Chemical Bonding And Molecular Structure Question 333
Question: The molecular orbital configuration of a diatomic molecule is $ \sigma 1s^{2}\sigma^{*}(1s)^2 \sigma 2s^{2},$ $\sigma^{*},2s^{2}\sigma ,2p_{x}^{2},$ $ \pi,{2p_z}^2,{\pi,{2p_y}^2} . $ Its bond order is
Options:
A) 3
B) 2.5
C) 2
D) 1
Show Answer
Answer:
Correct Answer: A
Solution:
B.O. $ =\frac{N_{b}-N_{a}}{2}=\frac{10-4}{2}=3 $ .
