A) \[s{{p}^{2}}\]
B) \[s{{p}^{3}}\]
C) \[s{{p}^{3}}d\]
D) sp
Correct Answer: A
Solution :
Keylder In \[-C-C-\,\,s{{p}^{3}}\] hybridisation \[-C=\] \[-\,\,s{{p}^{2}}\] hybridisation \[=C=C=sp\] hybridisation \[-C\equiv C-sp\] hybridisation \[\underset{s{{p}^{2}}}{\mathop{\overset{1}{\mathop{C}}\,{{H}_{2}}}}\,=\overset{2}{\mathop{\underset{sp}{\mathop{C}}\,}}\,=\underset{s{{p}^{2}}}{\mathop{\overset{3}{\mathop{C}}\,H}}\,-\underset{s{{p}^{3}}}{\mathop{\overset{4}{\mathop{C}}\,{{H}_{3}}}}\,\] In this molecule the hybrid state of carbon atoms. \[{{C}_{1}}\] and \[{{C}_{3}}\] is \[s{{p}^{2}},{{C}_{2}}\] is sp and \[{{C}_{4}}\] is \[s{{p}^{3}}\].
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