Chapter 7

COVALENT BONDING

COVALENT BONDING

7.1 Lewis Structures; The Octet Rule

7.2 Molecular Geometry Valence-Shell Electron-Pair Repulsion (VSEPR)

7.3 Polarity of Molecules

7.4 Atomic Orbitals; Hybridization

The steps in converting a molecular formula into a Lewis structure.

Molecular formula

Atom placement

Sum of valence e-

Remaining valence e-

Lewis structure

Place atom with lowest

EN in center

Add A-group numbers

Draw single bonds. Subtract 2e- for each bond.

Give each atom 8e-

(2e- for H)

Step 1

Step 2

Step 3

Step 4

Molecular formula

Atom placement

Sum of valence e-

Remaining valence e-

Lewis structure

For NF3

NFF

F

N 5e-

F 7e- X 3 = 21e-

Total 26e-

:

: :

::

: :

:: :

SAMPLE PROBLEM Writing Lewis Structures for Molecules with One Central Atom

SOLUTION:

PROBLEM: Write a Lewis structure for CCl2F2, one of the compounds responsible for the depletion of stratospheric ozone.

PLAN: Follow the steps outlined in Figure 10.1 .

Step 1: Carbon has the lowest EN and is the central atom.

The other atoms are placed around it.

C

Steps 2-4: C has 4 valence e-, Cl and F each have 7. The

sum is 4 + 4(7) = 32 valence e-.

Cl

Cl F

F

C

Cl

Cl F

FMake bonds and fill in remaining valence

electrons placing 8e- around each atom.

:

::

::

:

:

::

: ::

SAMPLE PROBLEM Writing Lewis Structure for Molecules with More than One Central Atom

PROBLEM: Write the Lewis structure for methanol (molecular formula CH4O), an important industrial alcohol that is being used as a gasoline alternative in car engines.

SOLUTION: Hydrogen can have only one bond so C and O must be next to each other with H filling in the bonds.

There are 4(1) + 4 + 6 = 14 valence e-.

C has 4 bonds and O has 2. O has 2 pair of nonbonding e-.

C O H

H

H

H

::

SAMPLE PROBLEM Writing Lewis Structures for Molecules with Multiple Bonds.

PLAN:

SOLUTION:

PROBLEM: Write Lewis structures for the following:

(a) Ethylene (C2H4), the most important reactant in the manufacture of polymers

(b) Nitrogen (N2), the most abundant atmospheric gas

For molecules with multiple bonds, there is a Step 5 which follows the other steps in Lewis structure construction. If a central atom does not have 8e-, an octet, then two e- (either single or nonbonded pair)can be moved in to form a multiple bond.

(a) There are 2(4) + 4(1) = 12 valence e-. H can have only one bond per atom.

CCH

H H

H

:

CCH

H H

H

(b) N2 has 2(5) = 10 valence e-. Therefore a triple bond is required to make the octet around each N.

N

:

N

:

. .

..

N

:

N

:

. . N

:

N

:

Resonance: Delocalized Electron-Pair Bonding

Resonance structures have the same relative atom placement but a difference in the locations of bonding and nonbonding electron pairs.

OO O

A

B

C

OO O

A

B

C

O3 can be drawn in 2 ways - OO O

OO O

Neither structure is actually correct but can be drawn to represent a structure which is a hybrid of the two - a resonance structure.

is used to indicate that resonance occurs.

O

O O

SAMPLE PROBLEM Writing Resonance Structures

PLAN:

SOLUTION:

PROBLEM: Write resonance structures for the nitrate ion, NO3-.

After Steps 1-4, go to 5 and then see if other structures can be drawn in which the electrons can be delocalized over more than two atoms.

Nitrate has 1(5) + 3(6) + 1 = 24 valence e-

N

O

O O

N

O

O O

N

O

O O

N does not have an octet; a pair of e- will move in to form a double bond.

N

O

O O

N

O

O O

N

O

O O

Formal Charge: Selecting the Best Resonance Structure

An atom “owns” all of its nonbonding electrons and half of its bonding electrons.

