1. Basic Concepts of Chemical Bonding 熊同銘 tmhsiung@gmail.com
Chapter 8
Basic Concepts of Chemical Bonding
熊同銘

2. Basic Types of Chemical Bonds
Metallic bond: Bonding, usually in solid metals, in which the bonding electrons are relatively free (delocalized) to move throughout the three-dimensional structure. (chapter 10)
Ionic bond: A bond between ions between oppositely charged ions (electrostatic forces).
Covalent bond: A bond formed between two atoms by a sharing of electrons.
Shared: 共用
Malleability:可塑性
Ductility: 延展性

3. Contents
Lewis Symbols and the Octet Rule
Ionic Bonding
Covalent Bonding
Bond Polarity and Electronegativity
Drawing Lewis Structures
Resonance Structures
Exceptions to the Octet Rule
Strengths and Lengths of Covalent Bonds

4. 1. Lewis Symbols and the Octet Rule
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1. Lewis Symbols and the Octet Rule
Lewis (dot) symbols: Combination of a chemical symbol for the element and valence electrons (the dots around the element symbol) of the atom.

5. Lewis dot symbols for mail-group elements:
* The first four dots are placed singly on each of the four sides of the chemical symbol, and pairing the dots as the next four are added.

6. H, Li, Be, B attain an electron configuration like He (a duet).
Octet Rule: When atoms bond, they tend to gain, lose, or share electrons to until they are surrounded by eight valence electrons (ns2np6, noble gas configuration).
Exceptions:
H, Li, Be, B attain an electron configuration like He (a duet).
Expanded octets for elements in period 3 or below, using empty valence d orbitals.
Octet Rule: 八隅體法則

7. Lewis structures:
A drawing that represents chemical bonds between atoms as shared or transferred electrons.
Metals and Nonmetals combine: valence electrons transferred from the metal to the non-metal atoms giving rise to ionic bonds.
Nonmetals and Nonmetals combine: one or more pairs of valence electrons are shared between the bonded atoms producing covalent bonds.
Transferred: 轉移

8. Lewis structure for ionic compounds KCl
2. Ionic Bonding
NaCl for example:
* Atoms tend to lose (metals) or gain (nonmetals) electrons to make them isoelectronic to the noble gases.
Lewis structure for ionic compounds
KCl
Na2S
CaCl2

9. Characteristic properties of ionic compounds:
Hard and brittle
High melting points (solids at room temperature)
Ionic solids do not conduct electricity. (Ionic compounds conduct electricity when dissolved in water).
Crystalline
The crystal structure of NaCl

10. Eel =  Q1Q2 d Lattice Energy
The energy required to completely separate one mole of a solid ionic compound into its gaseous ions.
NaCl(s) → Na+(g) + Cl–(g) ΔHlattice = +788 kJ
* Huge exothermic transition is the reverse of the lattice energy.
* The lattice energy associated with electrostatic interactions is governed by Coulomb’s law:
Eel = 
Q1Q2 d
Ionization energy/electronegativity/lattice energy
* lattice energy increases as
the charges on the ions increase
the radii of ions decrease
† depends more on ionic charge than on ionic radius

12. Sample Exercise 8.1 Magnitude of Lattice Energies
Arrange the ionic compounds NaF, CsI, and CaO in order of Increasing lattice energy.
Solution
NaF consists of Na+ and F− ions, CsI of Cs+ and I− ions, and CaO of Ca2+ and O2− ions. We expect the lattice energy of CaO, which has 2+ and 2− ions, to be the greatest of the three.
Ionic size: Cs+ is larger than Na+ and I− is larger than F−. Therefore, the distance between Na+ and F− ions in NaF is less than the distance between the Cs+ and I− ions in CsI. The lattice energy of NaF should be greater than that of CsI.
Therefore, we have CsI < NaF < CaO.

13. Born-Haber cycle: A hypothetical series of steps based on Hess’s law that represents the formation of an ionic compound from its constituent elements.

