All Chem Notes 1 9
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All Chem Notes 1 9 All Chem Notes 1 9 Presentation Transcript

  • Chemistry: The Study of Change Chapter 1 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.  
  • Introduction to Chemistry and the Scientific Method ask a question make observations and collect data design an experiment analyze data draw a conclusion do research
  • Chemistry: A Science for the 21 st Century
      • Sanitation systems
      • Surgery with anesthesia
      • Vaccines and antibiotics
      • Fossil fuels
      • Solar energy
      • Nuclear energy
    1.1 Health and Medicine Energy and the environment View slide
  • Chemistry: A Science for the 21 st Century
      • Polymers, ceramics, liquid crystals
      • Room-temperature superconductors?
      • Molecular computing?
      • Genetically modified crops
      • “ Natural” pesticides
      • Specialized fertilizers
    1.1 Materials Technology Food Technology View slide
  • 1.2 The Study of Chemistry Macroscopic Microscopic Chemists study the microscopic properties of matter, which in turn produce matter’s observable macroscopic properties – thus, we often switch back and forth between microscopic and macroscopic views of matter in this course.
  • The scientific method is a systematic approach to research. Although it is systematic, it is not a rigid series of steps that must be done in a particular order. ask a question make observations and collect data design an experiment analyze data draw a conclusion do research researcher’s hidden bias form a hypothesis
  • A theory is a unifying principle that explains a body of facts and/or those laws that are based on them. 1.3 Atomic Theory A hypothesis is a tentative explanation for a set of observations that can be tested . tested modified A law is a concise statement of a relationship between phenomena that is always the same under the same conditions. Force = mass x acceleration
  • Classification of Matter
  • Substances Matter is anything that has mass and occupies space. Matter that has a uniform and unchanging composition is called a (pure) substance. examples of pure substances include table salt, pure water, oxygen, gold, etc.
  • States of Matter Matter normally occupies one of three phases, or states. These are: P Solid P Liquid P Gas * Plasma is a 4 th state of matter in which the particles are at extremely high temperatures (over 1,000,000 °C).
  • As we shall see in more detail later, the phase (or state) of a substance is determined by the average kinetic energy of the particles that make up the substance, (i.e., temperature ) and the strength of the attractive forces holding the substance’s particles together. weak moderate strong States of Matter gas liquid solid
  • Solids
    • Solids have a definite shape and volume.
    • The particles of a solid cannot exchange positions.
    • Solids are incompressible.
    States of Matter
    • Liquids have definite volumes
    • Liquids do not have a fixed shape
    • Like solids, liquids are also incompressible
    Liquids States of Matter
    • Gases take on the shape and volume of their container
    • Unlike solids and liquids, gases are highly compressible
    Gases States of Matter
  • Technically, the word “gas” refers to a substance that is in the gas phase at room temperature. The word “vapor” refers to the gaseous state of a substance that is normally a solid or liquid at room temperature. States of Matter
  • 1. mixtures
    • homogeneous (solution) ‏
    • heterogeneous
    2. (pure) substances
    • compounds
    • elements
    Matter can be classified based on its characteristics into the following categories and subcategories: Classification of Matter
  • A pure substance is a form of matter that has a definite composition and distinct properties. examples: gold, salt, iron, pure water, sugar A mixture is a combination of two or more substances in which each substance retains its own distinct identity. examples: salt water, oil & vinegar dressing, granite, air Classification of Matter
  • see page 13 Classification Summary
  • Heterogeneous mixture : the composition is not uniform throughout. You can visibly see the different components. Mixtures can be heterogeneous or homogeneous . examples: cement, iron filings in sand, granite, milk, oil and water, etc. Classification of Matter Mixtures
  • Homogenous mixture (also called a solution ): The composition of the mixture is the same throughout. Classification of Matter Solutions are made up of two components : (1) the solute which is dissolved in (2) the solvent . If the solvent is water , the solution is called an aqueous solution which is symbolized: ( aq ).
  • We often think of a solution as being a solid dissolved in a liquid. However… In a solution, both the solvent and solute can be in any phase – solid, liquid or gas. air (O 2 dissolved in nitrogen) ‏ gas gas alloys (brass, bronze, etc.) ‏ solid solid gasoline (a mix of liquids) ‏ liquid liquid salt dissolved in water solid liquid example solute solvent Classification of Matter
  • In the case of liquids , we use a special term: If two liquids completely dissolve in each other, they are said to be miscible . If they do not, they are immiscible. Classification of Matter If a substance dissolves in another substance, we say the first substance is soluble in the second. If they do not dissolve, they are said to be insoluble. example: carbon dioxide is soluble in air gold is in soluble in water example: alcohol and water are miscible gasoline and water are immiscible
  • A mixture can be separated into its pure components by simple physical methods. Classification of Matter Filtration is a means of separating a solids from liquids. For example, we can filter out the sand from a mix of sand and water. Magnetic substances can be separated using a magnet.
  • Fractional crystallization is a means of separating two solids by adding a solvent that will dissolve one of the solids but not the other; the mixture is then filtered to separate out the insoluble solid. Finally, the solvent is evaporated off to recover the remaining solid. Separation of a Mixture For example, we can separate salt from sand by adding hot water to dissolve the salt, then filter off the sand. The water is then evaporated off, leaving the salt behind.
  • Distillation is a means of separating two liquids based on differences in their boiling points. Separation of a Mixture This method is only effective for substances that are liquids at room temperature with significant differences in their boiling points. The substance with the lowest boiling point “boils off” and is then cooled and condensed back into a liquid. The liquid is collected in a receiver flask. distillation apparatus
  • Chromotography is the separation of a mixture based on solubility in a “mobile” solvent coupled with an adherence to a “stationary phase” medium, such as paper or silica gel, etc. Separation of a Mixture column chromotography is a common means of separating components from a mixture Thin Layer Chromotography can be used to separate the components of chlorophyll from a crushed plant leaf.
  • Other means of separation: Other techniques of separating a mixture include sublimation, extraction, and leaching, etc. If you had a jar containing both nails and marbles, the only way to separate them would be by hand Separation of a Mixture speak to the hand…
  • You are given a test tube which contains a mixture of water, methanol, aspirin, acetanilide and aluminum oxide. (Acetanilide, aluminum oxide and aspirin are all white, powdery solids at room temperature and are thus visibly indistinguishable from each other. Water and methanol are both colorless liquids at room temperature and are also visibly indistinguishable from each other.) Assume your only source of heat is a Bunsen burner which can produce a maximum temperature of 600  C . Using the following information, devise a method to separate this mixture. Be specific and complete in your answer. Example: dissolves only in hot (50  C) water or warm (25 °C) methanol 304  C 114  C acetanilide does not dissolve in either methanol or water at any temperature 2980  C 2072  C aluminum oxide dissolves in methanol or water (if above 10  C) decomposes at 140  C 135  C aspirin dissolves in cold or hot water 65  C – 97  C methanol dissolves in cold or hot methanol 100  C 0  C water what it dissolves/does not dissolve in boiling point melting point substance
    • Mixtures are composed of two or more substances physically combined.
    • Recall that a (pure) substance is matter that has a uniform, unchanging composition
    • Pure substances may be elements or compounds
    Pure Substances: Elements and Compounds
  • see page 13 Classification Summary
  • An element is a substance that cannot be separated into simpler substances by chemical means.
      • 32 elements have been synthesized by scientists.
    • 114 elements have been identified
    • 82 elements occur naturally on Earth
    Classification of Matter examples include carbon, sulfur, copper, iron, and mercury
      • examples: technetium, americium, and seaborgium
    Elements carbon sulfur copper iron mercury
  • Some symbols are based on the Latin name eg , iron is Fe (for ferrum ) and sodium is Na (for natrium ) ‏ Symbols for Elements Elements are identified by a one or two-letter symbol. The first letter, which is ALWAYS capitalized, is typically the first letter in the name of the element. eg, C = c arbon, H = h ydrogen The second letter (which is only used if other elements have the same first letter) is NEVER capitalized. eg, Cl = c h l orine, He = he lium.
  • http:// Symbols for Elements
  • A compound is a substance composed of atoms of two or more different elements chemically bonded in fixed proportions. Water (H 2 O) ‏ Sucrose (C 12 H 22 O 11 ) ‏ Classification of Matter table salt (NaCl) ‏ Compounds sugar As such, they can be chemically decomposed into their component elements.
  • The properties of a compound are different from the properties of its component elements Classification of Matter For example, table salt is composed of sodium and chlorine. Sodium is a soft, silver colored metal that reacts violently with water, and chlorine is a pale-green poisonous gas – yet when chemically combined, they form table salt, a white crystalline solid you put on your eggs in the morning! = +
  • Compounds can only be separated into their pure components (elements) by chemical means. Water can be separated into its elements, hydrogen and oxygen, by passing an electric current through it, a process called electrolysis . For example: Iron is separated from iron ore (Fe 2 O 3 ) by heating the ore in a blast furnace and reacting it with carbon monoxide and elemental carbon (in the form of “coke”). Compounds
  • There are TWO kinds of compounds, depending on the nature of the chemical bond holding the atoms together. Molecules form when two or more neutral atoms form bonds between them by sharing electrons Note that some elements exist as molecules . For example,the following elements occur in nature as molecular diatomic elements : H 2 , N 2 , O 2 , F 2 , Cl 2 , Br 2 and I 2 They are molecules, but they are NOT compounds , because they have only one kind of element present. Cl 2 O 2 Compounds H 2 N 2
  • Ionic compounds are composed of ions, which are atoms that have a (+) or (-) charge. Ionic compounds form when cations and anions form electrostatic attractions between them (opposite charges attract) ‏ Compounds + + + + + + + + ─ ─ ─ ─ ─ ─ ─ ─ ─ + + ions are called cations and form when an atom l oses electrons
      • ions are called anions and form when an atom gains electrons
  • PURE SUBSTANCE can it be separated by physical means? compound element YES YES NO NO YES NO Classification Summary MATTER heterogeneous mixture is the mixture uniform throughout? can the substance be chemically decomposed into simpler substances? solution MIXTURE YES
  • Properties of Matter
  • Physical & Chemical Properties Physical Properties are measurable properties
    • mass $ density
    • boiling point $ solubility in water
    Chemical Properties describe how a substance reacts with other substances
    • flammability $ bonds with oxygen
    • reacts with water $ decomposes when heated
  • An extensive property of a material depends upon how much matter is being considered. Extensive properties are additive.
    • mass
      • length
        • volume
    Extensive and Intensive Properties Physical properties can be classified as being either extensive or intensive properties.
  • An intensive property of a material is independent of the amount of matter is being considered, and is not additive.
    • density • melting point
    • temperature •color
    Extensive and Intensive Properties Note that ALL chemical properties are intensive properties.
  • A physical change does not alter the composition or identity of a substance. Physical & Chemical Changes A chemical change (reaction) alters the identity or composition of the substance(s) involved. ice melting sugar dissolving in water hydrogen burns in air to form water
  • Physical & Chemical Changes Evidence of a chemical reaction include:
    • Heat and light (both) produced
    • Gas produced (bubbles)
    • Solid precipitate forms
    • Color changes occur
  • Measurement
  • Measurement The SI System of Measurement Scientists around the world use a unified system of measurement ( Le S ysteme I nternationale d’Unites, or SI for short). There are seven fundamental “quantities” that can be measured: Length Mass Time Temperature Electric Current Chemical quantity Luminous intensity
  • International System of Units (SI) page 16 Each base quantity is given a unit with a specific name and symbol
  • The SI units are based on metrics. Each power of ten change is given a special prefix used with the base unit. International System of Units (SI) You must know these prefixes see page 17
  • Measurements with SI Units
    • Length ( SI unit = meter ) The meter is often divided into cm and mm. (10 mm = 1 cm ).
    • Your little finger is about 1 cm in width. A dime is about 1 mm thick.
    • English/Metric equivalencies
        • 1 inch = 2.54 cm
        • 1 meter = 39.37 inches
  • Volume ( SI unit = m 3 ) Volume is the amount of space occupied by something. A more common unit is the dm 3 =1 liter. A smaller unit that we will use frequently is the cm 3 . Measurements with SI Units
        • 1 cm 3 = 1 ml
        • 1000 ml = 1 liter
    • English/Metric equivalencies
        • 1 liter = 1.057 quarts
        • 1 ml ~ 15 drops
  • liquids —use a graduated cylinder. To read the scale correctly, read the volume at the lowest part of the meniscus - the curve of the liquid’s surface in a container. Measuring Volume meniscus Measurements with SI Units Your eye should be level with the meniscus when reading the volume regular solids : volume = length x width x height
  • Measurements with SI Units Measuring Volume continued irregular solids : volume is found by displacement. 4 6 2 Begin with a known volume of water. 4 6 2 Add the solid. volume of solid = volume displaced : 6.0 – 4.0 = 2.0 cm 3 The amount of water displaced is the volume of the solid.
  • Measurements with SI Units Mass (SI unit = kilogram): the amount of matter. The mass of a given object is constant . A kilogram is about 2.2 pounds -- this is too large a unit for most chemistry labs, so we will use grams instead. Note that mass and weight are two different things…
  • Weight is a measure of the force due to gravity acting on a mass. The weight of an object changes , depending on the gravitational force acting on it. For example, on the moon you would weigh only 1/6th what you do on Earth, because the force of gravity on the moon is only 1/6th that of Earth . Measurements with SI Units
  • The Importance of Units On 9/23/99, the Mars Climate Orbiter entered Mar’s atmosphere 100 km (62 miles) lower than planned and was destroyed by heat because the engineers that designed the rocket calculated the force provided by the engines in pounds, but NASA engineers thought the force was given in the units of Newtons (N) when they determined when to fire the rockets… 1 lb = 1 N 1 lb = 4.45 N “ This is going to be the cautionary tale that will be embedded into introduction to the metric system in elementary school, high school, and college science courses till the end of time.”
  • Measurements with SI Units Measuring Mass We still use the term “weighing” even though we are finding the mass of an object, not its weight… Triple beam balance Electronic balance English/Metric equivalencies 1 kg = 2.203 lbs 1 paperclip  1 gram 1 lb = 453.6 grams
  • Temperature (SI unit = kelvin) is a measure of the average kinetic energy (energy due to motion) of the atoms and molecules that make up a substance. Measurements with SI Units There are three common temperature scales Fahrenheit ( o F) – English system, based on the freezing point of salt water. Centigrade ( o C) – metric system, based on the freezing and boiling points of pure water Kelvin (K) – SI unit , also called the “Absolute” scale; 0 K (Absolute Zero) is defined as the temperature at which all motion stops (kinetic energy = 0).
  • 32 o F = 0 o C 212 o F = 100 o C Temperature Conversions: K = o C + 273.15 273 K = 0 o C 373 K = 100 o C o C = ( o F – 32) 9 5 o F = ( o C) + 32 9 5
  • Temperature Examples A thermometer reads 12 o F. What would this be in o C ? The conversion formula from o F to o C is: o C = 5/9( o F – 32) Inserting the values gives: : o C = 5/9(12 o F – 32) o C = 5/9(-20) = -11.1 o C A thermometer reads 315.3 K. What would this be in o F ? The conversion formula from o C to o F is: o F = 9/5( o C) +32. Inserting the values gives: : o C = 9/5(42.15 o C) + 32 o C = (75.9) + 32 = 107.9 o F First convert K to o C: 315.3 K – 273.15 = 42.15 o C
  • page 21
  • Measurements with SI Units Time (SI unit = second) . This is the only non-metric SI unit. We still use 1 day = 24 hours, 1 hour = 60 minutes, 1 minute = 60 seconds Chemical Quantity ( SI unit = mole). Since atoms are so tiny, it takes a LOT of them to make even one gram. In fact, you would have to put 602,200,000,000,000,000,000,000 atoms of carbon (that’s 6.022 X 10 23 ) on a balance to get just 12 grams of carbon! We do use metric fractions of time, however, such as milliseconds (1/1000th of a second), etc.
  • That huge number (6.022 X 10 23 ) is given a special name; it is called “ Avogadro’s Number, ” symbolized N A , after the Italian physicist, Lorenzo Romano Amedeo Avogadro who lived between 1776-1856. Just like 1 dozen = 12 things, we define: Measurements with SI Units The Mole continued Avogadro 1 mole = 6.022 X 10 23 things
    • 1 mole of oranges would cover the surface of the earth to a depth of 9 miles!
    • If you stacked 1 mole of notebook paper, it would take you 5,800 years, traveling at the speed of light (186,000,000 miles per second) to reach the top of the stack!
    • If you were given 1 mole of dollar bills when the universe began 13 billion years ago, and you immediately began spending money at the rate of one million dollars per second , you would still have about 190 billion trillion dollars left !
    • but 1 mole of Hydrogen atoms would only mass about 1 gram!
  • 6.022 x 10 23 1.99 x 10 -23 N x 10 n N is a number between 1 and 10 n is a positive or negative integer Working with Numbers: Scientific Notation The number of atoms in 12 g of carbon: 602,200,000,000,000,000,000,000 The mass of a single carbon atom in grams: 0.0000000000000000000000199
  • Measurements with SI Units Derived Units Although we need only seven fundamental SI units, we can combine different units to obtain new units, called derived units. For example, speed is distance per unit time, so we must combine the unit for distance (m) and time (sec) to get the SI unit for speed: speed = meters per second (m/s) We will be working with many different derived units in this course. It is important to pay attention to the individual units that make up derived units!!
  • Derived Units SI derived unit for density is kg/m 3 . This is not a convenient unit in chemistry, so we usually use the units g/cm 3 or g/mL Density is the mass per unit volume of a substance. It is calculated using the equation: Every substance has a unique density . For example: You need to know the density of water. Any object that is more dense than water will sink in water; if it is less dense, it will float in water d = m V density = mass volume 0.70 g/cm 3 gasoline 11.35 g/cm 3 lead 2.70 g/cm 3 aluminum 1.00 g/cm 3 water density substance
  • page 18
  • page 18
  • Dimensional Analysis: A problem solving technique given unit x = desired unit desired unit given unit
  • In algebra, we learn that: u x u = u 2 u u = 1 (the u’s cancel!) and If we let “u” = units, then every measured quantity is a number x a unit. We can solve problems by setting them up so that the unit we do NOT want gets cancelled out by dividing u/u in the problem. Thus, if a/u is a conversion (say 100 cm/1 m) then we can convert cm to meters etc. using this conversion factor so that the cm cancel… a u u x = a (2u) 3 = 8 u 3 and… The Mathematics of Units
  • Dimensional Analysis Method of Solving Problems
    • Determine which unit conversion factor(s) are needed
    • Carry units through calculation
    • If all units cancel except for the desired unit(s) , then the problem was solved correctly.
    given quantity x conversion factor = desired quantity given unit x = desired unit desired unit given unit
  • Dimensional Analysis Method of Solving Problems Example: How many μ m are in 0.0063 inches? conversion factors needed: 0.0063 in = ? 1 inch = 2.54 cm 10 6 μ m = 1 m 1 m = 100 cm 0.0063 inch x x x = 160 μ m Begin with what units you have “in hand,” then make a list of all the conversions you will need. 2.54 cm 1 inch 1 m 10 2 cm 10 6 μ m 1 m
  • Example: The speed of sound in air is about 343 m/s. What is this speed in miles per hour? (1 mile = 1609 meters) 1 mi = 1609 m meters to miles seconds to hours 1.9 conversion units Dimensional Analysis Method of Solving Problems 1 min = 60 s 1 hour = 60 min 343 m s x 1 mi 1609 m 60 s 1 min x 60 min 1 hour x = 767 mi hour
  • page 29
  • page 30
  • Uncertainty, Precision & Accuracy in Measurements
  • Measurements with SI Units Uncertainty, Precision and Accuracy in Measurements When you measure length using a meterstick, you often have to estimate to the nearest fraction of a line. The uncertainty in a measured value is partly due to how well you can estimate such fractional units. http:// The uncertainty also depends on how accurate the measuring device, itself, is.
  • Accuracy – how close a measurement is to the true or accepted value To determine if a measured value is accurate, you would have to know what the true or accepted value for that measurement is – this is rarely known! Precision and Accuracy Precision – how close a set of measurements are to each other; the scatter of repeated measurements about an average. We may not be able to say if a measured value is accurate, but we can make careful measurements and use good equipment to obtain good precision, or reproducibility.
  • accurate & precise precise but not accurate not accurate & not precise A target analogy is often used to compare accuracy and precision. Precision and Accuracy
  • example: which is more accurate: 0.0002 g or 2.0 g? answer: you cannot tell , since you don’t know what the accepted value is for the mass of whatever object this is that you are weighing! example: which is more precise: 0.0002 g or 2.0 g? answer: surprisingly, the most precise value is 2.0 g, not the 0.0002 g. The number of places behind the decimal is not what determines precision! If that were so, I could increase my precision by simply converting to a different metric prefix for the same measurement: Which is more precise: 2 cm or 0.00002 km? They are, in fact, identical! Precision and Accuracy
  • Precision and Accuracy Precision is a measure of the uncertainty in a measured value. Any measured value is composed of those digits of which you are certain, plus the first estimated digit. 1 2 3 4 1 The length of the object is at least 1.7 cm, and we might estimate the last digit to be half a unit, and say it is 1.75 cm long. Others might say 1.74 or possibly 1.76 – the last digit is an estimate, and so is uncertain.
  • We always assume an uncertainty of ±1 in the last digit. The percent error in a measured value is defined as: The smaller the percent error, the greater the precision – the smaller the % error, the more likely two measurements will be close together using that particular measuring instrument. Thus: 2.0 ± 0.1 has a % error of (0.1/2.0) x 100 = ±5% but 0.0002 has a % error of (0.0001/0.0002) x 100 = ± 50% Precision and Accuracy % error = ± uncertainty measured value x 100
  • Percent Difference When determining the accuracy of an experimentally determined value, it must be compared with the “accepted value.” One common method of reporting accuracy is called the percent difference (  %) – this gives how far off your value is, as a percent, from the accepted value: Percent difference: experimental value – accepted value accepted value x 100  % =
  • example: In an experiment, a student determines the density of copper to be 8.74 g/cm 3 . If the accepted value is 8.96 g/cm 3 , determine the student’s error as a percent difference. = − 2.46 % The (-) sign indicates the experimental value is 2.46% smaller than the accepted value; a (+)  % means the experimental value is larger than the accepted value. experimental value – accepted value accepted value x 100  % = x 100 8.74 – 8.96 8.96  % =
  • Precision and Accuracy We will be doing math operations involving measurements with uncertainties, so we need a method of tracking how the uncertainty will affect calculated values – in other words, how many places behind the decimal do we really get to keep the answer? The method requires us to keep track of significant digits. Significant digits (or significant figures) are all of the known digits, plus the first estimated or uncertain digit in a measured value.
  • Significant Figures: Rules
    • Any digit that is not zero is significant
      • 1.234 kg 4 significant figures
    • Zeros between nonzero digits are significant
      • 606 m 3 significant figures
    • Zeros to the left of the first nonzero digit are not significant
      • 0.08 L 1 significant figure
    • If a number is greater than 1, then all zeros to the right of the decimal point are significant
      • 2.0 mg 2 significant figures
    • If a number is less than 1, then only the zeros that are at the end and in the middle of the number are significant
      • 0.00420 g 3 significant figures
  • How many significant figures are in each of the following measurements? 24 mL 2 significant figures 3001 g 4 significant figures 0.0320 m 3 3 significant figures 6.4 x 10 4 molecules 2 significant figures 560 kg You cannot tell!! Significant Figures: Rules
  • Significant Figures: Rules Suppose you wanted to estimate the number of jellybeans in a jar, and your best guess is around 400. Now – is the uncertainty in your estimate ±1 jellybean, or is it ±10 jellybeans, or maybe even ±100 jellybeans (if you weren’t very good at estimating jellybeans…) We need a way to write 400 and indicate in some way whether that was 400 ±1 vs 400±10 vs 400±100. The plain number “400” is ambiguous as to where the uncertain digit is. Use scientific notation to remove the ambiguity : 400 ± 1 = 4.00 x 10 2 = 3 sig figs 400 ± 10 = 4.0 x 10 2 = 2 sig figs 400 ± 100 = 4 x 10 2 = 1 sig fig
  • Rounding Numbers Given the number 6.82 and asked to round to 2 sig digits we would write 6.8. We write 6.8 because 6.82 is closer to 6.8 than it is to 6.9 Given the number 6.88 and asked to round to 2 sig digits, we would write 6.9. We write 6.9 because 6.88 is closer to 6.9 than it is to 6.8 You were taught this long ago. You were also probably taught that, given the number 6.85, and asked to round this to 2 sig digits, you would write 6.9. My question is, WHY did you round UP? 6.85 is JUST as close to 6.8 as it is to 6.9! Since it is in the middle, it could be rounded either way! And we should round it “either way.”
  • Rounding Numbers Since the rounding is “arbitrarily” up, this can introduce some round-off errors in chain calculations involving this number – the final value will be too large if you always round up when the next digit is exactly 5. Because rounding is “arbitrary” when the next digit is exactly 5, we introduce the following “ odd-even rounding rule : e.g. : 4.65 ≈ 4.6 but 4.75 ≈ 4.8 Note however, that 4.651 is closer to 4.7 than 4.6, so we round it to 4.7: only invoke the “odd-even rule” when the next digit is exactly 5. When the next digit is exactly 5, round up or down to make the number an even number .
  • Math Operations with Significant Digits We need a set of rules to determine how the uncertainty or error will “propagate” or move through a series of calculations and affect the precision of our final answer. There is one rule for addition and subtraction, and one rule for multiplication and division. Do not mix them and match them and confuse them!
  • Addition or Subtraction The answer cannot have more digits to the right of the decimal point than any of the original numbers. Significant Figures 89.332 1. 1 + 90 . 4 32 round off to 90.4 one digit after decimal point 3. 70 -2.9133 0. 78 67 two digits after decimal point round off to 0.79
  • Addition or Subtraction Significant Figures We often encounter two numbers that must be added that are in scientific notation. We cannot add them and determine the number of places “behind the decimal” unless they have the same power of 10 – we may have to convert! Example : What is the sum of 2.4 x 10 2 + 3.77 x 10 3 ? 3.77 x 10 3 0.24 x 10 3 4.01 x 10 3 Always convert the smaller power of 10 to the larger power of 10 The answer is good to 2 behind the decimal when written as x10 3 -- that is, the uncertain digit is in the “tens” place (± 10)
  • 10 n 10 n + + Decreasing the power of 10 means you must move the decimal to the RIGHT by one place for each power of 10 decrease. 10 n + Increasing the power of 10 means you must move the decimal to the LEFT one place for each power of 10 increase Moving the decimal determines both the magnitude and the +/- value of 10 n To determine the power of 10, visualize a see-saw when you move the decimal point: n n n n n n Significant Figures
  • Significant Figures Example: What is the answer to the following, to the correct number of significant digits? - 1.2 x 10 -4 - 0.012 x 10 -2 0 3.0268 x 10 -2 3.0148 x 10 -2 = 3.015 x 10 -2
  • Significant Figures Multiplication or Division The number of significant figures in the result is set by the original number that has the smallest number of significant figures 4.51 x 3.6666 = 16.5 36366 = 16.5 6.8 ÷ 112.04 = 0.060 6926 = 0.061 3 sig figs round to 3 sig figs 2 sig figs round to 2 sig figs
  • page 25
  • Significant Figures 1.8 Exact Numbers Numbers from definitions or numbers of objects are considered to have an infinite number of significant figures Because 3 is an exact number the answer is not rounded to 7, but rather reported to be 6.67 cm (three sig figures). Find the average of three measured lengths: 6.64, 6.68 and 6.70 cm. These values each have 3 significant figures Example: 6.64 + 6.68 + 6.70 3 = 6.67333 = 6.67 = 7
  • Atoms, Molecules and Ions Chapter 2 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.
  • Early Ideas Our understanding of the structure of matter has undergone profound changes in the past century. Nonetheless, what we know today did not arrive on a sudden inspiration. We can trace a fairly steady plodding towards our current understanding, starting as far back as 400 BCE…
  • Early Ideas The properties of matter could be explained by the shape and size of its atoms. Democritus (c.a. 400 BCE) All matter was composed of tiny, indivisible particles called atoms ( atomos = indivisible ) Each kind of matter had its own unique kind of atom – ie., there were water atoms, air atoms, fire atoms, bread atoms, etc. Fire atoms water atoms rolls & flows “ ouch!”
  • As a result, Democritus’ ideas were not very well receieved. It would be some 1200 years before the idea of atoms was revisited! Early Ideas Most importantly, Democritus believed atoms existed in a vacuum – that is, there was “nothing” in the spaces between atoms… Aristotle, among others, refused to believe in the existence of “nothingness” that still occupied space… Vacuum??
  • Early Ideas Aristotle Aristotle was the court philosopher to Alexander the Great . Because of this, Aristotle’s ideas were given a lot of weight . Aristotle believed that all matter was composed of four elements: earth, air, fire and water .
  • Heating WATER exchanged “hot” for “cold” which created “AIR” (which we see as steam…) WATER ( cold , wet ) AIR ( hot , wet ) These elements could be “inter-converted” into each other by exchanging the “properties” of hot, cold, dry and wet. example Early Ideas EARTH AIR WATER FIRE hot dry wet cold
  • Early Ideas This idea that one kind of element could be converted into another eventually led to the belief in Alchemy – that one could turn lead into gold by performing the right chemical reaction!
  • The “scientific method” of inquiry was developed during the 17 th and 18 th centuries. The invention of the balance and other instruments soon led to a new understanding about the nature of matter. Early Ideas The French chemist, Antoine-Laurent Lavoisier (1743-1794), presented two important ideas which would later help lead to a new, more developed atomic theory of matter…
    • The Law of Conservation of Matter: matter is not created or destroyed in chemical reactions . Any atomic theory would have to explain why matter is not gained or lost in reactions.
    • Lavoisier defined element as any substance that could not be chemically broken down into a simpler substance.
    Lavoisier Lavoisier was a meticulous experimenter. He also helped develop the metric system of measurement. He is often called the “ Father of Modern Chemistry ,” in recognition of his pioneering works.
  • Lavoiser experimenting with respiration
  • Joseph Proust, another 18 th century French scientist, proposed the Law of Definite Proportion , which states that the mass ratios of elements present in different samples of the same compound do not vary. Early Ideas For example, the percent by mass of the elements present in sugar are always found to be: 53.3% oxygen, 40.0% carbon and 6.7% hydrogen.
  • John Dalton (1766-1824) Dalton started out as an apothecary's assistant (today, we would call him a pharmacist). He was also interested in both meteorology and the study of gases. Dalton developed a new atomic theory of the nature of matter based on several postulates. His theory differed significantly from the early ideas of Democritus, but they both agreed that the simplest form of matter was the atom.
  • Dalton’s Atomic Theory (1808)
    • Elements are composed of extremely small particles called atoms .
    • All atoms of a given element are identical, having the same size, mass and chemical properties. The atoms of a given element are different from the atoms of all other elements.
    • Compounds are composed of atoms of more than one element. In any compound, the ratio of the numbers of atoms of any two of the elements present is either an integer or a simple fraction.
    • A chemical reaction involves only the separation, combination, or rearrangement of atoms; it does not result in their creation or destruction.
  • 2.1 Dalton’s Atomic Theory Law of Conservation of Matter and Definite Proportion Explained… + = 16 X 8 Y + 8 X 2 Y
  • Law of Multiple Proportions If Dalton’s ideas about atoms were correct, then he proposed that the mass of a compound containing different numbers of a given element (atom) would vary by the mass of that one whole atom – that is: If two elements can combine to form more than one compound, then the masses of one element that combine with a fixed mass of the other element are in ratios of small, whole numbers.
  • 16 12 = 1.33 = 2.67 2.67 / 1.33 = 2 Dalton’s Atomic Theory Consider the mass ratio of oxygen to carbon in the two compounds: CO and CO 2 32 12 Note that the mass of oxygen that combines with 12 g of carbon in carbon dioxide is 2 x greater than the mass of oxygen that combines with 12 g of carbon in carbon monoxide.
  • In the late 19 th and early 20 th centuries, three important experiments that shed light on the nature of matter were conducted:
      • J.J. Thomson’s investigation of cathode rays that led to the discovery of the electron.
      • Robert Millikan’s “Oil drop experiment” that determined the charge and mass of the electron.
      • Ernest Rutherford’s “Gold foil experiment” that finally gave us the current “nuclear” model of the atom.
    Modern Ideas
  • Cathode rays , discovered by William Crookes, are formed when a current is passed through an evacuated glass tube. Cathode rays are invisible, but a phosphor coating makes them visible.
  • J.J. Thomson The Electron is Discovered Sir Joseph John Thomson 1856-1940 J.J. Thomson helped show that cathode rays are made up of negatively charged particles (based on their deflection by magnetic and electric fields). N S
  • Thomson showed that all cathode rays are identical, and are produced regardless of the type of metals used for the cathode and anode in the cathode ray tube. Thomson was unable to determine either the actual electric charge or the mass of these cathode ray particles. He was, however, able to determine the ratio of the electric charge to the mass of the particles. J.J. Thomson
  • To do this, he passed cathode rays simultaneously through electric and magnetic fields in such a way that the forces acting on the cathode ray particles (now called electrons ) due to the fields cancelled out. The ratio of the electric field strength to the square of the magnetic field strength at this point was proportional to the charge to mass ratio of the electron. J.J. Thomson Magnetic field only Electric field only Both _ +
  • The value he obtained, −1.76 x 10 8 C/g*, was always the same, regardless of the source of the cathode rays. This value was nearly 2000 times larger than the charge to mass ratio of a hydrogen ion (H + )! This indicated that either the charge of the electron was very large, or that the mass of the electron was very, very small – much smaller than the mass of a hydrogen atom, which was the lightest atom known. J.J. Thomson *the SI unit of electric charge is the Coulomb (C)
  • Thomson proposed that these electrons were not just very small particles, but were actually a sub-atomic particle present in all atoms . We thus credit Thomson with the “discovery” of the electron because of his work in determining their physical characteristics, and his rather bold hypothesis that they were present in all atoms (which was later shown to be true). J.J. Thomson
  • Since the atom is neutrally charged, if it has (-) charged electrons, there must also be a (+) part to the atom to cancel the negative electrons. This showed that Dalton’s idea that atoms were indivisible is NOT correct – instead, the atom is composed of TWO oppositely charged parts. Thomson thought the atom was a diffuse (+) charged object, with electrons stuck in it, like raisins in pudding (the plum pudding model ). The Plum Pudding Model
  • Thomson’s Plum Pudding Model of the Atom
  • Millikan’s Oil Drop Experiment Robert Millikan (1911) designed an experiment to determine the actual charge of an electron. He suspended charged oil drops in an electric field. The drops had become charged by picking up free electrons after passing through ionized air.
  • F ELEC = E x q F GRAVITY = m x g when the downward force of gravity on the drop was balanced by the upward force of the electric field, then: Millikan’s Oil Drop Experiment E x q = m x g or q = mg/E Knowing the mass (m) of the oil drop, and the strength of the electric field (E), he was able to find the charge (q) on the oil drop .
