Chapter 8, Problem 8.155AP

(a)

Interpretation Introduction

Interpretation: The number of pairs of electrons that xenon share to form the given molecules or ions 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.

To determine: The number of pairs of electrons that xenon share to form ${\text{XeF}}_{\text{2}}$.

Answer to Problem 8.155AP

Solution

Xenon shares only one pair of electrons to form ${\text{XeF}}_{\text{2}}$.

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

Therefore, out of four pairs of electrons, xenon shares only one pair of electrons to form ${\text{XeF}}_{\text{2}}$.

(b)

Interpretation Introduction

To determine: The number of pairs of electrons that xenon share to form ${\text{XeOF}}_{\text{2}}$.

Answer to Problem 8.155AP

Solution

Xenon shares only two pairs of electrons to form ${\text{XeOF}}_{\text{2}}$.

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 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 ${\text{XeOF}}_{\text{2}}$ is $8+6+2\times 7=28$

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

Figure 2

Therefore, out of four pairs of electrons, xenon shares only two pairs of electrons to form ${\text{XeOF}}_{\text{2}}$.

(c)

Interpretation Introduction

To determine: The number of pairs of electrons that xenon share to form ${\text{XeF}}^{+}$.

Answer to Problem 8.155AP

Solution

Xenon shares only one pair of electrons to form ${\text{XeF}}^{+}$.

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.

A species on losing electron possesses a positive charge. The number of electrons gains or lost is equal to the charge raised on the structure of species.

Therefore, ${\text{XeF}}^{+}$ has one electron less than that of the valence electrons of xenon and fluorine.

Thus, the total number of valence electrons of ${\text{XeF}}^{+}$ is $8+7-1=14$

Therefore, the Lewis structure of ${\text{XeF}}^{+}$ is,

Figure 3

Therefore, out of four pairs of electrons, xenon shares only one pair of electrons to form ${\text{XeF}}^{+}$.

(d)

Interpretation Introduction

To determine: The number of pairs of electrons that xenon share to form ${\text{XeF}}_5{}^{+}$.

Answer to Problem 8.155AP

Solution

Xenon shares only one pair of electrons to form ${\text{XeF}}_5{}^{+}$.

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.

A species on losing electron possesses a positive charge. The number of electrons gains or lost is equal to the charge raised on the structure of species.

Therefore, ${\text{XeF}}_5{}^{+}$ has one electron less than that of the valence electrons of xenon and fluorine.

Thus, the total number of valence electrons of ${\text{XeF}}_5{}^{+}$ is $8+5\times 7-1=42$

Therefore, the Lewis structure of ${\text{XeF}}_5{}^{+}$ is,

Figure 4

Therefore, out of four pairs of electrons, xenon shares only one pair of electrons to form ${\text{XeF}}_5{}^{+}$.

(e)

Interpretation Introduction

To determine: The number of pairs of electrons that xenon share to form ${\text{XeO}}_4$.

Answer to Problem 8.155AP

Solution

Xenon shares only four pairs of electrons to form ${\text{XeO}}_4$.

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 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.

Thus, the total number of valence electrons of ${\text{XeO}}_4$ is $8+4\times 6=32$

Therefore, the Lewis structure of ${\text{XeO}}_4$ is,

Figure 5

Therefore, out of four pairs of electrons, xenon shares all four pairs of electrons to form ${\text{XeO}}_4$.

Conclusion

1. a. Xenon shares only one pair of electrons to form ${\text{XeF}}_{\text{2}}$.
2. b. Xenon shares only two pairs of electrons to form ${\text{XeOF}}_{\text{2}}$.
3. c. Xenon shares only one pair of electrons to form ${\text{XeF}}^{+}$.
4. d. Xenon shares only one pair of electrons to form ${\text{XeF}}_5{}^{+}$.
5. e. Xenon shares only four pairs of electrons to form ${\text{XeO}}_4$.