Chapter 8, Problem 8.166AP

(a)

Interpretation Introduction

Interpretation: The molecules that contain an atom with the expanded valence shell are to be identified.

Concept introduction: The Lewis symbols of the atoms indicate the total number of valence electrons surrounding the atom or the ions.

For an atom to contain an expanded valence shell, the total number of bonding and nonbonding domain should be more than four.

To determine: If ${\text{XeF}}_{\text{2}}$ contain an atom with the expanded valence shell.

Answer to Problem 8.166AP

Solution

Xenon atoms in ${\text{XeF}}_{\text{2}}$ contains an expanded valence shell.

Explanation of Solution

Explanation

The atomic number of xenon is $54$.

The electronic configuration of xenon is $\left[\text{Kr}\right]{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}$. Therefore, xenon has eight valence electrons.

The atomic number of fluorine is $9$.

The electronic configuration of fluorine is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^5$. Therefore, fluorine has seven valence electrons.

Thus, the total number of valence electrons of ${\text{XeF}}_{\text{2}}$ is $8+2\times 7=22$

Therefore, the Lewis structure of ${\text{XeF}}_{\text{2}}$ is,

Figure 1

For an atom to contain an expanded valence shell, the total number of bonding and nonbonding domain should be more than four. In the above Lewis structure, xenon atom has five bonding and nonbonding domains. Therefore, xenon atoms in ${\text{XeF}}_{\text{2}}$ contains an expanded valence shell.

(b)

Interpretation Introduction

To determine: If ${\text{GaCl}}_3$ contain an atom with the expanded valence shell.

Answer to Problem 8.166AP

Solution

None of the atoms in ${\text{GaCl}}_3$ contains an expanded valence shell.

Explanation of Solution

Explanation

The atomic number of gallium is $31$.

The electronic configuration of gallium is $\left[\text{Ar}\right]{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{1}}$. Therefore, gallium has three valence electrons.

The atomic number of chlorine is $17$.

The electronic configuration of chlorine is $\left[\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^5$. Therefore, chlorine has seven valence electrons.

Thus, the total number of valence electrons of ${\text{GaCl}}_3$ is $3+3\times 7=24$

Therefore, the Lewis structure of ${\text{GaCl}}_3$ is,

Figure 2

For an atom to contain an expanded valence shell, the total number of bonding and nonbonding domain should be more than four. In the above Lewis structure, none of the atoms have more than four bonding and nonbonding domains. Therefore, none of the atoms in ${\text{GaCl}}_3$ contains an expanded valence shell.

(c)

Interpretation Introduction

To determine: If ${\text{ONF}}_3$ contain an atom with the expanded valence shell.

Answer to Problem 8.166AP

Solution

None of the atoms in ${\text{ONF}}_3$ contains an expanded valence shell.

Explanation of Solution

Explanation

The atomic number of oxygen is $8$.

The electronic configuration of oxygen is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^4$. Therefore, oxygen has six valence electrons.

The atomic number of nitrogen is $7$.

The electronic configuration of nitrogen is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^3$. Therefore, nitrogen has five valence electrons.

The atomic number of fluorine is $9$.

The electronic configuration of fluorine is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^5$. Therefore, fluorine has seven valence electrons.

Thus, the total number of valence electrons of ${\text{ONF}}_3$ is $6+5+3\times 7=32$

Therefore, the Lewis structure of ${\text{ONF}}_3$ is,

Figure 3

For an atom to contain an expanded valence shell, the total number of bonding and nonbonding domain should be more than four. In the above Lewis structure, none of the atoms have more than four bonding and nonbonding domains. Therefore, none of the atoms in ${\text{ONF}}_3$ contains an expanded valence shell.

(d)

Interpretation Introduction

To determine: If ${\text{SeO}}_2{\text{F}}_{\text{2}}$ contain an atom with the expanded valence shell.

Answer to Problem 8.166AP

Solution

Selenium atoms in ${\text{SeO}}_2{\text{F}}_{\text{2}}$ contains an expanded valence shell.

Explanation of Solution

Explanation

The atomic number of selenium is $34$.

The electronic configuration of selenium is $\left[\text{Ar}\right]{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{4}}$. Therefore, selenium has six valence electrons.

The atomic number of oxygen is $8$.

The electronic configuration of oxygen is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^4$. Therefore, oxygen has six valence electrons.

The atomic number of fluorine is $9$.

The electronic configuration of fluorine is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^5$. Therefore, fluorine has seven valence electrons.

Thus, the total number of valence electrons of ${\text{SeO}}_2{\text{F}}_{\text{2}}$ is $6+2\times 6+2\times 7=32$

Therefore, the Lewis structure of ${\text{SeO}}_2{\text{F}}_{\text{2}}$ is,

Figure 4

For an atom to contain an expanded valence shell, the total number of bonding and nonbonding domain should be more than four In the above Lewis structure, selenium atom has five bonding and nonbonding domains. Therefore, selenium atoms in ${\text{SeO}}_2{\text{F}}_{\text{2}}$ contains an expanded valence shell.

Conclusion

1. a. Xenon atoms in ${\text{XeF}}_{\text{2}}$ contains an expanded valence shell.
2. b. None of the atoms in ${\text{GaCl}}_3$ contains an expanded valence shell.
3. c. None of the atoms in ${\text{ONF}}_3$ contains an expanded valence shell.
4. d. Selenium atoms in ${\text{SeO}}_2{\text{F}}_{\text{2}}$ contains an expanded valence shell.