Chapter #10 - 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( VB ) Theory and Molecular Shape

10.4 Molecular Shape and Molecular Polarity

Lewis Structures

• 1) Only the valence electrons appear in a Lewis structure.

• 2) The line joining two atoms represents a pair of electrons shared between two atoms.– single bond - two shared electrons, one line

– double bond - four shared electrons, two lines

– triple bond - six shared electrons, three lines

• 3) Dots placed next to an atom represent nonbonding electrons.

Lewis Structures of the Elements by Group in the Periodic Table

Group I II III IV V VI

H

Li

Na

.

.

.

.Be

Mg

Ca

. .

. .. .

. .B

Al

. .

. .

.

.

..

..

.

..

C

Si. .

.

.

..

..

..

N

P

. .

. .

. .O

..

..

..

S

VII VIII

F Cl . .

. .

. .

. . . .. . . .. .

. .He Ne Ar. . . . . .

. .

. .. .

Fig 10.1 (P 362)

Problem: Write a Lewis structure for the molecule CHCl3, Chloroform, a molecules that has been used to put people to sleep!Solution: Step 1: Place the atoms next to each other with carbon in the center, since it is the lowest element in a group with more that one electron. Place the others around the carbon in the four locations.Step 2: Count valence electrons.

Step 3: Draw single bonds between the atoms, and subtract 2 electronsper bond. 26 electrons - 8 electrons = 18 electrons.Step 4: Distribute the remaining electron in pairs beginning with the surrounding atoms.Check:

Writing Lewis Structures for molecules with one central atom.

C

Cl

Cl Cl

H

[1xC(4e-)] [1xH(1e-)] [3xCl(7e-)] = 26 electrons

1.

C

Cl

Cl Cl

H

C

Cl

Cl Cl

H3. 4.

......

........

....

Lewis Structures of Simple Molecules - I

H H

H H

H H

. .

. . . .. .

.

.

.

.

H H

F F . .

. .

. . . .. .

. .

.. . .H Cl

Sodium Chloride

Na+ Cl . .

. .

. .

-

Molecular FluorineHydrogen Chloride

MolecularHydrogen

Magnesium Chloride

Mg+2Cl Cl . .. .. . . .. . . .

. .

-

. .

-

C C

H H

H H

HH

. . . ...

. . ... ..

.

Ethane

F F. . . .

. .. .. . . .

. .H F . .

. .Hydrogen Fluoride

Writing lewis Structures for Molecules with more than one Central atom!

Problem: Write the Lewis structure for Hydrogen peroxide(molecular formul, H2O2 ) an important household bleech. Solution: Step 1. Place the atoms in the best geometry, with the hydrogen atoms having only one bond, they are on the ends or outside, and oxygen can have up to two bonds so put them in the middle.

H O O H

Step 2. Find the sum of electrons: [2 x H(1e-)] + [2 x O(6e-)] = 14e-

Step 3. Add single bonds and subtract 2e- for each bond:

H - O - O - H 14e- - 6e- = 8e-

Step 4. Add the remaning electron in pairs around the oxygen atoms as Hydrogen can only have two!

H - O - O - H...... ..

Check: oxygen has an octet of 8e-

and hydrogen ha sit’s two electrons.

Writing Lewis Structures for Molecules with Multiple Bonds.

Problem: Write lewis structures for Oxygen and Acetylene(C2H2): Plan: We begin with the first 4 steps we have done: placing atoms, counting electrons, placing single bonds, and completing octets, and if needed we finishas follows by placing multiple bonds in the molecules.Solution:

a) For oxygen: O2 O - O ..........

Change one of the lone pairs to a bonding pair. The oxygen on the right has an octet of electrons, while the oxygen aton on the left only has six electrons, so we convert the lone pair to another bondingpair between the two oxygen atoms.

O O..

....

..

b) For Acetylene: C2H2 H - C - C - H.. ..

Neither of the carbon atoms has an octet, or if they are placed around one atom, the other has only 4! so no octet! Therefore place both pairs into forming multiple bonds, a tripple bond! H - C C - H

Lewis Structures of Simple Molecules - II

CH4 Methane

C

H

H

H

H

H

C C HH

H

H

O

O

C2H4O2 Acetic Acid

C Cl

Cl

Cl

Cl....

......

.. ...... ..

...... ..

C

C O .CCl4 Carbon Tetrachloride

CO Carbon Monoxide O

O O

H

..