  • To find the charge of the electron, he found the smallest difference between the charges on any two oil drops. eg: Suppose you find three oil drops have the following charges: 12.4, 7.6, 10.8. The differences between the charges are: 12.4 – 10.8 = 1.6 10.8 – 7.6 = 3.2 12.4 – 7.6 = 4.8 4.8 – 3.2 = 1.6 You would conclude the charge of the electron was 1.6 charge units. Millikan’s Oil Drop Experiment
  • Millikan’s Oil Drop Experiment Using this technique, Millikan was able to determine the charge of an electron to be: Using Thomson’s charge to mass ratio and the charge for the electron, Millikan determined the mass of the electron to be 9.11 x 10 -31 kilogram. For his work, Millikan received the 1923 Nobel Prize in Physics. e = C 1.602 x 10 C 19 C
  • (Uranium compound) It was found that there are three distinct types of radiation: ( + ) alpha particles, ( - ) beta particles, and neutral gamma rays. Radioactivity was discovered in 1895
  • Rutherford’s Gold Foil Experiment Rutherford designed an experiment using these newly discovered alpha-particles to test if Thomson’s plum pudding model was correct. He fired (+) alpha particles at the gold foil. If the Thomson model was correct, most of the alpha particles would pass through the foil with little deflection. (1908 Nobel Prize in Chemistry)
  • R (+)  -particle Rutherford’s Experiment Expected Results of Rutherford’s Experiment θ The force of repulsion is directly proportional to the product of the charges of the alpha particle and nucleus and inversely proportional to the square of the distance between the center of the two charges. F = kQ 1 Q 2 /R 2 A large, diffuse positive charge is not able to repel a (+) alpha particle very strongly, because the alpha particle cannot make a close approach, so the angle of deflection, θ , would be fairly small.
  • However, some particles were deflected significantly, and perhaps one in 2000 were actually deflected nearly 180 degrees! Rutherford’s Gold Foil Experiment When Rutherford performed the experiment, nearly all the alpha particles passed through the foil without deflection, as expected…
  • Rutherford’s Experiment Rutherford was stunned. This would be like firing a machine gun at an apple, and having most of the bullets pass through -- but every once in a while one of the bullets would bounce off the apple and come back and hit you! Why would this happen??? something small and massive must be in there that deflects only those bullets that directly hit it… ?!? ? DUCK, ERNIE!
  • Only a positive charge with a very, very small radius would allow the alpha particle to approach close enough to experience a significant repulsion. Rutherford’s Experiment R Strong repulsion! nucleus  -particle By carefully measuring the angles of deflection, θ , Rutherford was able to determine the approximate size of this positive core to the atom. θ
  • Next, by measuring the kinetic energy of the alpha particle before and after the collision, Rutherford was able to apply conservation of momentum and determine the mass of the atom’s positive core. Putting it all together, he was able to conclude that all the positive charge -- and about 99.9% of the mass -- of an atom was concentrated in a very tiny area in the middle of the atom, which he called the nucleus. Rutherford’s Experiment
  • Only the very few (+) α -particles that passed very near this incredibly tiny (+) nucleus were strongly deflected; most α -particles never came near the nucleus and so were not deflected significantly. Rutherford’s Experiment *note carefully that the (+) α -particles never actually collide with the (+) nucleus – the repulsive force between the like charges is too great for that to occur!
  • atomic radius ~ 100 pm = 1 x 10 -10 m nuclear radius ~ 5 x 10 -3 pm = 5 x 10 -15 m Rutherford’s Model of the Atom “ If the atom is the Houston Astrodome, then the nucleus is a marble on the 50-yard line.” The estimated size of this nucleus was such a tiny fraction of the total volume of the atom, that at first Rutherford doubted his own conclusion.
  • As another size comparison, if the nucleus were the size of a basketball , placed at PHS, the atom would be over 20 km in diameter, reaching Martin to the North, and just missing the US 131 Business Loop exit to the South ! The basketball-sized nucleus would also mass about 70,000,000,000 tons! This is equivalent to about 100,000 cruise ship ocean liners!
    • Rutherford fired (+) charged alpha particles at thin sheets of gold foil and measured the angles at which the alpha particles were deflected.
    • Rutherford was testing the validity of Thomson’s plum pudding model. If this model were correct, the (+) alpha particles would not be deflected by the diffuse (+) charge of Thomson’s atom .
    • When Rutherford performed the experiment, he found that the majority of alpha particles did, in fact, pass without significant deflection. However, a small number were significantly deflected, and a very few were strongly deflected nearly 180 degrees.
    • By measuring the angles of deflection, Rutherford was able to calculate the size and mass of the (+) center that could produce the observed deflections. He found that all the (+) charge and about 99.9% of the atom’s mass was concentrated in a tiny region (about 1/100,000 the volume of the atom).
    • Only those alpha particles that passed very close to the nucleus experienced a strong enough repulsion to produced significant deflections – most particles never came near the nucleus, and so were not deflected.
    • AP Extras:
    • The repulsive force depends on 1/R 2 between the (+) alpha particle and the (+) charge of the nucleus.
    • He also relied on conservation of momentum to help him determine the mass of the nucleus which was repelling the alpha particles.
  • Chadwick’s Experiment (1932 ) (1935 Noble Prize in Physics)
    • H atoms have 1 p; He atoms have 2 p
    • ratio of mass He/mass H should be 2/1 = 2
    • measured ratio of mass He/mass H = 4 ???
    neutron (n) is neutral (charge = 0) n mass ~ p mass = 1.67 x 10 -24 g James Chadwick discovered that when 9 Be was bombarded with alpha particles, a neutral particle was emitted, which was named the neutron. Now the mass ratios can be explained if He has 2 neutrons and 2 protons, and H has one proton with no neutrons Discovery of the Neutron ??  + 9 Be 1 n + 12 C + energy
  • mass n  mass p  1,840 x mass e -
  • Atomic number (Z) = number of protons in nucleus Mass number (A) = number of protons + number of neutrons also called the nucleon number = atomic number (Z) + number of neutrons Isotopes are atoms of the same element (X) with different numbers of neutrons in their nuclei Mass Number Atomic number, Mass number and Isotopes X A Z Element Symbol Atomic Number H 1 1 H (D) 2 1 H (T) 3 1 protium deuterium tritium U 235 92 Uranium-235 C 14 6 Carbon-14 examples nuclide
  • 6 protons, 8 (14 - 6) neutrons, 6 electrons 26 protons, 33 (59 - 26) neutrons, 26 electrons How many protons, neutrons, and electrons are in How many protons, neutrons, and electrons are in Fe 59 26 ? Atomic number, Mass number and Isotopes Examples: C 14 6 ?
  • see page 50
  • We now understand that the number of protons in the nucleus of the atom is what “defines” the element and gives each element its unique properties. The Periodic Table of the Elements
  • The Periodic Table of the Elements Transition metals period group
    • Properties of Metals
    • malleable and ductile
    • lustrous
    • good conductors
    • lose e- to form cations
    • brittle
    • dull
    • poor conductors
    • gain e- to form anions
    Elements Properties of Non-metals
    • Properties of Metalloids
    • properties are intermediate between those of metals and nonmetals
    • semi-conductors
    1A = alkali metals 5A = pnictides 2A = alkaline earths 6A = chalcogens 3A = boron family 7A = halogens 4A = carbon family 8A = noble gases The chemical properties of elements within a Family or Group are similar Names of Families or Groups Elements
  • Elements Natural abundance of elements in the Earth’s crust Natural abundance of elements in the human body
  • Molecules and Ions
  • Molecules & Ions Note that some elements exist as molecules . For example,the following elements occur in nature as molecular diatomic elements : H 2 , N 2 , O 2 , F 2 , Cl 2 , Br 2 and I 2 They are molecules, but they are NOT compounds , because they have only one kind of element present. A molecule is an aggregate of two or more neutral atoms in a definite arrangement held together by chemical forces F 2 O 2 H 2 N 2
  • A polyatomic molecule contains more than two atoms O 3 , H 2 O, NH 3 , C 3 H 6 O An allotrope is one of two or more distinct molecular forms of an element, each having unique properties. For example, O 2 and O 3 are allotropes of oxygen; diamond, graphite and buckminster fullerene (C 60 ) are all different allotropes of carbon. Molecules & Ions
  • Ionic compounds are composed of ions, which are atoms that have a (+) or (-) charge. Classification of Matter Ionic compounds form when cations and anions form electrostatic attractions between them (opposite charges attract) + ions are called cations and form when an atom l oses electrons
    • ions are called anions and form when
    • an atom gains electrons
    + + + + + + + + + C C C C C C C C C C
  • A monatomic ion contains only one atom Examples: Na + , Cl - , Ca 2+ , O 2- , Al 3+ , N 3- Examples: ClO 3 - , NO 2 - , CN - , SO 4 2 - Molecules and Ions A polyatomic ion contains more than one atom note that the convention is to indicate the magnitude of the charge first , and then the sign: e.g. , Ca 2+ , not Ca +2
  • 13 protons, so there are 13 – 3 =10 electrons 34 protons, so there are 34 + 2 = 36 electrons How many electrons are in ? How many electrons are in ? Molecules and Ions Examples Al 27 13 3+ Se 78 34 2-
  • Charges of common monatomic ions Note that some atoms, especially transition metals, have multiple charge states Note also that metals typically form (+) charged ions, nonmetals form (-) charged ions. s ee page 54
  • Also note the relation between the magnitude of the charge and the group number (1A, 5A, etc) for most elements. The charge of representative metals (group 1A, 2A and 3A) is equal to the group number The charge of representative nonmetals (group 4A-7A) is equal to: (the group number – 8)
  • Chemical Nomenclature Determining the names and formulas of chemical compounds IUPAC = International Union of Pure and Applied Chemists . This is the group that determines the official rules of nomenclature for all chemical elements and compounds
  • Chemical Formulas A chemical formula is a combination of element symbols and numbers that represents the composition of the compound. Subscripts following an element’s symbol indicate how many of that particular atom are present. If no subscripts are given, it is assumed that only one of that atom is present in the compound. NH 3 C 3 H 6 S P 4 O 10 1 N + 3 H atoms 3 C + 6 H + 1 S atoms 4 P + 10 O atoms
  • A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance Chemical Formulas
  • An empirical formula shows the simplest whole-number ratio of the atoms in a substance H 2 O H 2 O molecular empirical C 6 H 12 O 6 CH 2 O N 2 H 4 NH 2 Chemical Formulas C 2 H 8 O 2 CH 2 O note that different molecular compounds may have the same empirical formula
  • For ionic compounds the formula is always the same as the empirical formula. The sum of the charges of the cation(s) and anion(s) in each formula unit must equal zero. Thus, the ratio of cations to anions can always be reduced to simple, whole number ratios. The ionic compound NaCl Ionic Formulas Na + 500 Cl - 500 = NaCl
  • Naming Binary Molecular Compounds
  • Naming Molecular Compounds We will only consider naming binary molecules . Binary molecular compounds typically form between two non-metals, or a non-metal and a metalloid. 1st element + root of 2nd element + “- ide ” Naming Molecules: e.g. : HCl = hydrogen chloride
      • If there is more than one of a given element, we use prefixes to indicate the number of each kind of atom present.
    Naming Molecular Compounds The prefix mono is only used for atoms that can form more than one compound with the second element. For this class, oxygen is the main element that does this. See page 62
  • HI hydrogen iodide NF 3 nitrogen tri fluoride SO 2 sulfur di oxide N 2 Cl 4 di nitrogen tetra chloride NO 2 nitrogen di oxide N 2 O Examples of naming molecules Naming Molecular Compounds di nitrogen mo noxide (laughing gas)
  • If the second element begins with a vowel, the terminal vowel of the prefix is allowed to be dropped. For example N 2 O 4 could be called dinitrogen tetroxide , rather than dinitrogen tetr a oxide. CO would be called carbon monoxide , not carbon mon oo xide Note, however, that the official IUPAC rule states that the vowel is only dropped for “compelling linguistic reasons.” Naming Molecular Compounds
  • Naming Compounds containing Hydrogen Compounds containing hydrogen can be named using the Greek prefixes, but most have common names that are accepted by IUPAC. The most common examples are: B 2 H 6 CH 4 SiH 4 NH 3 PH 3 H 2 O H 2 S diboron hexahydride diborane carbon tetrahydride silicon tetrahydride nitrogen trihydride phosphorus trihydride dihydrogen monoxide dihydrogen sulfide methane silane ammonia phosphine water hydrogen sulfide Naming Molecular Compounds
  • Determining the formula of molecules from the name The subscripts tell you the number of each type of element present, so naming molecules from the formula is straightforward. e.g. sulfur hexafluoride = SF 6 dichlorine heptoxide = Cl 2 O 7 Naming Molecular Compounds The order in which the atoms are listed in molecules is based on something called electronegativity. For now, we can predict the order using the chart on the next slide…
  • B Br Ge C N P As Sb O S Se Te F Cl Si I H Order of Elements in Writing Molecular Formulas Chemical Formulas
  • Organic chemistry is the branch of chemistry that deals with carbon compounds Carbon is unique among all the elements in its ability to catenate, or form long or branching chains of carbon atoms. C C C We usually write these chains as “condensed formulas” that assumes carbons are bonded to each other as follows: = CH 3 CH 2 CH 3 note that we could also write this as: C 3 H 8 H H H H H H H H
  • Organic molecules that contain only carbon and hydrogen are called hydrocarbons . The first 10 simple hydrocarbons Hydrocarbon compounds are named based on the number of carbon atoms in the “backbone” or chain of carbon atoms.
  • Naming Ionic Compounds
  • Naming Ionic Compounds
    • Ionic Compounds
    Ionic compounds are typically composed of a metal cation and a non-metal anion $ name of cation = simply the name of the element $ name of anion = root of element’s name + - “ide”
  • Naming Ionic Compounds BaCl 2 barium chloride K 2 O potassium oxide Binary ionic compounds are named: name of metal ion + root of non-metal + “-ide” e.g. Na 2 S Mg 3 N 2 Al 2 O 3 sodium sulfide magnesium nitride aluminum oxide
  • Determining the formula of ionic compounds from the name is a little more involved – unlike molecular compounds, the name does not give us the subscripts. These must be determined based on the charges of each ion. Remember that the total number of (+) and (-) charges in any ionic compound must sum to zero. Formula of Ionic Compounds
  • Formula of Ionic Compounds Al 2 O 3 2 x +3 = +6 3 x -2 = -6 Al 3+ O 2- CaBr 2 1 x +2 = +2 2 x -1 = -2 Ca 2+ Br - MgS 1 x +2 = +2 1 x -2 = -2 Mg 2+ S 2 - aluminum oxide calcium bromide magnesium sulfide
  • Formula of Ionic Compounds Note that if you take the magnitude of the charge of the cation, and make it the subscript on the anion, and take the magnitude of the anion’s charge and make it the subscript of the cation, the compound will always end up with a net neutral charge. Now, if possible, reduce the subscripts to a simpler ratio, and you have the correct formula for the compound! Al O Al 3+ O 2- Al 2 O 3 +3 -2 3 2
  • s ee page 58
  • Multivalent ions: The Non-Representative Atoms Cu W Mn Co Pb Fe
  • Most elements form only ions with one charge. However, most of the transition metals , as well as Pb and Sn, have more than one possible charge state. We say they are multi-valent. e.g. : copper can exist in either a +1 or +2 charge state: Cu + or Cu 2+ The formula or name of the compound must indicate which charge state the metal cation is in. Transition and other multi-valent metal ions
  • Transition and other multi-valent metal ions Cu + cuprous Fe 2+ ferrous Cu 2+ cupric Fe 3+ ferric Cr 2+ chromous Hg 2 2+ mercurous Cr 3+ chromic Hg 2+ mercuric Older method gives a common name for each valence state e.g. CuCl = cuprous chloride Hg 2 I 2 = mercurous iodide Fe 2 O 3 = ferric oxide PbO = plumbous oxide
  • Transition and other multi-valent metal ions To determine which charge state the cation is in, you must look at the anion, and calculate the charge of the cation… Fe 2 O 3 Subscript on O is the charge of the iron! Thus, Fe is +3 and this compound is ferric oxide. CuS S is always -2, and there is only one Cu to cancel this out, so copper must be +2. Thus, this is cupric sulfide .
      • Stock System:
      • We indicate charge on metal with Roman numerals
    FeCl 2 2 Cl - = -2 so Fe is 2+ iron(II) chloride FeCl 3 3 Cl - = -3 so Fe is 3+ iron(III) chloride Cr 2 S 3 3 S -2 = -6 so Cr is 3+ chromium(III) sulfide Transition and other multi-valent metal ions
  • Polyatomic Ions SO 4 2- C 2 O 4 2- C 2 H 3 O 2 2- NH 4 +
  • Naming Polyatomic Ions There are certain groups of neutral atoms that bond together, and then gain or lose one or more electrons from the group to form what is called a polyatomic ion. Most polyatomic ions are negatively charged anions. Examples: OH - = hydroxide ion CN - = cyanide ion NO 3 - = nitrate ion NH 4 + = ammonium ion SO 4 2- = sulfate ion SO 3 2- = sulfite ion
  • See page 60
  • Naming Polyatomic Ions Naming ionic compounds containing polyatomic ions is straightforward: Name the cation + name the (polyatomic) anion Examples: NaOH = sodium hydroxide K 2 SO 4 = potassium sulfate Fe(CN) 2 = iron (II) cyanide (NH 4 ) 2 CO 3 = ammonium carbonate
  • page 61
  • page 62
  • see page 64 Compound Summary
  • There are a different set of rules for naming acids. Some of the rules are based on a much older system of nomenclature, and so the rules are not as simple as they are for molecular and normal ionic compounds. NAMING ACIDS AND BASES
  • An acid can be defined as a substance that yields hydrogen ions (H + ) when dissolved in water. These H + ions then bond to H 2 O molecules to form H 3 O + , called the hydronium ion.
    • Many molecular gases , when
    • dissolved in water, become acids:
      • HCl (g) = hydrogen chloride
      • HCl (aq) = HCl dissolved in water
      • which forms (H 3 O + ,Cl - ) = hydrochloric acid
  • All acids have hydrogen as the first listed element in the chemical formula . For nomenclature purposes, there are two major types of acids: Oxoacids (also called oxyacids) = acids that contain oxygen. eg: H 2 SO 4 , HC 2 H 3 O 2 Non-oxo acids = acids that do not contain oxygen. eg: HCl (aq), H 2 S (aq) Acids
  • Acids Rules for naming non-oxoacids acid = “hydro-” + root of anion + “-ic acid” see page 65 *note that we add an extra syllable for acids with sulfur and phosphorus: it’s not hydrosulfic acid, but hydrosulf ur ic acid. Similarly, acids with phosphorus will end in phosph or ic, not phosphic acid. *
  • An oxoacid is an acid that contains hydrogen, oxygen, and another element – That is, oxoacids are the protonated form of those polyatomic ions that have oxygen in their formulas. HClO 3 chloric acid HNO 2 nitrous acid H 2 SO 4 sulfuric acid Acids examples:
  • If the name of the polyatomic anion ends in “ate,” drop the -ate and add “ic acid.” eg: SO 4 2- = sulfate anion H 2 SO 4 = sulfuric acid C 2 H 3 O 2 - = acetate anion HC 2 H 3 O 2 = acetic acid When naming oxoacids, NO “hydro” prefix is used. Instead, the acid name is the root of the name of the oxoanion + either “-ic” acid or “-ous” acid, as follows: If the name of the polyatomic anion ends in “ite,” drop the -ite and add “ous acid.” eg: SO 3 2- = sulfite anion H 2 SO 3 = sulfurous acid NO 2 - = nitrite anion HNO 2 = nitrous acid
  • Acids Naming Oxoacids and Oxoanions see page 66
  • Acids ic goes with ate because…. ”IC…I ATE it! ite goes with ous like……tonsil- ITE-OUS , senior-ITE-OUS As a mnemonic aid, I always use the following:
  • A base can be defined as a substance that yields hydroxide ions (OH - ) when dissolved in water. Bases NaOH sodium hydroxide KOH potassium hydroxide Ba(OH) 2 barium hydroxide
  • Hydrates are compounds that have a specific number of water molecules attached to them. BaCl 2 •2H 2 O LiCl•H 2 O MgSO 4 •7H 2 O Sr(NO 3 ) 2 •4H 2 O barium chloride dihydrate lithium chloride monohydrate magnesium sulfate heptahydrate strontium nitrate tetrahydrate CuSO 4 •5H 2 O cupric sulfate pentahydrate CuSO 4 anhydrous cupric sulfate Hydrates
  • Anhydrous: without water; this term describes hydrated compounds after “drying.” Hygroscopic: readily absorbs moisture directly from the air. Deliquescent: absorbs moisture from the air so readily, that these compounds can take on enough water to actually start to dissolve. Water of hydration: the water absorbed and incorporated into hygroscopic compounds Hydrates Other terms associated with hydrates
  • see page 68
  • Mass Relationships in Chemical Reactions Chapter 3 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.
  • By definition : 1 atom 12 C “weighs” 12 amu On this scale 1 H = 1.008 amu and 16 O = 16.00 amu Atomic mass is the mass of an atom in atomic mass units ( amu ). This is a relative scale based on the mass of a 12 C atom. Micro World atoms & molecules Macro World grams Relative Masses of the Elements
  • How do we find the relative masses of the other elements? Imagine we have 66.00 grams of CO 2 . The compound is decomposed and yields 18.00 grams of C and 48 grams of O. Since there are two oxygen atoms for every 1 carbon atom, we can say that This means that the relative mass of each oxygen atom is 1.333 x the mass of a carbon atom (12.00 amu) , or… mass of oxygen = 1.333 x 12.00 amu = 16.00 amu Relative Masses of the Elements
  • 7.42% 6 Li (6.015 amu) 92.58% 7 Li (7.016 amu) The Average atomic mass of lithium would be: Average Atomic Mass The average atomic mass of an element is the weighted average mass of that element, reflecting the relative abundances of its isotopes. example: consider lithium (Li), which has two isotopes with the following relative percent abundances: 7.42 100 6.015 92.58 100 7.016 6.941               amu amu amu
  • 6.941 The masses reported at the bottom of the “box” for each element in the Periodic Table is the average atomic mass for that element, (in amu). Average Atomic Mass
  • see page 79 Average Atomic Mass
  • The mole (mol) is the SI unit for the amount of a substance that contains as many “things” as there are atoms in exactly 12.00 grams of 12 C. 1 mol = N A = 6.022 x 10 23 “things” The Mole & Avogadro’s Number This number, called Avogadro’s number ( N A ), has been experimentally determined to be approximately 6.0221367 X 10 23 things. We can have 1 mole of atoms, or molecules, or even dump trucks. The mole refers only to a number, like the term “dozen” means 12.
    • If you stacked 1 mole of notebook paper, it would take you 5,800 years, traveling at the speed of light (186,000,000 miles per second) to reach the top of the stack!
    • If you were given 1 mole of dollar bills when the universe began 13 billion years ago, and you immediately began spending money at the rate of one million dollars per second , you would still have about 190 billion trillion dollars left !
    • 1 mole of oranges would cover the surface of the earth to a depth of 9 miles!
    • but 1 mole of Hydrogen atoms would only mass about 1 gram!
    JUST HOW BIG IS AVOGADRO’S NUMBER?? The Mole & Avogadro’s Number
  • Molar mass is the mass, in grams , of exactly 1 mole of any object (atoms, molecules, etc.) Note that because of the way we defined the mole : 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g 1 12 C atom = 12.00 amu For example: 1 mole 12 C atoms = 12.00 g 12 C 1 mole lithium atoms = 6.941 g of Li The Mole & Molar Mass Thus, for any element atomic mass (amu) = molar mass (grams)
  • One Mole of: C = 12.01 g S = 32.06 g Cu = 63.55 g Fe = 55.85 g Hg = 200.6 g The Mole & Molar Mass
  • The Mole & Molar Mass Solving Mole Problems We can now add the definitions of the mole, Avogadro’s number, and molar mass to our repertoire of conversion factors we can use in dimensional analysis problems. Thus, given the mass, we can use the molar mass to convert this to moles, and then use Avogadro’s number to convert moles to particles, and vice versa… N A = Avogadro’s number M = molar mass in g/mol
  • 0.551 g K How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 x 10 23 atoms K = 8.49 x 10 21 atoms K Solving Mole Problems conversion factors x 6.022 x 10 23 atoms K 1 mol K 1 mol K 39.10 g K x
  • Solving Mole Problems see page 81
  • Solving Mole Problems see page 82
  • Solving Mole Problems see page 82
  • Molecular mass (or molecular weight ) is the sum of the atomic masses of the atoms in a molecule. 1 molecule SO 2 = 64.07 amu 1 mole SO 2 = 64.07 g SO 2 Molecular Mass As was the case for atoms, for any molecule Example: consider SO 2 SO 2 1S 32.07 amu 2O + 2 x 16.00 amu SO 2 64.07 amu molecular mass (amu) = molar mass (grams) SO 2 64.07 amu
  • Molecular Mass see page 83
  • Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl Formula Mass For any ionic compound 1Na 22.99 amu 1Cl + 35.45 amu NaCl 58.44 amu NaCl
  • What is the formula mass of Ca 3 (PO 4 ) 2 ? 1 formula unit of Ca 3 (PO 4 ) 2 310.18 amu Formula Mass Since the formula mass, in grams (per mole), is numerically equal to the molar mass, in amu, we find that the formula mass of Ca 3 (PO 4 ) 2 = 310.18 grams per mole of Ca 3 (PO 4 ) 2 . 3 Ca 3 x 40.08 2 P 2 x 30.97 8 O + 8 x 16.00
  • Example : How many H atoms are in 72.5 g of C 3 H 8 O ? 1 mol C 3 H 8 O = (3 x 12) + (8 x 1) + 16 = 60 g C 3 H 8 O 1 mol H = 6.022 x 10 23 atoms H = 5.82 x 10 24 atoms H 1 mol C 3 H 8 O molecules = 8 mol H atoms 72.5 g C 3 H 8 O Molecular/Formula Masses Using Molecular/Formula Masses in Dimensional Analysis Problems We can now add molecular & formula masses to our list of conversion factors. They are used similarly to the way we used the molar mass of the elements as conversion factors. conversion factors 1 mol C 3 H 8 O 60 g C 3 H 8 O x 8 mol H atoms 1 mol C 3 H 8 O x 6.022 x 10 23 H atoms 1 mol H atoms x
  • Solving Mole Problems see page 84
  • Solving Mole Problems see page 85
  • The Mass Spectrometer Atomic and molecular masses of unknown compounds are determined using a mass spectrometer . A gaseous sample of the unknown is bombarded with electrons in an electron beam. This knocks electrons loose from the unknown to produce cations. These cations are then accelerated through perpendicular electric and magnetic fields. The charge:mass ratio ( e/m ) of the unknown ions determines the degree to which the particles are deflected. The greater the charge:mass ratio, the smaller the angle through which the beam is deflected.
  • The Mass Spectrometer We know the angle that a given e/m produces, so we can identify the unknown ion when it registers on a special screen. high e/m low e/m Mass Spectrometer
  • Percent composition of an element in a compound is the percent, by mass, of that element in the compound. It can be calculated as follows: where n is the number of moles of the element in 1 mole of the compound Percent composition Knowing the percent composition, one can determine the purity of a substance, (are there contaminants present in the sample?) and you can even determine the empirical formula of an unknown compound. n x molar mass of element molar mass of compound x 100%
  • check: 52.14% + 13.13% + 34.73% = 100.0% Percent composition Example: What is the percent composition of ethanol, which has the formula, C 2 H 6 O ? First, we find the molecular mass of ethanol. This is found to be: 2(12.01) + 6(1.008) + 1(16.00) = 46.07 grams/mole. % Composition: C 2 H 6 O %C = 2 x (12.01 g) 46.07 g x 100% = 52.14% %H = 6 x (1.008 g) 46.07 g x 100% = 13.13% %O = 1 x (16.00 g) 46.07 g x 100% = 34.73%
  • We can also determine the % by mass of groups of atoms present in a compound in the same manner. Example: what is the percent water in epsom salts, which has the formula: MgSO 4 • 7 H 2 O ? % H 2 O = 24.31 + 32.07 + 4(16.00) + 7(18.02) 7(18.02) mass of water mass of compound x 100 = = 246.52 g cmpd 126.14 g H 2 O x 100 = 51.17% H 2 O Percent composition this is the molar mass of water
  • Example: How many grams of CaCl 2 • 2 H 2 O must be weighed out to obtain 12.20 grams of CaCl 2 ? There are two ways of solving this problem: Method 1 : First determine the % CaCl 2 in CaCl 2 • 2 H 2 O: ii. 75.49% of ( X grams) of CaCl 2 •2 H 2 O = 12.20 g of CaCl 2  0.7549( X ) = 12.20 or X = 12.20/0.7549 = 16.16 grams i. % CaCl 2 = 110.98 g CaCl 2 147.02 g CaCl 2 • 2 H 2 O x 100 = 75.49% Then we note that the 12.20 g of CaCl 2 desired must be 75.49% of the mass of the hydrate used: Percent composition
  • Example: How many grams of CaCl 2 • 2 H 2 O must be weighed out to obtain 12.20 grams of CaCl 2 ? There are two ways of solving this problem: 12.20 g CaCl 2 x = 16.16 g note that, math-wise, both methods involve the exact same calculations (i.e., the ratio of the molar mass of the hydrate to the molar mass of the anhydrous form had to be determined) . The only difference was the “logic” you followed which led you to that calculation! Percent composition Method 2 : Use dimensional analysis and molar masses: 110.98 g CaCl 2 1 mole CaCl 2 x 1 mole CaCl 2 1 mole CaCl 2 • 2 H 2 O 147.02 g CaCl 2 • 2 H 2 O x 1 mole CaCl 2 • 2 H 2 O
  • Percent Composition and Empirical Formulas Knowing the percent composition of a compound, one can determine the empirical formula. It is essentially the same process as finding the percent composition – only you work backwards to find the molar mass of the compound…
    • First, you convert the % composition into grams. This is easily done – suppose you had 100 grams of the substance. Then, the mass, in grams, of each component element is numerically the same as its percent composition.
    example : a sample of an iron ore is found to contain 69.94% Fe and 30.06% O. In 100 grams of the ore, there would be 69.94 grams of Fe and 30.06 grams of oxygen.
  • Percent Composition and Empirical Formulas 2. Next, knowing the mass of each element (in your 100 gram sample), determine the number of moles of that element in your sample, by dividing the mass by the molar mass of the element. The number of moles of Fe and O in our sample of the iron ore would be: 69.94 grams Fe x = 1.252 mol Fe 30.06 grams O x = 1.879 mol O 55.847 g 1 mol Fe 16.00 g 1 mol O
  • Percent Composition and Empirical Formulas 3. To find the simplest mole ratio of the elements, divide each by the smallest number: in our iron ore sample, we would have: = 1.501 mol O per mole of Fe 4. If this ratio is a whole number , then you are done – if the ratio is NOT a whole number, it must be converted to a whole number ratio (we cannot have fractions of an atom!) Fe 1.00 O 1.50 = Fe O = Fe 2 O 3 1.879 mol O 1.252 mol Fe 3 2 2 2
  • Percent Composition and Empirical Formulas The process is summarized in Figure 3.5 on page 89 in your textbook.
  • Percent Composition and Empirical Formulas Example : Determine the empirical formula of a compound that has the following percent composition by mass: K = 24.75%, Mn = 34.77%, and O = 40.51% n Mn = 34.77 g Mn x = 0.6329 mol Mn 1 mol Mn 54.94 g Mn n O = 40.51 g O x = 2.532 mol O 1 mol O 16.00 g O n K = 24.75 g K x = 0.6330 mol K 1 mol K 39.10 g K
  • n K = 0.6330, n Mn = 0.6329, n O = 2.532 Percent Composition and Empirical Formulas The empirical formula for the compound is: KMnO 4 K : ~ ~ 1.0 0.6330 0.6329 Mn : 0.6329 0.6329 = 1.0 O : ~ ~ 4.0 2.532 0.6329 divide each element by the smallest mole value
  • Combust 10.0 g compound Collect 24.078 g CO 2 and 11.088 g H 2 O Experimental determination of a molecular formula Determination of Molecular Formulas sample
  • Determination of Molecular Formulas C 4 H 9 O empirical formula = mass of each element moles of each element mole ratios of the elements empirical formula
  • Determination of Molecular Formulas The molecular weight of the compound was determined experimentally to be 146.2 g/mol. To determine the true molecular formula, divide the molecular weight by the formula weight. This ratio gives the number each subscript must be multiplied by to give the molecular formula. Formula weight of C 4 H 9 O = 73.1 g/mol Molecular weight of compound = 146.2 g/mol thus, the true molecular formula is (C 4 H 9 O) 2 = C 8 H 18 O 2
  • Working with Chemical Equations
  • For example, there are several ways of representing the reaction of H 2 with O 2 to form H 2 O A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction Chemical Equations
  • How to “Read” Chemical Equations 2 Mg + O 2 2 MgO 2 atoms Mg + 1 molecule O 2 makes 2 formula units MgO 2 moles Mg + 1 mole O 2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O 2 makes 80.6 g MgO reactants form products IT DOES NOT IMPLY THAT 2 grams Mg + 1 gram O 2 makes 2 g MgO this is based on the molar masses of the species and the coefficients in the reaction…
  • Balancing Chemical Equations
    • Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.
    example: Ethane (C 2 H 6 ) reacts with oxygen to form carbon dioxide and water
    • Change the numbers in front of the formulas ( coefficients ) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.
    2 C 2 H 6 IS NOT = C 4 H 12 C 2 H 6 + O 2 CO 2 + H 2 O 
    • Start by balancing those elements that appear in only one reactant and one product.
    start with C or H but not O  multiply CO 2 by 2  multiply H 2 O by 3 Balancing Equations next balance H next we will balance O C 2 H 6 + O 2 CO 2 + H 2 O 2 carbon on left 1 carbon on right C 2 H 6 + O 2 2 CO 2 + H 2 O 6 hydrogen on left 2 hydrogen on right C 2 H 6 + O 2 2 CO 2 + 3 H 2 O
    • Balance those elements that appear in two or more reactants or products.
    = 7 oxygen on right to remove fraction multiply both sides by 2 Balancing Equations 2 oxygen on left 4 oxygen (2x2) C 2 H 6 + O 2 2CO 2 + 3H 2 O + 3 oxygen (3x1) multiply O 2 by 7 2 C 2 H 6 + O 2 2CO 2 + 3H 2 O 7 2 2 C 2 H 6 + 7 O 2 4 CO 2 + 6 H 2 O
    • Check to make sure that you have the same number of each type of atom on both sides of the equation.
    Balancing Equations  2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O Reactants Products 4 C 12 H 14 O 4 C 12 H 14 O 4 C (2 x 2) 4 C 12 H (2 x 6) 12 H (6 x 2) 14 O (7 x 2) 14 O (4 x 2 + 6)
  • example: balance the following equations: Fe + S -> Fe 2 S 3 K + H 2 O -> KOH + H 2 Al + S 8 -> Al 2 S 3 Pb(NO 3 ) 2 + KI -> KNO 3 + PbI 2 3 2 2 2 2 16 3 8 2 2
  • Stoichiometry One of the most important applications of balanced equations is in determining the amount of one reactant required to react completely with another, or in determining the theoretical amount of product that should be formed in a given reaction. These problems all follow the same set of “logic” steps – indeed, almost any problem involving balanced equations will always follow these same steps! Stoichiometry is the quantitative study of reactants and products in a chemical reaction.
