A) \[{{H}_{2}}C=CH-C\equiv CH\]
B) \[HC\equiv C-C\equiv CH\]
C) \[{{H}_{2}}C=C=C=C{{H}_{2}}\]
D)
Correct Answer: A
Solution :
\[s{{p}^{2}}\] hybridisation represents that C is attached with double bond and sp represents that C is attached with triple bond. [a] \[\underset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\,=\underset{s{{p}^{2}}}{\mathop{CH}}\,-\underset{sp}{\mathop{C}}\,\equiv \underset{sp}{\mathop{CH}}\,\] [b] \[\underset{sp}{\mathop{HC}}\,\equiv \underset{sp}{\mathop{C}}\,-\underset{sp}{\mathop{C}}\,\equiv \underset{sp}{\mathop{CH}}\,\] [c] \[\underset{sp}{\mathop{{{H}_{2}}C}}\,=\underset{sp}{\mathop{C}}\,=\underset{sp}{\mathop{C}}\,=\underset{sp}{\mathop{C{{H}_{2}}}}\,\] [d] \[\underset{s{{p}^{2}}}{\mathop{{{H}_{2}}C}}\,=\underset{s{{p}^{2}}}{\mathop{CH}}\,=\underset{s{{p}^{2}}}{\mathop{CH}}\,=\underset{s{{p}^{2}}}{\mathop{C\overset{\bigcirc -}{\mathop{H}}\,\overset{\oplus }{\mathop{N}}\,a}}\,\]You need to login to perform this action.
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