A) \[C{{H}_{2}}=C=C{{H}_{2}}\]
B) \[C{{H}_{3}}CH=CHCH_{2}^{+}\]
C) \[C{{H}_{3}}C\equiv CCH_{2}^{+}\]
D) \[C{{H}_{3}}-CH=CH-CH_{2}^{-}\]
E) \[C{{H}_{2}}=CHCH=C{{H}_{2}}\]
Correct Answer: C
Solution :
If there is four a bonds, hybridisation is\[s{{p}^{3}},\]if, three a bonds, hybridisation is\[s{{p}^{2}}\]and if two a bonds, hybridisation is sp. (a) \[\underset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\,=\underset{sp}{\mathop{C}}\,=\underset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\,\] (b) \[\underset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\,=\underset{s{{p}^{2}}}{\mathop{CH}}\,=\underset{s{{p}^{2}}}{\mathop{CH}}\,-\underset{s{{p}^{2}}}{\mathop{CH_{2}^{+}}}\,\] (c) \[\underset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\,-\underset{sp}{\mathop{C}}\,\equiv \underset{s{{p}^{2}}}{\mathop{CH}}\,-\underset{s{{p}^{3}}}{\mathop{CH_{2}^{-}}}\,\] (d) \[\underset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\,-\underset{s{{p}^{2}}}{\mathop{CH}}\,=\underset{s{{p}^{2}}}{\mathop{CH}}\,-\underset{s{{p}^{3}}}{\mathop{CH_{2}^{-}}}\,\] (e) \[\underset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\,=\underset{s{{p}^{2}}}{\mathop{CH}}\,-\underset{s{{p}^{2}}}{\mathop{CH}}\,=\underset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\,\] Hence, in\[C{{H}_{3}}-C\equiv C-CH_{2}^{+},\]all the three types of hybrid carbons are present.
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