Chemical Bonding I: The Covalent Bond

Chapter 9

Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H 2 S; and the NN bond in H 2 NNH 2 .

Cs – 0.7

H – 2.1

N – 3.0

Cl – 3.0

S – 2.5

N – 3.0

3.0 2.5 3.0 – 0.7 = 2.3

– 3.0 = 0 Ionic – 2.1 = 0.4 Nonpolar covalent Covalent 4/24/2020 2 9.5

Lewis structures

A Lewis structure is a representation of covalent bonding in which shared electrons pairs are shown either as lines or as pairs of dots between two atoms, and lone pairs are shown as pairs of dots on individual atoms.

H O H H O H 4/24/2020 3

Writing Lewis Structures

1. Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center.

2. Count total number of valence e . Add 1 for each negative charge. Subtract 1 for each positive charge.

3. Use valence e- determined in (2) to complete octets of atoms bonded to central atom (

except

hydrogen) 4. If central atom has less than an octet, form double and triple bonds on central atom as needed.

4/24/2020 4 9.6

Write the Lewis structure of nitrogen trifluoride (NF 3 ).

Step 1 – N is less electronegative than F, put N in center Step 2 – Count valence electrons N - 5 (2s 2 2p 3 ) and F - 7 (2s 2 2p 5 ) 5 + (3 x 7) = 26 valence electrons Step 3 – Draw single bonds between N and F atoms and complete octets on N and F atoms.

Step 4 - Does central atom have an octet? yes 3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons F N F F 4/24/2020 5 9.6

Write the Lewis structure of the carbonate ion (CO 3 2 ).

Step 1 – C is less electronegative than O, put C in center Step 2 – Count valence electrons C - 4 (2s 2 2p 2 ) and O - 6 (2s 2 2p 4 ) -2 charge – 2e 4 + (3 x 6) + 2 = 24 valence electrons Step 3 – Draw single bonds between C and O atoms and complete octet on O atoms.

Step 4 - Does central atom have an octet? no Step 5 - Form double bond to give central atom an octet O C O 4/24/2020 O 6 9.6

Write the Lewis structure of the carbonate ion (CO 3 2 ).

Step 1 – C is less electronegative than O, put C in center Step 2 – Count valence electrons C - 4 (2s 2 2p 2 ) and O - 6 (2s 2 2p 4 ) -2 charge – 2e 4 + (3 x 6) + 2 = 24 valence electrons Step 3 – Draw single bonds between C and O atoms and complete octet on O atoms.

Step 4 - Does central atom have an octet? no Step 5 - Form double bond(s) to give central atom an octet O C O 4/24/2020 O O C O O 7 9.6

Two possible skeletal structures of formaldehyde (CH 2 O) H C O H H H C O 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.

# bonds 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 The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.

4/24/2020 8 9.7

H -1 C +1 O H C – 4 e O – 6 e 2H – 2x1 e 12 e 2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 # bonds 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 C = 4 -2 ½ x 6 = -1 formal charge on O = 6 -2 ½ x 6 = +1 9 9.7

4/24/2020

H H 0 C 0 O C – 4 e O – 6 e 2H – 2x1 e 12 e 2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 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 C = 4 - 0 ½ x 8 = 0 formal charge on O = 6 -4 ½ x 4 = 0 10 9.7

4/24/2020

Formal Charge and Lewis Structures 1. For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present.

2. Lewis structures with large formal charges are less plausible than those with small formal charges.

3. 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.

Which is the most likely Lewis structure for CH 2 O?

H -1 C +1 O H H H 0 C 0 O 4/24/2020 11 9.7

Back to CO

2 O C O O Figure out which atoms have the charge.

4/24/2020 12

Write Lewis structures for the following:

1.

Draw skeletal structure of compound showing what atoms are bonded to each other. In general, least electronegative element goes in the center.

Hydrogen and fluorine usually occupy the periphery (terminal positions in the Lewis structure). Sometimes the higher symmetry structure is the correct one.

HCN BF 3 CHFO SnO 2 These correct Lewis structures should help you answer 9.23.

4/24/2020 13

A

resonance structure

is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure.

O O + O O + O O Expect one O-O bond to be longer than the other if ozone really does contain one double bond and one single bond.

O-O : 148 pm O=O : 121 pm Experimental evidence shows that bond oxygen-to-oxygen bonds are equal in length (128 pm).

