Question

The bond order of $CO\text{and}NO\text{is}$:

• A. 3 and 2
• B. 3 and 2.5
• C. 3 and 1.3
• D. 3 and 3.5

Answer 1

The correct option is B 3 and 2.5

Molecular Orbital of
$CO=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2P_x^2=\pi 2p_y^2\sigma 2P_z^2$
$\therefore Totalno.ofelectrons=14$
$\text{Bond order}=\frac{\text{No. of bonding electrons(10)-No. of electrons in ABMO (4)}}{2}=3$
$NO=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2P_z^2\pi 2P_x^2=\pi 2p_y^2{\pi }^{\ast }2P_x^1$
$\therefore Totalno.ofelectrons=15$
$\text{Bond order}=\frac{10-5}{2}=2.5$