14. Calculating lattice energy from Born-Haber cycle (NaCl for example):
Continued
Calculating lattice energy from Born-Haber cycle
(NaCl for example):
1. Enthalpy of sublimation:
Na(s) → Na(g) ΔH1 = +108 kJ
2. Bond energy:
½Cl2(g) → Cl(g) ΔH2 = +122 kJ
3. First ionization energy:
Na(g) → Na+(g) + e– ΔH3 = +496 kJ
4. Electron affinity:
Cl(g) + e– → Cl–(g) ΔH4 = – 349 kJ
5. Lattice energy:
Na+(g) + Cl–(g) → NaCl(s) – Lattice energy = ΔH5 = ?
Overall Enthalpy of standard formation:
Na(s) + ½ Cl2(g) → NaCl(s) ΔHfo = –411 kJ
ΔHfo = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5 = –411 kJ
Lattice energy = – ΔH5 = (ΔH1 + ΔH2 + ΔH3 + ΔH4) – ΔHfo = 788 kJ

15. Lewis Structure for Simple Molecules
3. Covalent Bonding
Lewis Structure for Simple Molecules
Lewis structure of molecule shows the bonded atoms with the electron configuration of a noble gas.
The shared electrons is counted for each atom that shares them.
The valence electrons of the bonded atoms obey the octet rule (H obeys the duet rule).
bonding pairs: The shared pairs of electrons (two electrons) in a molecule, commonly represented by a dash (—).
Lone pairs (nonbonding pairs): The electron pairs which are not shared.

16. Single Covalent Bonds
When two atoms share one pair of electrons, it is called a single covalent bond (two electrons), represented by (–).
Example: H2
Example: Cl2
Example: H2O

17. Double Covalent Bonds
When two atoms share two pairs of electrons the result is called a double covalent bond (four electrons), represented by (=).
Example: O2
Triple Covalent Bonds
When two atoms share three pairs of electrons the result is called a triple covalent bond (six electrons), represented by ().
Example: N2

18. Non-metals of the second period (except boron) form a number of covalent bonds equal to eight minus the group number:

19. Characteristic properties of covalently bonded molecular compounds:
fundamental units of covalently bonded compounds are individual molecules
found in all three states at room temperature.
low melting points and boiling points
do not conduct electricity in the solid or liquid state.
molecular acids conduct electricity when dissolved in water

20. 4. Bond Polarity and Electronegativity
Electronegativity (EN): The ability of an atom to attract electrons to itself.
EN is derived from ionization energy and electron affinity.
An atom with a very negative electron affinity and a high ionization energy, having highly electronegative.
The greater the EN of an atom, the more strongly it attracts the electrons in a chemical bond.
EN generally increases across a period in the periodic table.
EN generally decreases down a column in the periodic table.

21. Electronegativity values based on Pauling’s thermochemical data

22. Electronegativity Difference on Bond Type
Electronegativity Difference (ΔEN) : The difference in EN of bonded atom
ΔEN versus Bond type:
Pure nonpolar covalent bond: ΔEN = 0, ex. H─H.
metal-metal (metalic) bond, all atoms with low EN
Polar covalent bond: 極性共價鍵
Electronegativity Difference
Type of bond

23. Ways for expressing electron distributions of covalent bond
Dipole: A molecule with one end having a partial negative charge and the other end having a partial positive charge; a polar molecule. δ+ δ–
F—F
H—F
H—F

24. Dipole Moment (μ)
A measure of the separation and magnitude of the positive and negative charges in polar molecules.
A diatomic molecule with a polar covalent bond:
Q: magnitude of the partial charge (unit: coulomb, C)
r: distance that separates positive and negative charge (unit: m)
Unit for μ: Debye (D), 1 D = 3.34 x 10–30 C m

26. Sample Exercise 8.4 Bond Polarity
In each case, which bond is more polar? (a) B—Cl or C—Cl, (b) P—F or P—Cl..
Solution
(a) the B—Cl bond is more polar.
(b) the P—F bond is more polar.

27. Sample Exercise 8.5 Dipole Moments of Diatomic Molecules
The bond length in the HCl molecule is 1.27 Å. (a) Calculate the dipole moment, in debyes, that results if the charges on the H and Cl atoms were 1+ and 1– respectively. (b) The experimentally measured dipole moment of HCl(g) is 1.08 D. What magnitude of charge, in units of e, on the H and Cl atoms leads to this dipole moment?
Solution
(a)
(b)

28. 5. Drawing Lewis Structures of Molecular Compounds and Polyatomic Ions
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5. Drawing Lewis Structures of Molecular Compounds and Polyatomic Ions
Steps:
Determine the total number of valence electrons.
Write a skeletal structure and connect the atoms by single dashes (covalent bonds).
Place lone pairs electrons around the terminal atoms to give each terminal atom (except H) an octet.
Place the remaining electrons around the central atom to give an octet.
If necessary, move one or more lone pairs of electrons from terminal atom to form a multiple bond to the central atom.