Formal charge of atom =

# valence e- - (# unshared electrons + 1/2 # shared electrons)

OO O

A

B

C

For OA

# valence e- = 6

# nonbonding e- = 4

# bonding e- = 4 X 1/2 = 2

Formal charge = 0

For OB

# valence e- = 6

# nonbonding e- = 2

# bonding e- = 6 X 1/2 = 3

Formal charge = +1

For OC

# valence e- = 6

# nonbonding e- = 6

# bonding e- = 2 X 1/2 = 1

Formal charge = -1

Formal charge is the charge an atom would have if the bonding electrons were shared equally.

Resonance (continued)

Smaller formal charges (either positive or negative) are preferable to larger charges.

Avoid like charges (+ + or - - ) on adjacent atoms.

A more negative formal charge should exist on an atom with a larger EN value.

Three criteria for choosing the more important resonance structure:

EXAMPLE: NCO- has 3 possible resonance forms -

Resonance (continued)

N C O

A

N C O

B

N C O

C

N C O N C O N C O

formal charges

-2 0 +1 -1 0 0 0 0 -1

Forms B and C have negative formal charges on N and O; this makes them more preferred than form A.

Form C has a negative charge on O which is the more electronegative element, therefore C contributes the most to the resonance hybrid.

SAMPLE PROBLEM Writing Lewis Structures for Octet Rule Exceptions

PLAN:

SOLUTION:

PROBLEM: Write Lewis structures for (a) H3PO4 (pick the most likely structure); (b) BFCl2.

Draw the Lewis structures for the molecule and determine if there is an element which can be an exception to the octet rule. Note that (a) contains P which is a Period-3 element and can have an expanded valence shell.

(a) H3PO4 has two resonance forms and formal charges indicate the more important form.

O

PO O

O

H

H

H

O

PO O

O

H

H

H

-1

0

0

0

00

0

+10

0

0

0

0

0

0

0

lower formal charges

more stable

(b) BFCl2 will have only 1 Lewis structure.

F

BCl Cl

VSEPR - Valence Shell Electron Pair Repulsion Theory

Each group of valence electrons around a central atom is located as faraway as possible from the others in order to minimize repulsions.

These repulsions maximize the space that each object attached to thecentral atom occupies.The result is five electron-pair (group) arrangements of minimum energy seenin a large majority of molecules and polyatomic ions.

The electron-pairs, Np, are defining the object arrangement or electron pair geometry, but the molecular shape is defined by the relative positions of the atomic nuclei.

Because valence electrons can be bonding or nonbonding, the sameelectron-pair arrangement can give rise to different molecular shapes.

AXmEn

A - central atom X -surrounding atom E -nonbonding valence electron-group

integers

VSEPR

Electron-group repulsions and the five basic molecular shapes.

linear trigonal planar tetrahedral

trigonal bipyramidal octahedral

The single molecular shape of the linear electron-group arrangement. Number of Electron pairs, Np=2.

Examples:

CS2, HCN, BeF2

The two molecular shapes of the trigonal planar electron group arrangement. Np=3.

Class

Shape

Examples:

SO3, BF3, NO3-, CO3

2-

Examples:

SO2, O3, PbCl2, SnBr2

Factors Affecting Actual Bond Angles

Bond angles are consistent with theoretical angles when the atoms attached to the central atom are the same and when all electrons are bonding electrons of the same order.

C O

H

Hideal

1200

1200

larger EN

greater electron density

C O

H

H

1220

1160

real

Lone pairs repel bonding pairs more strongly than bonding pairs repel each other.

Sn

Cl Cl

950

Effect of Double Bonds

Effect of Nonbonding(Lone) Pairs

The three molecular shapes of the tetrahedral electron-group arrangement. Np=4.

Examples:

CH4, SiCl4, SO4

2-, ClO4-

NH3

PF3

ClO3

H3O+

H2O

OF2

SCl2

Lewis structures and molecular shapes.

The four molecular shapes of the trigonal bipyramidal electron-group arrangement. Np=5.

