the shapes of molecules

10-1

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

The Shapes of Molecules

10-2


The steps to follow in converting a molecular formula into LEDS.

1st: Place the atoms relative to each other. [Atom with the lowest electronegativity(EN)]

2nd: Determine the total number of valence electrons available. (Recall that the number of valence e- equals the A-group number)

3rd: Draw a single bond from each surrounding atom to the central atom and subtract two valence electrons for each bond.

4th: Distribute the remaining electrons in pairs so that each atom ends up with 8 electrons (or 2 for H). First place the lone pairs to the surrounding (more electronegative) atoms to give each an octet.

10-3


The steps to follow in converting a molecular formula into LEDS.

5th : CASES INVOLVING MULTIPLE BONDS If after step 4, a central atom still

does NOT have an octet, make MULTIPLE bond by changing a lone pair from one of the surrounding atoms into a bonding pair in the central atom.

10-4


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

10-5


Molecular formula

Atom placement

Sum of valence e-

Remaining valence e-

Lewis structure

NF3

NFF

F

N 5e-

F 7e-

X 3 = 21e-Total 26e-

:

: :

:

: :

:: ..

N 5e- X 1 = 5e- F 7e-

NF

F

F

:

10-6


SAMPLE PROBLEM 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.

:

::

::

:

:

::

: ::

10-7


SAMPLE PROBLEM 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

::

10-9


There are some compounds that undergo the process of chemical bonding that form more or less than 8 electrons are are considered EXCEPTIONS TO THE OCTET RULE.

1) Electron Deficient molecules

- gaseous compounds containing Be or B as the central atom.

Ex. BF3; BeCl2

2) Odd-Electron molecules

-most have central atoms from an odd-numbered group.

Ex. N (Group 5A -15); Cl (Group 7A -17)

NO2 ( free radical contain a lone electron.)

10-10


There are some compounds that undergo the process of chemical bonding that form more or less than 8 electrons are are considered EXCEPTIONS TO THE OCTET RULE.

3) Expanded Valence Shells

- molecules having MORE than 8 valence electrons around the central atom.

- occur around a central NON METAL atom from period 3 or higher, those in which d orbitals are available.

Ex. SF6

10-11


SAMPLE PROBLEM 3 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

10-12


• VSEPR focuses not only on electron pairs it also focus on electron groups as a whole.

• An electron group is an electron pair, a lone pair, a single unpaired electron, a double bond or a triple bond on the central atom.

The valence-shell electron-pair repulsion (VSEPR) theory states that electron pairs repel each other

whether they are in bond pairs or in lone pairs. Electron pairs will spread themselves as far from each other as

possible to minimize repulsion.

The valence-shell electron-pair repulsion (VSEPR) theory states that electron pairs repel each other

whether they are in bond pairs or in lone pairs. Electron pairs will spread themselves as far from each other as

possible to minimize repulsion.

10-13


VSEPR - Valence Shell Electron Pair Repulsion Theory has a its general formula.

A Xm En

A - central atom

X -surrounding atom

E -nonbonding valence electron-

group

integers

Understanding the molecular structure of a compound can help determine the polarity, reactivity, phase of matter, color, magnetism, as well as the biological activity.

10-14


The actual shape of a molecule can be determined by the location of the nuclei and the distribution of electrons.

CATEGORIES:

1) Electron-group geometry is determined by the NUMBER of ELECTRON GROUPS

2) Molecular Geometry depends on the NUMBER OF LONE PAIRS.

Number of electron groups

Name of electron group geometry

2 linear

3 trigonal-planar

4 tetrahedral

5trigonal-

bipyramidal

6 octahedral

ELECTRON GROUP GEOMETRY

10-15

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Number of

Electron Groups

Electron-Group

Geometry

# of Lone Pairs

VSEPR Notation(Type of Shape)

Molecular Geometry

Ideal Bond

Angles

Examples

2 linear

0 AX2 180° BeH2

3Trigonal planar

0 AX3 120° CO32-

1 AX2E 120° O3

10-16

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Number of

Electron Groups

Electron-Group

Geometry

# of Lone Pairs


Molecular Geometry

Ideal Bond

Angles

Examples

4

Tetrahe

-dral

0 AX4 109.5° S04

2-

1 AX3E 109.5° H3O+

2 AX2E2 109.5° H2O

10-17

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.No. of Electro

n Groups

Electron-Group

Geometry

# of Lone Pairs


Molecular Geometry

Ideal Bond

Angles

Examples

5

Trigonal-Bipyramidal

0

AX5 90°, 120° PF5

1

AX4E90°, 120° TeCl4

2 AX3E2 90°

3 AX2E3180°

I3-

10-18

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.No. of Electro

n Groups

Electron-Group

Geometry

# of Lone Pairs


Molecular Geometry

Ideal Bond

Angles

Examples

6 Octahedral

0

AX6 90°

PF6

-

1AX5E

90° SbCl52-

2 AX4E290° ICl4

-

10-19


Figure 10.2

Electron-group repulsions and the five basic molecular shapes.

linear trigonal planar tetrahedral

trigonal bipyramidal octahedral

10-20


Figure 10.3 The single molecular shape of the linear electron-group arrangement.

Examples:

CS2, HCN, BeF2

10-21


Figure 10.4 The two molecular shapes of the trigonal planar electron-group arrangement.

Class

Shape

Examples:

SO3, BF3, NO3-, CO3

2-

Examples:

SO2, O3, PbCl2, SnBr2

10-22


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

10-23


Figure 10.5 The three molecular shapes of the tetrahedral electron-group arrangement.

Examples:

CH4, SiCl4, SO4

2-, ClO4-

NH3

PF3

ClO3

H3O+

H2O

OF2

SCl2

10-24


Figure 10.6 Lewis structures and molecular shapes.

10-25


Figure 10.7 The four molecular shapes of the trigonal bipyramidal electron-group arrangement.

SF4

XeO2F2

IF4+

IO2F2-

ClF3

BrF3

XeF2

I3-

IF2-

PF5

AsF5

SOF4

10-26


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

SF6

IOF5

BrF5

TeF5-

XeOF4

XeF4

ICl4-

10-27


Figure 10.9 A summary of common molecular shapes with two to six electron groups.

10-28


Figure 10.10 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

See Figure 10.1

10-29


SAMPLE PROBLEM 10.6 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

10-30


SAMPLE PROBLEM 10.6 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

10-31


SAMPLE PROBLEM 10.7 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

the shapes of molecules

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