A) \[C{{H}_{3}}-C\equiv C-C{{H}_{3}}\]
B) \[C{{H}_{2}}=CH-C{{H}_{2}}-C\equiv CH\]
C) \[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}\]
D) \[C{{H}_{3}}-CH=CH-C{{H}_{3}}\]
Correct Answer: A
Solution :
\[{{H}_{3}}C-C\equiv C-C{{H}_{3}}\] is linear, because \[{{C}_{2}}\] and \[{{C}_{3}}\] are sp hybridised.
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