Bonding - ionic covalent & metallic

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Bonding and structure - ionic compounds, covalent compounds and metals. Relationship between intermolecular forces and physical properties. Allotropes.

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  • valence electrons, inner core electrons, effective nuclear charge and shielding.
  • Bonding - ionic covalent & metallic

    1. 1. STRUCTURE AND BONDING Prepared by Janadi Gonzalez-Lord
    2. 2. TABLE OF CONTENTS Prepared by JGL 8/21/2009  Syllabus requirements  Covalent bonding  Review: atoms and the  Covalent nomenclature periodic table  Properties of covalent  Elements and bonding compounds  Representing ions and  Metallic bonding molecules  Properties of metallic  Ion formation compounds  CATION formation  Allotropes  ANION formation  Atomic radius  Ionic Bond formation  Electronegativity  Representing ionic bonding  Thermodynamics of ion  Ionic nomenclature formation  Properties of ionic compounds 2
    3. 3. Prepared by JGL 8/21/2009 SYLLABUS REQUIREMENTS 3 BONDING
    4. 4. SYLLABUS REQUIREMENTS – IONIC BONDING Prepared by JGL 8/21/2009 The students should be able to :  state that atoms like to achieve a stable state by electron gain or loss  recognize the tendency for loss or gain based on the electronic configuration or the position in the periodic table ( for the first twenty elements ) 4  state that ions are formed by the gain or loss of the electrons
    5. 5. SYLLABUS REQUIREMENTS - BONDING Prepared by JGL 8/21/2009 The students should be able to :  Define anion and cation  Recognize that charge is equal to protons minus electrons  Identify the number of protons , electrons in and the electronic configuration of an ion . ( first twenty elements only )  Write symbols for ions and molecules  Identify the two main types of bonding as ionic / electrovalent and 5 covalent
    6. 6. SYLLABUS REQUIREMENTS - BONDING Prepared by JGL 8/21/2009 The students should be able to : •Explain metallic bonding •State the differences between ionic, covalent and metallic bonding •Identify the types of bonding present in substances based on their properties •Predict the properties of substances based on the bonding present •Draw diagrams to illustrate the types of bonding •Predict the types of bonding between elements 6
    7. 7. Prepared by JGL 8/21/2009 SYLLABUS REQUIREMENTS 7 STRUCTURE
    8. 8. SYLLABUS REQUIREMENTS - STRUCTURE Prepared by JGL 8/21/2009 Students should be able to 1.Define and give examples of ionic crystals, simple molecular structures and giant molecular crystals 2.Explain the term allotropy 3.Relate structures of sodium chloride, diamond and graphite to their properties 8 4.Distinguish between ionic and molecular solids
    9. 9. Prepared by JGL 8/21/2009 REVIEW: ATOMS AND THE PERIODIC TABLE 9
    10. 10. Prepared by JGL 8/21/2009 10
    11. 11. These columns are known as GROUPS are also known as GROUPS FAMILIES GROUPS IN THE PERIODIC TABLE 10 11 12 13 14 15 16 17 18 1 2 3 4 5 6 7 8 9 There are 18 GROUPS Prepared by JGL 11 8/21/2009
    12. 12. Elements within a group have similar All have the same number of physical and chemical properties electrons in their outermost or valence shells Example Na (2,8,1) and K (2,8,8,1) are both in Group 1 Prepared by JGL 12 8/21/2009
    13. 13. The rows are known as PERIODS. There are 9 periods MAIN PERIODS IN PERIODIC TABLE 1 2 3 4 5 6 7 8 9 Prepared by JGL 13 8/21/2009
    14. 14. Prepared by JGL 8/21/2009 ELEMENTS AND BONDING 14 Introduction to ionic, covalent and metallic bonding
    15. 15. 3/30/2010 Prepared by JGL PURE AND IMPURE SUBSTANCES 15 A review
    16. 16. 3/30/2010 Prepared by JGL Matter can be sub-divided into PURE and IMPURE SUBSTANCES or MIXTURES. PURE substances can be sub-divided into ELEMENTS and COMPOUNDS 16 IMPURE substances or MIXTURES can be sub-divided into HOMOGENOUS and HETEROGENOUS
    17. 17. 3/30/2010 Can be separated into Prepared by JGL Can be Can be separated separated into into 17 Source: www.mghs.sa.edu.au/Internet/Faculties/Science/Year10/Pics/elementsAndCompounds.gif
    18. 18. 3/30/2010 ELEMENTS VERSUS COMPOUNDS An element.... A compound...... 18  consists of only one kind  consists of atoms of two or more different elements bound of atom together,  cannot be broken down  can be broken down into a into a simpler type of simpler type of matter (elements) by chemical means matter by either physical (but not by physical means), or chemical means  has properties that are different  can exist as either atoms from its component elements, (e.g. argon) or molecules  always contains the same ratio of its component atoms Prep (e.g., nitrogen). ared by JGL
    19. 19. 19 Prepared by JGL 3/30/2010 AN ELEMENT Consists of only one kind of atom Ar Ar can exist as either atoms (e.g. argon) or molecules (e.g., nitrogen). N N cannot be broken down into a simpler type of matter by either physical or chemical means N N If you try to break apart an atom or molecule, you get an ATOMIC BOMB
    20. 20. 20 Prepared by JGL 3/30/2010 H H A COMPOUND O consists of atoms of two or more different elements bound together always contains the same ratio of its component atoms H H O Water (formula H2O) H H O O H H For every water molecule, there are 2 O Hydrogen atoms for every 1 Oxygen atom H H has properties that are different from its O component elements H H O For example, hydrogen and oxygen are gases but water is a liquid
    21. 21. 21 Prepared by JGL 3/30/2010 EXAMPLES OF ELEMENTS AND COMPOUNDS Elements Compounds Source: www.physicalgeography.net/fundamentals/images/compounds_molecules.jpg
    22. 22. WHY DO COMPOUNDS FORM IN THE FIRST PLACE? Prepared by JGL 8/21/2009 Scientists found that elements in Group 8 were very non- reactive. They also noticed that those in Groups 1,2,6 and 7 were extremely reactive. They also noticed that metallic substances had several properties that were very different from other elements. They could not at first understand why. Eventually they discovered that it had to do with ELECTRON CONFIGURATIONS and STABILITY 22
