Chapter 8, Problem 131QP

Describe the molecular shape of each of the following.

$\begin{array}{l}\text{(a)}{\text{SiCl}}_{\text{4}}\\ \text{(b)}{\text{GaCl}}_{\text{3}}\\ \text{(c)}{\text{NCl}}_{\text{2}}^{+}\\ \text{(d)}{\text{IO}}_{\text{3}}^{-}\\ \text{(e)}{\text{PCl}}_{\text{4}}^{\text{+}}\\ \text{(f)}{\text{OF}}_{\text{2}}\\ \text{(g)}{\text{GeH}}_{\text{4}}\\ \text{(h)}{\text{SOCl}}_{\text{2}}\\ \text{(i)}{\text{Br}}_{\text{2}}\text{O}\\ \text{(j)}{\text{ClO}}_{\text{2}}^{-}\end{array}$

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{SiCl}}_4$ is to be determined.

Explanation of Solution

The outermost valence electrons for silicon and each chlorine are ${\text{4 e}}^{-}$ and ${\text{7 e}}^{-}$ , respectively. Thus, each chlorine requires one electron and silicon requires four electrons to complete the octet by sharing electrons. The Lewis structure of ${\text{SiCl}}_4$ is given below.

${\text{SiCl}}_4$ contains four electron domains in which silicon bonds to four chlorine atoms. Therefore, the shape of ${\text{SiCl}}_4$ is tetrahedral and the bond angle is $109.5°$ .

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{GaCl}}_3$ is to be determined.

Explanation of Solution

The outermost valence electrons for each chlorine and gallium are ${\text{7 e}}^{-}$ and ${\text{3 e}}^{-}$ , respectively. Thus, gallium requires five electrons and each chlorine requires one electron to complete the octet by sharing electrons. The Lewis structure of ${\text{GaCl}}_3$ is given below.

${\text{GaCl}}_3$ contains three electron domains in which gallium bonds to three chlorine atoms. Therefore, the shape of ${\text{GaCl}}_3$ is trigonal planar and the bond angle is $120°$ .

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{NCl}}_2^{+}$ is to be determined.

Explanation of Solution

The outermost valence electrons for each chlorine and nitrogen are ${\text{7 e}}^{-}$ and ${\text{5 e}}^{-}$ , respectively. The Lewis structure of ${\text{NCl}}_2^{+}$ is given below.

In the above structure, nitrogen does not complete its octet after sharing the electrons with chlorine atoms. Thus, nitrogen in ${\text{NCl}}_2^{+}$ has one lone pair of electrons. Therefore, the shape of ${\text{NCl}}_2^{+}$ is bent with a bond angle of $120°$ .

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{IO}}_3^{-}$ is to be determined.

Explanation of Solution

The outermost valence electrons for each oxygen and iodine are ${\text{6 e}}^{-}$ and ${\text{7 e}}^{-}$ , respectively. The Lewis structure of ${\text{IO}}_3^{-}$ is given below.

Iodine contains one lone pair of electrons after sharing the electrons with oxygen. Therefore, the shape of ${\text{IO}}_3^{-}$ is trigonal pyramidal and the bond angle is $109.5°$ .

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{PCl}}_4^{+}$ is to be determined.

Explanation of Solution

The outermost valence electrons for each chlorine and phosphorus are ${\text{7 e}}^{-}$ and ${\text{5 e}}^{-}$ , respectively. The Lewis structure of ${\text{PCl}}_4^{+}$ is given below.

The shape of ${\text{PCl}}_4^{+}$ is tetrahedral and the bond angle is $109.5°$ .

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{OF}}_2$ is to be determined.

Explanation of Solution

The outermost valence electrons for oxygen and each fluorine are ${\text{6 e}}^{-}$ and ${\text{7 e}}^{-}$ , respectively. Thus, each chlorine requires one electron and each oxygen requires two electrons to complete the octet by sharing electrons. The Lewis structure of ${\text{OF}}_2$ is given below.

In ${\text{OF}}_2$ , the oxygen atom has two lone pairs of electrons. Therefore, the shape of ${\text{OF}}_2$ is bent with a bond angle of $109.5°$ .

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{GeH}}_4$ is to be determined.

Explanation of Solution

The outermost electrons for each hydrogen and germanium are ${\text{1 e}}^{-}$ and ${\text{4 e}}^{-}$ , respectively. Thus, each hydrogen requires one electron and germanium requires four electrons to complete the octet by sharing electrons. The Lewis structure of ${\text{GeH}}_4$ is given below.

The shape of ${\text{GeH}}_4$ is tetrahedral and the bond angle is $109.5°$ .

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{SOCl}}_2$ is to be determined.

Explanation of Solution

The outermost electrons for oxygen, sulfur and each chlorine are ${\text{6 e}}^{-}$ , ${\text{6 e}}^{-}$ , and ${\text{7 e}}^{-}$ , respectively. Thus, sulfur and chlorine require two-electrons and each chlorine requires one electron to complete the octet by sharing electrons. The Lewis structure of ${\text{SOCl}}_2$ is given below.

The shape of ${\text{SOCl}}_2$ is trigonal pyramidal with a bond angle of $109.5°$ .

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{Br}}_2\text{O}$ is to be determined.

Explanation of Solution

The outermost electron for each bromine and oxygen are ${\text{7 e}}^{-}$ and ${\text{6 e}}^{-}$ , respectively. Thus, each bromine requires one electron and oxygen requires two electrons to complete the octet by sharing electrons. Thus, the Lewis structure of ${\text{Br}}_2\text{O}$ is given below.

In ${\text{Br}}_2\text{O}$ , the oxygen atom has two lone pair of electrons. Therefore, the shape of ${\text{Br}}_2\text{O}$ is bent and the bond angle is $109.5°$ .

Interpretation Introduction

Interpretation:

The molecular shape of ${\text{ClO}}_2^{-}$ is to be determined.

Explanation of Solution

The outermost electron for each oxygen and chlorine are ${\text{6 e}}^{-}$ and ${\text{7 e}}^{-}$ , respectively. Thus, chlorine requires one electron and each oxygen requires two electrons to complete the octet by sharing the electrons. Thus, the Lewis structure of ${\text{ClO}}_2^{-}$ is given below.

In ${\text{ClO}}_2^{-}$ , the chlorine atom two lone pair of electrons. Therefore, the shape of ${\text{ClO}}_2^{-}$ is bent with a bond angle of $109.5°$ .