The bond lengths of C_2,N_2 and B_2 molecules | vedclass

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Question

(A) Bond length is inversely proportional to bond order. The bond orders of the given molecules are calculated using Molecular Orbital Theory $(MOT)$.

Vedclass · Accepted Answer

$X_3 > X_1 > X_2$

Vedclass · Answer

(A) Bond length is inversely proportional to bond order. The bond orders of the given molecules are calculated using Molecular Orbital Theory $(MOT)$.
$	ext{Bond Order} = \frac{	ext{Number of BMO } e^{-} - 	ext{Number of ABMO } e^{-}}{2}$
For $B_2$ ($10$ electrons): $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \pi 2p_{x}^1 = \pi 2p_{y}^1$. Bond Order $= \frac{6-4}{2} = 1$.
For $C_2$ ($12$ electrons): $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \pi 2p_{x}^2 = \pi 2p_{y}^2$. Bond Order $= \frac{8-4}{2} = 2$.
For $N_2$ ($14$ electrons): $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \pi 2p_{x}^2 = \pi 2p_{y}^2 \sigma 2p_{z}^2$. Bond Order $= \frac{10-4}{2} = 3$.
Since the bond order is $N_2 (3) > C_2 (2) > B_2 (1)$,the bond length order is $B_2 > C_2 > N_2$,which corresponds to $X_3 > X_1 > X_2$.

The bond lengths of $C_2$,$N_2$ and $B_2$ molecules are $X_1$,$X_2$ and $X_3$ $pm$ respectively. The correct order of their bond lengths is

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