Answer:
(i) \[\overset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\,=\overset{sp}{\mathop{C}}\,=O\]
(ii) \[\overset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\,-\overset{s{{p}^{2}}}{\mathop{CH}}\,=\overset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\,\]
(iii) \[\overset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\,-\underset{s{{p}^{2}}}{\overset{\overset{\text{O}}{\mathop{\text{
}\!\!|\!\!\text{ }\!\!|\!\!\text{
}}}\,}{\mathop{C}}}\,-\overset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\,\]
(iv)\[\overset{s{{p}^{2}}}{\mathop{C{{H}_{2}}}}\,=\overset{s{{p}^{2}}}{\mathop{CH}}\,-\overset{sp}{\mathop{C}}\,\equiv
N\]
(v) Each
carbon in benzene \[({{C}_{6}}{{H}_{6}})\] is \[s{{p}^{2}}\] -hybridized.
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