    23. 23. ELECTRON CONFIGURATION AND STABILITY Prepared by JGL 8/21/2009 Scientists’ research showed that in compounds, elements will combine so that the valence or outermost electrons will have the same electron configuration as the nearest noble gas 23 (in Group 8)
    24. 24. HOW CAN ELEMENTS COMBINE TO ACHIEVE THIS? An element can gain electrons from the Prepared by JGL element it combines with to have the same 8/21/2009 electron configuration as the nearest noble gas. Once an atom gains one or more electrons, it becomes a negatively charged particle known as an ANION An element can lose electrons to another An element can share element to have the valence electrons with same electron another element to configuration as the nearest noble gas. have the same Once an atom loses one There electron configuration are three as the nearest noble or more electrons, it gas. forms a positively (3) ways charged particle known as a CATION. 24
    25. 25. Prepared by JGL YOU MAY WELL 8/21/2009 BE ASKING WHAT DOES THIS MEAN? 25
    26. 26. Prepared by JGL LET’S TAKE A LOOK 8/21/2009 AT SOME EXAMPLES TO UNDERSTAND THIS CONCEPT MORE FULLY 26
    27. 27. Sodium’s atomic number Neon’s atomic number is is Z=11. Its electron Z=10. Its electron Let’s take sodium as an example configuration is therefore configuration is 2,8. It is 2,8,1 the nearest noble gas to sodium. Sodium will combine with another element so that it can change its electron configuration from 2,8,1 to 2,8. To do this, it must lose 1 electron and give it to Prepared by JGL the element with which it combines. 27 8/21/2009
    28. 28. Chlorine’s atomic number Argon’s atomic number is Z=17. Its electron is Z=18. Its electron Let’s take chlorine as an example configuration is therefore configuration is 2,8,8. It 2,8,7 is the nearest noble gas to chlorine. Chlorine will combine with another element so that it can change its electron configuration from 2,8,7 to 2,8,8. To do this, it must gain 1 electron from the Prepared by JGL element with which it combines. 28 8/21/2009
    29. 29. Prepared by JGL 8/21/2009 SODIUM AND CHLORINE UNDERGO WHAT IS KNOWN AS IONIC BONDING. 29
    30. 30. IN IONIC BONDING…… Prepared by JGL 8/21/2009  An element can lose or gain electrons to another element within the same compound to have the same electron configuration as the nearest noble gas.  These are known as IONS.  The electrostatic attraction between these ions is known as an IONIC bond 30
    31. 31. In order to form IONIC BONDING OF SODIUM CHLORIDE the compound sodium chloride, there are Prepared by JGL three (3) steps. 8/21/2009 First, the sodium atom loses one electron to form a positive sodium ion. (cation) Then the chlorine atom accepts the electron from the sodium atom to form a negative chloride ion (anion). Then the sodium cation and chloride anion become attracted to each due to their different charges, forming an ionic bond 31 Source: www.revisionworld.co.uk
    32. 32. Carbon’s atomic number Neon’s atomic number is is Z=6. Its electron Z=10. Its electron Let’s take carbon as an example configuration is therefore configuration is 2,8. It is 2,4 the nearest noble gas to carbon. Carbon will combine with another element so that it can change its electron configuration from 2,4 to 2,8. To do this, it must share 4 electrons with the element with which it combines as it is equally difficult to lose or gain four electrons. Prepared by JGL 32 8/21/2009
    33. 33. COVALENT BONDING Prepared by JGL 8/21/2009  An element can share valence electrons with another element to have the same electron configuration as the nearest noble gas.  This sharing of valence electrons is known as COVALENT bonding. 33
    34. 34. Covalent bond is the sharing of two COVALENT BONDING Prepared by JGL electrons between the 2 8/21/2009 atoms Two hydrogen atoms can share their valence electrons to attain the same electron configuration of the nearest Noble gas configuration, Helium 34
    35. 35. The valence (outermost) electrons are loosely held by METALLIC BONDING Prepared by JGL the metal ions, so much so that they move away from 8/21/2009 The electrons are free to move from the atom to form a positively one positively charged ION to the charged ION. next (i.e. They are DELOCALISED) However, metals like in covalent and are shared (just behave bonding among the various metallic differently. positively charged ions Metallic bonding is similar to both covalent and ionic bonding The number of electrons = the number of protons. Source: www.daviddarling.info/images/metallic_bond.jpg The metal is therefore 35 Syllabus requirement met: electrically NEUTRAL Explain metallic bonding
    36. 36. IN SUMMARY…… Prepared by JGL 8/21/2009 Ionic bonding 3 types of bonding Metallic Covalent bonding bonding 36
    37. 37. COMPARE AND CONTRAST TYPES OF BONDING Prepared by JGL Similarities Differences 8/21/2009  Metallic and ionic  Covalent bonding shares bonding involve electrons rather than electrostatic attractions having electrostatic between positive and charges. negatively charged particles.  Ionic bonding will form  Metallic bonding shares compounds whereas electrons among the ions covalent bonding can in a similar manner to form a compound or how electrons are shared element and metallic in covalent bonding. bonding is strictly found in elements Syllabus requirement met: 37 State the differences between ionic, covalent and metallic bonding
    38. 38. Prepared by JGL 8/21/2009 REPRESENTING IONS AND MOLECULES 38 Ionic notation, chemical formulae
    39. 39. Mass Number RECALL BASIC ATOMIC NOTATION Prepared by JGL = number of protons + 8/21/2009 number of neutrons A Atomic number = Number of protons Z X Element symbol 39
    40. 40. Ionic charge Prepared by JGL WE CAN ALSO = number of protons - number of electrons 8/21/2009 DISPLAY OTHER INFORMATION n+/- Number of atoms X m of element X in molecule or ion Element symbol 40
    41. 41. Mass Number Ionic charge ALTOGETHER of protons + = number Prepared by JGL = number of protons - number of neutrons number of electrons 8/21/2009 A n+/- Atomic X number Number of atoms = Number of of element X in protons molecule or ion Z m Element symbol 41
