# 1.E: Structure and Bonding (Exercises)

- Page ID
- 61688

These are the homework exercises to accompany the Textmap for McMurry's Organic Chemistry textbook.

## 1.2: Atomic Structure: Orbitals

### 1.2 Exercises

#### Questions

**Q1.2.1**

Label the following orbitals:

#### Solutions

**S1.2.1**

1= 3*p*_{x}; 2= 3*s* ; 3= 2*p*_{z}

## 1.3: Atomic Structure: Electron Configurations

### 1.3 Exercises

#### Questions

**Q1.3.1**

Give the electron configurations for Al, Br, Fe.

#### Solutions

**S1.3.1**

Al = 1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}

Br = 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}

Fe = 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}

## 1.4: Development of Chemical Bonding Theory

### 1.4 Exercises

#### Questions

**Q1.4.1**

List the bond angles for each of the following compounds: BH_{3}, CF_{4, }H_{2}O.

**Q1.4.2**

Why is sulfur dioxide a bent molecule (bond angle less than 180°)?

#### Solutions

**S1.4.1**

HBH = 120°

FCF = 109.5°

OHO = 104°

**S1.4.2**

This deviation is due to the lone pairs on the sulfur. These force the molecule to exhibit a “bent” geometry and therefore a deviation from the 180°.

## 1.5: The Nature of Chemical Bonds: Valence Bond Theory

### 1.5 Exercises

#### Questions

**Q1.5.1**

Draw an energy diagram for energy vs. intermolecular distance for a fluorine molecule (F_{2}) and describe the regions of the graph.

#### Solutions

**S1.5.1**

A - Repulsive Forces are present, p-orbitals are too close together

B - Optimal distance between the two p-orbitals to have a bond (the bond length)

C - Cannot form a bond, orbitals are too far away

## 1.7: sp3sp3 Hybrid Orbitals and the Structure of Ethane

### 1.7 Exercises

#### Questions

**Q1.7.1**

Draw pentane, CH_{3}CH_{2}CH_{2}CH_{2}CH_{3}, predict the bond angles within this molecule.

#### Solutions

**S1.7.1**

All the bond angles will be the same size.

## 1.8: sp2sp2 Hybrid Orbitals and the Structure of Ethylene

### 1.8 Exercises

#### Questions

**Q1.8.1**

Consider the following molecule:

At each atom, what is the hybridization and the bond angle? At atom A draw the molecular orbital.

#### Solutions

**S1.8.1**

A - sp^{2}, 120°

B - sp^{3}, 109°

C - sp^{2}, 120° (with the lone pairs present)

D - sp^{3}, 109°

## 1.9: spsp Hybrid Orbitals and the Structure of Acetylene

### 1.9 Exercises

#### Questions

**Q1.9.1**

1-Cyclohexyne is a very strained molecule. By looking at the molecule explain why there is such a intermolecular strain using the knowledge of hybridization and bond angles.

#### Solutions

**S1.9.1**

The alkyne is a sp hybridized orbital. By looking at a sp orbital, we can see that the bond angle is 180°, but in cyclohexane the regular angles would be 109.5°. Therefore the molecule would be strained to force the 180° to be a 109°.

## 1.10: Hybridization of Nitrogen, Oxygen, Phosphorus and Sulfur

### 1.10 Exercises

#### Questions

**Q1.10.1**

Identify geometry and lone pairs on each heteroatom of the molecules given.

#### Solutions

**S1.10.1**

Diethyl ether would have two lone pairs of electrons and would have a bent geometry around the oxygen.

Dimethyl amine would have one lone pair and would show a pyramidal geometry around the nitrogen.

## 1.12: Drawing Chemical Structures

### 1.12 Exercises

#### Questions

**Q1.12.1**

Below is the molecule for caffeine. Give the molecular formula for it.

#### Solutions

**S1.12.1**

C_{8}H_{10}O_{2}N_{4}