    • Write balanced chemical equation
    • Convert quantities of known substances into moles
    • Use coefficients in balanced equation to determine the mole:mole ratio between substances A and B, and from this determine the number of moles of the sought quantity
    • Convert moles of sought quantity into desired units
    Stoichiometry Flow Chart Stoichiometry
  • Example : Methanol burns in air according to the equation: If 209 g of methanol are used up in the combustion, what mass of water is produced? 209 g CH 3 OH = 235 g H 2 O The sequence of steps we follow in solving the problem are: Stoichiometry 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O grams CH 3 OH moles CH 3 OH grams H 2 O 1 mol CH 3 OH 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH x 18.0 g H 2 O 1 mol H 2 O x moles H 2 O moles CH 3 OH
  • The molarity of a solution is the concentration of that solution, expressed as the moles of solute present in 1 liter of a solution Solution Stoichiometry It is often easier to work with solutions, rather than solids. This means we also need a means of quantitatively working with reactions in solution. read as, for example: 2M NaCl = 2 “molar” solution of NaCl see pages 142-150 in Chapter 4 M = molarity = moles of solute liters of solution
  • M KI MW KI 500. mL = 232 g KI x 2.80 mol KI 1 L soln x x Solution Stoichiometry example: what mass of KI is needed to prepare 500 mL of a 2.80 M solution of KI? Solution plan: convert volume to moles using molarity, then moles to mass using molar mass: volume of KI solution moles KI grams KI 166 g KI 1 mol KI 1 L 1000 mL
  • known mass of solute dissolve solute dilute to mark
  • Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Solution Stoichiometry Dilution Add Solvent Moles of solute (concentrated) (c) Moles of solute after dilution (d) = M c V c M d V d =
  • example: How would you prepare 60.0 mL of 0.200 M HNO 3 from a stock solution of 4.00 M HNO 3 ? M c V c = M d V d M c = 4.00 M M d = 0.200 M V d = 0.0600 L V c = ? L = 0.003 L = 3 mL Thus, add 3 ml of acid to 57ml of water to form 60 ml of solution (dilute the 3 ml of acid to a total volume of 60 ml) Solution Stoichiometry V c = M d V d M c = 0.200 M x 0.0600 L 4.00 M
  • As with all stoichiometry problems, convert the starting unit to moles. Note that we now have THREE methods of converting to moles: Solving Solution Stoichiometry Problems 1. Use Avogadro’s number to convert particles to moles 2. Use the molar mass of the substance to convert grams to moles # particles x N A = moles grams x mole molar mass = moles
  • volume x = moles moles L sol’n 3. And now our third method is to use the molarity of the solution to convert volume* to moles. *the volumes must be in LITERS when converting to moles using the molarity of the solution. Solution Stoichiometry Note also that we can convert moles to volume by multiplying moles x 1/M
  • example : How many ml of 0.35 M Na 3 PO 4 are needed to react completely with 28.0 ml of a 0.42 M solution of Ba(NO 3 ) 2 , according to the balanced equation shown below: 3 Ba(NO 3 ) 2 + 2 Na 3 PO 4 -> Ba 3 (PO 4 ) 2 + 6 NaNO 3 0.0280 L x = 22.4 ml Solution plan: convert to moles, use mol : mol ratio from the balanced equation, convert moles to liters using molarity, then convert to mL. 0.42 mol Ba(NO 3 ) 2 Liter x 2 mol Na 3 PO 4 3 mol Ba(NO 3 ) 2 x 1 L 0.35 mol Na 3 PO 4 x 1000 mL I L
  • Gravimetric Analysis Gravimetric analysis is an analytical technique based on the measurement of the mass (usually of an ionic substance.) The substance of interest is typically reacted in solution and comes out as a precipitate. The precipitate is then filtered off, dried and weighed. Knowing the mass and chemical formula of the precipitate that formed, we can calculate the mass of a particular chemical component of the original sample.
  • Gravimetric Analysis
    • Dissolve unknown substance in water
    • React unknown with known substance to form a precipitate
    • Filter and dry precipitate
    • Weigh precipitate
    • Use chemical formula and mass of precipitate to determine amount of unknown ion
  • example: A 0.5662 gram sample of an unknown ionic compound containing chloride ions is dissolved in water and treated with an excess of AgNO 3 . If 1.0882 grams of AgCl precipitated , what is the percent chlorine in the original sample? First determine the mass of Cl‾ ions in the AgCl ppt: 1.0882 g AgCl x 1 mol AgCl 143.4 g AgCl x x = 0.2690 g Now determine the % Cl in the original sample: 47.51 % 1 mol Cl‾ ions 1 mol AgCl 35.45 g Cl 1 mol Cl 0.2690 g Cl % Cl = 0.5662 g unknown x 100 =
  • Limiting Reagents The limiting reagent is the reactant that gets used up completely -- that is, the one not present in excess. It is very rare that you mix reactants together in the exact stoichiometric ratio needed for each to react completely with the other. Usually, you have a little “extra” of one of the reactants compared to the other one, that is, one reagent is present in excess .
  • The maximum amount of product that can be formed is thus limited by the amount of the limiting reagent present. Limiting Reagents When the reaction is completed, there will be no limiting reagent left over. But there will be some of the reagents in excess left over. The reagents present in quantities greater than the minimum amount necessary are called the reagents in excess.
  • Limiting Reagents NO is the limiting reagent O 2 is the excess reagent Consider the reaction between NO and O 2 to form NO 2 . If we start with the mix shown at the top right, and end with the mix shown at the bottom right, we see that oxygen was present in excess (some is left over) which means that NO was the limiting reagent. 2NO + 2O 2 2NO 2
  • example: In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 according to the rxn: 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. 124 g Al 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g) so Al is limiting reagent Limiting Reagents g Al mol Al mol Fe 2 O 3 needed g Fe 2 O 3 needed OR g Fe 2 O 3 mol Fe 2 O 3 mol Al needed g Al needed 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al x 160. g Fe 2 O 3 1 mol Fe 2 O 3 x = Start with 124 g Al need 367 g Fe 2 O 3
  • Use limiting reagent (Al) to calculate amount of product that can be formed. 124 g Al 234 g Al 2 O 3 3.9 Now… Limiting Reagents g Al mol Al mol Al 2 O 3 g Al 2 O 3 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al x 102. g Al 2 O 3 1 mol Al 2 O 3 x = 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe
  • Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. Reaction Yields % Yield = Actual Yield Theoretical Yield x 100
  • Reactions in Aqueous Solution Chapter 4 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display. AP Inorganic Chemistry
  • Symbols used in Equations (s) solid (l) liquid (g) gas ( aq ) aqueous = (dissolved in H 2 O) solid precipitate given off as a gas yields equilibrium  heated 20º C specified rxn temperature MnO 2 MnO 2 is a catalyst in the rxn arrows are only used for products!
  • examples Na 2 CO 3 (s) Na 2 O (s) + CO 2 (g)  HCl (g) + H 2 O (l) H 3 O + (aq) + Cl¯ (aq) We often write H 2 O over the yields arrows when we show that something is being dissolved in water. NaCl (s) + AgNO 3 (aq) AgCl ( ) + NaNO 3 (aq) 2 KClO 3 (s) 2 KCl (s) + 3 O 2 ( ) MnO 2  HCl (g) H 3 O + (aq) + Cl¯ (aq) H 2 O or
  • General Properties of Aqueous Solutions
  • A solution is a homogenous mixture of 2 or more substances The solute is the substance(s) present in the smaller amount. The solvent is the substance present in the larger amount Solutions
  • During the solution process, the solute is first surrounded by the solvent molecules, and the attractions between the solvent and the solute help to pull the solvent particles apart. The isolated solute particles are in turn surrounded by a “sphere” of solvent particles in a process called solvation – in the case of aqueous solutions, the term hydration is used. General Properties of Aqueous Solutions
  • Hydration (or solvation ) : the process in which a solute particle, such as an ion or a neutral molecule, is surrounded by water molecules arranged in a specific manner. Hydration “spheres” Solutions
  • An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity . A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity . Solutions nonelectrolyte weak electrolyte strong electrolyte
  • Strong Electrolyte – 100% dissociation Weak Electrolyte – not completely dissociated To conduct electricity, a solution must contain charged particles, that is, cations (+) and anions (-) Solutions (all three species are present at equilibrium) NaCl ( s ) Na + ( aq ) + Cl - ( aq ) H 2 O CH 3 COOH CH 3 COO - ( aq ) + H 3 O + ( aq ) H 2 O
  • Nonelectrolytes do not produce charged particles in solution Solutions does not dissociate into ions when in solution C 6 H 12 O 6 ( s ) C 6 H 12 O 6 ( aq ) H 2 O
  • Classification of Reactions PbI 2
  • Classes of Reactions There are literally millions of known chemical reactions. It would be impossible to learn or memorize them all. Instead, we will learn five fundamental “classes” of reactions.
    • Double displacement
    • Single displacement
    • Decomposition
    • Synthesis
    • Combustion
  • Also called “metathesis” reaction . All double displacement reactions follow a pattern in which two elements “trade partners.” A X + B Y B X + A Y 1. Double Displacement
    • Double displacement reactions are typically one of two types:
    • precipitation reactions
    • neutralization reactions
    Classes of Reactions
  • Precipitation reactions involve the exchange of cations between two ionic compounds that results in the formation of an insoluble precipitate*: *we will learn to determine which substances are soluble and which are not later on in this chapter… Neutralization reactions involve an acid and a metal hydroxide; the acid’s H + ion is exchanged with the hydroxide’s metal cation to produce an ionic “salt” and water. Classes of Reactions (which is just H 2 O!) Ag NO 3 (aq) + Na Cl (aq) Ag Cl (s) + Na NO 3 (aq) eg: H Cl + K OH K Cl + H OH eg:
  • 2. Single displacement: A + B C B + A C Classes of Reactions Hydrogen Displacement Metal Displacement Halogen Displacement Single displacement is a reaction in which one element displaces another in a compound. The general pattern is: Single displacement reactions are typically one of three types: 2 Mg + Ti Cl 4 2 Mg Cl 2 + Ti Cl 2 + 2K Br 2K Cl + Br 2 Sr + 2 H 2 O Sr (OH) 2 + H 2
  • eg: 2 Al + 3 Br 2 2 AlBr 3 3. Synthesis: A + B AB Classes of Reactions H 2 O + SO 3 H 2 SO 4 A synthesis reaction is one in which two substances react and combine to form one substance. The general pattern of the reaction is: The reactants can be elements, compounds, or one of each.
  • eg: 2 KClO 3 2 KCl + 3 O 2 4. Decomposition: Decomposition is the opposite of synthesis: one substance decomposes (often by heating it) into two or more new substances. The general pattern is: 2 NaHCO 3 Na 2 O + 2 CO 2 + H 2 O Classes of Reactions AB A + B Δ Δ
  • Classes of Reactions 5. Combustion: eg: 2 Mg + O 2 2 MgO Technically, any reaction involving oxygen is a combustion reaction. Note that this reaction can also be classified as a synthesis reaction. A + O 2 A O x
  • C n H m + O 2 CO 2 + H 2 O Combustion reactions that involve hydrocarbon compounds (which may or may not contain O or N) reacting with oxygen gas will form CO 2 and H 2 O. The general reaction is: C 2 H 6 O + 3 O 2 2 CO 2 + 3 H 2 O Incomplete combustion forms CO (carbon monoxide). Classes of Reactions eg: CH 4 + 2 O 2 CO 2 + 2 H 2 O
  • Classify the following reactions:
    • 2 Na + 2 H 2 O 2 NaOH + H 2
    • Ba(NO 3 ) 2 + Na 2 CrO 4 BaCrO 4 + 2NaNO 3
    • C 4 H 10 + 13 O 2 8 CO 2 + 10 H 2 O
    • Ca(OH) 2 + H 2 S ( aq ) CaS + H 2 O
    • 4 Cu + O 2 2 Cu 2 O
    • K 2 CO 3 K 2 O + CO 2
    • NaHCO 3 + HCl ( aq ) NaCl + H 2 O + CO 2
    SD DD (P) C DD (N) S D DD/D Classes of Reactions
  • We are now going to look at each class of reaction in more detail. We will begin with the sub-categories of the double displacement reaction
  • Precipitation Reactions
  • Solubility is the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature. Precipitation Reactions If a substance dissolves in a solvent, it is said to be soluble ; if it does not, it is insoluble. Substances that are insoluble have a stronger attraction towards each other than they do towards the solvent (or sometimes, the solvent molecules have a stronger attraction towards themselves than towards the solute particles). Precipitate – insoluble solid that separates from solution One of the most common types of double displacement reaction is the precipitation reaction.
  • Precipitation Reactions PbI 2 precipitate of PbI 2 When aqueous solutions of Pb(NO 3 ) 2 and KI are mixed, a bright yellow precipitate of PbI 2 forms. Pb 2+ and I - form strong attractions Pb(NO 3 ) 2 ( aq ) + 2KI ( aq ) PbI 2 ( s ) + 2KNO 3 ( aq )
  • How do you know which substances are soluble and which form precipitates in aqueous solutions? There is a Table of general Solubility Rules on page 123 in your textbook. AP Chemistry students are required to memorize this list. (You need to know it for the AP Test!) Non-AP students are NOT required to memorize this list; it will be provided to you on tests and quizzes. Solubility Rules Precipitation Reactions
  • see page 123 Precipitation Reactions and acetates ( C 2 H 3 O 2 - )
  • While most double displacement reactions involve the formation of a precipitate, some double displacement reactions involve “dissolving” an insoluble compound by forming a soluble salt. These types of reactions most often involve the reaction of an insoluble ionic compound with an acid: AgCl (s) + HC 2 H 3 O 2 (aq) AgC 2 H 3 O 2 (aq) + HCl (aq) PbS (s) + 2 HNO 3 (aq) Pb(NO 3 ) 2 (aq) + H 2 S (g) Precipitation Reactions
  • Molecular and net ionic equations Precipitation Reactions The complete balanced equation, showing the formulas for each reactant and species is called the “molecular equation.” We can also show how each species in the reaction will dissociate or ionize when dissolved in water. Insoluble compounds do not dissociate, but soluble compounds will. We call this the “ionic equation.” molecular equation Pb(NO 3 ) 2 ( aq ) + 2NaI ( aq ) PbI 2 ( s ) + 2NaNO 3 ( aq ) ionic equation Pb 2+ + 2NO 3 - + 2Na + + 2I - PbI 2 ( s ) + 2Na + + 2NO 3 -
  • Molecular and net ionic equations Those species which appear unchanged on both sides of the yields sign did not “participate” in the reaction. They are said to be spectator ions. The net ionic equation shows only those species which actually participated in the reaction – all the spectator ions are cancelled out and not shown. net ionic equation In the above reaction, Na + and NO 3 - are the spectator ions – they do not participate in the net reaction. Pb 2+ ( aq ) + 2I − ( aq ) PbI 2 ( s ) be sure to include the phases, etc. in net ionic equations! Pb 2+ + 2NO 3 − + 2Na + + 2I − PbI 2 ( s ) + 2Na + + 2NO 3 −
  • Writing Net Ionic Equations
    • Write the balanced molecular equation.
    • Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions.
    • Cancel the spectator ions on both sides of the ionic equation
    • Check that charges and number of atoms are balanced in the net ionic equation
    Net Ionic Equations
  • example: Write the net ionic equation for the reaction of silver nitrate with sodium chloride. Net Ionic Equations molecular ionic net ionic AgNO 3 ( aq ) + NaCl ( aq ) AgCl ( s ) + NaNO 3 ( aq ) Ag + + NO 3 − + Na + + Cl − AgCl ( s ) + Na + + NO 3 − Ag + ( aq ) + Cl − ( aq ) AgCl ( s ) CuS (s) + 2 HC 2 H 3 O 2 ( aq ) Cu(C 2 H 3 O 2 ) 2 (aq) + H 2 S (g) CuS (s) + 2 H + (aq) Cu 2+ (aq) + H 2 S (g) CuS (s) + 2 H + + 2 C 2 H 3 O 2 − Cu 2+ + 2 C 2 H 3 O 2 − + H 2 S (g) example: Write the net ionic equation for the reaction of CuS with acetic acid molecular ionic net ionic
  • Note that if both reactants and products exist as solvated ions in solution, then NO REACTION has occurred – you began with a mix of hydrated ions, and you ended with the same mix of hydrated ions… K + + NO 3 − + Na + + Cl − K + + Cl − + Na + + NO 3 − They are ALL spectator ions! We write: Net Ionic Equations example KNO 3 ( aq ) + NaCl ( aq ) KCl ( aq ) + NaNO 3 ( aq ) KNO 3 ( aq ) + NaCl ( aq ) N.R.
  • Acid-Base Reactions H 3 O + = hydronium ion H H H O +
  • Arrhenius and Br  nsted-Lowry Definitions There are several “definitions” of acids or bases, from a chemical standpoint. The two most important definitions are those given by Svante Arrhenius in the late 19th century, and by J.N. Br  nsted and Thomas Lowry , who independently developed similar chemical descriptions of acids and bases in the 20th century. A very common type of double displacement reaction involves the neutralization of an acid with a base. Acid-Base Reactions
  • An Arrhenius acid is a substance that ionizes to produce H + ions in water An Arrhenius base is a substance that dissociates to produce OH - ions in water Acids and Bases i.e., Arrhenius bases are metal hydroxides that are soluble in water. H 2 O NaOH (s) Na + (aq) + OH ¯ (aq) H 2 O HCl (g) H + (aq) + Cl ¯ (aq) H 2 O
  • An H + ion is essentially a bare proton – this is an extremely reactive species! H + ions will instantly bond to a water molecule to form the polyatomic cation, H 3 O + , called the “hydronium ion.” Acids and Bases So, actually, an Arrhenius acid is a substance that produces H 3 O + ions in water .
  • A Br ø nsted-Lowry acid is a proton (H + ) donor A Br ø nsted-Lowry base is a proton (H + ) acceptor Acids and Bases Brønsted-Lowry made use of the fact that H + ions are essentially just a proton in their definition of acids and bases: B-L acid and base is a somewhat more “general” definition, since it does not require the presence of water as a solvent. However, one can certainly have an aqueous B-L acid or base!
  • acid base In the forward direction, water acts as the proton donor and NH 3 the acceptor… acid base Consider the reaction between NH 3 and H 2 O: … in the reverse direction, NH 4 + is the proton donor and OH - is the acceptor. Acids and Bases
  • Identify each of the following species as a Br ø nsted acid, base, or both. (a) HI, (b) OH - (c) HPO 4 2- acid base acid base Acids and Bases note that HPO 4 2- can act as both an acid or a base! Such substances are said to be amphoteric. HI (g) + H 2 O H 3 O + ( aq ) + I − ( aq ) OH − ( aq ) + H + ( aq ) H 2 O HPO 4 2− + H 2 O H 3 O + ( aq ) + PO 4 3− ( aq ) HPO 4 2− + H 3 O + H 2 PO 4 − ( aq ) + H 2 O
  • Monoprotic acids Acids and Bases Diprotic acids Triprotic acids Acids with only one ionizable H + are said to be monoprotic acids Acids with 3 ionizable H + are said to be triprotic acids . Acids with 2 ionizable H + are said to be diprotic acids HCl H + + Cl − H 2 SO 4 H + + HSO 4 − HSO 4 − H + + SO 4 2− H 3 PO 4 H + + H 2 PO 4 − H 2 PO 4 − H + + HPO 4 2− HPO 4 2− H + + PO 4 3− HCN H + + CN −
  • Neutralization Reaction A neutralization reaction is a special type of double displacement reaction in which an acid reacts with a hydroxide ion (base) to produce an ionic “salt” and water. Acids and Bases acid + base salt + water HCl ( aq ) + NaOH ( aq ) NaCl ( aq ) + H 2 O H + + Cl − + Na + + OH − Na + + Cl − + H 2 O note that the net ionic equation for all neutralization reactions will be H + + OH¯ H 2 O ! H + + OH − H 2 O net ionic =
  • Neutralization Reaction Examples: 2 HC 2 H 3 O 2 + Ca(OH) 2 2 H 2 O + Ca(C 2 H 3 O 2 ) 2 HCN ( aq ) + KOH H 2 O + KCN H 2 SO 4 + 2 NaOH 2 H 2 O + Na 2 SO 4 Acids and Bases
  • Not all reactions involving acids and bases are neutralization reactions . For example, ammonia (NH 3 ), a Br  nsted-Lowry base, can react with acids to form aqueous ammonium salts. Although this IS an acid base reaction, technically it is NOT a neutralization reaction: The aqueous NH 4 Cl formed can react with the water present in the solution to produce H 3 O + : Acids and Bases HCl ( aq ) + NH 3 NH 4 Cl ( aq ) NH 4 + + H 2 O NH 3 + H 3 O +  sol’n is still acidic!
  • examples: Na 2 CO 3 + 2 HCl ( aq ) 2 NaCl (aq ) + H 2 O + CO 2 ( g ) K 2 SO 3 + 2 HBr ( aq ) 2 KBr ( aq ) + H 2 O + SO 2 ( g ) PbS + 2 HI ( aq ) PbI 2 (s) + H 2 S ( g ) Acids produce gases when they react with certain salts containing carbonate, bicarbonate, sulfite and sulfide ions. Note that the reaction with PbS and HI is a simple double displacement reaction. The others involve both double displacement AND decomposition! Other Reactions with Acids
  • Basic and Acidic Oxides Basic oxides are metal oxides . The name “basic oxide” come from the fact that metal oxides react with water in a synthesis reaction to form hydroxides: Acidic oxides are non-metal oxides . The name “acidic oxide” comes from the fact that non-metal oxides react with water in a synthesis reaction to form oxoacids. eg: H 2 O + Na 2 O 2 NaOH eg: H 2 O + SO 3 H 2 SO 4
  • The reaction of water with SO 3 is how acid rain forms. Sulfur trioxide is a common pollutant from burning coal. The SO 3 reacts with water droplets in the air to form sulfuric acid. Basic and Acidic Oxides Basic oxides can react with acidic oxides in a type of acid-base reaction: base acid salt H 2 O + SO 3 H 2 SO 4 Na 2 O + SO 3 Na 2 SO 4
  • Some oxides are somewhat acidic and somewhat basic. They are said to be amphoteric oxides. Aluminum oxide is a good example: basic oxides amphoteric oxides acidic oxides Amphoteric oxides are oxides of metals at the far right end on the periodic table Basic and Acidic Oxides Al 2 O 3 + 6 HCl 2 AlCl 3 + 3 H 2 O Al 2 O 3 + 2 KOH 2 KAlO 2 + H 2 O
  • The reaction occurs because the metal oxide first reacts with water to form hydroxides, which in turn reacts with the acid in a neutralization reaction: 2 HC 2 H 3 O 2 ( aq ) + CaO ( s ) Ca(C 2 H 3 O 2 ) 2 ( aq ) + H 2 O ( l ) Aqueous acids react with basic oxides . Not surprisingly, since metal oxides are basic, they react with acids in a double displacement reaction that is essentially a neutralization reaction: Basic and Acidic Oxides CaO ( s ) + 2 HC 2 H 3 O 2 ( aq ) Ca(C 2 H 3 O 2 ) 2 ( aq ) + H 2 O ( l ) net: i. CaO + H 2 O Ca(OH) 2 ii. Ca(OH) 2 + 2 HC 2 H 3 O 2 Ca(C 2 H 3 O 2 ) + 2 H 2 O
  • 2 NaOH (aq) + N 2 O 5 (g) 2 NaNO 3 (aq) + H2O ( l ) Aqueous bases react with acidic oxides . Similarly, since non-metal oxides are acidic, they react with bases in a double displacement reaction that is essentially a neutralization reaction: The reaction occurs because the non-metal oxide first reacts with water to form an oxoacid, which in turn reacts with the hydroxide ion in a neutralization reaction: i. N 2 O 5 + H 2 O 2 HNO 3 ii. 2 HNO 3 + 2 NaOH 2 NaNO 3 + 2 H 2 O Basic and Acidic Oxides N 2 O 5 (g) + 2 NaOH (aq) 2 NaNO 3 ( aq ) + H 2 O ( l ) net:
  • Acid-Base Titrations
  • In a titration, a solution of accurately known concentration is gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color (this is called the end point ). Indicators must be carefully chosen so that the end point occurs at the equivalence point in the titration. Slowly add base to unknown acid UNTIL the indicator changes color Titrations
  • example: A 24.00 ml sample of an unknown acid is titrated to a phenolphthalein end point, which required 31.46 ml of a 0.104 M solution of NaOH. What is the molar concentration of the unknown acid? ii. M A V A = moles acid = moles base = M B V B i. At the end point (equivalence point) the moles of acid = moles of added base: M B V B 0.102 M x 31.46 ml V A 24.00 ml iii. M A = = = 0.134 M
  • What volume of a 1.420 M NaOH solution is required to titrate 25.00 mL of a 4.50 M H 2 SO 4 solution to the eq. point? Write the chemical equation: H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4 25.00 mL = 158 mL M A V A = moles acid* (H + ) = moles base (OH - ) = M B V B Note that H 2 SO 4 is diprotic, so moles H + = 2 x moles H 2 SO 4 volume acid moles acid moles base volume base 4.50 mol H 2 SO 4 1000 mL soln x 2 mol NaOH 1 mol H 2 SO 4 x 1000 ml soln 1.420 mol NaOH x M acid mol ratio M base
  • Oxidation–Reduction Reactions
  • In precipitation reactions, there is a transfer of ions between two compounds. In acid-base reactions , there is a transfer of protons between acids and bases. In the next grouping of reactions, called redox reactions , there is a transfer of electrons between reacting species.
  • The term redox is a shortened form of oxidation-reduction. Redox Reactions Oxidation is a process in which one atom loses or donates electrons to another. Reduction is a process in which one atom gains or accepts electrons from another. In any redox reaction, one species is ALWAYS oxidized and one is ALWAYS reduced. You cannot have oxidation without reduction also occurring.
  • Mnemonic Aid : to help remember these two definitions, use this aid: LEO = L oss of E lectrons is O xidation; GER = G ain of E lectrons is R eduction Redox Reactions LEO says GER
  • In fact, nearly every class of reaction we have already looked at, with the exception of double displacement reactions, is a form of a redox reaction. Redox reactions are a very important class of reactions. Everything from burning fossil fuels to the action of household bleach on stains is a redox reaction. In addition, most metals and nonmetals are obtained from their ores by redox chemistry. Redox Reactions
  • Redox reactions often involve ions, but despite the fact that we speak of electrons being transferred, redox reactions do not always involve ions. Redox Reactions In all cases, however, this loss or gain of electrons describes the difference in the electron density around a bonded atom in a compound, compared to the electron density in that atom’s elemental, un-bonded state.
  • The oxidation number of an atom is simply the “charge” the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. Note carefully that this is usually only a “pretend” complete electron transfer, and has meaning only as a book-keeping tool in accounting for shifts in the electron density around bonded atoms! We assign an oxidation number to an atom to indicate the relative electron density in its current state, compared to that in its elemental state. Redox Reactions
  • For ionic compounds , cations are assigned positive oxidation numbers because they really have lost electrons compared to the electrons present in their elemental state. Similarly, anions really have gained electrons, compared to their elemental state, and are assigned negative oxidation numbers. Redox Reactions
  • For non-ionic compounds , however, a negative oxidation number simply means the atom has a greater electron density in its bonded state than the atom has in its elemental, un-bonded state. A positive oxidation number means the atom has less electron density around it in its bonded state than when in its elemental, un-bonded state. Redox Reactions
  • Rules for Assigning Oxidation Numbers
    • The oxidation state of any neutral element in its naturally occurring state is zero.
    • The oxidation number of any cation or anion composed of
    • just one atom is that ion’s actual charge.
    Na, Be, K, Cl 2 , H 2 , O 2 , P 4 = 0 Li + = +1 ; Fe 3+ = +3 ; O 2- = -2
    • The oxidation number of oxygen is usually –2 . In H 2 O 2 and O 2 2- it is –1 .
    • The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1 .
    • The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.
    • Oxidation numbers do not have to be integers. Oxidation number of oxygen in the superoxide ion, O 2 - , is -½ .
    Rules for Assigning Oxidation Numbers continued
    • When you begin, assign the most electronegative element present the charge it would have if it were an anion. If oxygen is present, it has the higher priority and will always be –2.
    • The oxidation number of F is always -1 .
  • Example : Consider the compound PCl 3 : Cl is the more electronegative element, so assign each Cl an oxidation # equal to its charge as an anion (= –1). PCl 3 is a neutral molecule, so the sum of the oxidation numbers for P and Cl must add to 0. Let x = oxidation # of P: x + 3(–1) = 0  x = +3 PCl 3 : P = +3, Cl = -1 Redox Reactions
  • Since oxygen is present, but not as a peroxide ion, the oxygen is assigned an oxidation # of –2. The H is not bonded to a metal, so it must have an oxidation # of +1. Let x = oxidation number of Cl, and note that the sum of the oxidation #’s must equal zero since HClO 4 has a net charge of zero: 0 = +1 + x + 4(–2) solving for x gives x = +7 = Cl. HClO 4 : H = +1, Cl = +7, O = -2 Redox Reactions example: Consider HClO 4 . Note carefully that Cl is NOT a +7 cation in HClO 4 !! The +7 oxidation state simply tells us the electron density around Cl in HClO 4 is significantly lower than the electron density around Cl in its elemental state.
  • Example: Consider Cr 2 O 7 2– -2 = 2 x + 7(–2) or x = +6 Oxidation number for oxygen is: O = -2 The sum of the charges in the compound = the charge of the polyatomic ion = -2 Let x = oxidation number of Cr. Solving for x gives: Thus: O = -2 and Cr = +6 Redox Reactions
  • Common oxidation numbers of elements in their compounds
  • If an element’s oxidation number has increased in a reaction, this means the element has lost electrons and has been oxidized. If an element’s oxidation number has decreased , this means it has gained electrons in the reaction and has been reduced. Oxidizing and Reducing Agents Example: 2Mg + O 2 2MgO 2Mg 2Mg 2+ Mg has been oxidized O 2 2O 2- O has been reduced 0 0 Redox Reactions
  • The element that was oxidized “donated” its electrons to the element that was reduced (gained electrons). Thus, the species that contains the element being oxidized is said to be the reducing agent . The element that was reduced “stole” electrons from the element that was oxidized (lost electrons). Thus, the species that contains the element being reduced is said to be the oxidizing agent. Oxidizing and Reducing Agents The more readily an element is oxidized, the better it is as a reducing agent. Conversely, the more readily an element is reduced, the better it is as an oxidizing agent. Redox Reactions
  • 1. Zn ( s ) + CuSO 4 ( aq ) ZnSO 4 ( aq ) + Cu ( s ) Zn is oxidized Cu 2+ is reduced Zn is the reducing agent CuSO 4 is the oxidizing agent 2. 2 K ( s ) + 2 H 2 O ( aq ) 2 KOH ( aq ) + H 2 (g) H + is reduced H 2 O is the oxidizing agent Example: Identify the oxidizing agents and reducing agents in the following reactions: K is oxidized K is the reducing agent Oxidizing and Reducing Agents H + H o Zn Zn 2+ Cu 2+ Cu K K +
  • One element is simultaneously oxidized and reduced. 0 +1 -1 Disproportionation Reaction Redox Reactions One chlorine atom is being oxidized to a +1 oxidation number, and at the same time, the other Cl atom is being reduced to a -1 oxidation number. Note that the O and H are neither oxidized nor reduced in this reaction. Cl 2 + 2OH − ClO − + Cl − + H 2 O
  • Redox and Reaction Classification Almost every reaction that is not a double displacement reaction is a redox reaction. 0 0 2+ 2− 1+ 2− 0 0 4− 1+ 0 4+ 2− 1+ 2− Single Displacement: Mg + NiCl 2 MgCl 2 + Ni 0 2+ 1− 2+ 1− 0 Synthesis: Fe + S FeS Decomposition: 2 H 2 O 2 H 2 + O 2 Combustion: CH 4 + 2 O 2 CO 2 + 2 H 2 O
  • Predicting the products of redox reactions can be a little tricky. We will focus primarily on single displacement reactions In single displacement reactions, an element displaces another element in a compound. This means that one element must be oxidized, and the other reduced, in order to accomplish this.
  • 2 Ag (s) + CuCl 2 (aq) 2 AgCl (s) + Cu (s) Some elements are better oxidizing agents than others, and so not every element is able to oxidize another. We may be able to write an equation, but the reaction may not actually occur if we try it: This reaction does occur; the Cu 2+ ion can oxidize Zn This reaction does NOT occur; the Cu 2+ ion cannot oxidize Ag So…How can you tell if a given reaction will occur? Zn (s) + CuCl 2 (aq) ZnCl 2 (aq) + Cu (s)
  • To determine if a single displacement reaction can occur, we need to know the relative strengths of elements as oxidizing agents. This table of values is called an activity series. The table at right lists common metals in order of decreasing strength as reducing agents – or increasing strength as oxidizing agents. Activity Series of Metals strong reducing agents poor reducing agents
  • Hydrogen Displacement Reaction all metals above Cu will react with acids reactive metals above Mg can react with cold water metals below Cu do not react even with acids The Activity Series for Metals Single Displacement Rxns Ni + 2 HCl NiCl 2 + H 2 Ca + 2H 2 O Ca(OH) 2 + H 2 Au + HCl N.R.
  • Metal Displacement Reaction no reaction occurs if you try to displace a metal with a metal that lies above it in the table a metal can displace any metal that lies below it in the table Single Displacement Rxns The Activity Series for Metals Mg + CdS Cd + MgS Cu + NiCl 2 N.R.
  • Halogens , unlike metals, are poor reducing agents . However, they make good oxidizing agents. strongest oxidizing agent weakest oxidizing agent A halogen can displace any halogen below it in the activity series Halogen Displacement Reaction Activity Series of Halogens I 2 Br 2 Cl 2 F 2 Cl 2 + 2KBr 2KCl + Br 2 I 2 + 2KBr 2KI + Br 2
  • Predicting Products of Reactions
    • Precipitation reactions always involve reacting two ionic compounds (exchange cations).
    • Neutralization reactions always involve reacting an acid with a metal hydroxide (exchange H + and metal cation)
    In order to predict the products of a reaction, keep in mind the following points:
    • Synthesis reactions – put all the elements together – does the formula look like a compound you recognize? Try changing the subscripts by multiplying or dividing them by an integer – now does the formula look familiar?
    • Remember the synthesis reactions of acidic and basic oxides with water, and with each other!
      • Decomposition reactions – typically, there will be only one compound as the only reactant. Break it down into two compounds or two elements, etc.
      • Combustion reactions – look for hydrocarbons reacting with elemental O 2 to form CO 2 + H 2 O
      • Single displacement reactions – almost always involve a metal replacing another metal in a compound, or a metal replacing H in water or acids. Also, halogens can displace other halogens.
    Predicting Products of Reactions continued Be sure to use the activity series to determine if the rxn will occur at all. Also remember that alkali and alkaline earth metals displace H from water to form hydroxides , not oxides!
  • example: predict if a product will form. If so, complete and balance the reaction. 1. Na + Fe(NO 3 ) 3 2. Ni + CdBr 2 3. Pb + H 2 O 4. Ba + H 2 O 5. KCl + F 2 2 NaNO 3 + Fe N.R. N.R. Ba(OH) 2 + H 2 2 2 KF + Cl 2 2
  • The Behavior Of Gases Chapter 5 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.
  • 5.1
  • The Kinetic Molecular Theory of Gases
  • Kinetic Molecular Theory of Gases
    • A gas is composed of molecules that are widely separated from each other . The volume occupied by the molecules is but a tiny fraction of the total volume of the gas, to the extent that the gas molecules can be considered to be point masses ; that is, they possess mass but have negligible volume.
    • Gas molecules are in constant motion in random directions, and they frequently collide with one another.
  • Kinetic Molecular Theory of Gases continued
    • The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy.
    • There are neither attractive nor repulsive forces between gas molecules.
    • Collisions among molecules are perfectly elastic, that is, there is no net gain or loss of kinetic energy during these collisions.
