Chapter 7, Problem 7.88PAE

7.88 The ${\text{N}}_5{}^{+}$ cation has been synthesized and studied. Consider the possible Lewis structure below.
Indicate the hybridization expected for each nitrogen atom and the expected bond angles.

Assuming that the structure shown above is correct, how many of the five nitrogen atoms would always lie in the same plane?

Interpretation Introduction

Interpretation:

To find out the hybridization of ${\text{N}}_{\text{5}}^{\text{+}}$ cation and the bond angles

Concept introduction:

The valence bond theory and molecular orbital theory are used to explain the chemical bonding in the molecule.

Answer to Problem 7.88PAE

Solution: Pentazenium ion ${\text{N}}_{\text{2}}\text{and}{\text{N}}_{\text{4}}$ are lies the same plane. ${\text{N}}_{\text{(1)}}\text{and}{\text{N}}_{\text{(5)}}$ are $\text{sp}$ hybridized , ${\text{N}}_{\text{2}}\text{and}{\text{N}}_{\text{4}}$

${\text{sp}}^{\text{2}}$ hybridized and ${\text{N}}_{\text{3}}$ is $\text{sp}$ hybridized in the pentazenium ion.

Explanation of Solution

The ${\text{N}}_{\text{5}}^{\text{+}}$ cation is commonly called as pentazenium cation. It contains positively charged polynitrogen ion with chemical formula ${\text{N}}_{\text{5}}^{\text{+}}$.The ${\text{N}}_{\text{5}}^{\text{+}}$ cation, the Lewis structure is $\text{:N}\equiv \text{N-}\underset{\text{••}}{\overset{\text{••}}{\text{N}}}\text{-N}\equiv \text{N:}$. Based on valence bond theory, ${\text{N}}_{\text{5}}^{\text{+}}$ contains six resonance structures as follows:

$\begin{array}{l}{\left[\text{N}\equiv {\text{N}}^{\text{+}}{\text{-N}}^{\text{-}}{\text{-N}}^{\text{+}}\equiv \text{N}\right]}^{\text{+}}\leftrightarrow {\left[{\text{N}}^{\text{-}}{\text{=N}}^{\text{+}}{\text{=N-N}}^{\text{+}}\equiv \text{N}\right]}^{\text{+}}\leftrightarrow {\left[\text{N}\equiv {\text{N}}^{\text{+}}{\text{-N=N}}^{\text{+}}{\text{=N}}^{-}\right]}^{\text{+}}\leftrightarrow {\left[\text{N}\equiv {\text{N}}^{\text{+}}{\text{-N}}^{+}\equiv {\text{N}}^{\text{+}}{\text{-N}}^{2-}\right]}^{\text{+}}\leftrightarrow \\ {\left[{\text{N}}^{\text{2-}}{\text{-N}}^{\text{+}}\equiv {\text{N}}^{\text{+}}{\text{-N}}^{\text{+}}\equiv \text{N}\right]}^{\text{+}}\leftrightarrow {\left[{\text{N}}^{\text{-}}{\text{=N}}^{\text{+}}{\text{=N}}^{\text{+}}{\text{=N}}^{\text{+}}{\text{=N}}^{\text{-}}\right]}^{\text{+}}\end{array}$

Based on the x-ray experimental information, the ${\text{N}}_{\text{5}}^{\text{+}}$ cation is planar and symmetrical in nature. It has a V-shaped structure with bond angles $\text{111°}$ at the centre atom $\left(\text{N2-N3-N4}\right)$ and the bond angle for $\left(\text{N1-N2-N3}\text{and}\text{N3-N4-N5}\right)$ is calculated as $168°$. The bond lengths for $\text{N1-N2}\text{and}\text{N4-N5}$ are predicted as $\text{1}\text{.10}\stackrel{\circ }{\text{A}}$. The bond lengths for $\text{N2-N3}\text{and}\text{N3-N4}$ are calculated as $\text{1}\text{.30}\stackrel{\text{o}}{\text{A}}$.

Consider the structure, ${\text{:N}}_{(1)}\equiv {\text{N}}_{(2)}\text{-}\underset{\text{••}}{\overset{\text{••}}{{\text{N}}_{(3)}}}{\text{-N}}_{(4)}\equiv {\text{N}}_{(5)}$ : in this pentazenium ion ${\text{N}}_{\text{(1)}}\text{and}{\text{N}}_{\text{(5)}}$ are $\text{sp}$ hybridized it contains triple bond with other species and a lone pair of electrons. For ${\text{N}}_{\text{2}}\text{and}{\text{N}}_{\text{4}}$ are ${\text{sp}}^{\text{2}}$ hybridized it contains a single bond and the double bond. For ${\text{N}}_{\text{3}}$ it is ${\text{sp}}^{\text{3}}$ hybridized in nature. It contains two single bonds and a lone pair of electron. The pentazenium ion ${\text{N}}_{\text{2}}\text{and}{\text{N}}_{\text{4}}$ are lie in the plane.

Conclusion

Pentazenium ion ${\text{N}}_{\text{2}}\text{and}{\text{N}}_{\text{4}}$ are lies the same plane. ${\text{N}}_{\text{(1)}}\text{and}{\text{N}}_{\text{(5)}}$ are $\text{sp}$ hybridized , ${\text{N}}_{\text{2}}\text{and}{\text{N}}_{\text{4}}$

${\text{sp}}^{\text{2}}$ hybridized and ${\text{N}}_{\text{3}}$ is $\text{sp}$ hybridized in the pentazenium ion.