• Share
  • Email
  • Embed
  • Like
  • Save
  • Private Content
Chapter 8 Lecture- Basic Bonding
 

Chapter 8 Lecture- Basic Bonding

on

  • 4,838 views

Chapter 8 lecture for AP Chemistry on basic bonding (includes Lewis Dot Structures).

Chapter 8 lecture for AP Chemistry on basic bonding (includes Lewis Dot Structures).

Statistics

Views

Total Views
4,838
Views on SlideShare
4,826
Embed Views
12

Actions

Likes
1
Downloads
114
Comments
0

2 Embeds 12

http://www.slideshare.net 11
https://blackboard.muskegoncc.edu 1

Accessibility

Categories

Upload Details

Uploaded via as Microsoft PowerPoint

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    Chapter 8 Lecture- Basic Bonding Chapter 8 Lecture- Basic Bonding Presentation Transcript

    • Bellwork Question4-Style #4 1)Write a balanced net ionic equation for the reaction that occurs when sodium metal is added to pure water. a)How could you verify the identity of the gas evolved? b) What would be the pH of the resulting solution? Less than 7, near 7, or greater than 7?
    • Section 8.1 - 8.4 - Bonds
    • Chemical Bonds
      • Three basic types of bonds:
        • Ionic
          • Electrostatic attraction between ions
        • Covalent
          • Sharing of electrons
        • Metallic
          • Metal atoms bonded to several other atoms
    • Bonding always involves valence electrons which can be represented by Lewis dot symbols OCTET RULE- Atoms seek to gain a noble gas configuration, which is usually eight valence electrons. Remember Helium only has 2 valence electrons, so elements closest to He are most stable when they obtain 2 valence e - like H, Li, and Be. Hydrogen is also stable as a H + ion which has no electrons and is simply a proton + . Li 1 valence e - Be 2 valence e - B 3 valence e - C 4 valence e - N 5 valence e - O 6 valence e - F 7 valence e - Ne 8 valence e -
    •  
    •  
    • Ionic Bonding
    •  
      • Each magnesium atom loses two electrons and each fluorine atom gains one electron.
    • Energetics of Ionic Bonding
      • As we saw in the last chapter, it takes 495 kJ/mol to remove electrons from sodium.
    • Energetics of Ionic Bonding
      • We get 349 kJ/mol back by giving electrons to chlorine.
    • Energetics of Ionic Bonding
      • But these numbers don’t explain why the reaction of sodium metal and chlorine gas to form sodium chloride is so exothermic!
    • Energetics of Ionic Bonding
      • There must be a third piece to the puzzle.
      • What is as yet unaccounted for is the electrostatic attraction between the newly formed sodium cation and chloride anion.
    • Lattice Energy
      • This third piece of the puzzle is the lattice energy:
        • The energy required to completely separate a mole of a solid ionic compound into its gaseous ions.
      • The energy associated with electrostatic interactions is governed by Coulomb’s law:
      E el =  Q 1 Q 2 d
      • E el = 
      Q 1 Q 2 d Like charges produce a positive value equal to the energy of repulsion. Opposite charges produce a negative value equal to the energy of attraction. For an ionic solid this is called LATTICE ENERGY. Coulomb’s Law describes the energy of interactions between charges. Here we apply it to ions. First ion’s charge Second ion’s charge Distance between the two ions depends on ionic size!!! Constant. If I only ask you for RELATIVE energies then it doesn’t matter!
    • Lattice Energy
      • Lattice energy, then, increases with the charge on the ions.
      • It also increases with decreasing size of ions.
      The higher the charges on the ions the greater the energy holding them together. BIGGEST FACTOR The smaller the distance between the two ions the higher the lattice energy
    • Energetics of Ionic Bonding
      • By accounting for all three energies (ionization energy, electron affinity, and lattice energy), we can get a good idea of the energetics involved in such a process.
    • Arrange the following ionic compounds in order of increasing lattice energy: LiCl, NaCl, MgCl 2
      • NaCl < LiCl < MgCl 2
      • LiCl < NaCl < MgCl 2
      • MgCl 2 < NaCl < LiCl
    • Correct Answer:
      • NaCl < LiCl < MgCl 2
      • LiCl < NaCl < MgCl 2
      • MgCl 2 < NaCl < LiCl
      Lattice energy depends on size of the ions and their charges. MgCl 2 has the highest lattice energy because it has a more positive cation. LiCl is higher than NaCl because the interatomic distance between cation and ion is shorter .
    • Energetics of Ionic Bonding
      • These phenomena also help explain the “octet rule.”
      • Metals, for instance, tend to stop losing electrons once they attain a noble gas configuration because energy would be expended that cannot be overcome by lattice energies.