  • The average kinetic energy (KE) of a particle depends on its mass, m , and its average velocity, u: This means the higher the temperature, the faster a particle moves. At a given temperature, low mass particles move faster than higher mass particles. Kinetic Molecular Theory of Gases * a bar over a variable just means, “the average value.” KE = ½ mu 2
  • The kinetic molecular theory of gases can be used to explain several properties and behaviors of gases:
    • gases can be compressed
    • gases have low densities
    • gases are fluids
    • gases can undergo diffusion and effusion
    Kinetic Molecular Theory of Gases
  • Properties of Gases There is a lot of empty space between the particles in the gas state. This means that gases are very compressible . A liter of gas can be compressed down to about 1 milliliter. Compressibility
  • We use the units of grams per LITER instead of g/mL for the density of gases. Because the particles of a gas are so spread out, the density of gases is usually about 1000 x less than solids and liquids. Low Density Fluidity Since there are no attractive forces between gas particles, they can “slide” past one another freely. Properties of Gases
  • Effusion and Diffusion Diffusion is the spontaneous mixing of particles due to their random motion. Gas particles move randomly and have no attractive forces between themselves that would limit their ability to mix. Effusion is the process whereby particles under pressure escape through small openings. Gas particles are very tiny, and can pass through small holes at a rate that depends on their velocities. Properties of Gases
  • When discussing gases, we must be aware of how the temperature, the volume, the number of particles, and the pressure of the gas all affect each other. The way each of these affects the other is the subject of a set of gas laws , which were investigated in countless experiments over several centuries. But first, we need to get a better understanding of this new unit, pressure. Properties of Gases
  • Pressure Pressure is defined as the force pushing on one square meter of a surface: P = Force/Area. Normal atmospheric pressure is caused by the weight of a column of air pressing down on an object. Sea level 1 atm 4 miles 0.5 atm 10 miles 0.2 atm Properties of Gases
  • We measure pressure with a barometer. The air pushes down on a bowl of mercury, which in turn pushes the mercury up a column. The height of the column depends on the air pressure – higher pressures can support a taller column of mercury. mercury Pressure Evangelista Torricelli invented the barometer in 1643. vacuum weight of mercury column
  • 1 atmosphere = 760 mm Hg 1 torr = 1 mm Hg 1 atmosphere = 101.325 kPa Units of Pressure SI Unit = pascal (Pa) = N/m 2 The conditions 0 o C and 1 atm of pressure is called standard temperature and pressure (STP).
  • Manometers Used to Measure Gas Pressures Open to the atmosphere
  • Pressure in a container is caused by the force with which particles strike the walls of a container. The pressure increases with increased frequency of collisions , since more particles striking the wall means more force. Pressure also increases if the speed of the particles increases – faster speed = greater force of impact on the walls of the container. Pressure
  • The Gas Laws
  • The Gas Laws Boyle’s law At constant temperature , the pressure of a gas in a confined space will increase when the volume decreases -- and the pressure will decrease when the volume increases. volume increases volume decreases pressure decreases pressure increases P P P
  • We say that p ressure and volume are inversely proportional to each other. Boyle’s law P  , where k = a constant k V P V Pressure goes up, volume goes down P V Pressure goes down, volume goes up
  • Graphically , P vs V produces a curved line as shown below: Boyle’s law
  • Explanation At constant temperature, the average kinetic energy of gas particles (and therefore the average speed ) is constant. If the volume increases, the distance the particles must travel between collisions with the wall increases, which means the number of collisions/sec decreases (it takes longer for particles to reach the walls of the container). This means the pressure decreases. If you decrease the volume, then the number of collisions/second in creases and the pressure goes up. Boyle’s law
  • The Gas Laws Charles’ Law At a fixed pressure , the volume of a gas will increase if the temperature increases, and the volume will decrease if the temperature decreases. P P temp increases temp decreases volume increases volume decreases P
  • V  k T , where k = a constant We say that temperature and volume are directly proportional to each other. Charles’ Law V T Volume goes up, temperature goes up V T Volume goes down, temperature goes down
  • Charles’ Law A graph of Volume vs Temperature would look like the graph below. The Kelvin scale was developed based on this graph. When the volume of a gas is reduced to zero, the temperature was found to be -273.15 o C, which was defined as Absolute Zero on the Kelvin scale.
  • http:// Explanation: As the temperature increases the particles increase in kinetic energy, moving faster and faster. They then hit the walls of the container with a greater force, causing the walls to expand. As the temperature decreases the particles decrease in kinetic energy and hit the walls of the container with less force. The external air pressure causes the walls of the container to contract. Charles’ Law
  • At constant volume , the pressure of a gas will increase with increasing temperature and decrease with decreasing temperature. Gay-Lussac & Charles’ Law Gay-Lussac & Charles’ Law temp increases temp decreases pressure increases pressure decreases P P P
  • P  k T , where k = a constant We say that temperature and pressure are directly proportional to each other. Gay-Lussac & Charles’ Law P T Pressure goes up, temperature goes up P T Pressure goes down, temperature goes down
  • http:// Gay-Lussac & Charles’ Law Explanation: As temperature increases, the average kinetic energy increases and therefore the speed increases. The particles travel the same distance in a shorter period of time, which increases the number of collisions/sec, and the pressure goes up. As temperature decreases the average kinetic energy decreases therefore speed decreases. Particles travel the same distance in longer periods of time resulting in fewer collisions/sec and a decrease in pressure.
  • Avogadro’s Law Avogadro’s Law At a fixed pressure and temperature , the volume of a gas is directly proportional to the number of gas particles (i.e., the number of moles) present. A B Gas “A” has twice as many particles as gas “B” so gas “A” has twice the volume of gas “B,” when both gases are at the same temperature and pressure. http://
  • http:// Avogadro’s Law Explanation: If the temperature and pressure of two gases are the same, then the particles in both gases are moving at the same speed, and colliding with the container walls with the same frequency. If one container’s volume is greater than the other container, then there must be more particles to increase the frequency of collisions in order to maintain an equal pressure with the smaller volume container.
  • Charles’ law: V  T  (at constant n and P ) Avogadro’s law: V  n  (at constant P and T ) V  or V = R x R is the gas constant PV = nRT The Ideal Gas Equation where R = constant of proportionality This equation is usually rearranged to: T in kelvins! Boyle’s law: V  (at constant n and T ) 1 P nT P nT P
  • Solving Gas Law Problems To see how changing one variable affects the others, we look at the ratio of the ideal gas equations “before” and “after” the changes: We then cancel out those variables (if any) being held constant, enter the values for the other variables, and solve for the desired variable’s value. The temperature MUST be in Kelvins when using the gas law equations! Note: P 2 V 2 n RT 2 = P 1 V 1 n RT 1
  • Example : A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mm Hg. What is the pressure of the gas (in mm Hg) if the volume is reduced at a constant temperature to 154 mL? We note that the temperature is held constant, and presumably, no gas enters or escapes, so n is also constant. And, of course, R is also constant. P 1 V 1 P 2 V 2 = 1 P 1 V 1 V 2 = P 2 = 726 mm Hg x 946 ml 154 ml = 4460 mm Hg ii. iii. Gas Law Problems P 1 V 1 nRT 1 P 2 V 2 nRT 2 i. =
  • Example: Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 o C is heated to 85 o C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? P 1 V 1 nRT 1 n, V and R are constant P 1 T 1 P 2 T 2 = P 1 = 1.20 atm T 1 = P 2 = ? T 2 = = 1.48 atm P 2 V 2 nRT 2 Gas Law Problems = 291 K 358 K 18 o C 85 o C P 2 = P 1 x T 2 T 1 = 1.20 atm x 358 K 291 K
  • P 2 V 2 T 2 Example: A sample of carbon monoxide gas occupies 3.20 L at 125 o C and 740 torr. At what temperature, in o C, will the gas occupy a volume of 1.54 L if the pressure is increased to 800 torr? P 1 V 1 nRT 1 P 2 V 2 nRT 2 i. We assume that no gas enters or escapes, so n is constant. And, of course, R is also constant. All other variables are changing P 1 V 1 T 1 ii. = P 1 V 1 T 1 P 2 V 2 iii. = T 2 Solve for T 2 . Watch your algebra! = x 800 torr x 1.54 L 740 torr x 3.20 L 125 o C 398.15 K iv. T 2 = 207.1 K = − 66.0 o C Gas Law Problems =
  • see pages 185-186
  • PV = nRT R = 0.082057 L • atm / (mol • K) Experiments show that at 1.0 atm of pressure and a temperature of 0 o C (STP): We use this information to determine the value for R, the ideal gas constant: 1 mole of an ideal gas occupies 22.414 L The Ideal Gas Equation R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K)
  • Note that the value for R depends on the units used for pressure , as follows: Rather than memorize three different R-values, simply always convert the pressure to atmospheres first, then use 0.08206 L·atm/mol·K for all problems. 62.36 L · torr/mol · K torr (mm Hg) 8.314 L · kPa/mol · K kilopascals 0.08206 L · atm/mol · K atmosphere Gas Constant Pressure units
  • Example: What is the volume (in liters) occupied by 49.8 g of HCl (g) at STP? iii. PV = nRT iv. V = i. at STP, T = 0 0 C = 273.15 K, and P = 1.00 atm ii. n = 49.8 g x = 1.37 mol 1.00 atm 1.37 mol x 0.08206 x 273.15 K = 30.6 L The Ideal Gas Equation = nRT P 1 mol HCl 36.45 g HCl L•atm mol•K
  • Density of Gases at various T and P i. d = ii. but V = nRT/P from the ideal gas equation… iii. d = m nRT/P m x P nRT m n x P RT = = = But m/n = mass/mole which is the molecular weight (MW) of the gas. We can write: The Ideal Gas Equation m V d = x P RT MW
  • Molar Mass (MW) of a Gaseous Substance d is the density of the gas in g/L Finding the Molecular Weight of an Unknown Gas We can easily determine the density of an unknown gas experimentally. More importantly, knowing the density, we can rearrange the equation we just developed to solve for the molecular weight of the unknown gas dRT P MW =
  • Example: A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 o C. What is the molar mass of the gas? i. MW = ii. d = = 2.21 g L iii. MW = 1 atm x 0.08206 x 300.15 K Molar Mass of Unknown Gas dRT P m V 4.65 g 2.10 L = 2.21 g L L•atm mol•K MW = 54.4 g/mol
  • Gas Stoichiometry PV = nRT
  • We can make use of Avogadro’s Law in solving gas stoichiometry problems – we can measure volume ratios to determine mole ratios , if all the gases have the same temperature and pressure. Avogadro’s Law
  • Example: Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many liters of O 2 are required to completely react with 2.4 liters of ammonia at the same temperature and pressure? Recall: at constant T and P, n  V  4 volumes NH 3 5 volumes O 2  2.4 L NH 3 x 5 L O 2 4 L NH 3 = 3.0 L O 2 required Mole ratios: 4 mole NH 3 5 mole O 2 Avogadro’s Law rxn: 4 NH 3 + 5 O 2 4NO + 6H 2 O
  • We can also solve gas stoichiometry problems in a way very similar to the stoichiometry problems we have been doing – that is, convert to moles, use the mole:mole ratio from the balanced equation, etc. The ideal gas equation is simply yet another way of finding moles! Or converting moles to pressure, or moles to volumes, or… Gas Stoichiometry n = PV RT
  • Example: What volume of N 2 gas at STP conditions is required to react completely with 24.00 liters of H 2 at 0.974 atm and 24.5 o C? i. As always, convert to moles first, then use the mole:mole ratio, then convert to the desired unit. ii. n H2 = = 0.974 atm x 24.00 L 0.08206 x 297.65 K = 0.957 mol H 2 3 mol H 2 1 mol N 2 = 0.3190 mol N 2 = 7.15 L PV RT L atm mol K iii. From bal eqn: 0.957 mol H 2 x Rxn: N 2 + 3 H 2 2 NH 3 iv. V = = P nRT 0.3190 mol x 0.08206 x 273.15K L atm mol K 1.00 atm
  • Gas Stoichiometry Example: What is the volume of CO 2 produced at 37 o C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 ( s ) + 6O 2 ( g ) 6CO 2 ( g ) + 6H 2 O ( l ) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 i. 5.60 g C 6 H 12 O 6 = 0.187 mol CO 2 ii. V = = 4.76 L 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x nRT P 0.187 mol x 0.08206 x 310.15 K L•atm mol•K 1.00 atm =
  • Dalton’s Law of Partial Pressures P 1 P total = P 1 + P 2 P 2 The total pressure of a gas in a container is equal to the sum of the “partial pressures” of each component gas in the container. + V and T are constant
  • Collecting a Gas over Water A common method of collecting a gas is to collect it “over water” using a pneumatic trough . The gas collected will be “contaminated,” however, by a small amount of water vapor since the gas was collected over water. We use the Law of Partial Pressures to find the pressure of the “dry” gas. The volume of the water displaced is equal to the volume of the gas collected, and the temperature of the water is equal to the temperature of the gas. 2KClO 3 ( s ) 2KCl ( s ) + 3O 2 ( g )
  • After collecting the gas, the bottle is raised or lowered until the level of water in the collection bottle is even with the water in the pneumatic trough. At this point, the atmospheric pressure equals the total pressure of the gas + water vapor in the collection bottle. We now can find the pressure of the “dry” gas collected: P gas + P water = P total = P atm Collecting a Gas over Water continued gas + H 2 O = atm
  • Collecting a Gas over Water continued Example: A sample of KClO 3 was decomposed in the lab to produce 182 ml of oxygen gas: 2 KClO 3 2 KCl + 3 O 2 . The oxygen was collected over water. Atmospheric pressure in the room at the time was 731.0 torr. The temperature of the water (= temp of the gas) was 20.0 o C. According to the table on page 196, the vapor pressure of water at this temperature is 17.54 torr. How many moles of oxygen were collected?
    • n =
    • P gas = P atm – P water = 731.0 – 17.54 = 713.46 torr
    iii. P gas = 713.46 torr x 1 atm 760 torr = 0.9388 atm 0.9388 atm x 0.182 L 0.08206 x 293.15 K L atm mol K iv. n = = 7.10 x 10 -3 moles PV RT
  • see page 196
  • Mole Fraction The mole fraction ( X ) of a mixture is the ratio of the number of moles of one component to the total number of moles of particles in the mixture: Consider a mixture of two gases, A and B. Then i. P T = ii. P T = x (n A + n B ) n A = n A RT V multiply by X A gives: = P A  P A = X A P T n A n A + n B + … mole fraction ( X A ) = (n A + n B ) x RT V (n A + n B ) x RT V
  • Example : A sample of natural gas contains 8.24 moles of CH 4 , 0.421 moles of C 2 H 6 , and 0.116 moles of C 3 H 8 . If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = = 0.0132 P propane = 0.0132 x 1.37 atm = 0.0181 atm Mole Fraction i. ii. iii. 0.116 8.24 + 0.421 + 0.116
  • Example: Recall the sample problem in which KClO 3 was decomposed in the lab to produce 182 ml of oxygen gas. The oxygen was collected over water. Atmospheric pressure in the room at the time was 731.0 torr. The temperature of the water (= temp of the gas) was 20.0 o C. According to the table on page 196, the vapor pressure of water at this temperature is 17.54 torr. What was the mole fraction of oxygen in the collection bottle?
    • P gas = P atm – P water = 731.0 – 17.54 = 713.46 torr
    ii. P gas = P T X gas = 0.9760 Mole Fraction iii. X gas = P T P gas 713.46 torr 731.0 torr =
  • More Gas Mathematics u rms = 3 RT MW 
  • Distribution of Molecular Speeds James Maxwell (1831-1879) analyzed the proportion of gas particles in a sample moving at a given speed and obtained a graph similar to the one below, called a Maxwell Distribution Curve. The peak of each graph gives the most probable velocity of the particles – that is, the speed of the largest number of particles. Speed Distribution for Nitrogen at Different Temperatures Note that the most probable speed increases with increasing temperature , and as the temperature increases, the peaks shift and flatten out.
  • It can be shown that the total kinetic energy of one mole of gas is equal to ( )RT Root Mean Square Speed 3 2 u rms = 3 RT MW  Since the average KE of one molecule is ½ mu 2 , we can say that the total KE of a gas sample is given by: iii.  = ½ MW u 2 3RT 2 i. KE total = 3 2 RT = N A (½ mu 2 ) ; m = mass of one molecule ii. but N A m = # molecules mole x grams molecule = grams mole = MW iv.  = u 2 , OR 3RT MW
  • The distribution of speeds of three different gases at the same temperature Root Mean Square Speed continued Note that the speed is inversely proportional to the molar mass of the molecule. u rms = 3 RT MW 
  • Gas Diffusion and Effusion Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. The rate at which a gas diffuses depends on its velocity. Particles moving at a higher rate of speed will diffuse faster than slow moving particles. u rms = 3 RT MW  Recall that the velocity of a gas is inversely proportional to it’s molar mass:
  • If we compare the rate at which two gases diffuse at the same temperature, we obtain the relation: This reduces to Graham’s Law of Diffusion u rms A = 3 RT MW A  u rms B = 3 RT MW B  rate A rate B = 1 MW A  1 MW B  rate A rate B = MW B MW A 
  • NH 3 17 g/mol HCl 36 g/mol Example: Compare the rate at which NH 3 and HCl diffuse at the same temperature. put another way, NH 3 diffuses 1/0.687, that is, NH 3 diffuses 1.46 times faster than HCl. so HCl diffuses 0.687 times as fast as NH 3 … Graham’s Law of Diffusion NH 4 Cl rate HCl rate NH3 = MW NH3 MW HCl  36 g/mol  = 17 g/mol = 0.687
  • Gas effusion is the is the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening. As with diffusion, the rate of effusion is inversely proportional to the molar mass of the gas. r A r 2 MW B MW A  =
  • Example: Nickel forms a gaseous compound of the formula Ni(CO) x . What is the value of x, that is, what is the actual formula of the Ni compound, given that under the same conditions, methane (CH 4 ) effuses 3.26 times faster than the compound? Let A = CH 4 and B = Ni(CO) x ii. iii. 170.5 = Ni + x(CO) = 58.70 + x(28.01) iv. x = = 3.99 ~ 4  formula is Ni(CO) 4 MW B = 3.26 2 x 16.04 = 170.5 rate A rate B = MW B MW A  16.04 g/mol  = MW B 3.26 1 = i. 170.5 – 58.70 28.01
  • Non-Ideal Behavior
  • Deviations from Ideal Behavior Ideal gases have no attractive or repulsive forces acting between molecules, and ideal gas molecules have essentially zero volume. Real world gases do have attractive and repulsive forces and their molecules do occupy some space (albeit only a small space).
  • Effects of Intermolecular Attractions When intermolecular attractions come into play, the particles begin to “stick” together. This effectively reduces the total number of particles colliding with the container walls, which lowers the pressure. In addition, the attractive forces reduces the speed of the particles impacting the walls, which reduces the force at impact, which again lowers the pressure.
  • THUS, the measured pressure of a REAL gas will always be slightly LESS than the ideal pressure that gas should have. A corrective factor must be made in the gas equation for the pressure of a real world gas: P true = P measured + a is a constant that depends on the individual molecule. The likelihood of forming attractions increases with increasing particle density (moles/liter = n/V) which explains the second term. a n 2 V 2
  • Effects of Particle Volume Since a gas molecule does, in fact, occupy some volume, the total volume available for any one molecule will be LESS than the total volume of the gas container. measured volume true volume At sufficiently high concentrations, and for sufficiently small volumes, this can have an impact on the observed behavior of the gas.
  • For real gases, the TRUE volume the gas occupies is always slightly LESS than the container’s volume. A corrective term must be subtracted from the measured volume of an ideal gas: V true = V measured – nb n = # moles of the gas and b = a constant that is different for each molecule (essentially, it gives the volume per molecule)
  • Van der Waals Equation The Dutch physicist J.D. Van der Waals was the first to take into account the effects of these attractive forces and volume corrections. The “corrected” form of the ideal gas equation is thus: Van der Waals equation nonideal gas P + ( V – nb ) = nRT an 2 V 2 ( ) corrected pressure corrected volume
  • At sufficiently high temperatures (T ≥ 0 o C), the particles have enough kinetic energy to overcome the weak attractions that exist between gas phase molecules, to the point that these attractions can be ignored. What good is the Ideal Gas Equation? At sufficiently low pressures (< 5 atm) the particle density is low enough that the particles are rarely close enough for attractive forces to form, and so this effect can also be ignored.
  • Thus, gases at temperatures at or above 0 o C and with pressures below 5 atm do, in fact, behave as ideal gases, that is, the corrective terms are negligible (insignificant). Most gases that we work with under ordinary lab conditions do fall within these limitations, so most gases behave ideally! What good is the Ideal Gas Equation?
  • Chapter 6 Thermochemistry
  • Energy is the capacity to do work . To do work is to apply a force (F) to an object and cause the object to move through some distance (d): Energy Changes in Reactions units for work (energy): Nm = Joule (J) W = Fd
  • Other units for (thermal) energy: calorie (c) = (heat) energy required to raise the temperature of 1.00 gram of water by 1ºC. Food Calorie (C) –capital “C” is not the same as a little “c” calorie. There are 1000 calories in a food Calorie (that is, a food Calorie is a kilocalorie.) 1 calorie = 4.184 joules Energy
    • Radiant energy comes from the sun and is earth’s primary energy source
    • Thermal energy is the energy associated with the random motion of atoms and molecules ( a form of kinetic energy)
    • Chemical energy is the energy stored within the bonds of chemical substances
    • Nuclear energy is the energy stored within the collection of neutrons and protons in the atom
    • Potential energy is the energy available by virtue of an object’s position
    Energy is always conserved in any process. It is possible, however, to convert energy from one form to another. Energy
  • Thermochemistry is the study of heat change in chemical reactions. Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the average kinetic energy of the particles that make up a substance. Note, however, that higher temperature does NOT always mean greater thermal energy. example: A 90ºC cup of coffee is at a higher temperature, but a bathtub full of 40ºC water has more thermal energy, since there are more particles in the bathtub of water! Some Definitions
  • Thermodynamics “ Thermodynamics is a funny subject. The first time you go through the subject, you don’t understand it at all. The second time you go through it, you think you understand it, except for one or two small points. The third time you go through it, you know you don’t understand it, but by that time, you are so used to the subject that it doesn’t bother you anymore .”
    • Arnold Sommerfield (1868-1951) Well known German physicist and textbook author; response as to why he had never written a book on thermodynamics.
  • In order to observe and study energy changes and conversions, we must first define two terms: system and surroundings. Thermodynamics The surroundings is everything else in the universe outside the system. Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy. The system is the specific part of the universe that is of interest in the study .
  • open mass & energy What can be exchanged with surroundings? closed energy only isolated nothing There are three types of systems, based on whether mass, energy, both or neither can be exchanged with the surroundings. Three systems: Thermodynamics
    • The First Law of Thermodynamics deals with conservation of heat energy
    • The Second Law of Thermodynamics is a complex law dealing with ways that energy can be distributed within a system and its surroundings.
    There are 3 fundamental Laws of Thermodynamics. We will study and use the first two laws in this chapter: Laws of Thermodynamics
  • First Law of Thermodynamics – (heat) energy can be converted from one form to another, but cannot be created or destroyed. This is just the law of conservation of energy, applied to heat energy conversions Δ E system + Δ E surroundings = 0 Δ E system = – Δ E surroundings Thus, any energy lost by the system, must be transferred to the surroundings , and vice versa. Mathematically: or Laws of Thermodynamics
  • The Second Law of Thermodynamics deals with energy transformations and the distribution of energy within a system. There are many ways the Law can be stated, depending on what aspect of these energy distributions you want to emphasize. For our current purposes, the most important form of the second law states that heat energy always flows from the object at a higher temperature, to the object at the lower temperature. * In college, a friend who was a physics major had to use a text titled, “An Introduction to the Second Law” – which was about 1,000 pages in length! Laws of Thermodynamics
  • A fast moving particle can collide with a slow moving particle. During the collision, the fast moving particle transfers kinetic energy to the slower moving particle. As a result, the slower particle speeds up (and its temperature increases) and by conservation of energy, the faster particle must slow down ( so its temperature decreases). The Second Law of Thermodynamics The reason why heat must flow from hot objects to cooler objects relies on the definition of temperature :
  • The internal energy of a system (E) is the total of the kinetic plus potential energies of the particles making up that system. There are two ways that this internal energy can change:
    • Heat energy can be transferred to the system from the surroundings, or heat can be transferred from the system to the surroundings
    • Work can be done on the system by the surroundings , or work can be done by the system on the surroundings .
    Laws of Thermodynamics
  • Δ E system = q + w q = heat energy absorbed or given off w = work done on or by the system. Laws of Thermodynamics Mathematically, if Then the change in the internal energy ( Δ E) of a system is the sum of the heat energy changes plus the work done on or by the system: We assign (+/-) signs to q and w depending on whether energy is leaving the system or entering the system.
  • Sign Conventions in Thermochemistry
    • Energy is transferred INTO the system
    • Energy is transferred OUT OF the system
    We assign positive values to q and w when they represent a transfer of energy into the system from the surroundings, increasing the internal energy of the system. We assign negative values to q and w when they represent a transfer of energy out of the system to the surroundings, decreasing the internal energy of the system.
  • Summary of Sign Conventions in Thermochemistry  E system = q + w where q = heat energy absorbed or given off, and w = work done on or by the system. (energy leaves the system) (energy enters the system) (energy enters the system) (energy leaves the system)
  • Example: A beaker containing water is brought to a boil by absorbing 44.0 kJ of heat energy, while at the same time, the increased pressure of the water vapor as it expands does 11.5 kJ of work on the air above the beaker. What is Δ E of the system? i. We define our system as the water in the beaker. All else becomes the surroundings. ii. The water absorbs heat energy, so q is (+) 44 kJ. iii. The expanding water vapor does work on the surroundings, so we assign w = – 11.5 kJ iv. Thus, Δ E = q + w = (+44 kJ) + (–11.5 kJ) = + 32.5 kJ
  • It is important to recognize that there are literally an infinite number of combinations of heat and work that can lead to the same value for  E for the system. State Functions State functions are properties with a unique value, which is determined only by the current state of the system, regardless of how that condition was achieved. For example: We could bring about a change of 100 J in the system by allowing it to absorb 125 J of heat energy, and then do 25 J of work on the surroundings – or by absorbing 50 J of heat energy and having 50 J of work be done on it by the surroundings, or… We say that  E of the system is a state function.
  • At the summit, the potential energy of hiker 1 = hiker 2 , even though they took different paths to get there. Example: Potential energy is a state function. It does not matter what path is taken when you determine the change in gravitational potential energy -- all that matters is the change in altitude,  h:  P = P final − P initial  V = V final − V initial  T = T final − T initial Other examples of state functions include pressure, volume and temperature. State Functions  E = E final - E initial
  • Heat and work , on the other hand, are not state functions. These two functions can only describe how the system is changing, rather than a current state or condition. State Functions We cannot say a system “has” work or “has heat” – only that the system is doing work, or exchanging heat.
  • For example , if your car stalls and you need to push it off the street, the path you follow in pushing the car will determine the amount of work you do on the car ( W = F d ). The longer the distance you push the car, the more work you have to do. more work using this path! note that the net change in the car’s position is the same, so we can conclude that displacement is a state function – but the net work done in moving the car is different for each path -- thus, work is not a state function. w = Fd w =Fd
  • Even though the different quantities of work and heat applied or produced do depend on how the internal energy change occurs, the change in the internal energy itself, (the sum of q + w ), does not. To measure the internal energy changes of the system, we must be able to measure heat changes and the work done on or by the system. State Functions Work and heat are not state functions, but their sum, q + w =  E, is a state function.
  • Measuring heat and work done on a system We measure the heat exchanged between the system and the surroundings by measuring temperature changes in the system.
  • Measuring heat An exothermic process is any process that gives off heat – i.e., thermal energy is transferred to the surroundings from the system. The temperature of the surroundings increases eg: burning methane in a Bunsen burner, and condensing steam are both exothermic processes. CH 4 ( g ) + 2 O 2 ( g ) CO 2 (g) + 2H 2 O (g) + energy H 2 O ( g ) H 2 O ( l ) + energy
  • An endothermic process is any process that absorbs heat – i.e., thermal energy is transferred to the system from the surroundings. The temperature of the surroundings decreases eg: decomposing HgO by heating and melting ice are both endothermic processes. Measuring heat energy + 2HgO ( s ) 2Hg ( l ) + O 2 ( g ) energy + H 2 O ( s ) H 2 O ( l )
  • We can depict the energy changes in a reaction using a reaction coordinate diagram.  E released Energy reactants products  E absorbed Energy reactants products An Exothermic Rxn An Endothermic Rxn Energy Changes in Reactions CH 4 + 2 O 2 CO 2 + 2 H 2 O 2 HgO 2 Hg + O 2
  • Measuring P-V Work Recall: work = force applied through a distance, or: W = Fd . This is called mechanical work . One type of mechanical work is the expansion and compression of a gas. We call this “ P-V work ” because it depends on pressure and volume changes. We can derive a mathematical expression for this type of work as follows: iii. and P x V = x d 3 = F x d = Work ii. Volume = d 3 F d 2 i. Pressure = force area F d 2 =
  • We will only consider cases in which a gas expands or is compressed under constant pressure conditions (if the pressure is variable, you have to use differential calculus to solve the problem...) Under constant pressure conditions , the pressure of the gas is in equilibrium with, and must be equal to the external pressure (usually, this is atmospheric pressure). Because it is easier to measure, you are typically given the “external” pressure acting on the system in work problems. But remember, the external pressure = pressure of the system P-V Work In order to do work, however, regardless of the pressure, the volume of the gas must change…
  • If the system does work on the surroundings, then energy is leaving the system, and the internal energy of the system will be reduced. We say that (+) work is being done on the surroundings so negative work is done on the system.  w = –P  V If the gas expands , its change in volume,  V, will be positive. An expanding gas exerts a force on the surroundings and therefore must do work against the external pressure; that is, work is done ON the surroundings. P-V Work A gas expanding (+  V) AGAINST the external pressure, P, of the surroundings *we often say that work done BY the system is (-) work. since  V >0 we write: + Δ V
  • Gas is compressed (− Δ V) as the surroundings does work against the pressure of the gas. If the gas is compressed , its change in volume,  V, will be negative. And if the gas is compressed, the surroundings must have done work on the system. If the surroundings does work on the system, then energy is entering the system and the internal energy must increase . This means we assign the work done on the system as it is compressed to be positive work. P-V Work  w = –P  V Since  V < 0, we write: – Δ V
  • Thus, the formula for the P-V work done as a gas expands or is compressed that reflects the +/- sign convention is: P-V Work w = –P  V P = pressure exerted on the system  V = change in volume of the system
  • Calculations with w = -P  V The units for work are the Nm or joule. We need a way to convert pressure x volume to joules. It can be shown (see appendix 2) dimensionally, that if we measure volume in liters and pressure in kilopascals , the product unit (L-kPa) is equivalent to the unit: Nm = joule. Recall that there are 101.325 kPa in 1.00 atm. If a L-kPa = J, then the atm kPa pressure conversion is also a conversion to obtain work in joules: P  V = L-atm x 101.325 J L-atm = Joules
  • = w = – P  V initial final example : An 18-L sample of a gas is compressed to 12-L under a constant atmospheric pressure of 1.087 atm. What is the work done on the system? i. W = –P  V = –1.087 atm x (12L – 18L) = + 6.522 L-atm + 661 J of work was done ON the gas. P-V Work = ii. + 6.522 L-atm x 101.325 J L-atm
  • Example: A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done on the system, in joules, if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm? w = –P  V w = – (0 atm x 3.8 L) = 0 L•atm = 0 joules w = – (3.7 atm x 3.8 L) = –14.06 L•atm w = -14.06 L•atm x = –1424 J = work done on the system (gas) *we could also say that +1424 J of work was done on the surroundings. (a)  V = 5.4 L – 1.6 L = 3.8 L ; P = 0 atm (b)  V = 5.4 L – 1.6 L = 3.8 L ; P = 3.7 atm 101.3 J 1L•atm
  • Although there are many ways in which work can be done on or by the system (the mechanical work done in stirring a mixture, for example), PV work is perhaps the most common type, especially in an open container. If we focus on just the PV work ad the heat energy changes, we obtain a more useful equation with which we can measure the change in the internal energy of a system (  E) during any process :  E system = q + w Heat of Reaction   E system = q + (–P  V)
  • Since KE is part of the internal energy, if  E < 0, then  KE <0 as well, which means that the temperature drops! Chemistry in Action: Making Snow  E = q + (−P Δ V)  q = 0 and  E = −P  V  E < 0  T < 0 = SNOW! The compressed air + water vapor expands so fast, that no heat is exchanged with the surroundings: Since the compressed air + water vapor expands ( Δ V >0), it does work on the surroundings, lowering the internal energy of the system.
  • Heat of Reaction Most reactions are carried out under constant pressure conditions (the beaker or test tube is open to the atmosphere, etc.) However, reactions can also be carried out under constant volume conditions (within a sealed container). It is relatively easy to measure heat energy changes during a reaction. Solving the eqn for q we obtain, what we shall see, is a much more useful form: q = Δ E + P Δ V Δ E = q + (−P Δ V)
  • If a chemical reaction is run at a constant volume , then  V = 0 and no P-V work is done. This means that: 0  q v =  E Under constant volume conditions, the net change in the internal energy of a system ( Δ E) is equal to the heat energy (q) absorbed or given off by the system. the “v” subscript is used to remind us that this is under constant volume conditions. q =  E + P  V Heat of Reaction
  • Under constant pressure conditions, we write: the “p” subscript reminds us that this is at constant pressure q p =  E + P  V Heat of Reaction This new thermodynamic function is called the enthalpy ( H ) of the system: Since the majority of reactions are carried out under constant pressure conditions (for example, in an open beaker), we use this equation to define a new function whose change gives the heat of reaction, q p . H = E + PV
  • Enthalpy For any process under constant pressure conditions, the change in the enthalpy (  H) gives the heat energy absorbed or given off in the reaction: *Note that under constant pressure conditions, the change in enthalpy,  H, depends only on the change in internal energy,  E and/or the change in volume,  V. These are both state functions, which means that  H is also a state function!  H =  E + P  V
  • Comparing  E and  H From the equation:  H =  E + P  V, we can make the following statement: If  V is zero (no work done; no expansion or compression, etc.) or if  V is at least very small, then:  H ≈  E Thus, by measuring the enthalpy (heat energy) changes, which is relatively easy to do, we are able to determine the changes in the internal energy of the system (which would otherwise be very difficult, if not impossible, to do)!
  • Enthalpy of Reaction  H rxn = H (products) – H (reactants) We define the enthalpy of reaction ,  H rxn , as the difference between the enthalpies of the products and reactants:  H rxn can be (+) or (-), depending on the process. If  H rxn is ( + ) the process is endothermic If  H rxn is ( - ) the process is exothermic.
  • Thermochemical equations are equations that show enthalpy changes as well as mass relationships. Note that since heat energy is absorbed, the reaction is endothermic  Δ H > 0. example: 6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm. thermochemical equation: Enthalpy of Reaction H 2 O ( s ) H 2 O ( l )  H = + 6.01 kJ/mol
  • CH 4 ( g ) + 2O 2 ( g) CO 2 ( g) + 2H 2 O ( l )  H rxn = –890.4 kJ/mol Thermochemical Equations example: 890.4 kJ are released for every 1 mole of methane that is combusted at 25°C and 1 atm. thermochemical equation: Since heat energy is given off in the reaction, the reaction is exothermic and Δ H < 0.
  • In general, we often call the Δ H of a reaction the “heat of reaction .” Specific types of reactions are given their own specific names.
    • Δ H for vaporizing a liquid is called the “heat of vaporization”
    • Δ H for a combustion reaction is called the “heat of combustion.”
    • Δ H for a neutralization reaction is called the “heat of neutralization,” etc.
    Thermochemical Equations
  • Writing Thermochemical Equations
    • The physical states of all reactants and products must be specified in thermochemical equations.
