Question

Arrange the following in increasing order of bond length
(i) $N _2$
(ii) $N _2^{+}$
(iii) $N _2^{2+}$

• A. (ii), (i) and (iii)
• B. (ii), (iii) and (i)
• C. (iii), (ii) and (i)
• D. (i), (ii) and (iii)

Answer 1

Answer: (D)

As the bond order decreases, bond length increases Bond order
$=\frac{\text { No. of bonding } e ^{-} s -\text { No. of antibonding }e^{-} s}{2}$
For $N _2$, electronic configuration is
$\sigma 1 s^2<\sigma^* 1 s^2<\sigma 2 s^2<\sigma^* 2 s^2<\left(\pi 2 p_x^2=\pi 2 p_y^2\right)<2 p_z^2$
(i) Bond order of $N _2=\frac{10-4}{2}=3$
(ii) Bond order of $N _2^{+}=\frac{9-4}{2}=2.5$
(iii) Bond order of $N _2^{2+}=\frac{8-4}{2}=2$
Hence, order of Bond length will be,
$N _2< N _2^{+}< N _2^{2+}$