    42. 42. Mass Number EXAMPLE: SODIUM ION Prepared by JGL 8/21/2009 Element symbol Na 23 1+ Ionic charge Charge = +1 Atomic Number Z = 11 11 Na Mass number A = 23 Element symbol Atomic number Z 42
    43. 43. Mass Number EXAMPLE: HYDROGEN MOLECULE Prepared by JGL Number of 8/21/2009 Element symbol H 2 atoms of hydrogen in a Charge = 0 Atomic Number Z = 1 H molecule of hydrogen 1 2 Mass number A = 2 Element symbol Number of Hydrogen Atomic number Z atoms in a molecule of hydrogen = 2 43
    44. 44. Prepared by JGL 8/21/2009 ION FORMATION 44
    45. 45. IONS DEFINED Prepared by JGL 8/21/2009 An ion is an atom or molecule where the total number of electrons is not equal to the total number of protons, giving it a net positive or negative electrical charge. Syllabus objective met: 45 State that ions are formed by the gain or loss of the electrons
    46. 46. REVIEW – ELECTRON CONFIGURATIONS Prepared by JGL 8/21/2009  What is an electron configuration? Definition: Electron configuration is the arrangement of electrons in an atom, molecule or other body.  How do we represent electron configurations? By using Bohr-Rutherford diagrams Or electron configuration notation 2,8,1 11 p 10 n 46
    47. 47. REMEMBER – “CONTRAST” MEANS “LOOK AT THE DIFFERENCES” LET’S CONTRAST –F AND NE Prepared by JGL 8/21/2009 Fluorine Neon  Element symbol F  Element symbol Ne  Group 17  Group 18  Atomic Number Z = 9  Atomic Number Z = 10  Mass number A = 19  Mass number A = 20  Electron configuration: 2,7  Electron configuration: 2,8  Bohr-Rutherford diagram  Bohr-Rutherford diagram 9p 10 p 10 n 10 n 47
    48. 48. LET’S CONTRAST – NA & NE Prepared by JGL 8/21/2009 Sodium Neon  Element symbol Na  Element symbol Ne  Group 1  Group 18  Atomic Number Z = 11  Atomic Number Z = 10  Mass number A = 23  Mass number A = 20  Electron configuration: 2,8,1  Electron configuration: 2,8  Bohr-Rutherford diagram  Bohr-Rutherford diagram 11 p 10 p 12 n 10 n 48
    49. 49. Remember – “Compare” means “Look at COMPARE AND CONTRAST ALL 3 ELEMENTS Prepared by JGL 8/21/2009 Similarities Differences  F and Ne have the  Different atomic numbers (Z) and therefore protons same number of  Different mass numbers (A) electron shells and therefore different neutrons Scientists found that when  F needs to gain 1 electron elements from Group 1 and to have the same number of Group 7 combined, they lost or electrons as Ne gained an electron to have the same number of electrons as  Na needs to lose 1 electron the nearest Noble Gas. to have the same number of electrons as Ne i.e. F and Na form ions that Syllabus requirement met: 49 are ISO-ELECTRONIC with state that atoms like to achieve a stable Ne state by electron gain or loss
    50. 50. IN GENERAL Prepared by JGL Groups 1, 2 and 3 Groups 15, 16 and 17 8/21/2009  To become ISO-  To become ISO-ELECTRONIC ELECTRONIC with the with the nearest Noble Gas nearest Noble Gas (either (either within the same Period within the same Period or the or the Period just above) Period just above) 1. Group 15 elements gain 3 e- 1. Group 1 elements lose 1 e- 2. Group 16 elements gain 2 e- 2. Group 2 elements lose 2 e- 3. Group 17 elements gain 1 e- 3. Group 3 elements lose 3 e-  This only happens when combining or reacting with  This only happens when another element(s) from combining or reacting with Groups 1,2 or 3 another element(s) from Groups 15,16 or 17 50
    51. 51. Prepared by JGL 8/21/2009 CATION FORMATION 51
    52. 52. CATION FORMATION Prepared by JGL 8/21/2009 Let’s take an unknown Let’s assume that this element X that has an atom of X loses 1 atomic number Z = 10 electron. and a electrical charge of zero. Now # of protons (p) = 10 For an atom of X, # of electrons (e) = 10 -1 = 9 # of protons (p) = 10 Charge on ion = 10 – 9 = +1 # of electrons (e)= 10 Charge of atoms = 10 – 10 =0 52
    53. 53. IF WE THINK OF IT LIKE AN EQUATION, Prepared by JGL 8/21/2009 For an atom of X For an positive ion of X + 10 p + 10 p - 10 e - 9e 0 +1 In general, a positive ion formed by the loss of one or more electrons is known as a CATION. Syllabus objectives met: Define cation Recognize that charge is equal to protons minus electrons 53
    54. 54. Group 1 elements have 1 This is the same electron valence (outermost) electron configuration as Ne, the noble gas just above Na (Period 2, THE PERIODIC TABLE If Na loses 1 electron, its electron configuration Na ion and Ne Group 8). i.e. becomes (2,8) are ISO-ELECTRONIC If K loses 1 electron, its electron configuration This is the same electron Example Na configuration as Ar, the noble becomes (2,8,8) (2,8,1) and gas just above K (Period K (2,8,8,1) 2, Group 8). i.e. K ion and Ar are both in are ISO-ELECTRONIC Group 1 Prepared by JGL 54 8/21/2009
    55. 55. CATION FORMED Prepared by JGL 8/21/2009  If Na loses 1 p = +11 electron, its electron e- = - 10 configuration moves Charge = +1 from (2,8,1) to (2,8).  Since the number of protons remains the It therefore becomes a same (p=11) but the cation with an electrical number of electrons charge of +1 change (e- = 10), it is no longer electrically neutral 55
    56. 56. CATION FORMED Prepared by JGL 8/21/2009  If K loses 1 electron, its p = +19 electron configuration e- = - 18 moves from (2,8,8,1) to Charge = +1 (2,8,8).  Since the number of protons remains the It therefore becomes a same (p=19) but the cation with an electrical number of electrons charge of +1 change (e- = 18), it is no longer electrically neutral 56
    57. 57. IN GENERAL Prepared by JGL 8/21/2009 Li, Na and K are in group All group 1 1 elements So lose 1 e- to Li loses 1 e- to form Li+ form a Na loses 1 e- to form Na+ cation with an electrical K loses 1 e- to form K+ charge of +1 57
    58. 58. Group 2 elements have 2 This is the same electron valence (outermost) electrons configuration as Ne, the noble gas just above Mg (Period T HE PERIODIC TABLE If Mg loses 2 electrons, its electron configuration 2, Group 8). i.e. Mg ion and becomes (2,8) Ne are ISO-ELECTRONIC If Ca loses 2 electrons, its Example Mg electron This is the same electron (2,8,2) and configuration configuration as Ar, the noble Ca(2,8,8,2) becomes (2,8,8) gas just above Ca (Period are both in 2, Group 8). i.e. Ca ion and Group 2 Ar are ISO-ELECTRONIC Prepared by JGL 58 8/21/2009
