Chemical Bonding And Molecular Structure Question 210
Question: Bond order of $ O_2 $ is
[DPMT 2004]
Options:
2
1.5
3
3.5
Show Answer
Answer:
Correct Answer: A
Solution:
Electronic configuration of $ O_2 $ is $ O_2={{(\sigma 1s)}^{2}}{{({{\sigma }^{}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{}}2s)}^{2}}{{({{\pi }^{}}2p_{x})}^{2}}{{({{\pi }^{}}2p_{y})}^{2}}{{(\sigma 2p_{z})}^{2}} $
$ (\pi 2p_x^{2}\equiv \pi 2p_y^{2})\ ({{\pi }^{}}2p_x^{1}\equiv {{\pi }^{}}2p_y^{1}) $
Hence bond order $ =\frac{1}{2}[ N_{b}-N_{a} ] $
$ =\frac{1}{2}[10-6]=2 $ .
BETA
