A) \[HCOOH\]
B) \[N{{H}_{2}}CON{{H}_{2}}\]
C) \[{{(C{{H}_{3}})}_{3}}COH\]
D) \[C{{H}_{3}}CHO\]
Correct Answer: C
Solution :
\[\underset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\,-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \\ \,\,\,\,\,\,s{{p}^{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} \,\,\,\,\,s{{p}^{3}} \\ C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-OH\] In the above compound all bonds are\[\sigma \]bond and hence, carbon atom uses only\[s{{p}^{3}}-\]hybrid orbitals for bond formation.
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