    59. 59. CATION FORMED Prepared by JGL 8/21/2009  If Mg loses 2 p = +12 electrons, its electron e- = - 10 configuration moves Charge = +2 from (2,8,2) to (2,8).  Since the number of protons remains the It therefore becomes a same (p=12) but the cation with an electrical number of electrons charge of +2 change (e- = 10), it is no longer electrically neutral 59
    60. 60. CATION FORMED Prepared by JGL 8/21/2009  If Ca loses 2 p = +20 electrons, its electron e- = - 18 configuration moves Charge = +2 from (2,8,8,2) to (2,8,8).  Since the number of It therefore becomes a protons remains the cation with an electrical same (p=20) but the charge of +2 number of electrons change (e- = 18), it is no longer electrically neutral 60
    61. 61. IN GENERAL Prepared by JGL 8/21/2009 Be, Mg and Ca are in All group 2 group 2 elements So lose 2 e- to Be loses 2 e- to form Be2+ form a Mg loses 2 e- to form Mg2+ cation with an electrical Ca loses 2 e- to form Ca2+ charge of +2 61
    62. 62. Group 13 elements have 3 This is the same electron valence (outermost) electrons configuration as He, the noble gas just above B (Period THE PERIODIC TABLE 1, Group 8). i.e. B ion and He If B loses 3 electrons, its electron configuration becomes (2) are ISO-ELECTRONIC If Al loses 3 electrons, its Example electron B(2,3) and configuration This is the same electron Al(2,8,3) are becomes (2,8) configuration as Ne, the both in noble gas just above Al Group 3 (Period 2, Group 8). i.e. Al ion and Ar are ISO- ELECTRONIC Syllabus requirements met: Recognize the tendency for loss of electrons based on the electronic configuration or the position in the periodic table ( for metals) Prepared by JGL 62 8/21/2009
    63. 63. CATION FORMED Prepared by JGL 8/21/2009  If B loses 2 electrons, p = +5 its electron e- =-2 configuration moves Charge =+3 from (2,3) to (2).  Since the number of protons remains the It therefore becomes a same (p=5) but the cation with an electrical number of electrons charge of +3 change (e- = 2), it is no longer electrically neutral 63
    64. 64. CATION FORMED Prepared by JGL 8/21/2009  If Al loses 3 p = +13 electrons, its electron e- = - 10 configuration moves Charge = +3 from (2,8,3) to (2,8).  Since the number of protons remains the It therefore becomes a same (p=13) but the cation with an electrical number of electrons charge of +3 change (e- = 10), it is no longer electrically neutral 64
    65. 65. IN GENERAL Prepared by JGL 8/21/2009 B and Al are in group 13 All group 13 elements So B loses 3 e- to form B3+ lose 3 e- to form a Al loses 3 e- to form Al3+ cation with an electrical charge of +3 65
    66. 66. RECALL: METALS Group 1 to 13 and periods 8 and 9 are METALS Metals 10 11 12 13 14 1 2 3 4 5 6 7 8 9 Metals Prepared by JGL 66 8/21/2009
    67. 67. RECALL Prepared by JGL 8/21/2009 Basic Math equation Ionic half-equation For the basic mathematical We could write equation: Na – 1e- → Na+ X–3=0 If we treat the “→” as In order to solve for x , you “=“, we could take across take the number 3 across to the 1e- to the RHS and the right hand side (RHS) of the equation and change change the sign as well the sign. Na → Na+ +1e- 67 X = +3
    68. 68. SO FOR GROUP 1 ELEMENTS Prepared by JGL 8/21/2009 Statement Ionic half-equation Li loses 1 e- to form Li+ Or Li – 1e- → Li+ Li → Li+ +1e- Na loses 1 e- to form Na+ Or Na – 1e- → Na+ Na → Na+ +1e- K loses 1 e- to form K+ K → K+ +1e- Or K – 1e- → K+ 68
    69. 69. SO FOR GROUP 2 ELEMENTS Prepared by JGL 8/21/2009 Statement Ionic half-equation Be loses 2 e- to form Be2+ Or Be – 2e- → Be2+ Be → Be2+ + 2e- Mg loses 2 e- to form Mg2+ Or Mg – 2e- → Mg2+ Mg → Mg2+ +2e- Ca loses 2 e- to form Ca2+ Ca → Ca2+ +2e- Or Ca – 1e- → Ca2+ 69
    70. 70. SO FOR GROUP 13 ELEMENTS Prepared by JGL 8/21/2009 Statement Ionic half-equation B loses 3 e- to form B3+ Or B – 3e- → B3+ B → B3+ + 3e- Al loses 3 e- to form Al3+ Or Al – 3e- → Al3+ Al → Al3+ + 3e- 70
    71. 71. IN GENERAL METALS LOSE ELECTRONS TO Prepared by JGL FORM POSITIVE IONS 8/21/2009 Group 1 metals form mono-positive cations For example In words: sodium loses 1 electron to form sodium cation In chemical equation form: Na → Na+ + 1e- Group 2 metals form di-positive cations. For example In words: Magnesium loses 2 electrons to form magnesium cation In chemical equation form: Mg → Mg2+ + 2e- Group 13 metals form tri-positive cations In words: Aluminium loses 3 electrons to form aluminium cation In chemical equation form: Al → Al3+ + 3e- 71
    72. 72. WHAT ABOUT TRANSITION METALS? Prepared by JGL 8/21/2009 Recall: Transition metals are Groups 3 to 12 They also form cations However, they ccan form cations with multiple valences. For e In chemical equation form: Na → Na+ + 1e- Group 2 metals form di-positive cations. For example In words: Magnesium loses 2 electrons to form magnesium cation In chemical equation form: Mg → Mg2+ + 2e- Group 13 metals form tri-positive cations In words: Aluminium loses 3 electrons to form aluminium cation 72 In chemical equation form: Al → Al3+ + 3e-
    73. 73. SUMMARY – CATION FORMATION Prepared by JGL 8/21/2009 Cations are positive ions. The atoms lose electrons to achieve the They are formed by loss electron configuration of of electrons. the nearest noble gas Metals (Groups 1 to 13) form cations 73
    74. 74. Prepared by JGL 8/21/2009 ANION FORMATION 74
    75. 75. This is the same electron Group 15 elements have 5 configuration as Ne, the noble valence (outermost) electrons gas in the same period as N T HE PERIODIC TABLE (Period 3, Group 8). i.e. N ion If N gains 3 electrons, its electron configuration and Ne are ISO- becomes (2,8) ELECTRONIC If P gains 3 electrons, its Example electron This is the same electron N(2,5) and configuration configuration as Ar, the P(2,8,5) are becomes (2,8,8) noble gas in the same period both in as P (Period 3, Group 8). i.e. Group 5 P ion and Ar are ISO- ELECTRONIC Prepared by JGL 75 8/21/2009