    CH 4 ( g ) + 2O 2 ( g) CO 2 ( g) + 2H 2 O ( l )  H = -890.4 kJ/mol H 2 O ( s ) H 2 O ( l )  H = 6.01 kJ/mol
  •  H = 58.04 kJ/mol Thermochemical Equations
    • If you multiply both sides of the equation by a factor n, then  H must change by that same factor, n:
    = 12.0 kJ/mol x 2  H = +58.04 kJ/mol –
    • If you reverse the reaction, the sign of  H also reverses
    H 2 O ( s ) H 2 O ( l )  H = 6.01 2 2 N 2 O 4 ( g ) 2 NO 2 ( g ) 2 NO 2 ( g ) N 2 O 4 ( g )
  • Note the units for Δ H are kJ/mol. The “per mole” refers to the coefficients (moles) of a particular species present in the equation as written . : 2 C 4 H 10 ( g ) + 13 O 2 ( g) 8 CO 2 ( g) + 5 H 2 O ( l )  H = −5757 kJ/ mol Hence, in the thermochemical equation: The –5757 kJ/mol means that 5757 kJ of heat energy are released for every 2 moles of C 4 H 10 , or for every 13 mole of O 2 that reacts, and for every 8 moles of CO 2 or for every 5 moles of H 2 O that form. Thermochemical Equations
  • Thermochemical Equations Often, thermochemical equations are written with fractional coefficients so that the species of interest in the reaction is unimolar Note that the magnitude of Δ H changes whenever the coefficients in the balanced equation changes. Again, the “per mole” in kJ/mol refers to the molar ratios of the equation as written . 3 2 Al 2 O 3 (s) Al (s) + O 2 (g)  H rxn = 1676 kJ/mol  2 Al 2 O 3 (s) 2 Al (s) + 3 O 2 (g)  H rxn = 3352 kJ/mol 
  • We see that the amount of heat energy absorbed or released is fixed for a given reaction – that is,  H rxn is stoichiometrically equivalent to the specific number of moles of reactants and products for that particular thermochemical equation. Think of it this way: Think of “heat” as being a “compound,” and the magnitude of the heat energy as the “coefficient” for that “compound.” Then treat the Δ H value as you would any other species in the reaction when solving stoichiometry problems. Thermochemical Equations
  • Moles of Reactant Moles of Product Joules of Heat Energy  H / mol mol:mol ratio As always, convert the starting material to moles, then use the mole:mole ratios from the balanced equation. Now, however, one of the conversion factors is kJ of heat energy per mole(s) of reactant or product: Thermochemical Equations
  • Example: How much heat is given off when 266 g of white phosphorus (P 4 ) burns in air, given that: P 4 ( s ) + 5O 2 ( g ) P 4 O 10 ( s )  H = -3013 kJ/mol 266 g P 4 1 mol P 4 123.9 g P 4 x x = – 6470 kJ We see that -3013 kJ of heat energy are given off for every mole of P 4 that reacts; thus we need to determine the number of moles of P 4 , and then use the ratio: -3013 kJ released per mole of P 4 to determine how much heat is released: Thermochemical Equations -3013 kJ 1 mol P 4
  • Example: Al 2 O 3 undergoes thermal decomposition as follows: Al 2 O 3 (s) Al (s) + O 2 (g)  H rxn = 1676 kJ/mol  How many grams of O 2 will be formed if 1000 kJ of heat is absorbed by excess Al 2 O 3 ? = 28.65 g of O 2 1000 kJ x +1676 kJ 3/2 mole O 2 32.00 g mol O 2 x *note that we can solve this problem using the fractional coefficient 3/2. You do not have to convert all the coefficients to integer values. Thermochemical Equations 3 2
  • example: Ammonia is formed by reacting nitrogen and hydrogen at elevated temperature and pressures: What is the enthalpy change when 2.00L of H 2 gas at 600K and 4.00 atm are reacted with excess nitrogen gas? ii. moles H 2 = 0.1625 moles iii. Δ H = 0.1625 mole H 2 x alternately, we could say that +5.02 kJ of heat energy is released = – 5.02 kJ Thermochemical Equations N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Δ H rxn = -92.6 kJ/mol i. n H = PV/RT = (4.00 atm x 2.00 L) / (0.08206 x 600 K) 2 – 92.6 kJ 3 mol H 2
  • Measuring  E of reactions with gases We start with the equation for  H, and solve for  E:  H =  E + P  V  E =  H – P  V but P  V =  (PV) =  (nRT) from ideal gas equation and  (nRT) =  n (RT) at a fixed temperature
    •  E =  H –  nRT
    where  n = moles product gases – moles reactant gases AP Only note that  nRT has units of L-atm when R = 0.08206 L-atm/mol K. Thus, we must multiply this by 101.3 to convert to Joules. Note, too, that 0.08206 x 101.3 = 8.314 kPa-L/mol K. If we use 8.314 for R, then  nRT equals joules directly. if  n = 0 then  E =  H
  • If there is no net change in the number of moles of gas , then the system neither expands nor is it compressed. Thus,  V = 0 and no work is done. If the reaction results in a net decrease in the number of moles of a gas , then the system is compressed and work is done on the system by the surroundings . If the reaction results in a net increase in the number of moles of a gas , then some of the internal energy goes into doing work on the surroundings as the system expands . Measuring  E of reactions with gases
  • i.  E =  H -  nRT -- we must find  n and  H: v .   E = (-238.9 kJ) - [ (0.650)(8.314 kPa-L/mol K) x 298 K ] note that  H ≈  E AP Only = 1610 J ~ 1.61 kJ Example: Find  E for the reaction if 1.30 mol of Na react with excess water at 298 K, given that: iii.  n = mol product gas – mol reactant gas = (0.650 – 0) = 0.650 mol = - 238.9 kJ vi.   E = - 238.9 kJ – 1.61 kJ = - 240.5 kJ 2Na ( s ) + 2H 2 O ( l ) 2NaOH ( aq ) + H 2 ( g )  H = -367.5 kJ/mol ii. 1.30 mol Na x 1 mol H 2 2 mol Na = 0.650 mol H 2 produced iv.  H = -367.5 kJ 1 mol H 2 x 0.650 mol H 2
  • Calori metry measurement of heat energy changes
  • Heat Capacity and Specific Heat Capacity There are three things that determine the amount of heat energy that is lost or gained by an object: 1. The temperature change of the object 2. What the object is made of (its composition). 3. The quantity (mass) of matter present
  • The heat capacity ( HC ) of a substance is the amount of heat ( q ) required to raise the temperature of a substance by one degree Celsius. Heat Capacity and Specific Heat Capacity units for HC = J/ o C The heat capacity varies depending on what the sample is made up of (metal vs wood, for example), and the mass of sample used. Thus, it would be helpful if we had a term that takes into account the nature and mass of the substance being heated (or cooled). HC = q  t
  • The specific heat capacity ( c ) – usually shortened to just “specific heat” -- is the amount of heat energy ( q ) required to raise the temperature of one gram of the substance by one degree Celsius. units for c = J/g o C q = heat energy (joules) m = mass (grams)  t = temp change (ºC) Heat Capacity and Specific Heat Capacity Note that specific heat is an intensive property. Every substance has a unique specific heat value. c = q m x  t
  • Heat Capacity and Specific Heat Capacity Note that the specific heat of metals tends to be quite small – typically less than 1 J/g o C . Adding a small amount of heat energy will cause a relatively large increase in temperature – that is, the metal gets HOT quickly. Metals are good conductors of heat. On the other hand, metals cool off quickly as well: a small loss of heat energy = large change in temperature.
  • The specific heat of water , on the other hand, is by comparison very large -- 4.184 J/g o C . This means that water can absorb a lot of heat energy, yet its temperature will not increase very much. It can also lose a lot of heat energy, without cooling down. Water is an insulator. Note that the energy unit we call the calorie is actually another unit for the specific heat of water: 1 calorie = heat energy/g o C 1 calorie = 4.184 J
  • Lake Effect Weather and Specific Heat Because of water’s high specific heat, large bodies of water like Lake Michigan can absorb heat all summer without having any significant increase in temperature. In the summer, the hot air comes over Lake Michigan. Since the lake is cooler than the air, heat energy is transferred from the air to the water. This cools the air, but barely warms the water at all. The cool air now blows onshore. The hot sand then radiates away heat to the cooler air. Thus it is always cooler near the beach in the summer. heat heat cooler air cools land hot land
  • Lake Effect Weather and Specific Heat heat heat In the winter , the water has lost a lot of heat energy, but has not cooled down that much – water stays warmer than the air above it. When a cold air mass comes over the water, heat energy is transferred from the water to the air, warming the air. Warm air can hold more moisture than cold air. The warm, moist air now moves over the land. The land is colder than the air and so the air transfers heat to the land. The cooler air cannot hold its moisture, and so we get lake effect snow storms! warm, moist air cold air cold land
  • Solving Specific Heat Problems To determine the amount of heat energy absorbed or given off by a substance, start with the definition of specific heat, and solve for q: where  t = t final - t initial Example: How much heat is given off when an 869 g iron bar cools from 94 o C to 5 o C? specific heat of Fe = 0.444 J/g • 0 C  t = t final – t initial = 5 o C – 94 o C = -89 o C q = c m  t = 869 g x 0.444 J/g • o C x (–89 o C) = -34,000 J q = cm  t
  • Calorimeters Calorimeters are devices used to determine heat energy changes for a given process or reaction.
    • There are two types:
    • Constant-volume calorimeter (also called a bomb calorimeter ) used to measure heat energy changes when a substance undergoes combustion.
    • Constant-pressure calorimeter (often nothing more than two nested styrofoam coffee cups!) . This type is used for measuring heat changes for reactions in solution.
  • The constant-volume ( bomb ) calorimeter Bomb calorimeters are used to find heats of combustion. The bomb calorimeter uses an electric current to “burn” a substance of known mass under about 30 atm of O 2 gas. The heat given off during the combustion is transferred to a water jacket. The heat absorbed by the water is measured using a thermometer.
  • Bomb calorimeters have precisely known heat capacities, so that very precise enthalpy changes can be measured. A bomb calorimeter holds a known, fixed amount of water. As a result, the heat capacity of a bomb calorimeter typically includes both the “bomb” and the water in the jacket as part of the “calorimeter.” The Bomb Calorimeter HC cal = HC bomb + HC water
  • The Constant-Volume Bomb Calorimeter q sys = q cal + q rxn assuming no heat enters or leaves the bomb calorimeter, we can say that q sys = 0 !  - q rxn = +( q water + q bomb ) q water = c m  t  - q rxn = + q cal ▪ dimensionally, HC = J/ o C so q bomb = HC bomb x  t q cal = heat absorbed by the bomb + heat absorbed by the water jacket… ▪ finally, note that Δ t of the bomb = Δ t of the water, so: heat given off in the reaction = heat absorbed by the calorimeter  − q rxn = q cal = [ HC cal Δ t ] = + ( cm Δ t water + HC Δ t bomb )
  • The Bomb Calorimeter Example: A 1.922 g sample of methanol (CH 3 OH) is combusted in a bomb calorimeter which has a heat capacity of 10.4 kJ/ºC. The temperature of the water rose by 4.20 ºC. What is the molar heat of combustion of methanol? i. −q rxn = + q cal ii. −q rxn = HC x Δ t = 10,400 J/ºC x (+4.20 ºC) = 43,680 J iii.  q rxn = − 43.68 kJ for a 1.922 g sample of methanol. iv.  q per mole = x = −728 kJ/mol − 43.68 kJ 1.922 g 32.04 g mol CH 3 OH convert HC to joules /ºC !
  • see sample problem 6.6 on page 242
  • Constant Pressure Calorimetry Reactions under constant pressure conditions are much easier to work with. We can use a simple “coffee-cup” calorimeter to measure energy changes for many non-combustion reactions. In this type of calorimeter, the heat energy given off by a substance or a reaction is absorbed by the water in the calorimeter. We measure the temperature changes of the water to determine the heat given off. We can usually assume no heat is exchanged with the surroundings.
  • - q metal = + ( q water + q cal ) No heat enters or leaves! q sys = q water + q cal + q metal q sys = 0 and q cal = HC cal x  t cal Constant Pressure Calorimetry hot metal water Determining the specific heat of a metal c (metal) = -q metal m  t metal finally: where q water = [ cm  t ] H 2 O
  • i. q lost by the lead pellet = q gained by the water Example: Determining sp heat of a metal (see 6.7 pg 243) A lead pellet ( m = 26.47 g) was heated to 89.98 o C and then placed in a coffee-cup calorimeter containing 100.00 g of water at a temperature of 22.50 o C. The final temperature of the water was 23.17 o C. What is the specific heat of Pb? Assume HC of the calorimeter was 0 J ii.  - c m  t metal = +c m  t water iii.  c metal = cm  t water - m  t metal = 280.3 J/1768.5 g o C = 0.158 ~ 0.16 J/g o C watch your signs!! = 0.67 o C = 2 sig fig + = 4.184 J/g o C x 100.00 g x (23.17 – 22.50 o C) -26.47g x (23.17 – 89.98 o C) q water
  • Example: To determine the specific heat of titanium, a student placed a 32.872 gram sample of Ti in a beaker of boiling water for several minutes, and then transferred the metal to a calorimeter. The calorimeter contained 98.86 ml of water at an initial temperature of 20.10  o C . The final temperature of the system was 23.04 o C. The heat capacity of the calorimeter was determined to be 18.9 J/ o C . What was the reported value for the specific heat of Ti? i. First we note that t o for the metal was 100.0 o C = boiling pt of water ii. Next we note that the mass of water = 98.86 grams since the density of water is 1.00 g/ml iii. Since we are given HC for the calorimeter, we use the “expanded” form: –q metal = + (q water + q cal )
  • Solution: i. = - q metal = q water + q cal ii.  -c m  t metal = +c m  t water + HC  t cal iii.  -c(32.872 g)(23.04 – 100.00 o C) = +4.184 J/g o C x 98.86 g x (23.04 – 20.10 o C) + 18.9 J/ o C (23.04 – 20.10 o C) iv.  -c(-2529.8 g o C)= 1216.07 + 55.57J v.  c = -(-2529.8 g o C) 1216.07 + 55.57 J = 0.503 J/g o C
      • Give a possible source of experimental error to explain why the student’s value for the specific heat of Ti might be less than the accepted value?
      • What would explain why the student’s value for the sp heat of Ti might be greater than the accepted value?
    If sp heat is too small, then the water absorbed LESS heat energy than was lost by the hot metal. Thus, some heat energy was lost to the surroundings when the metal was transferred to the calorimeter. This is the opposite case – some excess heat energy from the surroundings was transferred to the water . c = + q water + HC  t cal m  t metal
      • If the accepted value for the specific heat of Ti is 0.550 J/g o C, determine the Δ% between the student’s value and the accepted value.
     % = experimental - accepted accepted  % = 0.503 -0.550 J/g o C 0.550 J/g o C = -8.5% since the percent difference was negative, the sp heat value is too small and heat energy was lost to the surroundings during the experiment.
  • Measuring  H rxn in aqueous solutions We can also use a coffee-cup calorimeter to measure  H rxn for reactions that take place in aqueous solutions. Example: 100.0 ml of 0.500M NaOH was placed in a coffee-cup calorimeter. To this, 100.0 ml of 0.500 M HCl was added. The temperature of the solution rose from 22.50 o C to a final temperature of 25.86 o C. Find  H rxn per mole of acid that reacts. i. we assume that sp heat of the aqueous sol’n = 4.184 J/g o C ii. we assume that the density of the sol’n is ≈ 1.00 g/ml so that the mass of the sol’n ≈ 100 + 100 = 200 g iii. we assume  V ≈ 0 for the sol’n so that  H ≈ q rxn
  • Solution: i. –q rxn = + q sol’n + q cal 0 assume a perfect insulator ii. –q rxn = c m  t soln iii. – q rxn = 4.184 J/g o C x 200.0g x (25.86 – 22.50 o C) iv. – q rxn = 2812 J ≈ 2.81 kJ vii. moles of HCl = M A V A = 0.500M x 0.100L = 0.0500 mol vii.   H rxn = q rxn = -2.81 kJ/0.0500 mol = -56.2 kJ/mol of HCl v. thus + q rxn = 2.81kJ
  • Standard Enthalpy of Formation (H f o ) Hess’s Law and
  • Hess’s Law of Heat Summation Hess’s Law of Heat Summation states that, the enthalpy change for an overall process or reaction is the sum of the enthalpy changes of its individual steps. You can determine  H rxn for any reaction that can be written as the sum of two or more other reactions for which thermodynamic data is known.  H rxn =  H 1 +  H 2 +  H 3 + … Recall that enthalpy is a state function, It doesn’t matter how you get there, it only matters where you start and end!
  • This means that, for any reaction that can be written as a sum of 2 or more steps, we can find  H rxn if we know the  H values for each step; or we can find  H for any given step, if we know the enthalpy changes for the other step(s) and the overall  H rxn . Since nearly every reaction can be written as the sum of two or more reactions, we can calculate the enthalpy change for nearly any reaction we might want to look at! Hess’s Law of Heat Summation
  • Consider the formation of carbon monoxide from its elements: C (graphite) + ½ O 2 ( g ) CO ( g ) Although we can write this reaction on paper, in the real world, you cannot synthesize CO in this way. But we can still determine  H rxn by combining the enthalpies of reactions that do occur in such a way that the net reaction is the desired reaction given above:  H rxn = −393.5 kJ/mol  H rxn = −283.0 kJ/mol 1. The following two reactions will occur and go to completion: I will have to give you the reaction steps – you cannot do this on your own – yet! Hess’s Law C (graphite) + O 2 ( g ) CO 2 ( g ) CO ( g ) + ½ O 2 ( g ) CO 2 ( g )
  • C (graphite) + O 2 ( g ) CO 2 ( g ) CO 2 ( g ) ½ O 2 ( g ) + CO ( g )  H rxn = −393.5 kJ/mol  H rxn = +283.0 kJ/mol 2. If we reverse the second reaction and add it to the first reaction, we obtain the desired reaction for the formation of CO. Remember when you reverse a reaction, you must reverse the sign of its enthalpy change! ½  H rxn = −110.5 kJ/mol C (graphite) + ½ O 2 ( g ) CO ( g ) Note that identical substances on opposite sides of the “ “ will cancel out, just like identical numbers or variables cancel when they appear on opposite sides of the “=“ sign in math problems.
  • Example: CO and NO are toxic gases found in car exhaust. One way of reducing these emissions is to convert them to less toxic gases. Determine the  H for the reaction: CO (g) + NO (g) CO 2 (g) + ½ N 2 (g)  H = ?? Given the following thermodynamic data: rxn A: CO (g) + ½ O 2 (g) CO 2 (g)  H A = -283.0 kJ/mol rxn B: N 2 (g) + O 2 (g) 2 NO (g)  H B = + 180.6 kJ/mol
    • We will need to use rxn A as written. Rxn B must be reversed to obtain N 2 as the desired product.
    • Next, note that we only want ½ N 2 as a product, so we must multiply rxn B by ½ as well.
    Hess’s Law
  • rxn B: ½ [2 NO (g) N 2 (g) + O 2 (g) ]  H B = ½ ( − 180.6 kJ/mol) net  H rxn : Δ H A = − 283.0 kJ/mol + Δ H B = ½ (−180.6) kJ/mol  H rxn = −373.3 kJ/mol Hess’s Law reverse the sign of Δ H when you reverse the reaction direction rxn A: CO (g) + ½ O 2 (g) CO 2 (g)  H A = −283.0 kJ/ml net rxn CO (g) + NO (g) CO 2 (g) + ½ N 2 (g)
  • Chemistry in Action: Bombardier Beetle Defense 1. C 6 H 4 (OH) 2 ( aq ) C 6 H 4 O 2 ( aq ) + H 2 ( g );  H º = 177 kJ/mol 3. H 2 O 2 ( aq ) H 2 O ( l ) + ½O 2 ( g );  H º = -94.6 kJ/mol 2. H 2 ( g ) + ½ O 2 ( g ) H 2 O ( l );  H º = -286 kJ/mol  H º rxn = Δ H 1 + Δ H 2 + Δ H 3 = 177 kJ + (– 286 kJ) + (– 9.4 kJ) = −204 kJ/mol The bombardier beetle ejects a hot chemical spray of quinone (C 6 H 4 O 2 ). Given the following thermodynamic data, determine  Hº rxn for the reaction that produces quinone shown below: C 6 H 4 (OH) 2 ( aq ) + H 2 O 2 ( aq ) C 6 H 4 O 2 ( aq ) + 2H 2 O ( l) C 6 H 4 (OH) 2 ( aq ) + H 2 O 2 ( aq ) C 6 H 4 O 2 ( aq ) + 2H 2 O ( l)
  • Standard Enthalpies of Formation Recall that  H rxn = H (products) – H (reactants) Unfortunately, there is no way to measure the absolute enthalpy, H, of a substance . We can only measure enthalpies relative to an arbitrary reference point. It is like measuring the height of a mountain – we always arbitrarily measure it relative to sea level. The reference point for measuring enthalpy changes is called the standard enthalpy of formation (  H º f ).
  • The standard enthalpy of formation of any element in its most stable form is zero.  H f o (O 2 ) = 0 but…  H f o ( O 3 ) = 142 kJ/mol  H f o (C, graphite) = 0  H f o (C, diamond ) = 1.90 kJ/mol Standard Enthalpies of Formation The standard enthalpy of formation (  H f o ) for a compound is the heat change that results when one mole of that compound is formed from its elements at a pressure of 1 atm . There is a table of standard enthalpies of formation for many inorganic compounds on page 247 in your textbook. A more extensive list in given in Appendix 3 at the back of the text.
  • see page 247 in the textbook
  • The standard enthalpy of reaction (  H º rxn ) is the enthalpy of a reaction in which reactants and products are in their standard states at 1 atm. Standard Enthalpy of Reaction (  H º rxn ) In order to use this equation, we must know  Hº f values for each species. If not already known, these must be calculated first. a A + b B c C + d D  H º rxn d  H º (D) f c  H º (C) f = [ + ] - b  H º (B) f a  H º (A) f [ + ]  H º rxn n  H º (products) f =  m  H º (reactants) f  -
  • Example: Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted (i.e., determine  Hº rxn ) ? The standard enthalpy of formation of benzene is 49.04 kJ/mol. See also table 6.4. this is for 2 mol C 6 H 6   Hº rxn / mol = 2C 6 H 6 ( l ) + 15O 2 ( g ) 12CO 2 ( g ) + 6H 2 O ( l )  H º rxn n  H º (products) f =  m  H º (reactants) f  -  H º rxn 6  H º f (H 2 O) 12  H º f (CO 2 ) = [ + ] - 2  H º f (C 6 H 6 ) [ + 15  H º f (O 2 )]  H º rxn = [ 12(–393.5) + 6(–187.6) ] – [ 2(49.04) ] = -5946 kJ = - 2973 kJ/mol C 6 H 6 -5946 kJ 2 mol C 6 H 6 0
  • Example: It is found, experimentally, that the production of NH 3 from its component elements releases 92.6 kJ, according to the equation below. Calculate  H f º of NH 3 .  H o rxn = 2  H f º(NH 3 , g ) – [  H f º(N 2 , g) + 3  H f º(H 2 , g )] But  H f º(N 2 , g ) = 0 and  H f º(H 2 , g ) = 0  H rxn º = -92.6 kJ  Hº rxn = -92.6 kJ = 2  H f º(NH 3 , g ) – (0 + 0)  H f º(NH 3 , g ) = -92.6 kJ / 2 = -46.3 kJ/mol NH 3 N 2 (g) + 3 H 2 (g) 2 NH 3 (g)
  • To produce a table that listed the heats of formation for every compound known would be impossible. Luckily, we can use Hess’s Law to determine  Hº f for virtually any compound from even a limited list of enthalpy of formation values. Equipped with a means of finding  Hº f values for virtually any compound, we can also then calculate  Hº rxn values for virtually any reaction – even reactions that are not possible to perform! Hess’s Law of Heat Summation Revisited
  • Example : Calculate  H º f of CS 2 ( l ) given that: C (graphite) + O 2 ( g ) CO 2 ( g )  H º rxn = −393.5 kJ/mol S (rhombic) + O 2 ( g ) SO 2 ( g )  H º rxn = −296.1 kJ/mol CS 2 ( l ) + 3O 2 ( g ) CO 2 ( g ) + 2SO 2 ( g )  H º rxn = −1072 kJ/mol 1 . Write the enthalpy of formation reaction for CS 2 2. Add the given rxns so that the result is the desired rxn.  Hº f =  Hº rxn = −393.5 + 2(−296.1) + 1072 = + 86.3 kJ Δ Hº f = ? Δ Hº f = ? C (graphite) + 2S (rhombic) CS 2 ( l ) + C (graphite) + 2S (rhombic) CS 2 ( l ) rxn rxn C (graphite) + O 2 ( g ) CO 2 ( g )  H º = −393.5 kJ/mol 2 S (rhombic) + 2 O 2 ( g ) 2 SO 2 ( g )  H º = −296.1 x 2 kJ/mol CO 2 ( g ) + 2SO 2 ( g ) CS 2 ( l ) + 3O 2 ( g )  H º = +1072 kJ/mol rxn
  • The enthalpy of solution (  H soln ) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack?  H rxn = H products − H reactants  H soln = H soln − H components
  • Dissolving ionic solutes There are two enthalpy steps involved in dissolving a solid ionic solute in water:
    • all the ions in the solute must be separated. The energy for this step is called the lattice energy, U. This is always an endothermic step.
    • the solute must be hydrated by the solvent water molecules. The energy for this step is called the enthalpy of hydration, Δ H hyd . This is always an exothermic step.
    The enthalpy of solution is the sum of these two steps
  • The Solution Process for NaCl  H soln = U + Δ H hyd = +788 – 784 = +4 kJ/mol
  • Quantum Theory & the Electronic Structure of Atoms Chapter 7
  • Electromagnetic Waves & Energy Recall that around the turn of the century (20th century that is), scientists were beginning to explore the nature of the atom. In a flurry of discoveries over a relatively brief time, our understanding of matter changed from Dalton’s “tiny billiard ball-atoms” to Thompson’s plum-pudding model, and finally in 1919, Rutherford showed that the atom was composed of a very tiny, dense, positively charged “nucleus” around which a swarm of negatively charged “electrons” were located.
  • The initial assumption was that these electrons followed circular orbits, not unlike the planets in orbit around the sun. Electromagnetic Waves & Energy The electrostatic forces between the (+) charged nucleus and the (–)charged electrons -- rather than gravity – was what kept the electrons in orbit.
  • Unfortunately, there was a major problem with this “planetary” model. Electromagnetic Waves & Energy According to the laws of physics, a charged particle in orbital motion should radiate away energy . And by conservation of energy, this loss of energy would result in a decrease in the kinetic energy – and hence, the speed of the electron. As the electron slowed down, it would be pulled ever closer to the nucleus.
  • A quick calculation showed that the time it would take for an electron to radiate away sufficient energy to spiral into and collide with the nucleus was less than one microsecond. Electromagnetic Waves & Energy Clearly, this does not happen in the real world… Ouch!
  • Strangely enough, the solution to this conundrum came out of other scientist’s investigation into the nature of light...We were on the verge of a fundamental overhaul of all our macroscopic assumptions about matter and light. Electromagnetic Waves & Energy
  • “ We were about to discover that matter is not as “solid” and “particle-like” as we would like to think; and that light is not as “insubstantial” and “wave-like” as we thought. “ On the atomic scale of things, the difference between matter and electromagnetic waves begins to blur rather alarmingly...” Electromagnetic Waves & Energy -Michael Munowitz Principles of Chemistry
  • Wave-Particle Duality Light and matter -- on a sub-atomic scale – exhibits both wave-like and particle-like behaviors, which we call a wave-particle duality. To better understand this duality, we need to review just what it is that makes something a wave or a particle. We will start with waves. Because of this duality, we have come to refer to light as a “ wave-packe t” – a little bundle of energy that generally behaves as a wave, but can interact with matter on a sub-atomic scale as if it were a particle.
  • Wave-like Properties
    • Waves are vibrations that transfer energy . These vibrations can be in the form of the troughs and crests of ocean waves, or high-low pressure changes in the air which we call sound waves, or even oscillating electric and magnetic fields. Waves are thus a series of of maximas and minimas.
    Properties of Waves 2. Waves travel through a medium by displacing the medium: the medium either vibrates parallel with the passing waves, called longitudinal waves , or the medium vibrates perpendicular to the direction of the passing waves, which is called a transverse wave .
  • A Transverse Wave  A direction of wave propagation Properties of Waves 3. Wavelength,  (lambda) is the distance over which the wave pattern repeats (the distance between two crests, for example). 4. Amplitude ( A ). The distance between the “rest” position of a wave and the point of maximum displacement .
  • 5. How often the wave pattern repeats is called the frequency of the wave, and is given the symbol, v (nu). The units for frequency are waves per second, or simply sec -1 , which is called a hertz (Hz). Properties of Waves note that higher frequency means shorter wavelengths! v = 4 waves/sec = 4 Hz v = 2 waves/sec = 2 Hz
    • How fast the wave moves forward (or propagates ) is the speed of the wave. Speed ( u ) has units of distance/time. Looking at the units for frequency and wavelength, we find that the speed of a wave is given by:
    Properties of Waves Note that the wavelength and the frequency of a wave are inversely proportional to each other. u =  meters wave x wave sec = meters sec units:  = u   = u 
  • Example: Ultrasound waves are high frequency sound waves with a wavelength of about 1.2 cm. If the speed of sound in air is 340 m/s, what is the frequency of these sound waves?  = 340 m/s 0.012 m = = 2.83 x 10 4 Hz Example: What is the wavelength of a radio wave with a frequency of 102 MHz? (u = 3 x 10 8 m/s.) = 2.94 m Properties of Waves u   = u  = 3 x 10 8 m/s 1.02 x 10 8 Hz
  • Properties of Waves
    • The energy of a light wave, as we shall see later, is directly proportional to its frequency.
    • The intensity of a wave is the power output (energy delivered per second) of the wave per square meter of the medium through which the wave is moving.
  • destructive interference occurs when the two waves meet out of phase and their amplitudes subtract to yield a wave with an amplitude of A – B at the point of interference. The energy of the wave at that point is (A – B) 2 constructive interference occurs when the two waves meet in phase and their amplitudes add to yield a wave with an amplitude of A+B at the point of interference. The energy of the wave at that point is (A+B) 2 . When two particles collide, they bounce off each other; when two waves collide, they undergo interference . Wave Interactions
  • before during after Destructive Interference Constructive Interference
  • Note that it is possible for two waves to interfere destructively to the point that the net amplitude at the point of interference is zero (said to be complete destructive interference ). Properties of Waves On the other hand, if the amplitudes undergo complete constructive interference , then the wave amplitude ( and energy ) has a maximum at the point of interference, which is called an anti-node. If the amplitude is zero, then the energy of the wave is also zero at that point . Such a point is called a node.
  • Properties of Waves Standing Waves If a wave is limited to a particular path length, such as when a guitar string is plucked, the waves undergo constructive and destructive interference to produce what is called a standing wave – it appears as if the wave is stationary: A standing wave must be some integer multiple of  2 nodes
    • Waves come in trains – for a wave, the transfer of energy is not a “one shot” affair, as it is with a particle; rather, waves “pump” energy in a cumulative fashion, over time . ( For example, You can start a fire using the sun’s rays and a magnifying glass to focus them, but it takes a few minutes for the tinder to absorb enough heat energy to reach its combustion temperature.)
    Properties of Waves 11. Waves can undergo diffraction – that is, they bend when they encounter a barrier. A diffraction pattern is built up when the diffracted waves interfere with each other.
  • If light is a wave, then how does it travel through the vacuum of space -- there is no medium to vibrate! In 1873, James Clerk Maxwell proposed that light is a type of wave which he called electromagnetic radiation (or EM radiation) for short. Light As Electromagnetic Waves: As we shall see, EM radiation is emitted by oscillating charged particles (mainly electrons, but certain high energy EM radiation is produced by the vibration of protons.) Moving charged particles also produce a magnetic field. Thus, EM radiation has both an electric and a magnetic component.
  • Maxwell showed that electromagnetic waves are traveling magnetic and electric fields that oscillate at right angles to each other. Since these fields vibrate perpendicular to both each other and the direction of wave propagation, EM waves are transverse waves. Electromagnetic Waves EM radiation is thus its own medium – and only EM radiation can travel through a vacuum.
  • What we call “visible light” is only a tiny fraction of a much broader range of waves called the electromagnetic spectrum, which includes not only visible light waves, but radio waves, microwaves, infrared, ultraviolet, x-rays, and gamma rays. The regions of the spectrum are defined somewhat loosely, and depend on the wavelength and/or frequency of the wave. Every type of electromagnetic wave travels at the same speed in a vacuum, namely, the “speed of light,” to which we give the symbol, c . The speed of light is 2.997924 x 10 8 m/s (which is rounded to 3.00 x 10 8 m/s for our purposes). Electromagnetic Waves
  • The Electromagnetic Spectrum see page 271
  • Particle-Like Properties
  • Light also exhibits several distinctly particle-like behaviors . One of the most obvious is the fact that light casts distinct shadows, and the image of a window cast onto the floor as sunlight passes through it has well-defined edges, meaning light does not appear to diffract as it encounters large barriers, nor as it passes through large openings. Particle-Like Behaviors
  • One of the most intriguing, and clearly particle-like behaviors of light is something called the photoelectric effect. The Photoelectric Effect This is the process by which light, striking a metal, causes the metal to emit an electron (the emitted e– is called a photoelectron and the process is called photoemission ). However, there are some strict limitations on just when such photoemission can occur.
  • If light was, in fact, a wave, then even low energy light waves should be able to slowly pump an electron up to a high enough energy state that the electron is ionized, or removed from the atom, just as light focused by a magnifying lens can slowly increase the kinetic energy of the tinder to cause it to burst into flame. However, this is not what happens... Instead, if the frequency of light waves used is too low (and hence, the wavelength is too long) no photoemission occurs at all, regardless of the intensity of the light or how long the light shines on the metal. The Photoelectric Effect On the other hand, if light of the correct minimum frequency is used , then photoemission occurs instantly, even if low intensity light is used .
  • Furthermore, if light had the correct frequency to cause photoemission, then it was found that increasing the intensity of the light merely increased the number of electrons emitted, but did not increase the kinetic energy of the emitted electrons. The photoelectric effect was explained by Albert Einstein. He recognized that light was apparently transferring momentum to the electrons, which is clearly a particle-like behavior, but the amount of momentum somehow depended on frequency, which is a wave-like property… Einstein’s explanation relied heavily on work done by Max Planck… The Photoelectric Effect
  • Max Planck About 1900, Planck was studying the radiation of heat energy from high temperature objects, called “black-body radiation.” He discovered that heat did not radiate in a continuous fashion as “classical” physics predicted. Instead, Planck found that: Quantized Energy Changes The energies of the vibrating molecules that make up the heated substance were quantized , meaning, the energy could only change by small, discrete amounts of energy. These allowed energy states were called quantum states .
  • Quantized Energy Changes “ Classical” physics said that an object’s energy should change as it absorbs or emits heat energy in a smooth, continuous fashion, like a ball rolling up or down a hill. Instead, Planck showed that these energy changes were more like a stair-step, that is, they were quantized. continuous changes quantized changes 1 quantum
  • The molecules of the heated object, in turn, could only emit energy in discrete units of light energy called quanta (now called photons ). They do so by “jumping” from one quantum state to another. The minimum energy change between these quantum states was found to be proportional to the frequency of the emitted photon: Thus, a quantum = minimum energy that an atom can absorb or emit. Quantized Energy Changes E = h  where ν = frequency (Hz) and h = Planck’s constant = 6.626x 10 -34 Js
  • Einstein argued that light acted as a particle (i.e., a photon ) with an energy that depended on its frequency, according to Planck’s idea: E = h  . Einstein called the energy required to remove an electron from the metal its work function, Φ , (phi). The Photoelectric Effect Explained Albert Einstein received his only Nobel prize in Physics for explaining the photoelectric effect. Einstein applied Planck’s idea that energy is quantized, but said this quantization was also true for electromagnetic energy, not just molecular energies. *your text calls Φ the “binding energy.”
