• # question_answer The compound in which carbon uses only its $s{{p}^{3}}$ hybrid orbitals for bond formation is [IIT-JEE 1989] A) $HCOOH$ B) ${{(N{{H}_{2}})}_{2}}CO$ C) ${{(C{{H}_{3}})}_{3}}COH$ D) ${{(C{{H}_{3}})}_{3}}CHO$

$\overset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\,-\underset{\underset{\underset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\,\,\,\,\,\,}{\mathop{|\,\,\,\,\,\,\,\,\,\,\,\,\,}}\,}{\overset{\overset{\overset{s{{p}^{3}}}{\mathop{C{{H}_{3}}}}\,\,\,\,\,\,\,}{\mathop{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\,}{\mathop{{{C}^{s{{p}^{3}}}}-}}}\,OH$ All the carbon atoms are $s{{p}^{3}}$ hybridized.