    • When they form ions, transition metals lose valence electrons first and d-electrons second (if at all).
    •  
      • Rh
    • The electron configuration of Fe 3+ is which of the following?
      • 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 6
      • 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3d 3
      • 1 s 2 2s 2 2 p 6 3 s 2 3 p 6 3 d 5
      • 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 1 3 d 4
    • Correct Answer:
      • 1 s 2 2 s 2 2 p 5 3 s 2 3 p 6 4 s 2 3 d 6
      • 1 s 2 2 s 2 2 p 5 3 s 2 3 p 6 4 s 2 3d 3
      • 1 s 2 2s 2 2 p 5 3 s 2 3 p 6 3 d 5
      • 1 s 2 2 s 2 2 p 5 3 s 2 3 p 6 4 s 1 3 d 4
    • Covalent Bonding
      • In these bonds atoms share electrons.
      • There are several electrostatic interactions in these bonds:
        • Attractions between electrons and nuclei
        • Repulsions between electrons
        • Repulsions between nuclei
      • Attractive forces are between each electron and either nucleus; repulsive forces are those between the two nuclei and those between the two electrons.
      • The repulsive forces are greater than the attractive forces.
    • Polar Covalent Bonds
      • Although atoms often form compounds by sharing electrons, the electrons are not always shared equally.
      • Fluorine pulls harder on the electrons it shares with hydrogen than hydrogen does.
      • Therefore, the fluorine end of the molecule has more electron density than the hydrogen end.
    • Electronegativity:
      • The ability of atoms in a molecule to attract electrons to itself.
      • On the periodic chart, electronegativity increases as you go…
        • … from left to right across a row.
        • … from the bottom to the top of a column.
      • EA measures the energy released when an isolated atom gains an electron to form a 1 – ion;
      • EN measures the ability of an atom to hold onto its own electrons and attract electrons from other atoms.
    • Polar Covalent Bonds
      • When two atoms share electrons unequally, a bond dipole results.
      • The dipole moment ,  , produced by two equal but opposite charges separated by a distance, r , is calculated:
      •  = Q r
      • It is measured in debyes (D).
      • IF
    • Polar Covalent Bonds
      • The greater the difference in electronegativity, the more polar is the bond.
      EN difference < 0.5 (non-polar covalent) ≥ 2.0 (ionic)
    •  
      • In HI and HBr the electronegativity differences are too small to lead to large charge separations in the molecules.
    • Boron (B) has an electronegativity of 2.0; chlorine (Cl) has an electronegativity of 3.0. Which of the following BEST describes a B  Cl bond?
      • Ionic
      • Polar covalent
      • Nonpolar covalent
      • Ionic
      • Polar covalent
      • Nonpolar covalent
      Correct Answer: The difference in electronegativities is 1.0 because Cl = 3.0 and B = 2.0. This is too low to be an ionic bond but too high to be a nonpolar covalent bond.
    • Based on the electronegativities of the atoms, which of the following bonds would you predict is most polar?
      • C  C
      • C  N
      • C  O
      • C  Si
      • C  Br
    • Correct Answer: The difference in electronegativities is greatest between O (3.5) and C (2.5) for a net difference of 1.0. Each of the other combinations yields a smaller value for electronegativity difference.
      • C  C
      • C  N
      • C  O
      • C  Si
      • C  Br
    • Nomenclature for ionic & covalent binary compounds
      • Both name least electronegative element first and end in -ide.
      • IONIC- No prefixes (only one formula possible)
      • May need to denote charge if variable
      • ex. Copper(I) oxide
      • COVALENT- Uses prefixes
      • Ex. Dinitrogen pentoxide
      • Note if compound contains a metal in a high oxidation state (>4+) it may act like a molecule and be named accordingly.
    •  
      • MoO 3
      • OsO 4 is more likely to behave as a covalent compound due to the high oxidation state of Os (8+)
    • Section 8.5 - 8.7- Lewis Dot Structures
    • Lewis Structures
      • Lewis structures are representations of molecules showing all electrons, bonding and nonbonding.
    • Writing Lewis Structures
      • Find the sum of valence electrons of all atoms in the polyatomic ion or molecule.
        • If it is an anion, add one electron for each negative charge.
        • If it is a cation, subtract one electron for each positive charge.
      • PCl 3
      5 + 3(7) = 26
    • Writing Lewis Structures
      • The central atom is the least electronegative element that isn’t hydrogen. Connect the outer atoms to it by single bonds.
      Keep track of the electrons: 26  6 = 20