  • To produce photoemission, a single photon strikes the electron. If the photon’s energy is less than Φ , then the electron cannot be “kicked loose” from the surface of the metal (the photon has insufficient momentum ). If that one photon’s energy is equal to Φ , then the electron would absorb that energy and photoemission occurs. If the photon’s energy was greater than Φ , then the “excess” energy imparted to that one electron would simply increase its kinetic energy – a single photon would not be able to knock two electrons loose. The Photoelectric Effect Explained
  • If the intensity of the light used is increased, then the number of photons striking the metal is increased, but the energy of each photon is still hv, so each photoelectron would have the same kinetic energy. The Photoelectric Effect Explained Putting it all together, Einstein said the kinetic energy of an emitted photon was equal to the photon’s energy minus the work function required to emit the electron: Kinetic Energy of a Photoelectron: KE = hv – Φ
  • The unit for energy is the joule. The energy for a typical photon of visible light is only about 10 -17 J. The Photoelectric Effect Explained This is such a small number that scientists often use a unit called the electron volt (eV) instead of the joule for these small energy changes. The electron volt is the energy an electron gains being accelerated across a 1-Volt potential. It’s relation to the joule is given by: 1 eV = 1.602 X 10 -19 J
  • Example: When cesium metal is illuminated with light of wavelength 300 nm, the photoelectrons emitted have a maximum kinetic energy of 2.23 eV. What is the work function for Cs? i. KE = hv - Φ ii.  Φ = hv - KE v.  Φ = (6.626 x10 -34 Js x 1.00 x 10 15 Hz) – 3.57 x 10 -19 J = 3.06 x 10 -19 J or 1.91 eV iii. KE = 2.23 eV x 1.602 x 10 -19 J 1 eV = 3.57 x 10 -19 J iv. v = c  = 3.0 X 10 8 m/s 3.00 x 10 -7 meter = 1.00 x 10 15 Hz
  • Emission Spectra and Bohr’s Model of the Hydrogen Atom
  • Line Emission Spectra It was found that when a substance in the gas phase was energized (by heating or passing a current through it) the substance gave off a unique spectrum of EM radiation. But rather than being a continuous spread of energies or wavelengths, these spectra were composed of individual wavelengths, or lines, when dispersed by a prism and cast upon a screen. Line emission spectrum for Hydrogen Emitted light
  • Line Absorption Spectra It was also found that when all wavelengths of light (given off by a so-called “black body radiator” such as an incandescent lightbulb, or the sun) were passed through a gas, certain wavelengths were absorbed, and the rest passed through. emits all  gas prism black-body radiator black lines = those  that were absorbed
  • Each element produces a unique emission or absorption spectrum. In fact, He was first discovered by examining the absorption spectrum of light coming from the sun! Furthermore, the line emission and line absorption spectra for a given element were “negative copies” of each other. Line Spectra emission spectrum absorption spectrum
  • Niels Bohr In 1913, Niels Bohr, a Danish physicist, proposed an explanation for these line spectra, and in so doing, presented a new model for the atom. The Bohr model of the atom worked well in explaining and predicting the spectrum of hydrogen and other one-electron atoms, such as He + , but his model could not explain the behavior of multi-electron atoms. Still, his model served as an important foundation for our current understanding of the atom.
  • The Bohr Model of the Atom 1. Atoms consist of a positively charged nucleus around which the electrons orbit like planets orbiting the sun. 2. The potential energy of the electron depends on the radius of the orbit in which an electron is found. The greater the radius, the greater the potential energy of the electron.
    • There are only certain allowed energy states that an electron can have (and hence, only certain allowed orbits). This is due to Planck’s restriction that energy can only change in small discrete (or quantum ) steps.
    *This same “quantum restriction” also prevents electrons from spiraling into the nucleus, as classical physics would require an orbiting electron to do !
    • The atom has only certain allowable energy levels or “shells.” Each shell has a fixed radius.
    • The energy of the shell depends on its distance from the nucleus – larger shells have higher energy.
    The Bohr Model of the Atom The lowest energy shell is called the “ground state.” Higher energy shells are said to be “ excited states.”
  • (Shells are designated with the letter “n”) The Bohr Model of the Atom 1 2 3 4 5 6 increasing energy first shell is the ground state (n = 1)
  • Energy vs Shell Size in the Bohr Model Note that, although the energy of the shell depends on the radius, it does not increase in direct proportion to the increased shell size. Rather, the energy states get closer and closer together as the shell size increases . n = 1 n = 2 n = 3 n = 4 n = 5 n = 6 5 2 3 4 1 6 shell number Energy
    • An electron moves to a lower energy shell by giving off energy in the form of light.
    The Bohr Model of the Atom
    • An electron moves to a higher energy shell by absorbing energy.
    An electron can move from one shell to another, but it cannot occupy the space between shells. energy energy
  • 6. This means that the energy absorbed or given off must be exactly equal to the difference in energy ( Δ E) between the initial and final shells. The Bohr Model of the Atom - Δ E + Δ E Photon absorbed given off E E
  • This means that the electron can absorb or emit only those photons which have the correct energy (i.e., specific wavelength and frequency) which exactly corresponds to the energy difference between two shells. This explains the line emission and absorption spectrum of an atom! http:// The Bohr Model The energy of a photon depends on its frequency: E = h  . The frequency, in turn, determines the wavelength: λ = c /  . Now, recall:
  • In the case of an absorption spectrum , only those photons of light with an energy exactly equal to an allowed transition from one shell to a higher energy shell are absorbed by the atom’s electrons. The other photons are not absorbed and pass through. As a result, most of the photons pass through the gas sample, but those that are absorbed are now “missing” and leave a black band in the spectrum. This produces the observed emission line. The Bohr Model
  • 1 2 3 The Bohr Model In the absorption spectrum, only certain wavelengths have the correct energy to be absorbed.  E 3-2 Δ E 3-2 Δ E 3-1 Δ E 3-1
  • In the case of an emission spectrum , the applied voltage excites all the electrons in an atom to the highest possible shell. The electrons then relax to lower shells. Any drop in shells is a drop in energy, so the electrons can drop in multiple steps by moving from shell to shell to shell until they arrive at the ground state, or they can drop to the ground state in one step. Each transition releases a photon with an energy exactly equal to the Δ E between shells. This produces the observed emission line. The Bohr Model
  • Excited electrons relax to lower shells and give off photons with an energy equal to  E between the shells. This creates the emission line spectrum. The Bohr Model 4 3 2 1 E 3-4 E 1-3 E 1-4
  • Bohr’s model was able to predict the energy and wavelengths of the emission lines for hydrogen. These emission lines were named by their discoverers. Each grouping of lines fell in a different region of the EM spectrum visible region n f = 2 n f = 1 n f = 3 n f = 4 The Balmer series, in which e- relax from excited states to the second shell ( n f = 2) is the only group with lines in the visible region. The Bohr Model
  • Mathematical Description of Bohr’s Model n (principal quantum number) = 1,2,3,… R H (Rydberg constant) = 2.18 x 10 -18 J The energy of an electron in a specific shell, n is given by: Z = nuclear charge (atomic number) Note the (−) sign in the equation. We define the reference as being a “free electron” no longer attached to the atom ( n = ∞), An electron with n < ∞ is at a lower potential state, and hence E is negative. E n = −R H Z 2 n 2
  • This equation only holds for a one-electron atom. For this reason, the Z term often seems to be “left out” (as it is in your text), since the only neutral atom with one electron is hydrogen, for which Z = 1. Bohr’s equation for the energy of an electron in hydrogen is then: So the Z term is there – it just happens to be “1.” However, this equation also holds for He + or Li 2+ ions, (Z= 2 and Z=3, respectively) since these, too, are one-electron atoms, and you must include the Z 2 term for the energy of the electron in these ions! E n = −R H Z 2 n 2 = −R H 1 2 n 2 = −R H  1 n 2
  • The energy of a photon absorbed or emitted during an allowed electron transition between two shells, n final and n initial is given by: Δ E = E f – E i 1 n 2 f Mathematical Description of Bohr’s Model = −Z 2 R H − 1 n 2 i − Z 2 R H = 1 n 2 i + Z 2 R H 1 n 2 f Z 2 R H − 1 n 2 i − 1 n 2 f  Δ E = h  = + Z 2 R H – (–) = +
  • E photon = (2) 2 x 2.18 x 10 -18 J x (1/25 - 1/9)  = 6.63 x 10 -34 (J•s) x 3.00 x 10 8 (m/s)/ ( + 6.20 x 10 -19 J)  = 3.208 x 10 -7 m or ~ 321 nm Example: Calculate the wavelength (in nm) of a photon emitted by a He + atom when its electron drops from the n = 5 state to the n = 3 state. E photon = h x c /   = h x c / E photon i f Δ E = Z 2 R H ( ) E photon = note that Δ E is negative since e- is relaxing to a lower energy shell… but always use a + Δ E when determining the wavelength using this equation… E photon = Δ E = − 6.20 x 10 -19 J 1 n 2 1 n 2
  • See page 278
  • We have seen that photons of electromagnetic radiation can posses both distinctly wave-like properties and particle-like properties. In 1924, the French physicist Louis deBroglie made the rather daring proposal (as part of his doctoral thesis) that matter could also exhibit this wave-particle duality. Louis deBroglie If one made this rather large leap of faith assumption, then a lot of the paradoxes and problems with the nature of the electrons in atoms began to make some sense…
  • The Wave Nature of an Electron We saw that waves can undergo destructive and constructive interference in such as way as to produce standing waves. In standing waves, an integral number of half-waves can “fit” in a given length, L, such that: L = ½ n λ , where n = 1,2,3,… n = 1 n = 2 n = 3 What would happen if this standing wave were to form a circle, such that L = circumference of the circle?
  • The Wave Nature of an Electron For such a circular standing wave to persist, a whole number of wavelengths would have to fit into the circumference, such that: circumference = 2 π r = n λ, where n = 1,2,3,4, etc. That is, there are only certain “allowed” circumferences for a given wavelength – or put another way, there are only certain allowed wavelengths for a given circumference. Not allowed Allowed Furthermore, the circumference can only change in quantized steps of n λ . This is exactly what Bohr’s model required!
  • In this simplified version of de Broglie’s theory of the atom, the waves are shown only in circular paths around the nucleus. In an actual atom, the standing waves make up spherical and ellipsoidal shells rather than flat, circular ones. Standing Waves
    • There are only certain allowed orbital radii for the electron to occupy (limited by the requirements that the circumference be an integer multiple of the wavelength of the electron’s matter-wave.)
    • The energy of the electron is quantized , (which follows from the fact that the circumference can only change in integer multiples of λ , and E = hc/ λ . )
    • Being a standing wave , the electron is actually not moving – it is in a stationary state – and since it is not moving, it will not radiate energy and thus will not spiral into the nucleus, as would happen otherwise according to the laws of classical physics.
    The Wave Nature of an Electron Properties of electron matter-waves
  • DeBroglie was able to derive an expression for the wavelength of such an electron “matter-wave”: The Wave Nature of an Electron In physics, the quantity mv is called momentum and is usually given the symbol, p . Hence, deBroglie’s expression is often given as: The wavelength of an electron matter-wave is thus inversely proportional to its momentum; the constant of proportionality is simply Planck’s constant, h.  = h mv  = h p
  • So why haven’t we ever seen evidence of this wave-like characteristic of macroscopic matter? Well...consider, for example, the matter-wave of a 0.17 kg baseball after it has been thrown at 30 m/s (~70 mph)... From deBroglie’s matter-wave equation, we have: This is an exceedingly small wavelength! For comparison, this is 1/10,000,000,000,000,000,000 th (that’s 19 zeroes!) the radius of a proton. Thus, the wavelength for ordinary, macroscopic objects is so small that we cannot even measure it, more or less notice it! Louis deBroglie
  • On the other hand, consider the wavelength of an electron’s matter-wave : Example: The mass of an electron is 9.11 × 10 – 31 kg. A typical value for the velocity of an electron about the nucleus of an atom is 1.0 × 10 6 m/s. What would be the wavelength of the electron? This wavelength is about the same as the diameter of a an atom of oxygen, i.e., the wavelength of the electron is about the same size as an atom itself – still a small number, but at least it is measurable! Louis deBroglie
  • Werner Heisenberg and the Uncertainty Principle To locate something, we need to “bounce” something off the object, e.g., airport air traffic control towers “bounce” radar beams (radio waves) off of airplanes to locate and guide them. But note that when any two objects collide, a transfer of momentum always occurs. One object ends up “losing” some momentum, and the other object “gains” an equal amount of momentum (thus the total amount of momentum does not change – momentum, like energy, must be conserved). Just as things started to go well, a new problem arose…
  • Ideally, the object we are trying to locate should not be affected by the particle we are “bouncing” off it. It would not make sense to locate a moving airplane by throwing, for example, a dump truck at it! The dump truck would transfer sufficient momentum to the plane to dramatically alter the plane’s momentum (hence its position, as well)... So we locate the plane using something which cannot transfer a significant amount of momentum to it – say, a photon of electromagnetic radiation in the radio region of the spectrum – i.e., radar! Heisenberg’s Uncertainty Principle
  • In 1927, Werner Heisenberg realized that we run into a real problem when we are trying to locate very tiny objects, like an electron. It seems that the momentum of something as insignificant as a photon of light is sufficient to alter the position of the electron (as is evidenced in the photoelectric effect, among other things)! The very act of trying to find the electron would move it as the photon transferred momentum to the electron! Heisenberg’s Uncertainty Principle
  • Unfortunately, this presents yet another problem: Heisenberg’s Uncertainty Principle To avoid this problem we must use very low momentum photons. But, as deBroglie showed us, to reduce the momentum, we must increase the wavelength of the photons used, since p = h /λ To locate an object precisely, the wavelength of light used must be smaller than the object being located. Not a problem for finding an airplane, but to locate a tiny electron means a very short wavelength is needed. But a shorter wavelength means we have greater momentum. There is a trade-off here: we cannot have both low momentum and short wavelength.
  • Heisenberg reasoned that, if we let the minimum uncertainty in the position, Δ x , of the electron correspond to the wavelength of light (λ) used to locate it, and Δ p be the minimum uncertainty in the momentum of the photon that would be transferred to the electron, then: Since p = h /λ, according to deBroglie’s equation, the combined minimum uncertainty is given by: This is called Heisenberg’s Uncertainty Principle . This means that we cannot simultaneously determine both the momentum and position of an electron with a high degree of precision. Heisenberg’s Uncertainty Principle  x  p > (  ) x ( h/  ) or  x  p > h* *actually, Heisenberg’s equation is  x  p > h/4 π
  • Suppose the speed of an electron in an atom is on the order of 10 6 m/s. It’s momentum would be: This is ~ 1½ times larger than the diameter of a hydrogen atom! Wow! The uncertainty in the position of the electron in a hydrogen atom is larger than the whole atom! This is not good news... mv = (9.11 x 10 – 31 kg)(10 6 m/s) = 9.11 x 10 – 25 kgm/s The uncertainty in the position of that electron can be estimated using Heisenberg’s equation: Δ x Δ p ≥ h/4 π Δ x ≥ h / 4 π Δ p  x ≥ 6.626 x 10 – 34 Js / 1.114 x 10 – 23 kgm/s = 5.79 x 10 – 11 m. Heisenberg’s Uncertainty Principle bad is this, really?
  • To use the Bohr model of the atom to predict the allowed energy states of the electrons we must know both the position and momentum of the electron. Heisenberg just showed us we cannot do this! How do we get around this problem? We need to find a way of understanding the energy of the electron without having to know its position. The solution came when scientists took a closer look at what a matter-wave meant.
  • The Development of the Quantum Mechanical (or Wave Mechanical) Model of the Atom
  • The Quantum Wave Mechanical Model In 1926, Erwin Schrödinger developed a model of the atom, called the Quantum Wave Mechanical Model of the Atom , by plotting the position and energy of an electron based on its wave-like properties. Instead of plotting the electron’s position to find its energy, he plotted the wave’s amplitude (intensity) to find its position. Erwin Schrödinger
  • The intensity of a light wave is proportional to the square of its amplitude: I = kA 2 Schrödinger’s Wave Equation (ψ) The brightest band is always the central one because that’s where the greatest degree of constructive interference occurs. From a particle viewpoint, we could say that this is where the greatest concentration of photons occurs. High intensity means a brighter light. Think of the interference pattern when light is passed through two slits: we get bright bands (constructive interference) separated by dark bands (destructive interference).
  • This means we can relate the amplitude 2 (intensity) with the particle concentration. We could also say there is a greater likelihood, if you will, of finding a photon at the point of greatest intensity. Schrödinger’s Wave Equation (ψ) Schrödinger reasoned that if an electron behaved as a matter-wave, we can develop an equation that describes the energy of the electron in terms of its amplitude. This equation is called a wave function symbolized as ψ (psi).
  • Think of ψ as a “pointer” that gives us the amplitude of a wave at any given point in space, such that ψ ( x,y,z ) gives the amplitude at the 3-dimensional Cartesian coordinates, x,y, and z . Schrödinger’s Wave Equation (ψ) Since ψ is a function of the amplitude of the matter-wave, then ψ 2 is a measure of the intensity of the matter-wave, that is, a measure of the probability of finding an electron at that point. e-
  • As an analogy: Imagine a beehive with only one bee. If you take a time exposure photo of the one bee as it buzzes around, you might get an image like this: Quantum Wave Mechanical Model
  • If I ask you where the bee is right now – you could not say for certain. But you could be about 90% certain that the bee was most likely somewhere inside the circle shown around the bee hive… Quantum Wave Mechanical Model
  • When the probability (using ) of finding an electron vs distance from the nucleus is plotted , we obtain “shells” just like Bohr’s model. Quantum Wave Mechanical Model ψ 2
  • All of the orbitals in all of the shells combined is called the electron cloud. The shells can be sub-divided into regions of highest probability called orbitals. Quantum Wave Mechanical Model
  • The shell is a more specific probability region. This is like asking, “Which floor of the house is the person most likely on, between 3:00 and 5:00 pm? Finding an e- in an atom is a little like guessing where a person might be in a house at any given time. The electron cloud is the broadest, or most general probability region of the atom. This is like saying that at sometime during the day, we know the person will be at home. The orbital is a very specific probability region. This is like asking, “Which room of the house is the person most l likely in, at exactly 1 am?” Note that the person might be in the kitchen getting a late night snack, but the most likely location of the person at that time is in bed. Quantum Wave Mechanical Model
  • Schrödinger’s actual equation was: When Schrödinger’s equation is solved, it turns out that there are 3 terms in the solution. We call these terms “quantum numbers,” and we use these numbers to more fully describe the electron’s energy state. (A 4 th quantum number, unrelated to Schrödinger’s numbers, popped up in later studies.) Keep in mind as we discuss these quantum numbers that they are mathematical solutions to a “probability density plot,” and have no physical reality to them. Schrödinger’s Wave Equation (ψ)
  • Schrödinger’s Wave Equation (ψ) Thus, the atom is no longer viewed as a nucleus with orbiting particles called electrons. We now understand the atom as being a nucleus surrounded by a diffuse “cloud” of electrons in which the electrons act more as a “probability wave” than as a particle.
  • The work of Louis DeBroglie and Erwin Schrödinger in the 1920's showed that the electron could be treated as a wave, specifically a standing wave , rather than as a particle. The electron wave must have a whole number of wavelengths “encircling” the nucleus, or else destructive interference could occur, which effectively destroys the electron-wave...and thus the electron! To increase the number of waves by whole wavelengths means that there are only certain allowed radial distances the electron wave can be from the nucleus and have the waves “fit” without interference --- these are the same restrictions as those arbitrarily imposed by Bohr on his model to create the shells. b Electron Configurations
  • We can describe the matter-wave by its wave function ( ψ ) , which is a three dimensional pattern of “ups” and “downs” and nodes analogous to the standing wave on a vibrating string. It is a function of the amplitude of the “electron wave,” and thus is partly determined by the energy of the electron . Thus, the quantum model, like the Bohr model, also has shells. However, these “shells” are not of exact dimensions, as was the case with the Bohr model – they are merely “probability regions.”
  • The square of the wave function, ψ 2 , gives the probability distribution mapping of the electron in space (as an analogy, the square of the amplitude of a light wave equals the intensity of that wave, which essentially is where there is the greatest density of photons). The “point by point tabulation” of the most probable location of an electron of a specified energy is called an orbital . Note that the orbital is simply a “mapping” of the most likely position for the electron – like mapping the most likely location of a bee in a time lapsed photo of a bee-hive. An orbital is a mathematical construct, and has no physical reality to it.
  • To describe the location of any given point, we need three dimensions. Using polar coordinates , these dimensions are radial distance, r , longitudinal angle, Φ , and latitudinal angle, θ ( see drawing below ). There are three quantum numbers , which are essentially the solutions for the three dimensions, r , θ , and Φ for an electron of a specified energy state: θ Φ
    • Principal quantum number (n) determines the radial value, r , for an electron. The potential energy of the e – increases with increasing distance, r , from the (+) nucleus. The “direction” of r (north, south, east, etc.) is immaterial – the potential energy depends only on how far away the electron is from the nucleus.
    The Quantum Numbers Thus, the principle quantum number gives the shell number for the electron, and has values n = 1,2,3,4... This is by far the most important factor in determining the energy of the electron.
      • 2. Angular (azimuthal) quantum number ( l ) determines the longitudinal angle, θ . This angle depends on the electron’s angular momentum .
      • Angular momentum is a bit abstract – you are most familiar with its conservation in rotating systems: it is conservation of angular momentum that prevents gyroscopes and bicycles from tipping over when the gyroscope or tires are rotating fast, but when the gyroscope or bicycle tires are not rotating, there is no angular momentum and they fall over.
    The Quantum Numbers
  • The values for l depend on the values for the principle quantum number, n, as follows: l = 0…(n -1) Thus
    • when n = 1 l = 0…(1-1) = 0, that is, when n = 1, l = 0
    • when n = 2, l = 0, … (2-1) = 1, so when n = 1, l has two possible values: 0 and 1
    • when n = 3, l = 0, 1, … (3-1) = 2, so when n = 1, l has three possible values: 0, 1 and 2
    … etc. The Quantum Numbers
  • s -subshell is made up of s-orbitals which are spherical shape p -subshell is made up of p-orbitals which are &quot;peanut&quot; shaped d -subshell is made up of d-orbitals which are &quot;double peanut&quot; or &quot;donut peanut&quot; shaped f -subshell is made up of f-orbitals which are &quot;flower&quot; shaped The angular quantum number defines the subshell of n , and determines the SHAPE of the orbital . The Quantum Numbers
  • Remember, the “shape” of these orbitals is just the probability distribution mapping of the most likely location of an electron with the specified angular momentum ( l -value) – the orbital has no other physical “reality” to it beyond this. There are no little spheres or peanuts, etc. inside the atom! a p-orbital electron distribution plot The Quantum Numbers
  • The energy of the electron's shell puts limits on the angular momentum of the electron, hence, as we noted earlier, different shells can contain different types of subshells (orbital types): The Quantum Numbers s, p, d, f 0, 1, 2, 3 4 s, p, d 0, 1, 2 3 s, p 0, 1 2 s 0 1 orbital types l -values shell (n-value)
    • Magnetic quantum number ( m l ): a charged particle in motion creates a magnetic field. These magnetic interactions between the electron and the nucleus adds a second “directional” component, Φ , to the orbital’s angular momentum.
    What this means is that the magnetic quantum number gives the spatial orientation of each orbital. Different orbitals have different numbers of possible orientations in space. Thus, each subshell is made up of one or more orbitals, and each orbital is oriented differently than the others. The Quantum Numbers
  • m l = (– l …0..+ l ) The magnetic quantum number (and hence, the number of orientations possible for each orbital type) depends on the angular number, l, and can have the following values: The Quantum Numbers 7 -3, -2, -1, 0, +1, +2, +3 f = 3 5 -2, -1, 0, +1, +2 d = 2 3 -1, 0, +1 p = 1 1 0 s = 0 # of orientations m l values for orbital type orbital ( l -value)
  • The three orientations of p-orbitals The five orientations of d-orbitals The Quantum Numbers
  • 4 . Spin quantum number (m s ): In addition to locating the electron based on its energy, a fourth quantum number was required to account for a property known as electron spin . A moving electron creates a magnetic field. The random motion of the electron as it moves around the nucleus produces random magnetic fields that tend to cancel each other out. However, another way an electron can move is to spin -- although the e – is not actually spinning like a top, it produces a consistent magnetic field as if it were. The Quantum Numbers
  • N if the e- spins clockwise, it produces a magnetic field pointing up (we call this spin up ). N if the e- spins counter clockwise, it produces a magnetic field pointing down (we call this spin down ). The Quantum Numbers
  • The spin quantum number can have one of two values: it is either + ½ or – ½ , depending on whether the electron’s magnetic field has spin “up” or spin “down.” From a chemical point of view, the spin of the electron makes no difference in how that atom behaves. However, what is important for our purposes is that, in a given orbital, no two electrons can have the same spin. The Quantum Numbers
  • We know two electrons in a given orbital will repel each other due to their like charges. But if the electrons are also repelling because their magnetic fields both point in the same direction, that is just too much repulsion to handle. Instead, the electrons will pair up with their spins opposite (we say they are “spin paired” ). This means that, magnetically, the two electrons are actually attracting each other, which helps off-set the electron-electron repulsion and stabilizes the electrons in the orbital. The Quantum Numbers
  • This is called the Pauli Exclusion Principle, which states that no two electrons in an atom can have the exact same four quantum numbers. At least one (typically the spin) must be different. This means that an orbital can have a maximum of two electrons, and those two electrons must be spin-paired – that is, their spin quantum numbers cannot both be the same value. eg: only one electron can be in the first shell (n = 1), in the p x -orbital, (m l = -1) of the p-subshell, ( l = 1), with its spin up (s = + ½). The Quantum Numbers
  • Thus, since there are at most two electrons per orbital, we can determine the maximum number of electrons per subshell or shell. The Quantum Numbers 1 s-subshell x 1 s-orbital x 2 e- per orbital = 2 e- 1 s-subshell x 1 s-orbital x 2 e- per orbital = 2 e- 1 st shell: 2 nd shell: 1 p-subshell x 3 p-orbitals x 2 e- per orbital = 6 e- maximum number of electrons = 8 e-
  • 1 s-subshell x 1 s-orbital x 2 e- per orbital = 2 e- 3 rd shell: 1 p-subshell x 3 p-orbitals x 2 e- per orbital = 6 e- maximum number of electrons = 18 e- 1 d-subshell x 5 d-orbitals x 2 e- per orbital = 10 e- 1 s-subshell x 1 s-orbital x 2 e- per orbital = 2 e- 4 th shell: 1 p-subshell x 3 p-orbitals x 2 e- per orbital = 6 e- maximum number of electrons = 32 e- 1 d-subshell x 5 d-orbitals x 2 e- per orbital = 10 e- 1 f-subshell x 7 f-orbitals x 2 e- per orbital = 14 e- The Quantum Numbers
  • Mathematically, the maximum number of electrons in any given shell follows the formula: # of electrons = 2n 2 Thus n = 1 holds 2 x (1) 2 = 2 e- n = 2 holds 2 x (2) 2 = 8 e- n = 3 holds 2 x (3) 2 = 18 e- n = 4 holds 2 x (4) 2 = 32 e- The Quantum Numbers
  • Among the Elements Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display. Periodic Relationships Chapter 8
  • In the latter part of the 19 th century, chemists began trying to organize the elements into some kind of pattern. John Newlands (1838-1898) noted that, if the elements were arranged in order of increasing atomic mass, every eighth element had similar properties. He called this the Law of Octaves. However, this law only worked for the first 20 elements, and then broke down. Most chemists did not pay much attention to it. The Search for Patterns Among the Elements
  • Dmitri Mendeleev (1834-1907) The first chemist to develop a meaningful pattern to the elements was Dmitri Mendeleev.
  • 1. First, he arranged the elements in order of increasing atomic mass. 3. The properties were more important than the mass – this resulted in several “blank” spaces. Mendeleev realized that these blanks represented missing elements . Mendeleev 2. Elements that had similar properties were placed in the same column or group.
  • For example… Arsenic (As) followed zinc (Zn) by mass – but arsenic belonged in the group with phosphorus (P) based on its properties -- this left two blanks below Al and Si… Mendeleev
  • Mendeleev
    • Mendeleev noticed trends in the properties of the elements in his table.
    For example: he noticed that the reactivity of some elements, like Na, Li and K, located on the left side of his table, increased as you moved down the column in which they were located. He also noted that all of these elements formed compounds in a one to one ratio with chlorine: LiCl, NaCl, KCl, RbCl, etc.
  • On the other hand, oxygen was more reactive than sulfur, which was more reactive than selenium, etc. – for these elements, located at the right side of the table, reactivity seemed to increase as you moved up the column. In addition, these elements all formed compounds with Na in a two to one ratio: Na 2 O, Na 2 S, Na 2 Se, etc. Mendeleev
  • He also noted that the density and melting points of the elements generally increased as you moved down a column, and that they increased, and then decreased again as you moved across a period. Mendeleev these elements have the highest densities and melting points within their rows increasing density and melting points
  • Once we knew what the properties of these missing elements might be, it was easier to search for and find the elements. As a result, these two elements, (which we now know as gallium and germanium) were discovered within a short time. Mendeleev
    • Based on the observed trends, Mendeleev was able to predict the properties of the missing elements below Al and Si in his table.
  • Predicted and actual properties of the missing element eka-silicon (now called germanium) based on Mendeleev’s periodic table trends: Mendeleev GeO 2 XO 2 formula with oxygen 5.6 g/cm 3 5.5 g/cm 3 density 72.6 amu 72 amu atomic mass actual predicted property
  • When the Elements Were Discovered
  • Modern Periodic Table Mendeleev believed the properties of the elements were a periodic function of their atomic mass. There were some elements that were “ glitches ” to this hypothesis. For example , based on its properties, iodine clearly goes in the same group as F, Cl and Br -- but iodine’s mass of 126.9 amu is LESS than the atomic mass of Te (127.6 amu). When ordered by mass, iodine should follow Te, rather than precede it. This would place I in the same group as S and O, whose properties are not at all similar to iodine.
  • Modern Periodic Table In 1912, Henry Moseley proposed that the elements be listed in order of increasing atomic number rather than increasing atomic mass. When he did this, the glitch with iodine and tellurium disappeared, since iodine’s atomic number is greater than the atomic number of tellurium. The Modern Periodic Law states that the chemical and physical properties of the elements are a periodic function of their atomic number.
  • Organization of the Periodic Table
  • Recall we learned earlier that the periodic table is “broken down” into periods (rows) and groups (columns): period group
  • As we have seen, the periodic table can be divided into s, p, d and f “blocks” based on which subshell the highest energy electron in the atom occupies: Organization of Periodic Table
  • Periodic Classification of the Elements Depending on the type of subshell being filled, we can categorize the elements into several sub-groups: Representative elements (also called the main group elements) are elements in Group 1A through 7A. These elements have incompletely filled s or p subshells in their valence ( outermost ) shell. The representative elements can be further broken down into families, each with similar but unique valence electron configurations, as follows:
  • Families of Representative Elements 1A = alkali metals: ns 1 electron configuration 2A = alkaline earths: ns 2 electron configuration 3A = boron family: ns 2 np 1 electron configuration 4A = carbon family: ns 2 np 2 electron configuration 5A = pnictides: ns 2 np 3 electron configuration 6A = chalcogens: ns 2 np 4 electron configuration 7A = halogens: ns 2 np 5 electron configuration Classification of the Elements note that for representative elements, the group A number gives the number of valence electrons for atoms within that family.
  • Classification of the Elements The transition metals are the elements in groups 3-11 and are characterized by having incompletely filled d-subshells, or readily form cations with incompletely filled d-subshells. Note that Zn, Cd and Hg are technically not transition metals, although many texts label them as such. The noble gases are actually not representative elements; these elements have a completely filled s and p-subshell (ns 2 np 6 ). Recall that this is the most stable electron arrangement for any atom.
  • The inner-transition metals or the rare earths as they are often called, are those elements in which the f -subshell is being filled . Classification of the Elements There are two rows:
    • The lanthanides, which are the elements following lanthanum (La)
    • 2. The actinides, which are the elements following actinium (Ac).
  • The Major Classes of Elements
  • Periodic Trends It turns out that the electron configuration (shell and subshell) of an atom, especially its valence shell configuration, is the most important factor in determining many of the physical, and most of the chemical properties of a given atom. We can identify certain trends that arise which also relate to the electron configuration of the atoms. So, before we look at the specific characteristics of specific atoms or families, lets look at some of the general trends which we find within the periodic table as a whole.
  • Within a family, the trends are dependent on which shell the valence electrons are in. Periodic Trends For most trends, there are two main factors that must be examined: Across a period, the trends depend on the number of electrons in that shell, and on the number of protons in the nucleus for that particular shell.
  • Isoelectronic series Atoms and ions that have the same number of electrons (hence, the same ground state electron configuration) are said to be isoelectronic. e.g.: Na + , Al 3+ , O 2- and N 3- all have a total of 10 electrons – that is, they all have the same electron configuration as [Ne]. Example: what would be an isoelectronic series for elements # 19 – 25? K + Ca 2+ Sc 3+ Ti 4+ V 5+ Cr 6+ Mn 7+ = [Ar]
  • Effective Nuclear Charge ( Z eff ) Electrons in interior filled shells (which occupy the space between the nucleus and the valence electron) act to “screen” the nucleus from the valence electrons. Periodic Trends Each interior or “screening” (-) electron partially cancels out or reduces the effective (+) charge acting on the valence electron, which weakens the attractive forces acting on that electron. In addition, electrons in the same shell can repel each other, which further limits the net attraction a given electron experiences towards the nucleus.
  • Effective Nuclear Charge Effective nuclear charge ( Z eff ) is the “net” positive charge felt by a valence electron, and is given by: Z eff = Z - σ Where Z = atomic number (nuclear charge) and σ = shielding constant. Electrons in the same shell do not shield each other nearly as effectively as those in interior shells, so to a very good approximation, the effective nuclear charge can be given as: Z eff  Z – (number of inner shell screening electrons)
  • Valence electron Screening electrons Consider sodium (Na). Na has 11 protons and 11 electrons. Of the 11 electrons, 1 is a valence electron, and the remaining 10 are screening electrons in interior filled shells, (2 + 8 = 10) Z eff = Z – interior e- Z eff = (11-10) = +1 For example: Effective Nuclear Charge
  • Finding the effective nuclear charge of an element 10 10 18 2 +1 +2 +6 +2 Z eff screen e- Z Element 2 +11 28 +6 Effective Nuclear Charge Mg 12 S 16 Ca 20 Li 3 Note: even though Ca has 8 more protons than Mg, they have the same Z eff (+2) . The valence electrons in both atoms “feel” as if only 2 protons are pulling on them. Al 3+ 13 Se 2- 34 pay careful attention to Z eff of cations and anions!