29. Skeletal structure (the arrangement of atoms):
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Skeletal structure (the arrangement of atoms):
Hydrogen atoms are usually terminal atoms
Either molecules or polyatomic ions usually have compact and symmetrical structures.
The central atom of a structure usually has the lowest electronegativity.
Oxoacids such as HClO4, HNO3, etc., hydrogen atoms are usually bonded to oxygen atoms
Compact: 緊密的

30. Example Write the Lewis Structure for CO2
Solution
Step 1
Step O C O
Step O:C:O (4 of 16 electrons used)
Step (16 of 16 electrons used)
Step 5

31. Sample Exercise 8.7 Draw the Lewis structure for HCN.
Solution
The total number of valence electrons = = 10

32. Sample Exercise 8.8 Draw the Lewis structure for the BrO3– ion.
Solution
total number of valence electrons = 7 + (3 ✕ 6) + 1 = 26.

33. Estimation expression:
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Formal Charge
Formal Charge: A fictitious charge that an atom in a Lewis structure would have if all the bonding electrons were shared equally between the bonded atoms.
Estimation expression:
Number of valence electrons in the unbonded atom
Number of lone-pair electrons on the bonded atom
Number of bonding electrons on the bonded atom – –
Simply expressed by: FC = V – LP – ½ BP
The sum of all formal charges in a neutral molecule must be zero, in a polyatomic ion must equal the charge of the ion.
Formal charge: 形式電荷
Fictitious charge: 虛擬電荷

34. Select the dominant Lewis structures by formal charge:
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Select the dominant Lewis structures by formal charge:
The dominant Lewis structures is one with no formal charges in all atoms, otherwise it should be as small as possible.
The negative formal charges should appear on the most electronegative atoms.
The adjacent atoms in a structure should not carry formal charges of the same sign.

35. * Structure A is the dominant Lewis structure.
Example: HCN
* Structure A is the dominant Lewis structure.

36. Sample Exercise 8.9
Three possible Lewis structures for the thiocyanate ion, NCS–, are
(a) Determine the formal charges in each structure.
(b) Based on the formal charges, which Lewis structure is the dominant one?
Solution
(a)
(b) the middle Lewis structure is the dominant one for NCS–.

37. *****
Resonance Structures
A molecule or a polyatomic ion is presented two or more valid Lewis structures that are shown with double-headed arrows between them.
O3 for example:
NO3– for example:
Resonance structures: 共振結構
Lewis structures are equally dominant and all three N-O bond lengths are the same.

38. Resonance in Benzene

39. 9. Exceptions to the Octet Rule
Odd number of valence electrons: The molecule or ion with an odd number of electrons in its Lewis structure, called free radical.
Examples: NO and NO2
The unpaired electron put on the N rather than the O in order to minimize formal charges.
* In general, free radical tend to be unstable and reactive.

40. Fewer than eight electrons: Elements in the second period before carbon can make stable compounds with fewer than eight electrons.
Example:
The positive formal charge on F, the most electronegative element, makes this an unfavorable structure. *

41. More than eight electrons: When an element is in period 3 or below in the periodic table (e.g., periods 3, 4, 5, etc.), it can use d-orbitals to make more than four bonds.
Example:

42. 8. Strengths and Lengths of Covalent Bonds
Bond Enthalpy: The energy required to break 1 mol of the bond in the gas phase.
* Diatomic molecule, definite BE.
Others, affected by the linked atoms or groups, average BE.

43. Using bond enthalpies estimate enthalpy changes for reactions
Hrxn = (bond enthalpies of all bonds broken) − (bond enthalpies of all bonds formed).

44. Sample Exercise
Using data from Table 8.4, estimate ΔH for the combustion reaction
Solution
ΔH = [12D(C-H) + 2D(C-C) + 7D(O=O)] – [8D(C=O) + 12D(O-H)]
= [12(413 kJ) + 2(348 kJ) + 7(495 kJ)] – [8(799 kJ) + 12(463 kJ)]
= 9117 kJ – kJ
= –2831 kJ
Check This estimate can be compared with the value of –2856 kJ calculated from more accurate thermochemical data; the agreement is good.

45. Bond Length: The length of a bond between two particular atoms.
* For a praticular pair of atoms:
Bond lengths: Single > Double > Triple
Bond enthalpies: Single < Double < Triple

46. End of Chapter 08