A) \[C{{H}_{3}}-C\equiv C-H\]
B) \[C{{H}_{3}}-CH=CH-C{{H}_{3}}\]
C) \[C{{H}_{2}}=CH-CH=C{{H}_{2}}\]
D) \[HC\equiv CH\]
Correct Answer: A
Solution :
(a) \[\underset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\,-\underset{sp}{\mathop{C}}\,\equiv \underset{sp}{\mathop{C}}\,-H\] (b) \[\underset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\,-\underset{s{{p}^{2}}}{\mathop{CH}}\,=\underset{s{{p}^{2}}}{\mathop{CH}}\,-\underset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\,\] (c) \[\underset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\,=\underset{s{{p}^{2}}}{\mathop{CH}}\,-\underset{s{{p}^{2}}}{\mathop{CH}}\,=\underset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\,\] (d) \[\underset{sp}{\mathop{HC}}\,\equiv \underset{sp}{\mathop{CH}}\,\] Hence,\[C{{H}_{3}}-C\equiv C-H\]has a bond formed by overlap of \[sp-s{{p}^{3}}\] hybrid orbitals.
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