    76. 76. This is the same electron Group 16 elements have 6 configuration as Ne, the noble valence (outermost) electrons gas in the same period as O T HE PERIODIC TABLE (Period 2, Group 8). i.e. O ion If O gains 2 electrons, its electron configuration and Ne are ISO- becomes (2,8) ELECTRONIC If S gains 2 electrons, its Example electron O(2,6) and configuration This is the same electron S(2,8,6) are becomes (2,8,8) configuration as Ar, the noble both in gas in the same period as S Group 6 (Period 3, Group 8). i.e. S ion and Ar are ISO- ELECTRONIC Prepared by JGL 76 8/21/2009
    77. 77. This is the same electron Group 17 elements have 7 configuration as Ne, the noble valence (outermost) electrons gas in the same period as O T HE PERIODIC TABLE (Period 2, Group 8). i.e. F ion If F gains 1 electron, its electron configuration and Ne are ISO- becomes (2,8) ELECTRONIC If Cl gains 1 electron, its electron Example configuration F(2,7) and This is the same electron becomes (2,8,8) Cl (2,8,7) configuration as Ar, the noble are both in gas in the same period as Cl Group 7 (Period 3, Group 8). i.e. Cl ion and Ar are ISO- ELECTRONIC Syllabus requirements met: Recognize the tendency for loss or gain based on the electronic configuration or the position in the periodic table ( for the first twenty elements) Prepared by JGL 77 8/21/2009
    78. 78. RECALL: NON-METALS Groups 14 to 18 are Non-metals Non- Metals 14 15 16 17 18 Atoms and The Periodic Table Prepared 78 by JGL 3/30/2010
    79. 79. IN GENERAL, NON-METALS GAIN Prepared by JGL ELECTRONS TO FORM NEGATIVE IONS 8/21/2009 Group 15 non-metals form tri-negative anions For example In words: nitrogen gains 3 electrons to form nitride anion In chemical equation form: N + 3e-→ N3- Group 16 non-metals form di-negative anions For example In words: oxygen gains 2 electrons to form oxide anion In chemical equation form: O + 2e-→ O2- Group 17 non-metals form mono-negative anions For example In words: fluorine gains 1 electron to form fluoride anion In chemical equation form: F + 1e-→ F- 79
    80. 80. ANION FORMATION Prepared by JGL 8/21/2009 Let’s take an unknown Let’s assume that this element Y that has an atom of Y gains 2 atomic number Z = 11 electron. and a electrical charge of zero. Now # of protons (p) = 11 For an atom of Y, # of electrons (e) = 11 +2 = 13 # of protons (p) = 11 Charge on ion = 11 – 13 = -2 # of electrons (e)= 11 Charge of atoms = 11 – 11 =0 80
    81. 81. IF WE THINK OF IT LIKE AN EQUATION, Prepared by JGL 8/21/2009 For an atom of Y For a negative ion of Y + 11 p + 11 p - 11 e - 13 e 0 -2 In general, a negative ion formed by the gain of one or more electrons is known as a ANION. Syllabus objectives met: Define anion Recognize that charge is equal to protons minus electrons 81 State that ions are formed by the gain or loss of the electrons
    82. 82. SUMMARY – ANION FORMATION Prepared by JGL 8/21/2009 Anions are negative ions Non-metals lose electrons to attain the Non-metals gain electron configuration of electrons to form anions the nearest noble gas. Non-metals are elements in Groups 14 to 18 82
    83. 83. Prepared by JGL 8/21/2009 IONIC BOND FORMATION 83
    84. 84. RECALL... Prepared by JGL 8/21/2009 Cations Anions  Are positive ions  Are negative ions  Are formed from metals  Are formed from non- (groups 1-13) metals (groups 14-18)  Are formed when  Are formed when non- metals lose electrons metals gain electrons The ionic bondattain the electron to attain the electron to is the configuration of the configuration of the electrostatic attraction gas nearest noble gas nearest noble between anion and cation 84
    85. 85. HOW TO RECOGNIZE AN IONIC COMPOUND Prepared by JGL 8/21/2009 Ionic Metal Non-metal compound 85
    86. 86. Prepared by JGL Metals Non- 8/21/2009 Group 1 to 13 are METALS Metals Groups 14 to 18 are Non- Metal Non-metal metals Ionic compound Periods 8 and 9 are METALS Metals 86
    87. 87. Prepared by JGL 8/21/2009 LET US EXAMINE THE BONDING BETWEEN MAGNESIUM AND FLUORINE 87
    88. 88. Fluorine Magnesium Prepared by JGL Metals Non- Noble Gases Group 8 8/21/2009 Group 1 to 13 are METALS Metals Neon 88
    89. 89. USING THE PERIODIC TABLE... Prepared by JGL 8/21/2009 Magnesium Flourine  Group 2  Group 17  Period 3  Period 2  Metal  Non-metal  Forms cation  Forms anion  Z = 12  Z=9  Electron configuration  Electron configuration (2,8,2) (2,7)  Nearest noble gas Ne  Nearest noble gas Ne  Electron configuration of  Electron configuration of Ne = (2,8) Ne = (2,8) 89
    90. 90. FORMATION OF IONIC BOND Prepared by JGL 8/21/2009 Cation formation Anion formation 2 electrons need to be 1 electron needs to be lost to go from (2,8,2) gained to go from (2,7) to (2,8) to (2,8) Mg → Mg2+ + 2e- F + 1e- → F- # of e- lost # of e- gained The Law of conservation of matter states that matter can neither created nor destroyed. Therefore # of e- lost HOW? # of e- gained 90
    91. 91. THE ONLY WAY THIS CAN HAPPEN…. Prepared by JGL 8/21/2009 Is if Then Another Fluorine atom (F) accepts the 2nd electron F + 1e- → F- Mg → Mg2+ + 2e- And F + 1e- → F- 2 Fluorine ions are needed to bond to each Magnesium ion. 91
    92. 92. IN TERMS OF IONIC EQUATIONS Prepared by JGL 8/21/2009 Cation equation Anion equation Mg → Mg2+ + 2e- F + 1e- → F- F + 1e- → F- 2F + 2e- → 2F- 92
    93. 93. ADDING THE 2 EQUATIONS Prepared by JGL 8/21/2009 Total ionic equation Ionic equation Mg → Mg2+ + 2e- Mg2+ + 2F- → MgF2 2F + 2e- → 2F- Mg + 2F → Mg2+ + 2F- 93
    94. 94. THEREFORE THE CHEMICAL FORMULAE Prepared by JGL 8/21/2009 FOR IONIC COMPOUND FORMED BETWEEN MAGNESIUM AND FLUORINE IS MGF2 A chemical formulae expresses the ratio of atoms of elements within a compound. 94
    95. 95. IONIC BOND Prepared by JGL FORMATION PROCESS 8/21/2009 CATION formation IONIC Compound formation Mg → Mg2+ + 2e- Mg2+ + 2F- → MgF2 ANION formation Electrostatic 2F + 2e- → 2F- attraction 95
    96. 96. IONIC BONDING EXERCISES Prepared by JGL 8/21/2009  Demonstrate the type 1. Sodium and fluorine of bond formed 2. Magnesium and between the two sets oxygen of elements below in 3. Lithium and Sulphur 1. Diagrammatic form 2. Using atomic notation 3. Using ionic equations 96