....

....CO2 Carbon Dioxide

H2O WaterHydrogen Oxide

.. Cl Cl.. ...... ..

Cl2 Chlorine

.

Lewis Structures of Simple Molecules - III Multiple Bonds

. .. .O O

. .. .

. . . .

O O. .

. .. .. .

H C N . .Hydrocyanic acid Hydrogen Cyanide : HCN

Molecular Oxygen : O2

H C C H

Acetylene : C2H2

C C

H

H

H

H

Ethylene : C2H4

. . N N . .

Nitrogen N2

Writing Lewis Structures - IV

Step 1) Place the atoms relative to each other: For compounds of formula ABn , place the atom with the lower group number in the center, the one that needs more electrons to attain an octet. In NF3 (nitrogen trifluoride), the N (Group 5A) has five electrons so it needs three, whereas F (Group 7A) has seven so it needs only one; thus, N goes in the center with the three F atoms around it.

Step 2) Determine the total number of valence electrons available: For molecules, add up the valence electrons of all atoms (the number of valence electrons equals the A-group number). In NF3, N has five valence electrons, and each F has seven. For polyatomic ions, add one e - for for each negative charge, or subtract one e - for each positive charge.

Writing Lewis Structures - V

Step 3) Draw a single bond from each surrounding atom to the central atom, and subtract two valence electrons for each bond. There must be at least a single bond between bonded atoms.

Step 4) Distribute the remaining electrons in pairs so that each atom obtains eight electrons (or two for H). Place lone pairs on the surrounding (more electronegative) atoms first to give each an octet. If electrons remain, place them around the central atom. Then check that each atom has 8e -.

Lewis Structures of Simple Molecules - VI

C

H

H H

H

Cl

O

O O

K+

KClO3

CF4

..

..H C O H

H

H

H

H

C

Ethyl Alcohol (Ethanol)

Potassium Chlorate Carbon Tetrafluoride

......

..

..

..

.. ...... ..

.... C

F

FF

F

......

..

.. ..

..

....

CH4

Methane

Lewis Structures of Simple Molecules - VII

N

H

H H. .

N

H

H H

H

+

Ammonia

Ammonium Ion

C

NN

O

H H

HH

. .

. .

. .

. .

Urea

(P 365)

Resonance: Delocalized Electron-Pair Bonding - I

Ozone : O3 ......

..

O O

O ..

........ ..OOO ....

I II

O

O

O

..

........

Resonance Hybrid Structure

One pair of electron’s resonances between the two locations!!

Resonance:Delocalized Electron-Pair Bonding - II

C

CC

CC

C

CC

C

C C

C

C

CC

CC

C

H

HHH

H H

H

H

H

H

H

H

H

HHH

H

HResonance Structure Benzene

Lewis Structures of Simple Molecules

Resonance Structures -III Nitrate

N

O

O O

N

O

O O

..

..

..

..

..

..

....

.... ..

.... ..

....

.... ..

N

O

O O......

.. ..

Lewis Structures of Simple Molecules -VIII• Determine the Lewis

structure of molecular Nitrogen, N2

• N2 is a covalent compound.

• There are 10 valence electrons.

• N-N use 2 e-, leaving 8 around the 2 atoms.

• Three pairs are placed around one atom, leaving 1 pair.

• Provisional structure:• N N Calculate FC • Formal Charge

• N = 5 valence -(1 bonding

• + 2 nonbonding) = +2

• N = 5 valence -(1 bonding

• + 6 nonbonding) = -2

• Move electrons in to make a triple bond.

• N N

. .

. .. .

. .

. . . .

Lewis Structures for Octet Rule Exceptions

Cl

F

F

F ....

..

..

....

..

.... ..

..

..

..

BCl

Cl

Cl......

..

......

Each Fluorine atom has 8 electrons associated. Chlorine has 10 electrons!

Each Chlorine atom has8 electrons associated. Boron has only 6!

Cl ClBe....

..

.. ....

Each Chlorine atom has8 electrons associated. The Beryllium has only 4 electrons.

NO O

... .... ..

..NO2 is an odd electron atom.The nitrogen has 7 electrons.

Resonance Structures - IVExpanded Valence Shells..

.. S

F

F

F

FF

F......

.. ......

.. .. .. ....

.. .. ....

Sulfur hexafluoride

....

.. PF

F

F

FF......

