Question

Bond order of which among the following molecules is zero ?

• A. $F_2$
• B. $O_2$
• C. $B{e}_2$
• D. $L{i}_2$

Answer 1

Answer: (C)

(a) $F_2$ molecule

$=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2$

$\pi 2p_x^2\approx \pi 2p_y^2,{\pi }^{\ast }2p_x^2\approx {\pi }^{\ast }2p_y^2$

Bond order $=\frac{1}{2}\left(N_b-N_a\right)=\frac{1}{2}(10-8)=1$

(b) $O_2$ molecule

$=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2$

$\pi 2p_x^2\approx \pi 2p_y^2,{\pi }^{\ast }2p_x^1\approx {\pi }^{\ast }2p_{y^{\prime }}$

Bond order $=\frac{1}{2}(10-6)=2$

(c) $B{e}_2$ molecule

$=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2$

Bond order $=\frac{1}{2}(4-4)=0$

(d) $L{i}_2$ molecule $=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2$

Bond order $=\frac{1}{2}(4-2)=1$

Therefore, $B{e}_2$ molecule has zero bond order.