SF4

XeO2F2

IF4+

IO2F2-

ClF3

BrF3

XeF2

I3-

IF2-

PF5

AsF5

SOF4

The three molecular shapes of the octahedral electron-group arrangement. Np=6.

SF6

IOF5

BrF5

TeF5-

XeOF4

XeF4

ICl4-

A summary of common molecular shapes with two to six electron groups.

The steps in determining a molecular shape.

Molecular formula

Lewis structure

Electron-group arrangement

Bond angles

Molecular shape

(AXmEn)

Count all e- groups around central atom (A)

Note lone pairs and double bonds

Count bonding and nonbonding e-

groups separately.

Step 1

Step 2

Step 3

Step 4

SAMPLE PROBLEM Predicting Molecular Shapes with Two, Three, or Four Electron Groups

PROBLEM: Draw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) PF3 and (b) COCl2.

SOLUTION: (a) For PF3 - there are 26 valence electrons, 1 nonbonding pair

PF F

F

The shape is based upon the tetrahedral arrangement.

The F-P-F bond angles should be <109.50 due to the repulsion of the nonbonding electron pair.

The final shape is trigonal pyramidal.

PF F

F

<109.50

The type of shape is

AX3E

SAMPLE PROBLEM Predicting Molecular Shapes with Two, Three, or Four Electron Groups

continued

(b) For COCl2, C has the lowest EN and will be the center atom.

There are 24 valence e-, 3 atoms attached to the center atom.

CCl O

Cl

C does not have an octet; a pair of nonbonding electrons will move in from the O to make a double bond.

The shape for an atom with three atom attachments and no nonbonding pairs on the central atom is trigonal planar.C

Cl

O

Cl The Cl-C-Cl bond angle will be less than 1200 due to the electron density of the C=O.

CCl

O

Cl

124.50

1110

Type AX3

SAMPLE PROBLEM Predicting Molecular Shapes with Five or Six Electron Groups

PROBLEM: Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) SbF5 and (b) BrF5.

SOLUTION: (a) SbF5 - 40 valence e-; all electrons around central atom will be in bonding pairs; shape is AX5 - trigonal bipyramidal.

F

SbF

F F

FF Sb

F

F

F

F

(b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX5E, square pyramidal.

BrF

F F

F

F

SAMPLE PROBLEM Predicting Molecular Shapes with More Than One Central Atom

SOLUTION:

PROBLEM: Determine the shape around each of the central atoms in acetone, (CH3)2C=O.

PLAN: Find the shape of one atom at a time after writing the Lewis structure.

C C C

OH

H

H

HH

H

tetrahedral tetrahedral

trigonal planar

C

O

HC

HHH

CH

H>1200

<1200

The tetrahedral centers of ethane and ethanol.

ethane

CH3CH3

ethanol

CH3CH2OH

The orientation of polar molecules in an electric field.

Electric field OFF Electric field ON

SAMPLE PROBLEM Predicting the Polarity of Molecules

(a) Ammonia, NH3 (b) Boron trifluoride, BF3

(c) Carbonyl sulfide, COS (atom sequence SCO)

PROBLEM: From electronegativity (EN) values (button) and their periodic trends, predict whether each of the following molecules is polar and show the direction of bond dipoles and the overall molecular dipole when applicable:

PLAN: Draw the shape, find the EN values and combine the concepts to determine the polarity.

SOLUTION: (a) NH3

NH

HH

ENN = 3.0

ENH = 2.1N

HH

HN

HH

H

bond dipoles molecular dipole

The dipoles reinforce each other, so the overall molecule is definitely polar.

SAMPLE PROBLEM Predicting the Polarity of Molecules

continued

(b) BF3 has 24 valence e- and all electrons around the B will be involved in bonds. The shape is AX3, trigonal planar.

F

B

F

F

F (EN 4.0) is more electronegative than B (EN 2.0) and all of the dipoles will be directed from B to F. Because all are at the same angle and of the same magnitude, the molecule is nonpolar.

1200

(c) COS is linear. C and S have the same EN (2.0) but the C=O bond is quite polar(EN) so the molecule is polar overall.

S C O