NEET Chemistry Molecular Orbital Theory MCQ Question with Solution

For a diatomic molecule with a molecular orbital configuration of $\sigma_{1s}^2 \sigma_{1s}^*2 \sigma_{2s}^2 \sigma_{2s}^*2 \pi_{2p_x}^2 \pi_{2p_y}^2$ , what is the bond order?

A

0.5

B

1.0

C

1.5

D

2.0

Correct Answer

Option B

Detailed Explanation

Bond order is calculated as $\text{Bond Order} = \frac{N_b - N_a}{2}$ , where $N_b$ is the number of electrons in bonding orbitals and $N_a$ is the number in antibonding orbitals. Here, $N_b = 6$ and $N_a = 4$ , so the bond order is 1.0.

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