A) 1
B) 2
C) 0.5
D) 1.5
Correct Answer: D
Solution :
According to the molecular orbital theory electronic configuration of \[C_{2}^{+}({{e}^{-}}=11)\] \[=\sigma {{(1s)}^{2}}{{\sigma }^{*}}(1{{s}^{2}})\sigma {{(2s)}^{2}}{{\sigma }^{*}}{{(2s)}^{2}}\] \[\pi {{(2{{p}_{y}})}^{2}}\pi {{(2{{p}_{z}})}^{1}}\] \[\therefore \] \[B.O.=\frac{{{n}_{C}}-{{n}_{A}}}{2}=\frac{7-4}{2}=1.5\]
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