chapter 10

Chapter 10

The Shapes of Molecules

The Shapes of Molecules

10.1 Depicting Molecules and Ions with Lewis Structures

10.2 Using Lewis Structures and Bond Energies to Calculate Heats of Reaction

10.3 Valence-Shell Electron-Pair Repulsion (VSEPR) Theory and Molecular Shape

10.4 Molecular Shape and Molecular Polarity

Figure 10.1

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 10.1 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 10.2 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 10.3 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 e- 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.

OO O

SAMPLE PROBLEM 10.4 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

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 important 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 10.5 Writing Lewis Structures for Exceptions to the Octet Rule.

PLAN:

SOLUTION:

PROBLEM: Write Lewis structures for (a) H3PO4 and (b) BFCl2. In (a) decide on the most likely structure.

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

10.2 Using Lewis Structures and Bond Energies to Calculate Heats of Reaction

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

H2 (g) H (g) + H (g) H0 = 436.4 kJ

Cl2 (g) Cl (g)+ Cl (g) H0 = 242.7 kJ

HCl (g) H (g) + Cl (g) H0 = 431.9 kJ

O2 (g) O (g) + O (g) H0 = 498.7 kJ O O

N2 (g) N (g) + N (g) H0 = 941.4 kJ N N

Bond Energy

Bond Energies

Single bond < Double bond < Triple bond

Average bond energy in polyatomic molecules

H2O (g) H (g) + OH (g) H0 = 502 kJ

OH (g) H (g) + O (g) H0 = 427 kJ

Average OH bond energy = 502 + 427

2= 464.5 kJ

Bond Energies (BE) and Enthalpy changes in reactions

H0 = total energy input – total energy released= BE(reactants) – BE(products)

Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products.

Use bond energies to calculate the enthalpy change for:H2 (g) + F2 (g) 2HF (g)

H0 = BE(reactants) – BE(products)

Type of bonds broken

Number of bonds broken

Bond energy (kJ/mol)

Energy change (kJ)

H H 1 436.4 436.4

F F 1 156.9 156.9

Type of bonds formed

Number of bonds formed

Bond energy (kJ/mol)

Energy change (kJ)

H F 2 568.2 1136.4

H0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ

Figure 10.2 Using bond energies to calculate H0rxn

H0rxn = H0

reactant bonds broken + H0product bonds formed

H01 = + sum of BE H0

2 = - sum of BE

H0rxn

Ent

halp

y, H

SAMPLE PROBLEM 10.6 Calculating Enthalpy Changes from Bond Energies

SOLUTION:

PROBLEM: Use Table 9.2 (button at right) to calculate H0rxn for the

following reaction:

CH4(g) + 3Cl2(g) CHCl3(g) + 3HCl(g)

PLAN: Write the Lewis structures for all reactants and products and calculate the number of bonds broken and formed.

H

C H

H

H + Cl Cl3

Cl

C Cl

Cl

H + H Cl3

bonds broken bonds formed

SAMPLE PROBLEM 10.6 Calculating Enthalpy Changes from Bond Energies

continued

bonds broken bonds formed

4 C-H = 4 mol(413 kJ/mol) = 1652 kJ

3 Cl-Cl = 3 mol(243 kJ/mol) = 729 kJ

H0bonds broken = 2381 kJ

3 C-Cl = 3 mol(-339 kJ/mol) = -1017 kJ

1 C-H = 1 mol(-413 kJ/mol) = -413 kJ

H0bonds formed = -2711 kJ

3 H-Cl = 3 mol(-427 kJ/mol) = -1281 kJ

H0reaction = H0

bonds broken + H0bonds formed = 2381 kJ + (-2711 kJ) = - 330 kJ

Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest?

Establish an arbitrary scale with the standard enthalpy of formation (H0) as a reference point for all enthalpy expressions.

f

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.

f

The standard enthalpy of formation of any element in its most stable form is zero.

H0 (O2) = 0f

H0 (O3) = 142 kJ/molf

H0 (C, graphite) = 0f

H0 (C, diamond) = 1.90 kJ/molf

The standard enthalpy of reaction (H0 ) as the enthalpy of a reaction carried out at 1 atm.

rxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn nH0 (products)f= mH0 (reactants)f-

Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

Calculate the standard enthalpy of formation of CS2 (l) given that:C(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJrxn

S(rhombic) + O2 (g) SO2 (g) H0 = -296.1 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1. Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)

2. Add the given rxns so that the result is the desired rxn.

rxnC(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJ

2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -296.1x2 kJrxn

CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)

H0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJrxn

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. What is the heat released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn nH0 (products)f= mH0 (reactants)f-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ

-5946 kJ2 mol

= 2973 kJ/mol C6H6

The enthalpy of solution (Hsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.

Hsoln = Hsoln - Hcomponents

Which could be used for melting ice?

Which could be used for a cold pack?

