Question

The molecule which has highest bond order is

• A. $C_2$
• B. $N_2$
• C. $B_2$
• D. $O_2$

Answer 1

The Correct Option is B

$(a) C_2(12)=s 1s^2, s^* 1s^2, s2s^2, s^* 2s^2,\pi 2p^2_x \approx \pi 2p^2_y$
$\, \, \, \, \, \, \, Bond\, order=\frac{N_b-N_a}{2}=\frac{8-4}{2}=2$
$(b) N (14)=s 1s^2, s^* 1s^2, s2s^2, s^* 2s^2,\pi 2p^2_x \approx \pi 2p^2_y,\pi^* 2p^2_x$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, Bond\, order=\frac{10-4}{2}=3$
$(c) B2(10)=s 1s^2, s^* 1s^2, s2s^2, s^* 2s^2,\pi 2p^1_x \approx \pi 2p^1_y$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, Bond\, order=\frac{6-4}{2}=1$
$(d) 02 (16)=s 1s^2, s^* 1s^2, s2s^2, s^* 2s^2,\pi 2p^2_x \approx \pi 2p^2_y,\pi 2p^1_x \approx \pi 2p$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, Bond\, order=\frac{10-6}{2}=2$
$(d) 02 (16)=s 1s^2, s^* 1s^2, s2s^2, s^* 2s^2,\pi 2p^2_x \approx \pi 2p^2_y,\pi 2p^2_x$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, Bond\, order=\frac{10-8}{2}=1$
Hence, $N_2$ has the highest bond order.