MCQ
The compound in which carbon uses only its $s{p^3}$ hybrid orbitals for bond formation is
- A$HCOOH$
- B${(N{H_2})_2}CO$
- ✓${(C{H_3})_3}COH$
- D${(C{H_3})_3}CHO$
✓
Answer
Correct option: C.
${(C{H_3})_3}COH$
(c)$\mathop {C{H_3}}\limits^{s{p^3}} - \mathop {{C^{s{p^3}}} - }\limits_{\mathop {|\,\,{\kern 1pt} \,\,\,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} }\limits_{\mathop {C{H_{3\,\,\,\,\,}}}\limits_{s{p^3}\,\,} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} } }^{\mathop {|{\kern 1pt} {\kern 1pt} \,\,{\kern 1pt} {\kern 1pt} \,\,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} }\limits^{\mathop {C{H_{3\,\,\,\,\,}}}\limits^{s{p^3}\,\,} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} } } OH$
All the carbon atoms are $s{p^3}$ hybridized.
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