Chapter 8, Problem 8.171AP

(a)

Interpretation Introduction

Interpretation: The justification for the isoelectronic behaviour of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ and ${\text{N}}_{\text{2}}\text{O}$; the Lewis structure of ${\text{N}}_{\text{2}}{\text{F}}^{+}$; the atom that has the $+1$ formal charge in the Lewis structure; the possibility of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ to have resonance forms and the possibility of the central atom of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ to be fluorine atom is to be stated.

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

To determine: The justification for the isoelectronic behaviour of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ and ${\text{N}}_{\text{2}}\text{O}$.

Answer to Problem 8.171AP

Solution

The total valence electrons of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ is same as that of ${\text{N}}_{\text{2}}\text{O}$. Therefore, ${\text{N}}_{\text{2}}{\text{F}}^{+}$ and ${\text{N}}_{\text{2}}\text{O}$ are isoelectronic.

Explanation of Solution

Explanation

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.

The atomic number of oxygen is $6$.

The electronic configuration of oxygen is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^4$. Therefore, oxygen has six 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, ${\text{N}}_{\text{2}}{\text{F}}^{+}$ has an electron less with that of the valence electrons of two nitrogen and fluorine atoms.

Thus, the total number of valence electrons of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ is $2\times 5+7-1=16$ and the total valence electrons of ${\text{N}}_{\text{2}}\text{O}$ is $2\times 5+6=16$.

Hence, the total number of valence electrons of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ and ${\text{N}}_{\text{2}}\text{O}$ is same. Therefore, ${\text{N}}_{\text{2}}{\text{F}}^{+}$ and ${\text{N}}_{\text{2}}\text{O}$ are isoelectronic.

(b)

Interpretation Introduction

To determine: The Lewis structure of ${\text{N}}_{\text{2}}{\text{F}}^{+}$.

Answer to Problem 8.171AP

Solution

The Lewis structure of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ shows $16$ valence electrons.

Explanation of Solution

Explanation

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.

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, ${\text{N}}_{\text{2}}{\text{F}}^{+}$ has an electron less with that of the valence electrons of two nitrogen and fluorine atoms.

Thus, the total number of valence electrons of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ is $2\times 5+7-1=16$.

Therefore, the Lewis structure of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ is,

Figure 1

(c)

Interpretation Introduction

To determine: The atom that has the $+1$ formal charge in the Lewis structure.

Answer to Problem 8.171AP

Solution

The atom that has $+1$ formal charge in the Lewis structure is the central nitrogen atom.

Explanation of Solution

Explanation

Formal charge is calculated by the formula,

$\text{Formal}\text{charge}=\left(\begin{array}{l}\text{Valence}\text{electrons}\\ -\left(\text{Number}\text{of}\text{nonbonding}\text{electrons}+\frac{\text{Number}\text{of}\text{bonding}\text{electrons}}{2}\right)\end{array}\right)$

Therefore, the formal charge of nitrogen containing six bonding electrons and two nonbonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{5}-\left(\text{2}+\frac{\text{6}}{2}\right)\right)\\ =5-5\\ =0\end{array}$

The formal charge of nitrogen containing eight bonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{5}-\left(0+\frac{\text{8}}{2}\right)\right)\\ =5-4\\ =+1\end{array}$

The formal charge of nitrogen containing four bonding electrons and four nonbonding electrons is,

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{5}-\left(\text{4}+\frac{4}{2}\right)\right)\\ =5-6\\ =-1\end{array}$

(d)

Interpretation Introduction

To determine: The possibility of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ to have resonance forms.

Answer to Problem 8.171AP

Solution

There are three resonance forms of ${\text{N}}_{\text{2}}{\text{F}}^{+}$.

Explanation of Solution

Explanation

The Lewis structure of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ is,

Figure 1

Therefore, the possible resonance forms of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ are,

• Nitrogen bonded to nitrogen with triple bond an fluorine bonded to nitrogen with single bond.
• Nitrogen bonded to nitrogen with double bond an fluorine bonded to nitrogen with double bond.
• Nitrogen bonded to nitrogen with single bond an fluorine bonded to nitrogen with triple bond.

Thus, the resonating structures of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ are,

Figure 2

Therefore, there are three resonance forms

(e)

Interpretation Introduction

To determine: The possibility of the central atom of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ to be fluorine.

Answer to Problem 8.171AP

Solution

It is possible for the central atom of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ to be fluorine.

Explanation of Solution

Explanation

Principle quantum number of fluorine is $2$. Therefore, fluorine has one $\text{s}$ and three $\text{p}$ orbitals to form total four bonds. Therefore, being the central atom of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ fluorine forms minimum two bond and maximum four bonds with the two nitrogen atoms.

Conclusion

1. a. The total valence electrons of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ is same as that of ${\text{N}}_{\text{2}}\text{O}$. Therefore, ${\text{N}}_{\text{2}}{\text{F}}^{+}$ and ${\text{N}}_{\text{2}}\text{O}$ are isoelectronic.
2. b. The Lewis structure of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ shows $16$ valence electrons.
3. c. The atom that has $+1$ formal charge in the Lewis structure is the central nitrogen atom.
4. d. There are three resonance forms of ${\text{N}}_{\text{2}}{\text{F}}^{+}$.
5. e. It is possible for the central atom of ${\text{N}}_{\text{2}}{\text{F}}^{+}$ to be fluorine.