Question

The molecular orbital configuration of a diatomic molecule is Its bond order is $\sigma$ 1 $s^2$ ${\sigma }^{+}$ $1s^2$ $\sigma$ $2s^2$ ${\sigma }^{+}$ $2s^2$ $\sigma$ $2p_x^2\{\begin{array}{c}\pi 2p_y^2\\ \pi 2p_z^2\end{array}$ Its bond order is

• A. 3
• B. 2.5
• C. 2
• D. 1

Answer 1

The correct option is A

3

Bond Order = $\frac{No.ofBondingelectrons-No.ofantibondingelectrons}{2}$