......

..

...... ..

..

Phosphorous pentafluoride

O

S

O

O OH H

..

....

.... ..

.. ......

O

S

O

O OH H.. .... ..

........

Sulfuric acid

S = 12e- p = 10e-

S = 12e-

Resonance Structures

Lewis Structures of Simple Molecules

Resonance Structures-VSO O

O

O

SO O

O

O

. .

. . . .

. .. .

. .. .. .

. .

. .

. .

. .

. .

-2

. .

. .

. .. .. .

-2 Sulfate

S

O

O

O Oxx

x = Sulfur electrons o = Oxygen electrons

o o

o o

o o

o o

o o

x o

x x

x o

o o

o o

o o

o o

o *

o *-2

o o

Plus 4 othersfor a total of 6

. .

. .

Fig 10.2 (P 371)

Fig 10.3 (P 372)

Calculating H from Bond Energies - I

Problem: Using the bond energies in Table 9.2, calculate the H of the reaction between methane and chlorine and fluorine to give freon-12 (CCl2F2).Plan: Look up the bond energies of the reactants and products in table 9.2, and subtract the product bands from the reactant bonds.Solution:CH4 (g) +2 Cl2 (g) + 2 F2 (g) CF2Cl2 (g) + 2 HF(g) + 2 HCl (g)

Reactant bonds broken: For Methane : 4 mol C - H bonds For Molecular Chlorine : 2 mol Cl - Cl bonds For Molecular Fluorine : 2 mol F - F bonds Product bonds formed: For Freon - 12 : 2 mol C - F bonds , 2 mol C - Cl bonds For HF : 2 mol H - F bonds For HCl : 2 mol H - Cl bonds

Reactant bonds broken: 4 mol C - H bonds = 4 mol x 413 kJ/mol = 1652 kJ 2 mol Cl - Cl bonds = 2 mol x 243 kJ/mol = 486 kJ 2 mol F - F bonds = 2 mol x 159 kJ/mol = 318 kJ H0

bonds broken = _________ Product bonds formed: 2 mol C - F bonds = 2 mol x 453 kJ/mol = 906 kJ 2 mol C - Cl bonds = 2 mol x 339 kJ/mol = 678 kJ 2 molH - F bonds = 2 mol x 565 kJ/mol = 1130 kJ 2 mol H - Cl bonds = 2 mol x 427 kJ/mol = 854 kJ H0

bonds formed = ________

H0rxn = H0

bonds broken - H0bonds formed

H0rxn = _______ - _________ = _________

Calculating H from Bond Energies - II

Solution cont.

Two Three Four Five Six

Number of Electron Groups

Fig 10.4 (P 375)

Fig 10.5 (P 375)

Fig 10.6 (P 375)

AX2 Geometry - Linear

Cl ClBe

..

.. ....

..

.. ..1800

BeCl2

GaseousBeryllium Chloride is an example of a molecule in which the central atom - Be does not have an octet of electrons, and is electron deficient.Other alkaline earth elements also have the same valence electron configuration, and the same geometry for molecules of this type. Therefore this geometry is common to group II elements.

Molecular Geometry = Linear Arrangement

CO O..

..

..

1800

Carbon Dioxide also has the same geometry, and is a linear molecule, but in this case, the bonds between the carbon and oxygens are double bonds.

CO2

Fig 10.7 (P 376)

AX3 Geometry - Trigonal Planar

BF3B

F

F F..

..

......

.... ..

..N

O

O O

1200

1200

1200

NO3-

Boron Trifluoride

Nitrate Anion

All of the Boron Family(IIIA)elements have the same geometry. Trigonal Planar !

AX2E SO2

....

.... ....

....

....

....

..S

O O

The AX2E molecules have a pair ofElectrons where the third atom would appear in the space around the central atom, in the trigonal planargeometry.

-

(P 376)

(P 376)

(P 376)

(P 376)

(P 377)

Fig 10.8 (P 377)

Fig 10.9 (P 377)

The effect of lone pair replusion on bond angle.

(P 378)

(P 378)

Fig10.10 (P 378)

(P 378)

(P 379)

(P 379)

(P 379)

Fig 10.11 (P 379)

(P 379)

(P 380)

(P 380)

Using VSEPR Theory to Determine Molecular Shape

1) Write the Lewis structure from the molecular formula to see the relative placement of atoms and the number of electron groups.

