A) 2
B) 1.5
C) 3
D) 3.5
Correct Answer: A
Solution :
Key Idea: Bond order is calculated by using following formula after writing electronic configuration of molecule according to molecular orbital theory. \[{{O}_{2}}=8+8=16=\sigma 1{{s}^{2}},\overset{*}{\mathop{\sigma }}\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\mathop{\sigma }}\,2{{s}^{2}}\],\[\sigma 2p_{z}^{2},\pi 2p_{x}^{2},\pi 2p_{y}^{2},\overset{*}{\mathop{\pi }}\,2p_{x}^{1},\overset{*}{\mathop{\pi }}\,p_{y}^{1}\] Bond order \[=\frac{{{N}_{b}}-{{N}_{a}}}{2}\] Where, \[{{N}_{b}}=\] no. of electron in bonding molecular orbitals Na = no. of electron in antibonding molecular orbitals. Bond order \[=\frac{10-6}{2}=2\]
You need to login to perform this action.
You will be redirected in
3 sec