  • Trends in Z eff : Within a family Consider the trend within the Group 1A family H (Z = 1) has 1 valence electron, with no inner shell screening electrons, so it has a Z eff of (1-0) = + 1 Li (Z = 3) has 1 valence electron, with 2 inner shell screening electrons, so it has a Z eff = (3-2) = + 1 Na (Z = 11) has 1 valence electrons, with 10 inner shell screening electrons, so it’s Z eff = (11 -10) = + 1 Effective Nuclear Charge
  • We see that the effective nuclear charge stays constant within a family. Even though we are adding protons, we are also adding interior shell screening electrons at the same time, and so the Z eff does not change. Effective Nuclear Charge
  • Trends in Z eff : Across a period As you move across a period, the number of protons (Z) increases steadily. The added electrons are entering the same shell, so the number of interior shell screening electrons does not change . Na (Z = 11) Al (Z = 13) P (Z =15) Cl (Z = 17) [Ne] 3 s 1 [Ne] 3 s 2 3p 1 [Ne] 3 s 2 3p 3 [Ne] 3 s 2 3p 5 Effective Nuclear Charge
  • Na (Z = 11) Al (Z = 13) P (Z =15) Cl (Z = 17) Since Z eff = Z – core electrons, we see that the effective nuclear charge steadily increases from left to right across a period. Trends in Z eff : Across a period Effective Nuclear Charge = +1 Z eff = (11-10) = +3 Z eff = (13-10) = +5 Z eff = (15-10) = +7 Z eff = (17-10 )
  • Trends in Z eff across the transition metals For transition elements, the electron configuration is: ns 2 (n-1)d x . All the added electrons as you move across the period are entering the (n-1) d-subshell – that is, they are entering an interior shell and they therefore act as screening electrons. As a result, although the number of protons increases across the period, the Z eff stays nearly constant, just as it does down a family. Thus Z eff will be ≈ +2 for most transition metals, and +1 for Groups 6 and 11 (why?) Effective Nuclear Charge
  • Atomic Radius Defining the size of an atom is difficult, because the electron cloud is diffuse. We use the following definitions for atomic radius:
    • two adjacent metal atoms, or
    • b) two adjacent atoms in a diatomic molecule
    atomic radius = ½ the distance between the nuclei in:
  • Trends in the radius (size) of the elements Size Trends Note that, like Z eff , there are two distinct trends in the size of the atoms: 1. Within a family 2. Across a period The radius of the elements ranges in size from 31 picometers (He) to about 280 picometers (Fr). A picometer is 10 -12 m.
  • Size Trends Size trend within a family The size of the atoms gets larger as you move down a family valence shell # 1 2 3 The valence electrons are entering successively larger shells – larger shell means larger atomic radius. recall that Z eff ≈ constant within a family, so it has no significant effect on size trend within a family. H Li Na
  • Size Trends Size trend across a Period As we move across a period from left to right, the electrons are entering the same shell, and the number of protons is increasing. We have already seen that this causes the Z eff to increase. This increase in Z eff pulls more strongly on the electrons. The electron cloud is pulled inward, and the size of the atom gets smaller as you move across the period. Na (Z = 11) Al (Z = 13) P (Z =15) Cl (Z = 17) Z eff = +1 Z eff = +3 Z eff = +5 Z eff = +7
  • Increasing size Increasing size General Size Trends for Representative Elements
  • Size Trends So far, we have only looked at the “general” trends in size. There are some “adjustments” in this general trend that we need to include. The “good news” is the fact that the quantum model of the atom predicts both the general trend AND these adjustments. This is one piece of physical evidence that the quantum model is the correct model.
  • Size Trends Although the atoms tend to get smaller due to the increasing Z eff as you move from left to right across a period, the rate at which they get smaller tends to level off. As you move from left to right, the number of valence electrons increases. This causes an increase in electron-electron repulsion, which tends to limit how much the electron cloud can be pulled inwards, despite the increasing Zeff. Electron-electron repulsion effects
  • Thus, the trend in decreasing size as you move across a period “levels off” as you move farther to the right. increasing electron-electron repulsion Size Trends atomic number radius
  • Size trends across the Transition Metals As you move across the transition metals (and the inner-transition metals, as well) we have already seen that Z eff ≈ constant. Thus, since all the added electrons are entering the same (n-1) shell as you go across a given the period, the size of the elements across a given period among the transition metals is roughly constant . Size Trends
  • Putting it all together, we obtain the following graph of atomic radius vs atomic number… Size Trends
  • Size Trends The Atomic Radii of Elements 1-86 *note: unfortunately, the graph in your textbook has several mis-plotted points, so it does not look like the graph above…but it should . Note the jump in size as you move to a new shell… … and the fairly steady decrease in size moving across a period, which levels off at the far right. … and the roughly constant size across the transition metals
  • Cation is always smaller than the atom from which it is formed. If we compare the size of ions with the neutral atom from which the ion was formed, we find that the… Anion is always larger than the atom from which it is formed. Size of Ions + -- Size trends of ions
  • Explanation: Size of the cation A cation is formed when an atom loses valence electrons. When an atom loses its valence electrons, it tends to lose the entire valence shell – so the cation is much smaller than the original atom. Na : [Ne]3s 1 … only the 2 inner shells remain When this 3 rd shell e- is removed… Size of Ions + Na + : [Ne] 3s 0 Note also that the Zeff acting on the remaining valence shell electrons is now (11-2) = +9, so the 2 nd shell is pulled inward, making the cation even smaller.
  • Explanation: size of the anion An anion is formed when an atom gains electrons in its valence shell. But the atom gains electrons without gaining any protons – there is no increase in Z eff to off-set the electron-electron repulsion of the extra electrons in the valence shell. This causes the electron cloud to “swell” and as a result, the anion becomes larger than the original atom. Size of Ions This electron is added to the valence shell Cl Cl —
  • Radii of some common cations and anions cations anions neutral atoms neutral atoms
  • Ionization Energy (IE) Ionization energy is the energy required to remove a valence electron from an atom (in the gas phase, to insure that there are no interactions with neighboring atoms that might influence the IE.). Ionization Energy 1st IE: energy + X ( g )  X + ( g ) + e - Multiple electron atoms will have a different IE for each electron removed: 2nd IE: energy + X + ( g )  X 2 + ( g ) + e - 3rd IE: energy + X 2+ ( g )  X 3 + ( g ) + e -
  • The force of attraction for an electron depends on the Z eff and the distance of the electron from the nucleus . *k is just a constant of proportionality To understand the trends in ionization energies among the elements, we must recognize that the strength of the attraction between the nucleus and the electron is what determines the energy required to remove that electron. Trends in Ionization Energies Force of attraction = Z eff r 2 k
  • The greater the Z eff , and the closer the electron is to the nucleus, the stronger the attraction. The stronger the attraction, the more energy will be required to pull the electron loose. Ionization Energy high Z eff IE close to nucleus Unnhh…! Let go already!! this is my electron! e-
  • 1st Ionization Energy within a Family Ionization Energy
    • Within a family, the Z eff remains constant
    • As you move down a family, the valence electrons are farther from the nucleus (larger shells), which weakens the attraction.
    The 1st Ionization energy DECREASES as you go down a family
  • 1st Ionization Energy Across a Period Ionization Energy
    • Across a period, the Z eff steadily increases from left to right.
    • Across a period, the size of the atoms gets steadily smaller as you move from left to right.
    • Both of these factors increases the attraction for the valence electrons.
    1st Ionization energy INCREASES as you move from left to right across a period.
  • General Trend in 1st Ionization Energy Increasing IE Increasing IE
  • Variation of Ionization Energy with Atomic Number Note that there are several irregularities, but the general trend shows IE increasing across a period and decreasing down a family
  • Irregularities in the General Trends for 1st IE It turns out that the “irregularities” in the first ionization energy trends are actually predicted by the quantum model! Consider the apparent “glitch” in the 2A family: The general trend would have the IE of Al greater than that of Mg, but in fact, Mg is greater than Al… Na = 495.9 kJ/mol Mg = 738.1 kJ/mol Al = 577.9 kJ/mol ??
  • The explanation for this apparent glitch is as follows:
    • The electron being removed from group 2A atom is an ns 2 electron – meaning, you are breaking up that very stable filled sub-shell arrangement, so it takes a little more energy than expected.
    • In addition, the electron being removed from a group 3A atom is an ns 2 np 1 electron. It turns out that s-orbitals have greater electron density near the nucleus than do p-orbitals (we say the s-orbital electron is is able to “penetrate” better.)
    Irregularities in IE Trends
  • As a result, s-orbital electrons can partially screen a p-orbital electrons , even though they are in the same shell. Thus, it takes less energy to remove the partially screened p-orbital electron. The combination of slightly higher IE for Mg and slightly lower IE for Al makes a marked “glitch” in the trend. Irregularities in IE Trends
  • A second “hiccup” in the general trend of increasing IE across a period occurs among members of the Group 5A elements. Consider the apparent “glitch” in the first IE of P: The general trend would have the IE of S greater than that of P, but in fact, P is greater than S… Irregularities in IE Trends Si = 786.3 kJ/mol P = 1012 kJ/mol S = 999.5 kJ/mol ??
  • The explanation for this apparent glitch is as follows: 1. In group 5A atoms there are 3 p-orbital electrons – one each in the p x , p y and p z orbital, following Hund’s rule. group 5A Irregularities in IE Trends Removing an electron from a group 5A atom involves breaking up the special stability of exactly ½ filled sub-shells, so its IE is higher than expected .
  • 2 . O n the other hand, group 6A atoms have one set of spin-paired electrons in one p-orbital: The spin-paired electron in group 6A has a lower than expected IE , since removing one of these electrons would eliminate the electron-electron repulsion effects on the remaining electron, lowering its potential energy state. Irregularities in IE Trends Group 6A
  • Again, the higher than expected IE for group 5A elements, coupled with lower than expected IE for group 6A elements makes a marked “spike” in the graph of IE . Finally, we note that the IE is relatively constant across the transition metal elements, reflecting the fact that these atoms have relatively constant size and Zeff . Irregularities across the Transition Metals Irregularities in IE Trends
  • However, the IE of group 13 (3A) is much lower than that of group 12 for the same reason that the IE of group 3A is lower than that of group 2A – the electron being removed is coming from a higher energy p-subshell. The IE of group 12 is also higher than expected , because you are breaking up the special stability of a filled s,d subshell combination. This produces the marked “glitch” between groups 12 and 13. Irregularities in IE Trends
  • Irregularities in 1st Ionization Energy Trends
  • Another important trend appears when we examine multiple ionization energies ( i.e., 1 st , 2 nd , 3 rd , etc.) for atoms in a given family Multiple Ionization Energies As you remove an electron, the amount of electron-electron repulsion in the atom’s remaining valence shell electrons is reduced. This, in turn, reduces the shielding effect, σ , and increases the attraction for whatever valence electrons remain. As a result, successive ionization energies are higher: IE 1 < IE 2 < IE 3 < …
  • Multiple Ionization Energies Removing an electron may leave the atom with a filled or half-filled subshell. Removing the next electron will require a somewhat larger than expected jump in energy, due to the special stability of such filled and exactly half-filled subshells. Example: consider the successive IE for oxygen: 13,000 11,000 7,470 5,300 3,390 1,314 +2,000 +3,530 +2,170 +1,910 +2,076 increase: O 5+ O 4+ O 3+ O 2+ O + O [He]2s 1 [He]2s 2 [He]2s 2 2p 1 [He]2s 2 2p 2 [He]2s 2 2p 3 [He]2s 2 2p 4
  • If removing an electron leaves the atom with a noble gas core (s 2 p 6 valence configuration) then removing the next electron will require a much larger jump in energy, since a noble gas core electron configuration is the most stable possible configuration. Compare the first four ionization energies for Na and Al: Multiple Ionization Energies [He]2s 2 2p 4 [He]2s 2 2p 5 [Ne] [Ne]2s 1 Na [Ne] [Ne]2s 1 [Ne]2s 2 [Ne]2s 2 2p 1 Al 11,600 2,750 1,820 578 9,540 6,900 4,560 496 4 th IE (kJ/mol) 3 rd IE (kJ/mol) 2 nd IE (kJ/mol) 1 st IE (kJ/mol)
  • see page 331
  • Electron Affinity (EA) Electron affinity is a measure of the tendency of a single neutral atom (in the gas phase) to gain an electron in its valence shell. The force of attraction for an “added electron” will also depend on the Z eff pulling on it, and how far from the nucleus the added electron will be (i.e., which shell the electron goes in).. Electron Affinity *k is just a constant of proportionality Force of attraction = Z eff r 2 k
  • Just as was the case for ionization energy, the greater the Z eff , and the closer the electron is to the nucleus, the stronger the attraction. The stronger the attraction, the more likely an atom will gain the extra electron and form an anion. Electron Affinity Electron Affinity high Z eff IE close to nucleus high Z eff close to nucleus I’ll take you! Sob…no one wants me… e-
  • Measuring Electron Affinity Experimentally, EA is measured by first adding an electron to the neutral, isolated atom – in the gas phase -- then measuring how much energy it takes to “pull” the electron off again. That is, EA is the ionization energy of an anion . e.g., F − (g) F (g) + e−  H = +328 kJ/mol As a result, EA has a positive  H. The larger the +  H for the process, the more stable the anion, which means it has a high electron affinity.
  • Electron Affinity Some atoms do not readily accept an electron, so it is difficult to measure their electron affinity. These atoms are assigned a (-) EA, which cannot be precisely determined (what is the energy required to remove an electron from an atom that will not gain an electron?) EA values for some representative elements
  • Electron Affinity Down a Family
    • Within a family, the Z eff remains constant
    • As you move down a family, the added electron would enter a valence shell farther from the nucleus (larger shells), which weakens the attraction.
    Electron Affinity DECREASES as you go down a family Electron Affinity
  • Electron Affinity Across a Period Electron Affinity
    • Across a period, the Z eff steadily increases from left to right.
    • Across a period, the atom gets steadily smaller as you move from left to right, so the added electron would be ever closer to the nucleus.
    • Both of these factors increase the attraction for an added electron.
    Electron Affinity INCREASES as you move from left to right across a period.
  • Increasing EA Increasing EA Electron Affinity (General Trend)
  • Irregularities in EA Trends There are more irregularities in EA trends than for IE trends. Some of these irregularities can, however, be explained. If an atom gains a stable filled or half-filled subshell when it gains an electron, it will require more energy to remove it, and hence it’s EA is greater as well. Thus, the EA of Group 1A, 4A and 7A are higher than might otherwise be expected.
  • If an atom already has a stable filled or half-filled subshell, then it’s tendency to gain another electron is very low. As a result, the EA of Group 2A, 5A and 8A elements tends to be lower than expected . Coupled with the higher than expected EA values for Group 1A and 7A, we can account for the “glitches” in the general trend, so that the EA of carbon is greater than that for nitrogen, even though nitrogen is a smaller atom with a higher Z eff . Irregularities in EA Trends
  • Electron Affinity for the Group 8A Family The EA values for the noble gases deserves a special note. As we might expect, since the noble gases have a filled s and p subshell, their tendency to gain another electron is quite low. If we examine the Z eff acting on any electron that might be gained, we see a secondary reason for the low EA value for noble gases… Irregularities in EA Trends
  • added electron must go to a new shell Z eff = 10 - 10 = 0;  EA < 0 Since the valence shell is already filled for all noble gases, the added electron would have to enter the next larger shell. The Z eff acting on the added electron would be zero because all the other electrons in the atom would now act as screening electrons: Irregularities in EA Trends Neon (Z=10) filled shell! all electrons are now screening e-
  • Irregularities in Electron Affinity Trends Glitches: EA of 2A, 5A and 8A are much lower than expected EA of 1A, 4A and 7A are much higher than expected Be N Ne
  • The observed trends in the radius of the atoms, the first ionization energies, multiple ionization energies, and the electron affinities for the atoms are all in agreement, including the “glitches,” with what the quantum mechanical model of the atom would predict. This is strong evidence in support of our quantum mechanical model of the atom. Finally, we note that…
  • Variation in the Physical & Chemical Properties of the Representative Elements
  • General Trends in Physical Properties Before we examine the chemical properties of individual groups in detail, there are some general trends in the physical properties of the elements which we can point out. The periodic table is broken down into three major groups: metals, non-metals and metalloids. Note that metals are on the left; non-metals are on the right.
  • Classification of the Elements
    • General Properties of Metals
    • malleable : can be hammered into thin sheets
    • ductile: can be pulled into long, thin wires
    • lustrous: shiny (has a metallic “sheen.”)
    • good conductors: allow heat and electricity to easily pass through them
    • tend to lose e- due to their low IE, they tend to lose electrons in reactions to form cations
    Classification of Elements
  • The density & melting point of metals tends to increase as you go down a family (atomic mass increases faster than atomic radius). Across a period, these values tend to increase to a maximum among the d-block transition metals, and then slowly decrease again. (Transition metals are relatively small and have a close-packed structure.) Li is the lightest metal known (0.53 g/cm 3 ), and osmium has the highest density of any element (22.6 g/cm 3 ). Tungsten has the highest melting point of any known element (3,410 ºC); gallium will melt in your hand. Metals
  • Metals sodium copper gallium gold beryllium magnesium
    • brittle (break into fragments when stressed)
    • dull (not lustrous)
    • poor conductors (that is, good insulators )
    • gain e- to form anions (have high EA )
    General Properties of Non-metals Classification of Elements
  • For example, fluorine and chlorine are gases, bromine is the only liquid non-metal, and iodine is a solid (which readily sublimes). The density & melting points of non-metals is somewhat more varied than that for metals. However, like metals, these values tend to increase moving down a family. Non-Metals
      • $ properties are intermediate between those of metals and nonmetals
    • $ many are semi-conductors
    As Sb Si Ge General Properties of Metalloids Classification of Elements
  • Trends in the Metallic Properties of the Elements increasingly metallic in properties…(or less non-metallic)
  • The general trend is for elements to become more and more metallic as you move from right to left across a period. Across the 3 rd period, if we begin at the far right and move to the left, we have the non-metals argon (Ar), chlorine (Cl) sulfur (S) and phosphorus (P); moving farther left we reach the metalloid silicon (Si), then a true metal, aluminum (Al) and continuing, we have the metals magnesium (Mg) and sodium (Na). Classification of Elements
  • increasingly metallic Trends in the Metallic Properties of the Elements
  • Within the Group 4A family, moving downward, we see the first element is carbon (C), a non-metal, followed by silicon (Si) and germanium (Ge), both metalloids, and finally, there are two metals, tin and lead (Sn and Pb). The general trend is for elements in the same family to become more and more metallic as you move downward. Classification of Elements
  • Note that the trend in increased metallic properties follows the same general trend as that for decreasing IE . Because one of the most important chemical properties of metals is that they tend to lose electrons in reactions, it follows that the lower the IE, the more metallic the element! Classification of Elements Across a period , the most reactive metals are those at the far left (Group 1A). In a family , the most reactive metals are at the bottom of the family.
  • Note that the trend in increased non-metallic properties follows the same general trend as that for increasing EA . Because one of the most important chemical properties of non-metals is that they tend to gain electrons in reactions, it follows that the greater the EA, the more non-metallic the element! As a result: Across a period , the most reactive non-metals are those at the far right (halogens); Within a family , the most reactive non-metals are at the top of the family. Classification of Elements
  • A Brief Guided Tour of the Elements
  • Elements with similar valence electron configurations tend to have similar chemical properties. Thus, members of the same group tend to have similar properties. Chlorine reacts very similarly to how bromine reacts; sodium behaves much the same as potassium, etc. Since the electron configuration of an atom – especially its valence shell configuration -- is the most important factor in determining most of the chemical properties of a given atom, it is not surprising that: Group Properties
  • Note also that all the transition metal elements have similar valence electron configurations – nearly every one has an ns 2 (n-1)d x configuration. Thus, as a group, many transition elements are fairly similar in their chemical behaviors. However, the presence of an incomplete d-subshell does affect the properties of the transition metals, so they are different from those of, say Group 2A metals, which also have a valence ns 2 configuration, but do not have any (n-1) d-subshell electrons. Group Properties
  • For example, carbon, silicon, lead and tin all have an s 2 p 2 valence electron configuration, yet carbon is a non-metal, silicon is a metalloid, and lead is a metal! The first element in a given family also tends to be somewhat different from the rest of the elements in the family . This is due to the fact that elements at the top of a family tend to be smaller, with higher IE and EA compared to other elements in the family. Group Properties There are exceptions to this “similar e- configuration means similar properties” group rule.
  • Finally, similarities exist between pairs of elements in different adjacent groups and periods, known as the diagonal rule. Specifically, the first three members of the second period exhibit many similarities to those elements located diagonally below them in the periodic table. The reason for this has to do with the fact that “diagonal elements” have similar charge densities (charge per unit volume); this makes them react similarly with anions. Group Properties
  • Now let us examine specific groups
  • Hydrogen: Its Own Special Group Hydrogen is unique among the elements. It has only one electron, which it loses quite readily. For this reason, we place it in the same vertical row as the alkali metals, which also only have one electron -- but H is not a metal. e.g., HCl + H 2 O -> H 3 O + Cl- H can also gain one electron, like the halogens, to form the hydride ion, H‾. For this reason, some periodic tables place H at the top of both groups 1A and 7A. e.g., 2 NaH + 2 H 2 O -> 2 NaOH + H 2 Probably the most important compound containing H is water 2 H 2 ( g ) + O 2 ( g ) -> 2 H 2 O ( l )
  • Hydrogen is highly flammable . Because it is also lighter than air, it was used in making dirigibles (blimps) for many years. Hydrogen A spectacular accident occurred in New Jersey in 1937 when a spark ignited the hydrogen in the Hindenburg dirigible as it was “docking,” killing many people aboard. Today, we use helium instead, which is not flammable!
  • Group 1A Elements: The Alkali Metals
    • The alkali metals are all metals with an ns 1 valence electron configuration (n >1).
    Increasing reactivity
    • soft, low density metals
    • lose 1 electron to form +1 ions in reactions
    • very reactive; not found in nature in elemental state. Alkali metals are found only in compounds, such as NaCl and KI.
    • react with water to form alkaline (basic) solutions, (that is, they form hydroxides); hence the name:
    The Alkali Metals
      • e.g., 2 Na + 2 H 2 O -> 2 H 2 + NaOH
    • all react with oxygen to form oxides; all but Li can also form peroxides:
    • e.g., Na + O 2 -> Na 2 O 2
  • Group 1A Elements: Alkali Metals 8.6
  • Group 2A Elements: The Alkaline Earths
    • The alkaline earths are all metals with an ns 2 valence electron configuration.
    Increasing reactivity
    • lose 2 electrons to form +2 ions in reactions
    • relatively soft, low density metals (but harder and more dense than the group 1A metals)
    • very reactive, but not as reactive as alkali metals. Alkaline earths are still too reactive to be found in nature in elemental state (they, too, are only found in compounds).
    • Ca, Ba and Sr react with water (Mg reacts with steam) to form alkaline (basic) solutions (that is, hydroxides) – hence the name. Be does not react, however, even with steam.
    • Sr-90, a radioactive isotope of Sr, was released during the Chernobyl accident. This toxic isotope, which is chemically very similar to Ca, is readily taken up by bones, and led to a sharp increase in cases of leukemia.
    The Alkaline Earths
  • Group 2A Elements : Alkaline Earths
  • The Transition Metals Fe Co Cu Ni Sc Ti V Cr Mn 3 4 5 6 7 8 9 10 11 12
    • The transition metals are the elements that make up the “B” groups, or groups #3-11 using the newer numbering system.
    • tend to be harder, with higher melting points than the Group 1A and 2A metals (their smaller size allows them to pack more tightly together).
    • most form colorful compounds (most compounds with representative elements are white)
    • not as reactive as 1A and 2A metals; many found in elemental state, such as gold and silver .
    • like Al, most react with oxygen and form a protective oxide layer that prevents further oxidation (e.g., chrome plating).
    The Transition Metals
    • most can form +2 or +3 ions (often both). The +3 charge state is most stable at the left side, and the +2 state is more common on the right side.
    Charge states in red are the most common. Note that the oxidation state climbs to a maximum at Mn, and then drops back down. The Transition Metals
  • Some Properties & Trends among the 1 st Row Transition Metals
  • transition metal compounds in aqueous solutions Transition Metals Ti 3+ Mn 2+ Cr 3+ Fe 3+ Co 2+ Ni 2+ Cu 2+
  • A Comparison of Group A and Group B metals Consider the electron configurations of Groups 1A and Groups 2A with their B group counterparts (groups 11,12): Note that the A groups and B groups differ only in the presence or absence of a filled d-subshell of electrons. [noble] (n-1)d 10 ns 2 group 2B (12) metals [noble] ns 2 group 2A metals [noble] (n-1)d 10 ns 1 group 1B (11) metals [noble] ns 1 group 1A metals electron configuration Group number
  • The d-subshell electrons are not as effective at screening the s-orbital electrons . As a result, the 1B (11)and 2B (12) metals have higher IE than their A-group counterparts, which makes them less reactive. As a result, although alkali metals are never found in nature in their elemental state, the 1B elements of copper, silver and gold are very commonly found in their elemental state; indeed, they are sufficiently inert to be used in making coins (they are often called the coinage metals for this reason.) Similarly, the alkaline earths are too reactive to be found in their elemental states, but Zn, Cd and Hg (group 2B) are less reactive and can be found in their elemental states. Compare A & B Metals
  • The Rare Earths Elements in the bottom two rows on the periodic table are called the rare earth (metals). The rare earths are broken down into two groups…
    • The Lanthanides : Elements # 58-71 immediately following lanthanum (La).
    • The Actinides : Elements #90-103 immediately following actinium (Ac).
    • Elements with atomic numbers greater than 92 are called the trans-uranium elements . None of these elements are naturally occurring – they are all man-made .
    The Rare Earths
  • Group 3A Elements: The Boron Family
    • These elements all have an ns 2 np 1 configuration. Boron is a metalloid. The rest of the elements in this family are true metals.
    • Aluminum is the most abundant metal in the earth’s crust. It is usually found in the ore, bauxite (Al 2 O 3 ). About 200 years ago, it was so expensive to extract pure Al from bauxite that aluminum was worth more than gold! We have since found cheaper methods of extracting the Al.
    • Al is actually quite reactive, but quickly forms a thin coating of Al 2 O 3 when exposed to air. This protects the Al from further corrosion, making it a good, lightweight, structural metal.
    Group 3A Elements
    • In addition to forming ionic compounds, these metals can also form molecular compounds.
  • Group 3A Elements: The Boron Family 8.6
  • Group 4A Elements: The Carbon Family
    • These elements all have a ns 2 np 2 electron configuration. Carbon is a non-metal, silicon (Si) and germanium (Ge) are metalloids; tin (Sn) and lead (Pb) are true metals.
  • Group 4A Elements
    • Carbon comes in several allotropes. Recall that an allotrope is one of two or more distinct molecular forms of an element, each having unique properties. The allotropes of carbon are: graphite, diamond, and buckminster fullerene (also called “buckyballs” for short!)
    graphite and diamond buckminster fullerene (C 60 ) – its structure looks like a soccer ball
    • Carbon is also unique among the elements in its ability to bond to itself to form long chains. This process is called catenation . This allows carbon compounds to have a wide range of structures, etc., which enables them to have biological roles. The study of carbon based compounds is called organic chemistry .
    • Silicon is the second most abundant element in the earth’s crust (SiO 2 is the chemical formula for sand). Si is also important in the semi-conductor industry, and is an important ingredient in making computer chips.
    • The most stable oxidation state for both C and Si is 4 + . (Recall that this does NOT mean they form +4 ions!)
    Group 4A Elements
    • Lead is a toxic metal. It was once used in paints and also as an “anti-knock agent” in gasoline. Exposing young children to lead can lead to mental retardation and neurological problems, so its use has been cut back drastically in the past 20 years. In ancient Rome, the wealthy used lead pipes to carry water to their homes. Some historians believe the effects of lead poisoning on the people of Rome contributed to the decline and fall of the Roman Empire!
    • Elements at the top of the family are most stable in the +4 oxidation state , while lead, at the bottom of the family, is most stable in the +2 oxidation state (although the Pb 4+ also exists.)
    Group 4A Elements
  • Group 4A Elements: The Carbon Family
  • Group 5A Elements: the Pnictides These elements all have an ns 2 np 3 electron configuration. Elements in the nitrogen family include non-metals, metalloids (As and Sb) and one true metal (bismuth).
    • Nitrogen in its diatomic form (N 2 ) is a chemically inert gas (will not react) which makes up about 70% of our air. Nitrous oxide (N 2 O), or laughing gas, is used by many dentists as an analgesic. Ammonium nitrate (NH 4 NO 3 ) and other nitrates (TNT) are explosives. Ammonia (NH 3 ) is a common cleaning agent.
    • Phosphorus has two allotropes: white (yellow) phosphorus exists as P 4 molecules and must be stored under water because it burns in air with an intense, hot flame. Red phosphorus, the other allotrope, is less reactive.
    The Pnictides
    • Arsenic is used as a pesticide. There is evidence that Napoleon died of arsenic poisoning; it was first believed he was poisoned by a rival, but we now believe the arsenic actually came from his wallpaper, which used arsenic in its dyes!
    The oxoacids of P and N are formed by reacting the oxides with water: N 2 O 5 + H 2 O -> 2 HNO 3 P 4 O 10 + 6 H 2 O -> 4 H 3 PO 4 The Pnictides
  • Group 5A Elements: the pnictide family 8.6
  • Group 6A Elements: The Chalcogens These elements all have an ns 2 np 4 valence electron configuration. Most of the members of this family are non-metals, except for Te and Po. Oxygen is the most reactive element in the family. reactivity increases
    • Oxygen is a diatomic element. It is the most abundant element in the earth’s crust – mainly as silicate rocks (SiO 2 ) -- and makes up about 20% of the air we breathe . Ozone (O 3 ) is an allotrope of oxygen that is present in the upper atmosphere that screens out harmful UV rays. The ozone layer now has “holes” in it, caused partly by pollutants.
    • Most oxides of non-metals form acids in water (the word oxygen means “acid former”). SO 3 is a common pollutant from burning coal that reacts with water in the air to form sulfuric acid (H 2 SO 4 ) – or acid rain:
    The Chalcogens SO 3 + H 2 O -> H 2 SO 4
    • Sulfur comes in several allotropic forms. Crown sulfur (S 8 ), the most common allotrope of sulfur, is used in making matches; it is also added to latex rubber in a process called vulcanization , which makes the rubber less “tacky” and harder (the inventor of this process used the vulcanized rubber to make tires – his name was Charles Goodyear.)
    Group 6A Elements: the chalcogen family The Chalcogens
  • Group 7A Elements: the Halogens Increasing reactivity These elements all have a ns 2 np 5 valence electron configuration. The word halogen means “salt former.” All the halogens are diatomic elements. Bromine is a liquid, iodine and astatine are solids, and fluorine and chlorine are gases.
    • Fluorine is the most reactive of all the non-metals. It is a pale yellow-green gas at room temperature. F is used in making non-stick teflon, and compounds containing F - ions are used to fight tooth decay.
    • Chlorine is a pale green gas that is also quite reactive. Chlorine bleach is actually the compound, NaClO (sodium hypochlorite). Never mix bleach with other household cleaning agents such as toilet bowl cleaners and ammonia – they react with bleach to produce toxic fumes!
    The Halogens
    • Bromine is a corrosive, reddish brown liquid. Compounds containing bromine or chlorine are often used as antibacterial agents in swimming pools and hot tubs.
    • Iodine is a purple solid that sublimes very easily. Tincture of iodine (iodine dissolved in alcohol) used to be a common antiseptic to put on cuts, etc.
    • All the halogens react with water to form binary halic acids:
    • e.g., Cl 2 + H 2 O -> 2 HCl ( aq )
    • Organic compounds containing Cl and F are called chlorofluorocarbons or CFC’s. These compounds, used in making refrigerants, have been shown to destroy the ozone in the earth’s upper atmosphere.
    The Halogens
  • Group 7A Elements: the Halogens
  • Group 8A Elements: The Noble Gases The noble gases all have a filled s 2 p 6 valence shell (except He which is filled with only a 1s 2 configuration). Most are chemically inert (they will not react with anything). As their name implies, these elements are all gases at room temperature -- they have some of the lowest boiling points of any element (e.g., He boils at -269 o C!)
  • The Noble Gases
    • Helium has the distinction of being first discovered in the spectrum of the sun before it was discovered here on earth. The name “helium” comes from “Helios,” the Greek god of the sun.
    • Neon is often used in lighted signs.
  • Properties of Oxides Across a Period There is one last comparison we will make of the properties of the representative elements will help distinguish the behavior of metals from that of non-metals. We will examine the properties of the oxides of the 3 rd period: Na 2 O, MgO, Al 2 O 3 , SiO 2 , P 4 O 10 , SO 3 , and Cl 2 O 7
  • Oxygen tends to form -2 ions, especially when bonded to metals with low IE, such as Na, Mg and Al. These compounds form ionic bonds (as attested to by their very high melting points and extensive 3-dimensional lattice structures.) Oxides Across a Period As the ionization energy of the elements increases across a period, the ionic character of the oxides decreases, while the molecular nature of the oxides increases. Most molecular oxides exist as small, discrete molecules, except for SiO 2 , or quartz, which has a three dimensional structure very similar to that of diamond.
  • ionic oxides molecular oxides Trends in Ionic vs Molecular Nature of Oxides Oxides Across a Period
  • Acid-Base Properties of Oxides Oxides Across a Period Basic oxides are metal oxides that react with water to form bases (hydroxides) and/or react with acids. e.g., Na 2 O + H 2 O -> 2 NaOH ( aq ) e.g., MgO + 2 HCl ( aq ) -> MgCl 2 + H 2 O Acidic oxides are non-metal oxides that react with water to form acids, and/or react with bases to form salts and water. e.g., P 4 O 10 + 6 H 2 O -> 4 H 3 PO 4 ( aq ) e.g., SO 3 + 2 NaOH -> Na 2 SO 4 + H 2 O
  • Amphoteric oxides can display both acidic and basic properties. Al 2 O 3 + 6 HCl ( aq ) -> 2 AlCl 3 + 3 H 2 O Al 2 O 3 will also react with bases , which is what we would expect of an acidic oxide … Al 2 O 3 + 2 NaOH + 3 H 2 O -> 2 NaAl(OH) 4 ( aq ) Al 2 O 3 will react with acids to form salts and water, which is what we would expect of basic oxides : Oxides Across a Period
  • basic oxides amphoteric oxides acidic oxides Trends in Basic/Acidic Nature of Oxides Oxides Across a Period
  • Finally, note also that because the metallic character of the elements increases as you move down a family, oxides at the top of the family tend to be more acidic than those at the bottom of the family oxides at the bottom of the family are more basic than those at the top. Oxides Within a Family
  • Chapter 9 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display. Chemical Bonding: Basic Concepts AP Inorganic Chemistry
  • Why do atoms form bonds? We now know that the most energetically stable electron arrangement for an atom is to have a filled valence shell, or to have an s 2 p 6 valence electron configuration (the so-called octet rule ) – that is, to be isolectronic with a noble gas. Gilbert Lewis suggested that atoms obtain this stable valence electron configuration by sharing, gaining or losing electrons, which results in the formation of chemical bonds.
  • 1. ionic bond: electrostatic attraction between cations and anions, which are created by the transfer of electrons. The type of bond formed depends on the ionization energy and electronegativity of the atoms involved 2. covalent bond: a bond formed by the equal sharing of valence electrons between the nuclei of two atoms. Types of Chemical Bonds 3. polar covalent bond: a bond formed by the unequal sharing of valence electrons between two atoms.
  • Electronegativity (EN) = a measure, developed by Linus Pauling, of the tendency for an atom to attract or pull electrons towards itself in a chemical bond. Chemical Bonding Electronegativity Electron affinity refers to an isolated, gas phase atom’s attraction for electrons Note the difference between EN and EA: Electronegativity involves shifts in electron density between atoms that are bonded to each other.
  • Electronegativity The electronegativity of an atom is a relative value that can only be measured in relation to the EN of other elements. Thus, EN is a calculated value , not an experimentally determined value. In general, however, we can say that the factors that increase EA also increase EN, so we may predict periodic trends in EN. In general, elements with high EA and high IE also have high electronegativities.