2) Assign an electron-group arrangement by counting all electron groups around the central atom, bonding plus nonbonding.

3) Predict the ideal bond angle from the electron-group arrangement and the direction of any deviation caused by the lone pairs or double bonds.

4) Draw and name the molecular shape by counting bonding groups and non-bonding groups separately.

Fig 10.12 (P 380)

Predicting Molecular Shapes

Problem: Determine the molecular shape and ideal bond angles for: a) NCl3 b) COCl2 Solution:

Predicting Molecular Shapes

Problem: Determine the molecular shape and ideal bond angles for: a) NCl3 b) COCl2 Solution: a) for NCl3

1) Write the Lewis structure:

..

.... .. ..N

Cl

Cl Cl

..

..

..

..

..

..

2) Assign the electron arrangement: Four electron groups around N, ( three bonding, and one non-bonding), so we have the tetrahedral arrangement. 3) For the tetrahedral arrangement, the ideal angle is 109.50. Since there is one lone pair, the actual bond angle should be less than 109.50

4) Draw and name the molecular shape:

N

Cl

ClCl..

........

....

..

..

NCl3 Has a trigonal pyramidal shape

Solution: b) for COCl2

1) Write the Lewis structure: C

O

Cl Cl ..

..

........

....

2) Assign the electron-group arrangement: three electron groups around carbon ( two single, and one double bond ) which gives the trigonal planar arrangement.3) Predict the bond angles: the ideal angle is 1200, but the double bond between the Carbon and the oxygen should compress the Cl - C - Cl bond angle by repelling the chlorine atoms, and the bonds between them and the carbon atom.4) Draw and name the molecular shape:

CCl Cl

O

..

.. ....

.. ..

....124.50

1110

Predicting Molecular Shapes with five or six Electron Groups

Problem: Determine the molecular shape and predict the bond angles( relative to the ideal angles ) of (a) AsI5 (b) BrF5

Solution:(a) 1)Lewis structure for AsI5:

As

II I

I I ..

..

..

..

..

..

..

..

..

..

..

....

.. .. ...... ..

2) Electron group arrangement with five groups, this is the trigonal bipyramidal arrangement.

3) Bond angles: since all the groups and surrounding atoms are indentical, the bond angles are ideal: 1200 between equatorial groups and 900 between axial and equatorial groups.

As

I

I

I

I

I..

..

..

..

........

..

....

1200

900

4) Molecular arrangement: Trigonal bipyramidal

b) BrF5

1) Lewis structure for BrF5:

2) Electron Group arrangement 6 electron groups- octhedral!

3) Bond angles: Lone pair should make all angles less than 900.

4) Molecular shape: one lone pair, and five bonding pairs give square pyrimidal:

Br

F

F

F F

FF ........

....

..

..

....

........ ..

..

Br

FF

FF

......

..

..

..

..

..

..

..

.... ..

.. ....

Fig 10.13 (P 382)

(P 383)

(P 383)

(P 383)

(P 383)

Fig 10.14 (P 384)

(P 384-385)

(P 386)

Predicting The Polarity of Molecules -I

Problem: From electronegativity, and their periodic trends, predict whether each of the following molecules is polar and show the directionof bond dipoles and the overall molecular dipole.(a) Phosphine, PH3

(b) Carbon Disulfide, CS2 (atom sequence SCS)(c) Auminum Chloride, AlCl3

Plan: First we draw and name the molecular shape. Then using relative EN values, we decide on the direction of each bond polarity. Finally wedecide upon the polarity of the molecule based upon the geometry.Solution:

(a)

PH

H

H H

..P P

HH

H HH

.. ..

Molecular shape Bond dipoles Molecular Dipole

(b) Carbon Disulfide, CS2..

S C S....

..S C S.... S C S....

.. .... ..

Molecular shape Bond dipoles No -Molecular dipole

(c) Aluminum Chloride, AlCl3

Al Al Al

Cl Cl Cl ClCl

Cl

Cl

Cl Cl..

.... ..

....

..

.. .. ..

..

....

........

............

..

......

Molecular shape Bond dipoles No - Molecular dipole

(II)

Seven Primary Odors or Olfactory Receptor Sites

1. Camphor - like

2. Musky

3. Floral

4. Pepperminty

5. Etheral

6. Pungent

7. Putrid

Fig 10A (P 387)

Fig 10B (P 387)

Fig 10C (P 388)