A) sp and \[s{{p}^{2}}\] hybridised C-atoms
B) sp, \[s{{p}^{2}}\] and \[s{{p}^{3}}\] hybridized C-atoms
C) Only \[s{{p}^{2}}\] hybridised C-atoms
D) Only sp hybridised C-atoms
Correct Answer: C
Solution :
\[\underset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\,=\underset{s{{p}^{2}}}{\mathop{CH}}\,-\underset{s{{p}^{2}}}{\mathop{CH}}\,=\underset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\,\]
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