Question

Bond order of $O_2$ is:

• A. 2
• B. 1.5
• C. 3
• D. 3.5

Answer 1

Answer: (A)

Bond order is calculated by using following formula after writing electronic configuration of molecule according to molecular orbital theory.
$O_2=8+8=16=\sigma 1s^2,\stackrel{\ast }{\sigma }1s^2,\sigma 2s^2,\stackrel{\ast }{\sigma }2s^2$,
$\sigma 2p_z^2,\pi 2p_x^2,\pi 2p_y^2,\stackrel{\ast }{\pi }2p_x^1,\stackrel{\ast }{\pi }{p}_y^1$
Bond order $=\frac{N_b-N_a}{2}$
Where, $N_b=$ no. of electron in bonding molecular orbitals
$Na=$ no. of electron in antibonding molecular orbitals.
Bond order $=\frac{10-6}{2}=2$