    97. 97. IONIC BONDING BETWEEN SODIUM now p= 11 and e-=10. There are AND FLUORINE – protons, p=11 CATION FORMATIONcharge of +1 and The number of Since each p has a STEP 1: Prepared by JGL each e- has a charge of -1, the overall 8/21/2009 (because Z=11) charge becomes +11-10=+1.  Using Bohr-Rutherford diagrams atom The nearest noble gas to Na is therefore forms a positive ion Na neon, Ne. The atomic number for Ne 1+ The number of or cation of monopositive charge +1 is Z=10. Its electronic configuration neutrons, n=12 is therefore 2,8 (n = A – Z) 11 p - 1 e- 11 p + 1 e- 12 n 12 n To achieve this electronic Na atom has electronic configuration, Na must lose 1 configuration e- of 2,8,1 because there are 2 e- in the 1st shell, 8 2,8,1 the 2nd shell and 1 e- in e- in 2,8 Using atomic notation, the 3rd or outermost (valence) shell Now the electronic configuration becomes 2,8 which is iso-electronic Na Na + + 1e - with Ne. Sodium has mass number A=23 and atomic number Z=11. 97 The atomic notation for sodium atom is therefore 2311Na. The number of electrons, e-=11 (because the number of protons is equal to the number of electrons in an atom)
    98. 98. There are now p= 9 and e-=10. I ONIC BONDING BETWEEN SODIUM AND Since each p has a charge of +1 and FLUORINE TheTEP of ANION FORMATIONof -1, the overall –S number 2: each e- has a charge Prepared by JGL protons, p=9 charge becomes +9-10=-1. Na atom therefore forms a negative ion or 8/21/2009 (because Z=9)  Using Bohr-Rutherford diagramsof mono-negative charge -1 anion The nearest noble gas to F is 1- The number of neon, Ne. The atomic number for Ne neutrons, n=10 is Z=10. Its electronic configuration (n = A – Z) is therefore 2,8 9p + 1 e- 10 n 9p 10 n To achieve this electronic configuration, F must gain 1 e-. F atom has electronic configuration of Law of According to the 2,8 2,7 conservation of matter, matter can 2,7 because there are 2 e- in the 1 st Now the electronic configuration shell and 7 e- in the 2ndneither be created nor destroyed. Te Using atomic notation, or outermost electron must therefore come 2,8 which is iso-electronic becomes from - (valence) shell F + 1e - the Na atom. with Ne. F Fluorine has mass number A=19 and atomic number Z=9. 98 The atomic notation for fluorine atom is therefore 199F. The number of electrons, e-=9 (because the number of protons is equal to the number of electrons in an atom)
    99. 99. IONIC BONDING BETWEEN SODIUM AND FLUORINE – STEP 3: IONIC BOND FORMATION Prepared by JGL 8/21/2009  Using Bohr-Rutherford diagrams 1+ 1- 11 p + 9p 10 n 12 n IONIC bond 2,8 2,8 Using atomic notation, Na+ + F- NaF The fluorine anion is attracted to the sodium cation as opposites attract. The electrostatic (“electro” meaning “electrically” or “coming from electrons” and “static” 99 meaning “not moving”) attraction between the anion and cation is the IONIC bond.
    100. 100. IN SUMMARY, FOR SODIUM AND FLUORINE Prepared by JGL Na Na+ + 1 e- F + 1 e- F- 8/21/2009 9p 10 n 11 p 12 n 1- 1+ 9p 11 p 10 n 12 n 100 Na+ + F- NaF
    101. 101. SIMILARLY, FOR MAGNESIUM AND OXYGEN, Prepared by JGL Mg Mg2+ + 2 e- O + 2 e- O2- 8/21/2009 8p 8n 12 p 12 n 1- 1+ 9p 11 p 10 n 12 n 101 Mg2+ + O2- MgO
    102. 102. SIMILARLY, FOR LITHIUM AND SULPHUR, Prepared by JGL Li Li+ + 1 e- S + 2 e- S2- 8/21/2009 16p 16n 3p 3p 4n 4n 1- 1+ 1+ 3p 9p 3p 10 n 4n 4n 102 2Li+ + S2- Li2S
    103. 103. Prepared by JGL 8/21/2009 REPRESENTING IONIC BONDING 103 Lewis structures (Dot and cross diagrams), chemical formulae, ionic equations, ionic notation
    104. 104. EXAMPLE: ALUMINIUM OXIDE Prepared by JGL 8/21/2009 CATION information ANION information Aluminium  Oxygen Symbol Al  Symbol O Group 3  Group 16 Period 3  Period 2 Metal  Non-metal Forms cation  Forms anion Z = 13  Z=8 Electron configuration (2,8,3)  Electron configuration (2,6) Nearest noble gas Ne  Nearest noble gas Ne Electron configuration of Ne =  Electron configuration of Ne (2,8) = (2,8) Needs to lose 3 electrons to  Needs to gain 2 electrons to achieve electron configuration of achieve electron 104 Ne configuration of Ne
    105. 105. Prepared by JGL 8/21/2009 HOW DO WE REPRESENT THIS INFORMATION MORE SIMPLY? 105
    106. 106. Prepared by JGL 8/21/2009 An ionic equation expresses what happens at an ionic level 106
    107. 107. CATION FORMATION: ALUMINIUM Prepared by JGL 8/21/2009 In words: An aluminium atom loses three electrons to form aluminium cation 107
    108. 108. CATION FORMATION: ALUMINIUM Prepared by JGL Aluminium atom is neutrally charged. This is represented Ionic equations must 8/21/2009 In by the element symbol ions in them Aluminium cation is ionic equation form: have Al alone. represented by the element symbol and the charge on the ion. In this case it is 3+ Al → Al3+ + 3 e - There is no =. Instead there is an RHS = Right hand arrow. This means side LHS = “changes into”. It os The RHS is electrically Left called an equation neutral because the Hand because the LHS is number of electrons (3-) Side equivalent to RHS. cancels out the positive 108 charge (3+)
    109. 109. The 3 The charge is electrons that represented by the cations superscript. For loses is CATION FORMATION Prepared by JGL aluminium represented cation, it is 3+ here 8/21/2009 Using Bohr-Rutherford diagrams 3+ 12 p 12 p 3 e- 12 n 12 n The brackets indicate that this is part of a The outermost or larger entity. A cation valence electrons are no usually has an anion to longer on the outermost balance it off 109 shell electrically.
    110. 110. Prepared by JGL 8/21/2009 ANOTHER WAY TO REPRESENT BONDING IS LEWIS (DOT AND CROSS) DIAGRAMS 110
    111. 111. WHAT ARE LEWIS (DOT AND CROSS) DIAGRAMS? Prepared by JGL 8/21/2009  A Lewis structure is a simplified Bohr-Rutherford diagram  Since chemical reactions take place among the valence and outermost electrons only  Only these electrons are represented in the diagram. 111
    112. 112. COMPARISON OF BOHR-RUTHERFORD DIAGRAM TO LEWIS STRUCTURE Prepared by JGL 8/21/2009 Bohr-Rutherford diagram – Lewis structure – Al atom Al atom 13 p 14 n Al The 3 valence 112 electrons are represented here.
