ABC26IN0066 · Chemical Bonding
Subject: Inorganic Chemistry · Chapter: Chemical Bonding · Topic: Chemical Bonding – General · Exam: CSIR-NET DEC 2015 · Marks: 2 · Difficulty: Easy
The correct statement among the following is
(a)$\mathrm{N_2}$ has higher bond order than $\mathrm{N_2^{+}}$ and hence has larger bond length compared to $\mathrm{N_2^{+}}$.
(b)$\mathrm{N_2^{+}}$ has higher bond order than $\mathrm{N_2}$ and hence has larger bond length compared to $\mathrm{N_2}$
(c)$\mathrm{N_2}$ has higher bond order than $\mathrm{N_2^{+}}$ and hence has higher dissociation energy compared to $\mathrm{N_2^{+}}$
(d)$\mathrm{N_2}$ has lower bond order than $\mathrm{N_2^{+}}$ and hence has lower dissociation energy compared to $\mathrm{N_2^{+}}$ energy.
Answer
Explanation
The MO configuration of $\mathrm{N_2}$ and $\mathrm{N_2^{+}}$ are \[ \mathrm{N_2} = \sigma_{2s}^{2}\,\sigma_{2s}^{*2}\,\pi_{2px}^{2}\,\pi_{2py}^{2}\,\sigma_{2pz}^{2} \quad \text{B.O.} = \frac{8-2}{2} = 3 \]
\[ \mathrm{N_2^{+}} = \sigma_{2s}^{2}\,\sigma_{2s}^{*2}\,\pi_{2px}^{2}\,\pi_{2py}^{2}\,\sigma_{2pz}^{1} \quad \text{B.O.} = \frac{7-2}{2} = 2.5 \]
$\mathrm{N_2} \rightarrow \mathrm{N_2^{+}}$ (Bond order decreases) Open in whiteboard · Browse this chapter in the app