Chemistry-Molecular Orbital Theory

NEET Chemistry Molecular Orbital Theory MCQ Question

Type: MCQ-numerical-Hard-Class 11

For a diatomic molecule with a molecular orbital configuration of σ1s2σ1s2σ2s2σ2s2π2px2π2py2\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 Bond Order=NbNa2\text{Bond Order} = \frac{N_b - N_a}{2}, where NbN_b is the number of electrons in bonding orbitals and NaN_a is the number in antibonding orbitals. Here, Nb=6N_b = 6 and Na=4N_a = 4, so the bond order is 1.0.

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