A) \[10\sigma 3\pi \]
B) \[11\sigma 2\pi \]
C) \[9\sigma 3\pi \]
D) \[10\sigma 2\pi \]
Correct Answer: A
Solution :
\[H\overline{\sigma }C\underset{\pi }{\overset{\pi }{\mathop{\equiv }}}\,C\overset{\sigma }{\mathop{\_}}\,\underset{\begin{smallmatrix} |\sigma \\ C \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ \sigma | \end{smallmatrix}}{\mathop{C}}}\,\overset{\sigma }{\mathop{-}}\,\underset{{}}{\overset{\begin{smallmatrix} H \\ |\sigma \end{smallmatrix}}{\mathop{C}}}\,\underset{\pi }{\overset{\sigma }{\mathop{=}}}\,\underset{{}}{\overset{\begin{smallmatrix} H \\ \sigma | \end{smallmatrix}}{\mathop{C}}}\,\overline{\sigma }H\] Therefore, \[10\sigma \] and \[3\pi \] bonds are present
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