4/24/2020 14 9.8

What are the resonance structures of the carbonate (CO 3 2 -) ion?

O C O O O C O O O C O O 4/24/2020 15 9.8

Exceptions to the Octet Rule The Incomplete Octet BeH 2 Be – 2e 2H – 2x1e 4e H Be H BF 3 B – 3e 3F – 3x7e 24e F B F F 3 single bonds (3x2) = 6 9 lone pairs (9x2) = 18 Total = 24 4/24/2020 16 9.9

Exceptions to the Octet Rule Odd-Electron Molecules NO N – 5e O – 6e 11e N O Odd-electron molecules are sometimes called radicals.

Many radicals are highly reactive.

The Expanded Octet (central atom with principal quantum number n > 2) SF 6 4/24/2020 S – 6e 6F – 42e 48e F F S F F F F 6 single bonds (6x2) = 12 18 lone pairs (18x2) = 36 Total = 48 17 9.9

Write Lewis structures for the following:

HCN BF 3 CHFO SnO 2 4/24/2020 18

Exceptions to the Octet Rule The Incomplete Octet BeH 2 BX 3 AlX 3 Odd-Electron Molecules NO NO 2 The Expanded Octet (central atom with principal quantum number n > 2) SF 6 PF 6 4/24/2020 19 9.9

Stability of molecules based on their covalent bond energies… 4/24/2020 20

The enthalpy change required to break a particular bond in one mole of gaseous molecules is the

bond energy

.

H Cl HCl (

g

) O 2 (

g

) N 2 ( 2 ( 2 (

g g g

) ) ) H (

g

) + H (

g

) Cl (

g

) + D

Bond Energy

H 0 = 436.4 kJ/mol Cl (

g

) D H 0 = 242.7 kJ/mol H (

g

) + Cl (

g

) D H 0 = 431.9 kJ/mol O (

g

) + O (

g

) D H 0 = 498.7 kJ/mol N (

g

) + N (

g

) D H 0 = 941.4 kJ/mol O N O N Bond Energies Single bond < Double bond < Triple bond 4/24/2020 21 9.10

Average

bond energy

in polyatomic molecules H 2 O (

g

) H (

g

) + OH (

g

) D H 0 = 502 kJ/mol OH (

g

) H (

g

) + O (

g

) D H 0 = 427 kJ/mol Average OH bond energy = 502 + 427 = 464 kJ/mol 2 4/24/2020 22 9.10

Bond energies in thermochemistry

• Energy is always required to break a bond.

• Energy is always released when a bond is formed.

• Whether a reaction is exothermic depends on the relative strengths of the bonds in the product(s) compared to those in the starting material(s). 4/24/2020 23

Bond Energies (BE) and Enthalpy changes in reactions Imagine a reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products.

D H 0 = total energy input – total energy released = S BE(reactants) – S BE(products) endothermic exothermic 4/24/2020 24 9.10

H 2 (

g

) + Cl 2 (

g

) 2HCl (

g

) 2H 2 (

g

) + O 2 (

g

) 2H 2 O (

g

) 4/24/2020 25 9.10

Use bond energies to calculate the enthalpy change for: H 2 (

g

) + F 2 (

g

) 2HF (

g

) D H 0 = S BE(reactants) – S BE(products) Type of bonds broken H H F F Type of bonds formed H F Number of bonds broken 1 1 Number of bonds formed 2 Bond energy (kJ/mol) 436.4

156.9

Bond energy (kJ/mol) 568.2

D H 0 = 436.4 + 156.9 – 1136.4 = -543.1 kJ Energy change (kJ/mol) 436.4

156.9

Energy change (kJ/mol) 1136.4

4/24/2020 26 9.10

Ex. 9.11

Estimate the enthalpy change for the combustion of hydrogen gas.

-469 kJ/mol 4/24/2020 27

9.50

(a) For the reaction 2C 2 H 6(g) + 7O 2(g) -> 4CO 2(g) + 6H 2 O (g) Predict the enthalpy of reaction from the average bond energies in Table 9.2. (b) Calculate the enthalpy of reaction from the standard enthalpies of formation (see Appendix 2) of the reactant and product molecules, and compare the result with your answer for (a).

(a) -2759 kJ/mol (b) -2855 kJ/mol slightly different b/c (a) used

average

bond energies 4/24/2020 28