    • Writing Lewis Structures
      • Fill the octets of the outer atoms.
      Keep track of the electrons: 26  6 = 20  18 = 2
    • Writing Lewis Structures
      • Fill the octet of the central atom.
      Keep track of the electrons: 26  6 = 20  18 = 2  2 = 0
    • Writing Lewis Structures
      • If you run out of electrons before the central atom has an octet…
      • … form multiple bonds until it does.
    • Writing Lewis Structures
      • Then assign formal charges.
        • For each atom, count the electrons in lone pairs and half the electrons it shares with other atoms.
        • Subtract that from the number of valence electrons for that atom: The difference is its formal charge.
    • Writing Lewis Structures
      • The best Lewis structure…
        • … is the one with the fewest charges.
        • … puts a negative charge on the most electronegative atom.
    •  
    •  
      • There must be a better Lewis structure since F is not expected to carry a formal charge of +1, being the most electronegative element.
    • Which of the following BEST describes the formal charges on the atoms in the cyanide ion, CN  ?
      • C = +1, N =  1
      • C =  1, N = +1
      • C = 0, N =  1
      • C =  1, N = 0
    • Correct Answer: The formal charges are calculated from the difference between number of valence electrons and number of electrons assigned to each atom in the Lewis structure, thus: C: 4 valence e   5 e  =  1 N: 5 valence e   5 e  = 0
      • C = +1, N =  1
      • C =  1, N = +1
      • C = 0, N =  1
      • C =  1, N = 0
    • Considering formal charge, what is the preferred Lewis structure of NCO - ?
    • Considering formal charge, what is the preferred Lewis structure of NCO - ?
    • Resonance
      • This is the Lewis structure we would draw for ozone, O 3 .
      - +
    • Resonance
      • But this is at odds with the true, observed structure of ozone, in which…
        • … both O — O bonds are the same length.
        • … both outer oxygens have a charge of  1/2.
    • Resonance
      • One Lewis structure cannot accurately depict a molecule such as ozone.
      • We use multiple structures, resonance structures, to describe the molecule.
    • Resonance
      • Just as green is a synthesis of blue and yellow…
      • … ozone is a synthesis of these two resonance structures.
    •  
      • Yes
    • How many resonance structures can be drawn for NO 3 - ?
      • 1
      • 2
      • 3
      • 4
      • 5
    • How many resonance structures can be drawn for NO 3 - ?
      • 1
      • 2
      • 3
      • 4
      • 5
    •  
      • one-and-a-third bonds
    • Resonance
      • In truth, the electrons that form the second C — O bond in the double bonds below do not always sit between that C and that O, but rather can move among the two oxygens and the carbon.
      • They are not localized , but rather are delocalized .
    • Resonance
      • The organic compound benzene, C 6 H 6 , has two resonance structures.
      • It is commonly depicted as a hexagon with a circle inside to signify the delocalized electrons in the ring.
    •  
      • No, because double bonds cannot be moved to give equivalent structures due to fixed hydrogens.
    • Exceptions to the Octet Rule
      • There are three types of ions or molecules that do not follow the octet rule:
        • Ions or molecules with an odd number of electrons.
        • Ions or molecules with less than an octet.
        • Ions or molecules with more than eight valence electrons (an expanded octet).
    • Odd Number of Electrons
      • Though relatively rare and usually quite unstable and reactive, there are ions and molecules with an odd number of electrons.
      • Like ClO 2 , NO, NO 2 , & O 2 -
    • Fewer Than Eight Electrons
      • Consider BF 3 :
        • Giving boron a filled octet places a negative charge on the boron and a positive charge on fluorine.
        • This would not be an accurate picture of the distribution of electrons in BF 3 .
    • Fewer Than Eight Electrons
      • Therefore, structures that put a double bond between boron and fluorine are much less important than the one that leaves boron with only 6 valence electrons.
    • Fewer Than Eight Electrons
      • The lesson is: If filling the octet of the central atom results in a negative charge on the central atom and a positive charge on the more electronegative outer atom, don’t fill the octet of the central atom.
    • Famous Lewis Acid/Base reaction
    • More Than Eight Electrons
      • The only way PCl 5 can exist is if phosphorus has 10 electrons around it.
      • It is allowed to expand the octet of atoms on the 3rd row or below.
        • Presumably d orbitals in these atoms participate in bonding.
