Question

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

• A. \(X_3 \gt X_1 \gt X_2\)
• B. \(X_2 \gt X_3 \gt X_1\)
• C. \(X_1 \gt X_2 \gt X_3\)
• D. \(X_1 \gt X_3 \gt X_2\)

Hint:

In molecular orbital theory: \[ \text{Higher bond order} \Rightarrow \text{Shorter bond length} \] and \[ \text{Higher bond order} \Rightarrow \text{Greater bond strength} \]

Answer 1

The Correct Option is A

Step 1: Recall the relation between bond order and bond length.
Bond length is inversely proportional to bond order.
Higher bond order means stronger bonding and hence smaller bond length.
\[ \text{Bond length} \propto \frac{1}{\text{Bond order}} \]

Step 2: Determine the bond order of \(B_2\).
Electronic configuration of \(B_2\) according to molecular orbital theory is
\[ (\sigma1s)^2(\sigma1s^*)^2(\sigma2s)^2(\sigma2s^*)^2(\pi2p_x)^1(\pi2p_y)^1 \] Bond order is
\[ \text{B.O.}=\frac{N_b-N_a}{2} \] \[ =\frac{6-4}{2} \] \[ =1 \] Thus, \(B_2\) has bond order \(1\).

Step 3: Determine the bond order of \(C_2\).
Electronic configuration of \(C_2\) is
\[ (\sigma1s)^2(\sigma1s^*)^2(\sigma2s)^2(\sigma2s^*)^2(\pi2p_x)^2(\pi2p_y)^2 \] Bond order is
\[ =\frac{8-4}{2} \] \[ =2 \] Thus, \(C_2\) has bond order \(2\).

Step 4: Determine the bond order of \(N_2\).
Electronic configuration of \(N_2\) is
\[ (\sigma1s)^2(\sigma1s^*)^2(\sigma2s)^2(\sigma2s^*)^2(\pi2p_x)^2(\pi2p_y)^2(\sigma2p_z)^2 \] Bond order is
\[ =\frac{10-4}{2} \] \[ =3 \] Thus, \(N_2\) has bond order \(3\).

Step 5: Compare the bond lengths.
Since bond length decreases with increase in bond order,
\[ B_2 \gt C_2 \gt N_2 \] Therefore,
\[ X_3 \gt X_1 \gt X_2 \]

Step 6: Final conclusion.
Hence, the correct order of bond lengths is
\[ \boxed{X_3 \gt X_1 \gt X_2} \]