    113. 113. COMPARISON OF BOHR-RUTHERFORD DIAGRAM TO LEWIS STRUCTURE Prepared by JGL 8/21/2009 Bohr-Rutherford diagram Lewis structure Al cation Al cation 3+ 3+ 13 p 12 n Al No valence electrons are represented here all 113 have been lost to form cation.
    114. 114. FOR OXYGEN Prepared by JGL 8/21/2009 Bohr-Rutherford diagram Lewis structure O anion O anion All 8 valence 2- electrons are 2- represented here to form anion. 13 p 12 n O 114
    115. 115. ACCORDING TO THE LAW OF CONSERVATION OF MATTER, MATTER CAN NEITHER BE CREATED NOR Prepared by JGL DESTROYED. 8/21/2009 Since Al needs to give up And since O needs to gain 3e- only 2e- Al → Al3+ + 3e- O + 2e- → O2- There is a need to balance the number of e- on both sides so that the Law is not violated! The only way to do this is to find the lowest common multiple for the number of electrons 115 for the above.
    116. 116. EXCUSE ME?? WHAT IS THE LOWEST COMMON MULTIPLE? Prepared by JGL 8/21/2009  The Lowest Common Multiple (LCM) is the lowest number that is divisible among the members of a set of two or more numbers. For example,  For a set of numbers (2,3,6), 6 is the LCM as 6 is the lowest number that can be divided by 2, 3 and 6  For a set of numbers (2,4,6), 12 is the LCM as 12 is the lowest number that can be divided by 2, 4 and 6  For a set of numbers (9,3,6), 18 is the LCM as 18 is the lowest number that can be divided by 9, 3 and 6 116
    117. 117. SO FOR ALUMINIUM AND OXYGEN Prepared by JGL 8/21/2009 In order to balance this side, I would need to multiply by a factor/number that would give me the LCM. For this ionic half- For this ionic half- equation, the factor would equation, the factor would be 2 since be 3 since 2 x 3e- = 6 e- 3 x 2e- = 6 e- Al → Al3+ + 3e- O + 2e- → O2- x2 2Al → 2Al3+ + 6e- x3 3O + 6e- → 3O2- The set of numbers in this case would be (3,2) The LCM would therefore be 6 as 6 is the lowest 117 number that can be divided by both 2 and 3.
    118. 118. SO FOR ALUMINIUM AND OXYGEN Prepared by JGL 8/21/2009 Since the number of electrons is balanced on both sides, the net effect is zero 2Al → 2Al3+ + 6e- + 3O + 6e- → 3O2- 2Al+ 3O → 2Al3+ + 3O2- Since it requires 3 oxygen anions to bond with 2 aluminium cations, the ratio of Al:O is 2:3. The chemical formulae is therefore Al2O3 118
    119. 119. 2Al3+ + 3O2- USING BOHR-RUTHERFORD DIAGRAMS Prepared by JGL 2- 8/21/2009 Al3+ 3+ Al3+ 3+ 18p 8n 13 p 13 p 14 n 14 n O2- 2- 2- 8p 8p 8n 8n 119 O2- O2-
    120. 120. USING LEWIS (DOT-AND-CROSS) DIAGRAMS Prepared by JGL 2 - 8/21/2009 [Al] 3+ O [Al] 3+ 2- 2- O O 2Al3+ + 3O2- 120
    121. 121. A SIMPLER METHOD FOR DETERMINING CHEMICAL Prepared by JGL FORMULAE IS BY CROSSING CHARGES 8/21/2009 To cross charges, 1. Write the symbol for first the cation and then the anion 2. Take the number of the charge of the anion and bring it to the bottom right hand corner of the cation. 3. Do the same for the cation. 4. If the charges are equal, then leave the formula as a 1:1 ratio 5. If the charges are unequal but are both even numbers then divide by the LCM. 6. If the charges are unequal but both are uneven 121 numbers, then leave as is.
    122. 122. DETERMINING CHEMICAL FORMULA OF IONIC Prepared by JGL COMPOUNDS 8/21/2009 From the previous examples, when sodium and chlorine react:  one sodium atom gives 1 electron to a chlorine atom.  The loss of 1 electron transforms sodium atom into sodium cation  The gain of 1 electron from sodium atom transforms chlorine atom into chlorine anion  An ionic compound forms between sodium cations and chlorine cations  The ratio of sodium ion to chlorine ions involved in 122 ion formation is 1:1
    123. 123. DETERMINING CHEMICAL FORMULAE OF IONIC Prepared by JGL COMPOUNDS. EXAMPLE SODIUM CHLORIDE 8/21/2009 Step 1: Form cation • Na → Na+ + 1e- Step 2: Form anion • Cl + 1e- → Cl- Step 3: Write chemical symbols for cation and anion • Na 1+ + Cl1- Step 4: Cross charges of anion and cation • Na Cl 123
    124. 124. IN CLASS EXERCISE Prepared by JGL 8/21/2009 Determine the chemical formulae of the Answer following:  Magnesium and sulphur  MgS  Potassium and nitrogen  K3N  Aluminum and chlorine  AlCl3  Calcium and phosphorous  Ca3P2  Sodium and oxygen  Na2O  Beryllium and Fluorine  BeF2  Boron and nitrogen  BN  Magnesium and bromine  MgBr2 124
    125. 125. Prepared by JGL 8/21/2009 IONIC NOMENCLATURE 125 IUPAC naming rules for ionic compounds
    126. 126. (International Union of Pure and Applied Chemistry) IUPAC RULES FOR IONIC BONDING Prepared by JGL 8/21/2009 RULE #1: 1. The name of the metal comes first followed by the non- metal 2. Use lowercase letters throughout RULE #2: The metal’s name remains unchanged RULE #3: 1. The non-metal’s name changes to end with suffix – ide. 2. For oxygen, sulphur and nitrogen, remove “ygen”, “ur” and “ogen” respectively and replace with “ide” 3. For all others, remove the last 3 letters and replace with “ide” 4. It does not matter the number of non-metal ions in the 126 compound – they are not included in the name
    127. 127. (International Union of Pure and Applied Chemistry) IUPAC RULES FOR IONIC BONDING Prepared by JGL 8/21/2009  Using the first 3 rules stated in the previous slide, name the ionic compounds formed assuming that the elements undergo ionic bonding 1. Oxygen, magnesium 2. Aluminum, nitrogen magnesium oxide 3. Chlorine, sodium aluminum nitride 4. Fluorine, potassium sodium chloride 5. Sulphur, calcium potassium fluoride calcium sulphide 127
    128. 128. IUPAC RULES FOR IONIC BONDING Prepared by JGL 8/21/2009 RULE#4: For transition metals which can have more that one type of charge (oxidation state or valency), the valence number of the metal must follow the name using roman numerals in brackets  Name the ionic compounds formed given the following information: 1. Sn+ , oxygen tin(I) oxide 2. Iodine, Pb+ lead (I) iodide 3. Cu2+, chlorine copper (II) chloride 4. Fluorine, Ag+ silver (I) fluoride 5. Mn7+, oxygen manganese (VII) oxide 6. Sulphur, Fe3+ iron (III) sulphide 128
    129. 129. Prepared by JGL 8/21/2009 COVALENT BONDING 129 Covalent bonding, polar covalent bonds, Lewis structures
    130. 130. Recall that Prepared by JGL Metals Non- 8/21/2009 Group 1 to 13 are METALS Metals Groups 14 to 18 are Non- Metal Non-metal metals Ionic compound Periods 8 and 9 are METALS Metals 130
    131. 131. RECALL: NON-METALS Groups 14 to 18 are Non-metals What happens Non- when elements Metals that are non- metals want to 14 15 16 17 18 combine? Atoms and The Periodic Table Prepared 131 by JGL 3/30/2010
    132. 132. COVALENT BONDING Prepared by JGL 8/21/2009  If we look at group 14 elements , there are 4 valence electrons.  The question arises – what is the best way to achieve 8 valence electrons?  Should 4 electrons be lost or gained?  In this case, the choice to achieve a stable octet is by sharing electrons  This is known as Definition: A covalent bond is a bond formed between 2 atoms in which a pair of electrons are shared so that both atoms can 132 achieve the electron configuration of the nearest noble gas.