    • More Than Eight Electrons
      • Even though we can draw a Lewis structure for the phosphate ion that has only 8 electrons around the central phosphorus, the better structure puts a double bond between the phosphorus and one of the oxygens.
    • More Than Eight Electrons
      • This eliminates the charge on the phosphorus and the charge on one of the oxygens.
      • The lesson is: When the central atom is on the 3rd row or below and expanding its octet eliminates some formal charges, do so.
    • Section 8.8- Bond Enthalpies
    • The stability of a molecule is related to the strength of the bonds in the molecule.
    • Covalent Bond Strength
      • Most simply, the strength of a bond is measured by determining how much energy is required to break the bond.
      • This is the bond enthalpy .
      • The bond enthalpy for a Cl — Cl bond,
      • D (Cl — Cl), is measured to be 242 kJ/mol.
      • kJ per MOLE!!
    • Average Bond Enthalpies
      • This table lists the average bond enthalpies for many different types of bonds.
      • Average bond enthalpies are positive, because bond breaking is an endothermic process.
    • Average Bond Enthalpies
      • NOTE: These are average bond enthalpies, not absolute bond enthalpies; the C — H bonds in methane, CH 4 , will be a bit different than the
      • C — H bond in chloroform, CHCl 3 .
    • Bond enthalpies can be measured through the “enthalpy of atomization”, which measures the energy required to break a molecule into free gaseous atoms. D(C-H) = 1660kJ divided by 4 because there are 4 C-H bonds in methane.
      • The enthalpy of atomization – 6 · D (C  H) = a good estimate of D (C  C).
    • Breaking bonds = endothermic energy must be provided to break a bond Bond formation = exothermic a stable bond is always lower energy than the non-bonded atoms
    • Molecules with strong (stable) bonds tend to be less reactive. Naturally abundant compounds are usually very stable with strong bonds. Ex. O=O is 495kJ/mol O-O is 146 kJ/mol O 2 is stable & abundant H 2 O 2 is not
    • Enthalpies of Reaction
      • Yet another way to estimate  H for a reaction is to compare the bond enthalpies of bonds broken to the bond enthalpies of the new bonds formed.
      • In other words,
      •  H rxn =  (bond enthalpies of bonds broken) 
      •  (bond enthalpies of bonds formed)
    • Enthalpies of Reaction
      • CH 4 ( g ) + Cl 2 ( g ) 
      • CH 3 Cl ( g ) + HCl ( g )
      • In this example, one
      • C — H bond and one
      • Cl — Cl bond are broken; one C — Cl and one H — Cl bond are formed.
    • Enthalpies of Reaction
      • So,
      •  H rxn = [ D (C — H) + D (Cl — Cl)  [ D (C — Cl) + D (H — Cl)
      • = [(413 kJ) + (242 kJ)]  [(328 kJ) + (431 kJ)]
      • = (655 kJ)  (759 kJ)
      • =  104 kJ
      • Using  H° f calculations provides a better estimate of  H° rxn than bond enthalpies.
      • Bond enthalpies are an average value and they are calculated for gaseous molecules.
      • Sometimes the  H° f is not available for all of the molecules involved so this method is good to know.
      •  467 kJ
      • +467 kJ
      • +17 kJ
      •  983 kJ
      Calculate  H from the following bond enthalpies for the reaction: 2 NCl 3  N 2 + 3 Cl 2 D(N  Cl) = 200, D(N  N) = 941, D(Cl  Cl) = 242 (all kJ/mol).
    • Correct Answer:  H =  (bond enthalpies of bonds broken)   ( bond enthalpies of bonds formed)  H = 6(200)  [941 + 3(242)]  H = 1200  (1667)  H =  467
      •  467 kJ
      • +467 kJ
      • +17 kJ
      •  983 kJ
    • Bond Enthalpy and Bond Length
      • We can also measure an average bond length for different bond types.
      • As the number of bonds between two atoms increases, the bond length decreases.
    • The higher the bond order the shorter & stronger the bond Bond order = 1 Longest & weakest Bond order = 2 Med length & strength Bond order = 3 shortest & strongest
    •  
      • C–O in carbon monoxide is a triple bond.
    • Which of the following carbon – carbon atom bonds would be expected to be the shortest?
      • C  C
      • C  C
      • C  C
      • C  C
      • C  C
      • C  C
      Correct Answer: The bond length decreases as the bond enthalpy increases; in general, as the number of bonds between two atoms increases, the bond grows shorter and stronger.
    • Which is predicted to have the longest carbon – oxygen bonds: CO, CO 2 , or CO 3 2  ?
      • CO 3 2 
      • CO 2
      • CO
    • Correct Answer: In CO, the carbon – oxygen bond is a triple bond, while in CO 2 both carbon – oxygen bonds are double bonds. In CO 3 2  , there are three resonance structures and the average carbon – oxygen bond is 1 1/3.
      • CO 3 2 
      • CO 2
      • CO
    •