Valence shell electron pair repulsion (VSEPR) model:

Valence shell electron pair repulsion (VSEPR) model:

Predict the geometry of the molecule from the electrostatic repulsion between the electron (bonding and nonbonding) pairs.

AB2 2 0

Class

# of atomsbonded to

central atom

# lonepairs on

central atomArrangement of electron pairs

MolecularGeometry

linear linear

B B

Examples:

CS2, HCN, BeF2

Cl ClBe

2 atoms bonded to central atom0 lone pairs on central atom

AB2 2 0 linear linear

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB3 3 0trigonal planar

trigonal planar

Examples:

SO3, BF3, NO3-, CO3

2-


Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar

AB4 4 0 tetrahedral tetrahedral

Examples:

CH4, SiCl4, SO42-, ClO4

-


Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar


AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

Examples: PF5

AsF5

SOF4


Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar


AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB6 6 0 octahedraloctahedral

SF6

IOF5

Figure 10.5

Electron-group repulsions and the five basic molecular shapes.

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 Pairs

bonding-pair vs. bondingpair repulsion

lone-pair vs. lone pairrepulsion

lone-pair vs. bondingpair repulsion> >

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


trigonal planar

AB2E 2 1trigonal planar

bent

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB3E 3 1


tetrahedraltrigonal

pyramidal

NH3

PF3

ClO3

H3O+

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


AB3E 3 1 tetrahedraltrigonal

pyramidal

AB2E2 2 2 tetrahedral bent

H

O

HH2O

OF2

SCl2

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB4E 4 1trigonal

bipyramidaldistorted

tetrahedron

SF4

XeO2F2

IF4+

IO2F2-

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB4E 4 1trigonal


tetrahedron

AB3E2 3 2trigonal

bipyramidalT-shaped

ClF

F

F

ClF3

BrF3

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR

AB5 5 0trigonal

bipyramidaltrigonal

bipyramidal

AB4E 4 1trigonal


tetrahedron

AB3E2 3 2trigonal

bipyramidalT-shaped

AB2E3 2 3trigonal

bipyramidallinear

I

I

I

H2O

OF2

SCl2

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


AB5E 5 1 octahedral square pyramidal

Br

F F

FF

F

BrF5

TeF5-

XeOF

4

Class

# of atomsbonded to

central atom

# lonepairs on


MolecularGeometry

VSEPR


AB5E 5 1 octahedral square pyramidal

AB4E2 4 2 octahedral square planar

Xe

F F

FFXeF4

ICl4-

Predicting Molecular Geometry1. Draw Lewis structure for molecule.

2. Count number of lone pairs on the central atom and number of atoms bonded to the central atom.

3. Use VSEPR to predict the geometry of the molecule.

What is the molecular geometry of SO2 and SF4?

SO O

AB2E

bent

S

F

F

F F

AB4E

distortedtetrahedron

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 Pairs

Figure 10.4

A balloon analogy for the mutual repulsion of electron groups.

Figure 10.7 The two molecular shapes of the trigonal planar electron-group arrangement.

Class

Shape

Examples:

SO3, BF3, NO3-, CO3

2-

Examples:

SO2, O3, PbCl2, SnBr2

Figure 10.8 The three molecular shapes of the tetrahedral electron-group arrangement.

Examples:

CH4, SiCl4, SO4

2-, ClO4-

NH3

PF3

ClO3

H3O+

H2O

OF2

SCl2

Figure 10.9 Lewis structures and molecular shapes

Figure 10.10 The four molecular shapes of the trigonal bipyramidal electron-group arrangement.

SF4

XeO2F2

IF4+

IO2F2-

ClF3

BrF3

XeF2

I3-

IF2-

PF5

AsF5

SOF4

Figure 10.11 The three molecular shapes of the octahedral electron-group arrangement.

SF6

IOF5

BrF5

TeF5-

XeOF4

XeF4

ICl4-

Figure 10.12The 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

See Figure 10.1

SAMPLE PROBLEM 10.7 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 10.7 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 10.8 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

Figure 10.13

The tetrahedral centers of ethane.

Figure 10.13

The tetrahedral centers of ethanol.

SAMPLE PROBLEM 10.9 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

Dipole Moments and Polar Molecules

H F

electron richregion

electron poorregion

= Q x rQ is the charge

r is the distance between charges

1 D = 3.36 x 10-30 C m

Electric field OFF Electric field ON

Which of the following molecules have a dipole moment?H2O, CO2, SO2, and CH4

O HH

dipole momentpolar molecule

SO

O

CO O

no dipole momentnonpolar molecule

dipole momentpolar molecule

C

H

H

HH

no dipole momentnonpolar molecule

SAMPLE PROBLEM 10.10 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 10.10 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

chapter 10

Documents

pair of e

valence en

pair of nonbonding e

writing lewis structures

remaining valence electrons

formal charge of atom

lewis structure construction

lewis structures10