  • The Electronegativities of Common Elements The trend for EN is very similar to the trend for EA: EN increases up a family, and L->R across a period. *note the EN values for some of the common elements
  • Variation of Electronegativity with Atomic Number Electronegativity Note that there are fewer glitches in EN compared to EA!
  • Bond Type and  EN By comparing the difference in electronegativity,  E, between two atoms, we can estimate the degree to which electrons shared or transferred between the atoms. K Na Li Al B H C N O F 0.8 0.9 1.0 1.5 2.0 2.1 2.5 3.0 3.5 4.0 Δ EN = 3.1 Δ EN = 1.0 Δ EN = 0.4
  • A large Δ EN between the two atoms ( Δ EN ≥ 1.7), indicates the electrons will be shifted or transferred to the more electronegative atom to a significant extent (greater ionic character to the bond). A smaller Δ EN between the two atoms indicates that each atom’s pull on the electrons is more equal, which means the electrons will be shared between them (greater covalent character to the bond). The smaller the Δ EN, the more equally the electrons are shared. Bond Type and  EN
  • Bond Type and Δ EN Covalent roughly equal sharing of e - Polar Covalent unequal sharing or partial transfer of e - Ionic complete transfer of e - % ionic character Δ EN 50% 5% ionic polar covalent covalent 1.7 0.3 100% 3 Approximate Δ EN values and bond type:
  • The Ionic Bond  EN ≥ 2 +
  • Ionic Bonding The Ionic Bond If an atom with a low IE collides and reacts with an atom which has a high EA, the first atom can transfer electrons to the second atom. The atom that gives up the electron becomes a positively charged cation , and the atom that gains the electron becomes a negatively charged anion . An ionic bond is the “electrostatic” attraction formed between the oppositely charged cation and anion.
  • Ionic Bonding This electron transfer most often occurs between fairly reactive metals (very low IE) such as group 1A or 2A metals, and reactive non-metals (high EN) such as the halogens and oxygen. + metal (low IE) loses electron non-metal (high EN) gains electron this creates oppositely charged cations and anions Δ EN ≥ 1.7
  • Please note: you may have learned in a previous science class that “an ionic bond is a transfer of electrons.” This is incorrect . An ionic bond is simply the electrostatic attraction that occurs between oppositely charged ions. The cations and anions involved in an ionic bond, as often as not, were already ions before the reaction. But we can break apart and recombine these ions to form new ionic bonds. For example, consider the reaction between Ag + and Cl − to form the ionic compound, AgCl in the reaction: Ag NO 3 + Na Cl  AgCl + NaNO 3 No electron transfer occurs between Ag and Cl – the Ag + and Cl‾ ions were already ions before the reaction began! But the bond in AgCl is an ionic bond. Ionic Bonding
  • Ionic Bonding Note that compounds involving alkali metals and halogens always form ionic bonds between them -- For this class, we will generally assume that any compound involving a metal and a nonmetal will be primarily ionic in its characteristics. Bond is at least 50% ionic, and Δ EN ≥ 1.7
  • Lewis Dot Notation To keep track of the valence electrons involved in bonding, we use Lewis dot symbols which consists of the element’s symbol, with one dot for each valence electron. Thus, alkali metals have 1 dot, chalcogens have 6 dots, etc.
  • Lewis Dot Symbols We can represent the ionic bond that forms between two atoms using Lewis dot symbols, as follows: = [He] = [Ne] We often use a curved arrow to depict shifts in the e- during reactions. The arrow always points in the direction of e- movement. 1s 2 2s 1 1s 2 2s 2 2p 5 1s 2 1s 2 2s 2 2p 6
  • Ionic Bonding Arrangement of ions in ionic compounds Ionic compounds do not form individual “molecules,” composed of just two or three atoms. Instead, ionic compounds are composed of literally billions of cations and anions electrostatically attracted to each other. The empirical formula for an ionic compound is just the simplest ratio of the cations to the anions in compound.
  • Ionic Bonding The regular, repeating arrangement of ions in an ionic compound is called the crystal lattice . expanded view of the crystal lattice of NaCl actual crystal lattice structure of NaCl
  • Ionic Bonding Factors that Determine the Structure of an Ionic Compound’s Crystal Lattice
    • The charge of the ions : this determines the ratio of cations to anions
    • The size of the ions : larger ions will not fit in the same arrangement as smaller ions will.
    • Ions are arranged within the lattice in such a way as to maximize the separation of like charged ions, and minimize the separation of un-like charged ions . (Like charges repel and opposite charges attract.)
  • Ionic Bonding Lattice Energy The same factors that determine the structure of the lattice also determine the stability of a solid ionic compound, which is measured by its lattice energy. Lattice energy = energy required to completely separate 1 mole of a solid ionic compound into its gaseous ions. Thus, this is the same amount of energy released when those ions come together to form the solid compound.
  • Knowing the structure and formula of the compound, we can determine it’s potential energy, and hence its lattice energy using Coulomb’s Law: r = distance between the ions Note that the lattice energy will be greatest when r is small and the charges are large and opposite in sign. +/- Q = charges of the ions Lattice Energy E = k Q + Q - r
  • The Born-Haber Cycle The lattice energy of an ionic compound cannot be measured directly. Instead, we measure it indirectly by assuming the formation of an ionic compound takes place in a series of steps , involving the ionization energy of the cations and the electron affinities of the anions, and then use Hess’s Law! This process was developed by Max Born and Fritz Haber, hence, it is referred to as the Born-Haber cycle.
  • But this is the rxn for the enthalpy of formation for LiF. From the Appendix we find that, for LiF: 6. when rxns 1-5 are added, we obtain the net reaction: Li (s) + ½ F 2 (g) -> LiF (s) The Born-Haber Cycle 1. convert: Li (s) -> Li (g) Δ Hº 1 = 155 kJ/mol example: determine the lattice energy for LiF 2. dissociate F 2 : ½ F 2 (g) -> F (g) Δ Hº 2 = 75 kJ/mol 3. ionize Li: Li (g) -> Li + (g) + e – Δ Hº 3 = 520 kJ/mol 4. add e- to F: F (g) + e – -> F‾ (g) Δ Hº 4 = -328 kJ/mol Δ H° f = –594 kJ/mol 5. Li + (g) + F – (g) -> LiF (s) Δ H° 5 = U = ?? add lattice energy
  • Δ H 5 = U = Δ H° f − ( Δ Hº 1 + Δ Hº 2 + Δ Hº 3 + Δ Hº 4 ) U = −594.1 −[(155 + 75 + 520 + (−328)] kJ/mol U = −594 − (422) = −1016 kJ/mol LiF That is, −1016 kJ of heat energy is released in forming 1.0 mole of LiF from its component ions (or +1016 kJ of energy are needed to separate 1 mole of LiF into its gaseous ions). The Born-Haber Cycle Solving for Δ Hº 5 = lattice energy (U) , we obtain: By Hess’s Law, we have: Δ Hº 1 + Δ Hº 2 + Δ Hº 3 + Δ Hº 4 + Δ Hº 5 = Δ H° f
  •  H overall =  H° 1 +  H° 2 +  H° 3 +  H° 4 +  H° 5 The Born-Haber Cycle Lattice Energy =
  • Forming a cation, even if forming the cation results in an s 2 p 6 electron configuration, requires energy. Indeed, it requires more energy for each electron removed, so an Al 3+ ion requires more energy to form than an Al + ion, even though Al 3+ has a noble gas configuration. So where does the energy come from to form cations? Lattice Energy It is the huge release of the lattice energy that supplies the energy required to remove the electrons from all the cations.
  • If the lattice energy is not sufficient to “off-set” the energy needed to form the cations and anions from the neutral atoms, then no electron transfer occurs, and no ionic bonding is possible between those two atoms. lattice energy Na + Cl 2 Na + -Cl - net release of energy Energy neutral Na atoms & Cl 2 molecules combined ions in lattice reaction progression ions formed + + Lattice Energy − − − +
  • If, for example, Mg obtains a noble gas electron configuration by losing 2 e- to form Mg 2+ , this gain in stability does not outweigh the energy required to remove the two electrons. Lattice Energy There is a “trade-off” between increasing the charge of the metal cations (which increases the lattice energy) and the greater ionization energy required to remove multiple electrons.
  • However, the increase in the lattice energy released by forming ionic attractions with +2 ions instead of +1 ions is more than enough to compensate for the extra energy needed to form the doubly charged Mg 2+ ion. Thus, Mg reacts with Cl‾ ions to form MgCl 2 , rather than MgCl: Lattice Energy Mg Mg 2+ Δ H = +2188 kJ/mol Mg 2+ + 2 Cl‾ MgCl 2 (s) Δ H lat = -2527 kJ/mol Δ H net = -339 kJ/mol net energy change
  • Lattice Energy By the same token, Na does not form Na 2+ ions in its compounds – the gain in lattice energy is not sufficient to make up the difference for the huge increase in ionization energy required to remove the second electron from the stable [Ne] electron configuration of the Na + ion. What is true about the cations also holds for the anions : there is a trade off in the (-) EA of forming higher charged anions ( eg, F 3− ) vs the lattice energy released when such higher charged ions combine to form compounds.
  • The lattice energy vs energy required to form the ions also helps explain why elements with higher EN do not form cations (and hence, do not form ionic bonds), even though such a cation could form a stable ionic bond with an anion. It is simply because the energy required to form a cation from an element with high EN is greater than the lattice energy that could be released if such a compound were to form. Lattice Energy
  • lattice energy P + Se 2 PSe this much energy is still needed! Energy neutral P atoms & Se 2 molecules combined ions in lattice reaction progression Lattice Energy energy required The energy needed to form P + and Se − ions is greater than the lattice energy that would be released if PSe formed. As a result, P + and Se − ions do not form, and the ionic compound PSe does not exist. ions formed + + − − − +
  • Lattice Energy Thus, there is always a balance between the IE or EA needed to form the highest charge that can reasonably be obtained on the atoms, and the increased lattice energy released when these higher charged ions come together to form a solid ionic compound.
  • Properties of Ionic Compounds Ionic Bonding
    • Definite crystalline structure
    • High melting points
    • Brittle as solids
    • Most dissolve in water
    • Conduct electricity only when in the molten state or when dissolved in water.
  • Definite Crystalline Structure: This is explained by the factors already discussed concerning the crystal lattice structure: charge of ions, size of ions, and attraction/repulsion of ions. Ionic Bonding
  • High Melting Points: To melt a substance requires that the particles of the substance be separated, which requires energy to accomplish. In ionic compounds, each particle is attracted to multiple other particles of opposite charge in a three-dimensional lattice network. Melting ionic compounds requires breaking multiple ion-ion attractions, which requires more energy = higher melting pt. Ionic Bonding this cation is held in place by attractions to three anions, all three of which must be overcome in order to melt the compound. +
  • Brittle as Solids: If you “hammer” on an ionic crystal, it shatters. Any attempt to distort the crystal lattice causes the “layers” to shift, which causes like-charged ions to become aligned. The resulting repulsive forces shatters the crystal. Ionic Bonding force applied to crystal + + + + + like charges become aligned + + + + + crystal shatters due to repulsive forces between like-charged ions + + + + +
  • O H H water molecule Ionic Bonding Dissolve in Water: As we shall see later, water is a polar molecule. Polar molecules are molecules which have a small degree of charge separation, so that one end of the molecule is slightly negatively charged, and the other end is slightly positively charged. red = region of higher electron density , which gives a partial negative charge to the oxygen atom in this area of the water molecule blue = region of reduced electron density , which gives a partial positive charge to the hydrogen atoms in this area of the water molecule
  • The partial positive charge on the hydrogen atoms in water are attracted to the anions in an ionic compound; the partial negative charge on the oxygen atom in water are attracted to the cations in an ionic compound. The water molecule is able to literally “pluck apart” the ionic compound, and carry the ions off into the solution. Dissolve in Water: continued = water = anion ( ) = cation (+) + + Ionic Bonding
  • Ionic Compounds Conduct Electricity when Molten or Dissolved in Water Ionic Bonding When a voltage is applied, a current will flow only if there are charged particles present that are free to move. These charge carriers can be electrons or ions . We refer to these as “mobile charge carriers.” A A
  • Ionic Bonding When an ionic compound is in the solid phase, the ions are “locked” in place and cannot move. Thus, ionic solids are poor conductors.
  • However, when an ionic compound is dissolved in water or in the molten (liquid) phase , the ions are separated and held less tightly. They are thus free to move, which allows a current to flow. e − e − Ionic Bonding − + − − − + + +
  • Electrolytes are compounds that, when dissolved in water, produce a solution that conducts electricity. All ionic compounds are electrolytes . Compounds can be classified as strong or weak electrolytes, depending on the extent to which they dissociate into ions in solution. Strong electrolytes dissociate 100% into ions. Non-Electrolytes are compounds that, when dissolved in water, produce a solution that does NOT conduct electricity. Most molecular, covalently bonded compounds are non-electrolytes Ionic Bonding
  • Covalent Bonds N N O C O O - - O C O O - - O C O O - -
  • The Covalent Bond Covalent bonds involve nonmetals with relatively high, but similar electronegativities . These atoms compete roughly equally for the electrons, and thus must obtain a stable octet of electrons by sharing electrons between them, rather than transferring electrons from one atom to another. Covalent Bonding This sharing of electrons by overlapping orbitals between two atoms is called a covalent bond.
  • As the orbitals of two atoms overlap, electrons are attracted to the nuclei of both atoms. This “traps” the electrons in a region between the two atoms, which holds the atoms together in the covalent bond . The trapped electrons are “shared” by both atoms in the molecule formed. RED = electron rich area BLUE = electron poor area Covalent Bonding overlap = covalent bond
  • Covalent Bonding Polar covalent bonds form when there is a slight to moderate difference in EN between the two bonded atoms ( 0.30 < Δ EN < 1.7). Pure covalent bonds occur only in diatomic nonmetal elements, such as Cl 2 or N 2 , which have identical electronegativities ( Δ EN = 0).
  • If an ionic bond is a complete transfer of electrons, and a pure covalent bond is an equal sharing of electrons, then we can say a polar covalent bond is an un-equal sharin g of electrons, or a “partial transfer” of electrons. In polar covalent bonds, the shared electrons are shifted towards the atom with the greater electronegativity. This electron shift can produce a charge separation within the molecule as a whole. Polar Covalent Bonds
  • H — F For example : F is more EN than H. As a result, the electrons are shifted away from H and towards F. We represent this shift with an arrow , pointing towards the more electronegative atom, with a cross-bar at the more electropositive end of the bond. Polar Covalent Bonds = δ - δ + H F
  • The region with increased e- density is symbolized δ - and the region with reduced e- density is symbolized δ + . Molecules like HF and H 2 O which have a net charge separation are said to be polar molecules. We will investigate the nature of polar molecules in more detail when we discuss intermolecular forces in Chapter 11 H δ + δ + H O Polar Covalent Bonds δ
  • Covalent Bonds Because molecules can exist as individual, isolated units, many of the properties of molecular compounds depend on the nature of the attractive forces that exist between two or more molecules. Note carefully that these attractive forces are NOT covalent bonds. We will investigate the nature of these so-called “intermolecular” attractive forces in more detail in the next chapter… Properties of Covalently Bonded Compounds
    • Molecular compounds have lower melting points than ionic compounds.
    • Molecular compounds are poor conductors of electricity in any phase.
    • Molecular compounds have a wide range of solubility in water – some dissolve easily in water, some do not dissolve at all.
    Covalent Bonds Properties of Covalently Bonded Compounds
  • Low Melting Points The bonding within a molecule is a covalent bond between two atoms – however, there are only very weak attractions between separate molecules. As a result, it does not take a lot of energy to separate two molecules (melting). weak or no attractions between molecules… Covalent Bonds Molecules are easily separated = low melt pt.
  • Poor Conductors of Electricity Recall that, in order to conduct a current of electricity, you must have charged particles that are some distance apart to create an electric field. Molecules are made up of neutral atoms. By sharing the electrons, neither atom has gained or lost electrons to become ions, even when the molecules are separated. Thus, there are no charged particles with which to create an electric field, and so they are non-conductors. Covalent Bonds
  • Solubility in Water Water is a polar molecule , in which there is a charge separation within the molecule so that one end is partially positive and other end is partially negative. Polar molecules can dissolve in water because they form attractions to water molecules, much like ionic compounds. Non-polar molecules do not dissolve in water because they cannot form attractions to the polar water molecules. Covalent Bonds
  • 9.4
  • Some “Special Case” Types of Covalent Bonding
    • Some molecular compounds do not exhibit the properties typical of most molecules. Some of these special molecules have extremely high melting and boiling points; some have an actual electric charge to them.
    • The two types of special molecules are:
        • network solids
        • polyatomic ions
  • Network Solids Network solids (also called covalent crystals ) are a type of crystal in which every atom is covalently bonded to its neighbors in a vast, complex interconnected “network.” Quartz crystals (SiO 2 ) and diamonds are examples of these network solids. Some “Special Cases”
  • Because every atom is covalently bonded to its neighbor, network solids can be thought of as giant “macro-molecules.” In order to melt network solids, you must break multiple covalent bonds, much as we have to overcome multiple ion-ion attractions in ionic solids. As a result, network solids tend to have extremely high melting points, and tend to be very hard. diamond: melting point = 3550 o C diamond is the hardest substance known Network Solids
  • Polyatomic Ions Polyatomic ions are groups of atoms covalently bonded together, which have a net charge . Most polyatomic ions are anions, in which the group of atoms picks up one or more electrons to fill the valence needs of one of the atoms in the group. sulfate anion (SO 4 2- ) cyanide anion (CN - ) nitrate anion (NO 3 - )
  • Lewis Structures O C O O - - O C O O - - O C O O - - F N F F
  • Lewis Structures We are frequently interested in which atoms are bonded to which in a molecule. To show this, we use Lewis structures, which were developed by Gilbert Lewis. In the Lewis structure (or structural formula, as it is sometimes called), we indicate all the valence electrons around each atom, and which electrons are shared in order to form the covalent bonds between two atoms.
  • 9.1 Lewis Dot Symbols for the Representative Elements and Noble Gases
  • Lewis proposed that a single covalent bond requires the sharing of a pair of electrons between two atoms. A double bond occurs whenever two atoms share two pair of electrons between them. A triple bond occurs when atoms share three pair of electrons between them. Lewis structure terminology Unshared electron pairs on an atom are called lone pair electrons.
    • Determine the TOTAL number of valence electrons supplied by ALL the atoms in the compound. If the compound has a net charge, add or remove e- from the total count accordingly.
    • Determine which element is the central element. This will be the element that is the LEAST electronegative element of those present in the compound, with the exception of hydrogen. Hydrogen can NEVER be the central element, since H can never form more than one bond.
    • Arrange the other atoms around the central atom.
  • Example: Consider the molecule CHF 3 CHF 3 = 1C + 1 H + 3 F = 1(4) + 1(1) + 3(7) = 26 valence electrons The least electronegative element is H, but H cannot be the central atom; carbon has the next lowest EN so the central atom will be carbon. Arranging the other elements around C we write: F H C F F Lewis Structure Rules the arrangement of the outer atoms is arbitrary – typically, if the molecule has 3 atoms we place them in a straight line, for 4 atoms we use a T-shape, and for 5 atoms we use a + sign pattern as shown here.
    • Form a single bond between the central element and all the other elements in the compound. Each single bond requires 2 electrons. Subtract the number of e- used in making the single bonds from the total number of electrons available, which you determined in step #1.
    Lewis Structure Rules F H C F F : : : : It requires 8 electrons to form the 4 single bonds needed to attach the outside elements to the carbon atom This leaves: (26 — 8) = 18 electrons
      • IF the number of electrons needed to make the single bonds is LESS than the total number of electrons available, then place electrons, IN PAIRS, around the outside elements until their valence requirements are met. IF the number of e- needed for this equals the number of e- available, you are finished
    : : : : : : add electrons in pairs around each F atom until each has a total of 8 electrons; none are added to H because it already has 2 electrons. Lewis Structure Rules : : This used an additional 18 electrons, which is all that were left – we are done. This is the Lewis structure for CHF 3 . : F H C F F : : : :
  • : : : F H C F F : : : : : : : : : : We often show single bonds as a single line, rather than as 2 dots. We still show all the lone pairs as dots, however. Lewis Structure Rules F H C F F : : : : : : : : : : : : : : : : : :
  • Example: consider the molecule OF 2 total number of valence e- = 2(7) + 1(6) = 20 e- Arranging the elements with O as the central atom and forming the 2 bonds required to attach the F’s to the O requires 4 electrons. This leaves (20 – 4) = 16 e- Lewis Structure Rules F O F : : : : : : : : Next we place 6 e- around each F to fill their valence needs. This uses 12 e- . So far we have used 4 + 12 = 16 electrons. This leaves us with 4 e- left over: (20 – 16) = 4
    • IF there are any electrons left over after filling the valence needs of the outer atoms, add these extra electrons to the central atom, in pairs .
    • If the valence requirements of all atoms has been met at this point, you are finished.
    F O F : : : : : : : : : : When we add the 4 remaining electrons to the central oxygen atom (in pairs), we have now met the valence needs for every atom in the compound. We are done. Lewis Structure Rules F O F : : : : : : : :
  • Example: consider the molecule CH 2 O total number of valence e- = 1(4) + 2(1) + 1(6) = 12 e- Arranging the elements with C as the central atom and forming the 3 bonds required to attach the H’s and the O to the C requires 6 electrons. This leaves (12 – 6) = 6 e- The 2 H’s now have 2 e- each, so their valences are filled. The remaining e- are placed on the oxygen atom. At this point, the oxygen’s valence needs are met. H C H O : : : : : : However, the carbon only has 6 electrons in its valence, and it needs 8 – and we are out of electrons! Lewis Structure Rules
  • Lewis Structure Rules H C H O : : : : : :
    • If there are insufficient electrons to meet the valence requirements of every element, you must share 2 or more pairs of electrons with the central atom.
    H C H O : : : : : : Now every element has a filled octet (or a filled shell, which only requires 2 e- in the case of H) shift this pair of electrons down so that there are now 4 electrons being shared between carbon and oxygen…
  • When an atom shares 4 electrons ( 2 pairs ) we say the bond is a double bond . We represent a double bond as 2 lines. Double bonds are stronger and shorter than single bonds. Lewis Structure Rules H C H O : : : : : : H C H O : :
  • Example: Consider HCN The compound contains 1 + 4 + 5 = 10 e-. We form the bonds needed to attach the H and N to the central C atom. Next we place the remaining 6 e- around the nitrogen atom. H C N : : : : : To meet the valence needs of the carbon, we will have to shift TWO pairs of electrons from the nitrogen, so that C and N share 3 electron pairs. Lewis Structure Rules H C N : : : : : H C N : : : : :
  • When an atom shares 6 electrons (3 pairs ) we say the bond is a triple bond . We represent a triple bond as 3 lines. Triple bonds are stronger and shorter than double bonds. Lewis Structure Rules H C N OR : H C N : : : : :
  • Example: Write the Lewis structure of nitrogen trifluoride (NF 3 ). Step 1 – N is less electronegative than F, put N in center F N F F Step 2 – Count valence electrons N = 5 and F = 7 5 + (3 x 7) = 26 valence electrons Step 3 – Draw single bonds between N and F atoms and complete octets on outside F atoms. This uses 24 e- Step 4 - Add remaining two electrons to central atom. Check, do all atoms have filled octets? N F F F
  • Resonance When drawing out Lewis structures, you may encounter a situation where a double bond is needed, but there are two or more different atoms that could donate the extra electrons. For example, consider the molecule O 3 (ozone). When you draw out its Lewis structure, you reach this structure after putting in all the available electrons: O O O The question becomes, which of the two oxygens will donate a second e- pair with the central atom with which the required double bond can be made? Does it make any difference? Lewis Structure Rules
  • The answer turns out to be – BOTH of the oxygens will donate electrons. If only one did, that would imply there is something unique about that particular O—O bond, when in fact, all O—O bonds should be identical. I’m an oxygen, they’re an oxygen, wouldn’t you like to be an oxygen, too? Lewis Structure Rules
  • Lewis Structure Rules The problem is, how do we show that both oxygens are simultaneously donating electrons? This would require that the oxygen atoms share 3 electrons – that is, we need to show there are two “1½ bonds” to the central atom… O O O ?? ?? this would imply that two orbitals on the central oxygen have 3 electrons in them – that is, it does not really have an s 2 p 6 e- configuration, and violates both Pauli’s and Hund’s rules, too! O O O
  • We rectify this by drawing both possible Lewis structures, in which the central oxygen forms a double bond with one oxygen, and then the other. We use a double arrow between the two to show that the true structure is a blending of the two possible Lewis structures: The double bond is NOT switching back and forth between the left and right hand oxygens! Lewis Structure Rules This “double” Lewis structure is called a resonance structure. O O O O O O
  • Lewis Structure Rules When your Lewis structure calls for resonance structures, you must show all possible resonance structures to receive full credit! Shown below are two compounds – one requires resonance and the other does not. Can you tell which one requires resonance? NOT resonance – the atoms are not the same! identical atoms – this requires resonance
  • Exceptions to the Octet Rule 2. Odd-Electron Molecules NO 1. The Expanded Octet only for central atoms in 3rd shell and larger; (uses available empty d-orbitals to form the bonds) SF 6 N only has 7 valence e-. Odd electron molecules always have multiple bonds to central atom. N = 5e - O = 6e - 11e - N O S = 6e - 6F = 42e - 48e - S F F F F F F S has 12 e-: 4 are in the 3 d-subshell, the rest are in 3s and 3p subshells.
  • 3. The Incomplete Octet BeH 2 BF 3 Exceptions to the Octet Rule H H Be Be = 2e - 2 H = 2e - 4e - B = 3e - 3 F = 21 e- 24e - F B F F only 2 valence e- on Be only 6 valence e- on B
  • When setting up the Lewis structure for CH 2 O (this is formaldehyde), we find that there are actually two possible skeletal structures: Formal Charge We use a concept called “ formal charge ” to predict which of several possible Lewis structures is the most likely structure for that compound. H C O H H C O H
  • An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure. The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion. Formal Charge formal charge on an atom in a Lewis structure = total number of valence electrons in the free atom - total number of nonbonding electrons + 1 2 total number of bonding electrons ( )
  • formal charge on Carbon = 4 - [ 2 + ½ x 6 ] = -1 formal charge on Oxygen = 6 - [ 2 + ½ x 6 ] = +1 -1 +1 Example: What is the formal charge on the possible CH 2 O structure shown below? Formal Charge Σ charge = 0  H C O H formal charge on an atom in a Lewis structure = total number of valence electrons in the free atom - total number of nonbonding electrons + 1 2 total number of bonding electrons ( )
  • formal charge on Carbon = 4 - [ 0 + ½ x 8 ] = 0 formal charge on Oxygen = 6 - [ 4 + ½ x 4 ] = 0 0 0 Example: What is the formal charge on the possible CH 2 O structure shown below? Formal Charge Σ charge = 0  H C O H formal charge on an atom in a Lewis structure = total number of valence electrons in the free atom - total number of nonbonding electrons + 1 2 total number of bonding electrons ( )
  • Formal Charge & Stability of Lewis Structures
    • For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present.
    • Lewis structures with large formal charges are less plausible than those with small formal charges.
    • Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms.
  • H C O H -1 +1 H C O H 0 0 We see that for CH 2 O, the second structure yielded the smallest valued formal charges (zero) on each atom. Thus, this is the most likely structure for CH 2 O. Formal Charge example: Use formal charges to determine the correct Lewis structure for PClO. +2 0 -2 +2 -1 0 0 0 possible Lewis structures and their formal charges: -1 0 -1 +1 P Cl O P Cl O Cl P O Cl P O
  • There is evidence that the correct Lewis structure for H 2 SO 4 involves an expanded octet around S. Compare the Lewis structures with and without an expanded octet. with expanded octet, all formal charges are 0 without expanded octet, S has +2 formal charge S O O O O H H S O O O O H H 0 0 0 0 0 0 0 -1 -1 0 0 0 0 +2
  • Metallic Bonding
  • Metallic Bonding Metallic bonding is the type of bonds that form between two or more metal atoms. This means that the electrons are only loosely held. When the energy levels of many adjacent metal atoms overlap, it creates a “common” energy level called a molecular orbital which the valence e- of any one of the atoms can occupy. Metals have low IE and they also have low EN. molecular orbital
  • We say the electrons are delocalized. Note that the electrons are NOT being shared between two nuclei as in covalent bonds – rather, the electrons are free to “associate” with any metal atom, and can and do move from atom to atom. Metallic Bonding
  • A metallic bond can be thought of as a group of cations in a “sea” of delocalized electrons, which are mutually attracted to and associated with the metal cations. The (-) electrons act as the “glue” that holds the (+) cations together. Metallic Bonding delocalized electrons cations + + + + + + + + + + + + + + + +
  • Properties of metals
    • Metals are good conductors of heat & electricity
    • Metals are malleable and ductile
    • Metals are lustrous
    Our model of the metallic bond can be used to explain why metals have these properties. Metallic Bonding
  • Metals are Good Conductors Metallic Bonding Recall that to be a conductor of electricity requires that there be charges that are free to move. The delocalized electrons shared by all the cations in metallic bonds are not locked in place between two specific cations, as they are with covalent bonds, and so they are free to move in the presence of an applied voltage. This makes metals very good conductors in the solid phase. 6 V + + + + + + + + + + + + + + + + + + + + − − − −
  • Band Theory of Conductivity see page 876-877 in text The band theory is a model to explain the ability of metals to conduct electricity, It gets its name from the idea that delocalized electrons freely move through “bands” formed by overlapping molecular orbital energy levels. Because of their small size, the atoms in metals are closely packed together, and their outer shells can interact with each other, creating “molecular orbitals” which are very closely spaced in energy – so close that we refer to them as bands rather than shells.
  • The closely spaced filled energy shells make up the valence band , and the higher energy, unoccupied “molecular orbitals” formed by the overlap of empty p-orbitals make up the conduction band . Conduction and valence bands for Mg. Note that the conduction band involves the overlap of empty 3 p orbitals. Band Theory Each band is actually made up of individual energy lines so close together they essentially merge
  • To conduct a current, electrons in the valence band (which are held to the atom) must be promoted to the conduction band where the electrons are now delocalized and free to move. The energy gap (called the forbidden zone ) between the valence band and the conduction band varies widely from metals to non-metals, and explains why non-metals do not conduct (large forbidden zone), but metals (very small forbidden zone) conduct easily. Band Theory
  • Heat is a measure of the transfer of kinetic energy between particles . The mobile e- are able to collide with other e- and cations, transferring kinetic energy as they do so. Thus, metals conduct heat very well. + Metallic Bonding + + + + + + + + +
  • Metals are Malleable and Ductile Malleability and ductility requires that the bonding between atoms be very “flexible” so that, even when the substance is distorted, the atoms are still bonded. When a force is applied, the cations are free to move – as long as there are some e- between the cations, the atoms are still bonded together… force Metallic Bonding + + + + + + + + + + + + + + + + + +
  • Metals are Lustrous An object that appears lustrous or shiny is reflecting or emitting multiple photons of light with very similar, but slightly different, energies. The delocalized e- in metals move between shells that are close together in energy levels. This e- shift produces photons of similar energy = “lustrous” quality of the metal. Larger shells are close together in energy states, so the photons emitted are also similar in energy – this makes it shiny. Metallic Bonding
  • Bond Enthalpy
  • Bonds & Energy As the orbitals of two atoms begin to overlap, the extra attraction of the electrons towards two nuclei lowers the potential energy of the atoms – as the atoms approach closer and closer, the shared e- are also closer to both nuclei and the potential energy of the atoms/electrons is lowered more. However, if the atoms approach too closely , the repulsion of the two (+) nuclei becomes a factor, increasing the energy state of the bonded atoms quite dramatically. The bond is now destabilized.
  • The bond energy is the Δ E associated with the point of lowest potential energy in forming the bond. Δ E Bonds & Energy There is a point of maximum overlap where the attraction of the e- for the two nuclei is greatest, and the repulsion of the nuclei is not too great, which yields a minimum potential energy for the two atoms. stable bond
  • Chemical bonds form because the potential energy of the bonded atoms is lower than the potential energy of the atoms in the free or unbonded state. THUS: Breaking bonds is always an endothermic process – that is, Δ H > 0 Bonds & Energy Forming bonds is always an exothermic process – that is, Δ H < 0
  • The enthalpy change required to break a bond in one mole of gaseous molecules is the bond energy .  H° = 436.4 kJ  H° = 431.9 kJ  H° = 498.7 kJ  H° = 941.4 kJ Bond Energy Bond Enthalpy (energy) in Binary Compounds molecule H 2 ( g ) H ( g ) + H ( g ) HCl ( g ) H ( g ) + Cl ( g ) O 2 ( g ) O ( g ) + O ( g ) O O N 2 ( g ) N ( g ) + N ( g ) N N H H H Cl
  • Single bond < Double bond < Triple bond The bond energies for multiple bonds is always greater than that for single bonds: Bond Energy C C Δ Hº = 347 kJ/mol C C Δ Hº = 620 kJ/mol C C Δ Hº = 812 kJ/mol
  • Bond Enthalpies for Polyatomic Molecules In compounds that contain more than one bond, the bond energies can differ for different bonds, even if the bonds involve the same two kinds of atoms. example: the bond energy to break the first O-H bond in water is greater than the energy needed to break the second O-H bond: OH (g) H (g) + O (g) Δ H = 427 kJ/mol H 2 O (g) H (g) + OH (g) Δ H = 502 kJ/mol
  • The structure of the molecule is different after the first O-H bond is broken , so it is not too surprising that the enthalpy change in breaking the second O-H bond is also different. Note that this also means that the bond energy for the O-H bond in water is different than the bond energy for the O-H bond in, say, methanol (CH 3 OH) or hydrogen peroxide (H 2 O 2 ) etc. For polyatomic molecules, we usually only refer to the average bond energy for a particular bond. Bond Enthalpies
  • Average O-H bond energy in water: H 2 O ( g ) H ( g ) + OH ( g )  Hº = 502 kJ/mol OH ( g ) H ( g ) + O ( g )  Hº = 427 kJ/mol Average OH bond energy = 502 + 427 2 = 464 kJ/mol
  • see page 386 Δ H values in red are for diatomic molecules
  • Bond Energies (BE) and Enthalpy changes in reactions Since all reactions involve making and breaking chemical bonds, we can analyze the enthalpies of reactions by examining the enthalpy changes in breaking the bonds of the reactant molecules and forming the bonds among the product molecules. This technique only gives an approximation of the actual enthalpy of reaction, since we are only using average bond energies, and seems to work best when Δ H rxn > 100 kJ/mol.
  • Δ Hº = total energy input – total energy released Δ Hº =  BE(reactants) –  BE(products) Mathematically, we imagine a reaction proceeding by breaking all the bonds in the reactants and then using the gaseous atoms to form all the bonds in the products. Then: Bond Energies and Enthalpy changes in Rxns Where Σ BE (react) = bond energy in breaking reactants’ bonds Σ BE (prod) = bond energy in forming product bonds
  • endothermic reaction exothermic reaction Bond Energies and Enthalpy changes in Rxns Net Δ H Net Δ H
  • Example: Use bond energies to calculate the enthalpy change for: H 2 ( g ) + F 2 ( g ) 2HF ( g )  Hº =  BE(reactants) –  BE(products)  H º = ( 436.4 + 156.9 ) – (2 x 568.2 ) = -543.1 kJ Type of bonds broken Number of bonds broken Bond energy (kJ/mol) Energy change (kJ) H H 1 436.4 436.4 F F 1 156.9 156.9 Type of bonds formed Number of bonds formed Bond energy (kJ/mol) Energy change (kJ) H F 2 568.2 1136.4