    133. 133. COVALENT BONDING Prepared by JGL 8/21/2009 7 8 8 6 7 5 1 6 2 4 5 3 3 4 What happens is that the electron shells overlap and the electrons are counted as if they belong to both nuclei. 133
    134. 134. EXAMPLE: CARBON AND OXYGEN Prepared by JGL 8/21/2009 Carbon information Oxygen information Carbon  Oxygen Symbol C  Symbol O Group 4  Group 16 Period 2  Period 2 Non-Metal  Non-metal Z = 6 Electron configuration (2,4)  Z=8 Nearest noble gas Ne  Electron configuration (2,6) Electron configuration of Ne =  Nearest noble gas Ne (2,8)  Electron configuration of Ne Needs to share 4 electrons to = (2,8) achieve electron configuration of  Needs to share 2 electrons to Ne achieve electron 134 configuration of Ne
    135. 135. USING BOHR-RUTHERFORD DIAGRAMS Prepared by JGL Oxygen requires 2 e-. Oxygen requires 2 e-. 8/21/2009 It shares 2 e- with the carbon It shares 2 e- with the carbon atom. atom. 1 2 5 6 35 1 3 8p 6p 8p 6n 8n 8n 2 4 46 C atom 7 8 8 7 O atom O atom However, carbon requires 4 e-. 135 Even after it shares 2 e- with an oxygen atoms, it still requires 2 more e- in order to achieve its stable octet. It does so by sharing e- with another oxygen atom
    136. 136. USING BOHR-RUTHERFORD DIAGRAMS Prepared by JGL Because the ratio of C atoms to O atoms is 1:2 8/21/2009 the chemical formula is CO2 6p 8p 6n 8p 8n 8n C atom O atom O atom The covalent bonding is represented as shown above. 136 Note that there are 4 covalent bonds because there are 4 pairs of shared electrons
    137. 137. USING LEWIS STRUTURES Prepared by JGL 8/21/2009 The covalent O C O bonding is represented here 137
    138. 138. USING LEWIS STRUTURES Prepared by JGL Instead of 8/21/2009 drawing the O C O “ Note that each “ represents a “dots” and pair of electrons. “crosses”, a covalent bond can be OC O shown using a“ ” 138
    139. 139. IN CLASS EXERCISE Prepared by JGL 8/21/2009  For the pairs of  Sulphur and sulphur elements listed, draw  Nitrogen and nitrogen the following:  Oxygen and oxygen  Bohr-Rutherford  Hydrogen and hydrogen diagram showing  Nitrogen and hydrogen bonding between elements  Nitrogen and phosphorous  Lewis structure showing bonding between elements  Chemical formula 139
    140. 140. SULPHUR AND SULPHUR S S Prepared by JGL 8/21/2009 S S 16p 16 n 16p 16 n S2 140
    141. 141. NITROGEN AND NITROGEN N N Prepared by JGL 8/21/2009 N N 7p 7n 7p 7n N2 141
    142. 142. OXYGEN AND OXYGEN O O Prepared by JGL 8/21/2009 O O O2 8p 8p 8n 8n 142
    143. 143. HYDROGEN AND HYDROGEN Prepared by JGL 8/21/2009 H H H H 1p 1p 0n 0n H2 143
    144. 144. NITROGEN AND HYDROGEN HN H Prepared by JGL 8/21/2009 H 1p 0n 7p 7n 1p 0n 1p HN H 0n H NH3 144
    145. 145. NITROGEN AND PHOSPHORUS Prepared by JGL 8/21/2009 PN 15p 7p NP 16 n 7n P N 145
    146. 146. Prepared by JGL 8/21/2009 DATIVE OR COORDINATE COVALENT BONDING 146
    147. 147.  Sometimes one of the elements involved in the bonding will give up or share both of its electrons Prepared by JGL  This type of covalent bond is called a 8/21/2009 DATIVE OR COORDINATE COVALENT BOND This usually occurs with elements that have lone pairs of electrons 147
    148. 148. WHAT IS A LONE PAIR OF ELECTRONS? Prepared by JGL 8/21/2009 A pair of valence electrons that is not directly involved in bonding 148
    149. 149. HOW MANY LONE PAIRS DO THE FOLLOWING HAVE? Prepared by JGL 8/21/2009 Atoms…. Ions  Oxygen  Oxygen ion in  Fluorine magnesium oxide  Fluorine in sodium  Nitrogen fluoride  Phosphorous  Nitrogen in calcium  Aluminium nitride  Sodium  Phosphorous in sodium phosphide  Aluminium in aluminium chloride 149
    150. 150. Prepared by JGL SOLUTIONS 3 2 8/21/2009 1 3 2 2 2 O 3 1 F 3 1 N Lewis structure for Lewis structure for Lewis structure for oxygen atom fluorine atom nitrogen atom 2 2 1 0 1 P 1 Al Na Lewis structure for Lewis structure for 150 Lewis structure for Phosphorous atom fluorine atom sodium atom

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