Question
The compound in which $$\mathop C\limits^* $$ uses its $$s{p^3}$$ hybrid orbitals for bond formation is :
A.
$$H\mathop C\limits^* \,OOH$$
B.
$${\left( {{H_2}N} \right)_2}\mathop C\limits^* O$$
C.
$${\left( {C{H_3}} \right)_3}\mathop C\limits^* OH$$
D.
$$C{H_3}\mathop C\limits^* HO$$
Answer :
$${\left( {C{H_3}} \right)_3}\mathop C\limits^* OH$$
Solution :
TIPS/Formulae:
$$4\sigma \,{\text{bonds}} - s{p^3}\,{\text{hybridisation}}$$
$$2\sigma \,{\text{and}}\,2\pi \,{\text{bonds}} - s{p^2}\,{\text{hybridisation}}\,$$
$$1\sigma \,{\text{and}}\,3\pi \,{\text{bonds}} - sp\,\,\,{\text{hybridisation}}$$
[For hybridization only $$\sigma {\text{ - bonds}}$$ are considered]
$$\left( {\text{a}} \right)3\sigma ,1\pi \,\,\left( {\text{b}} \right)3\sigma ,1\pi \,\,\left( {\text{c}} \right)4\sigma \,\,\left( {\text{d}} \right)3\sigma ,1\pi $$
$$\therefore {\left( {C{H_3}} \right)_3}COH$$ has $$4\sigma $$ bonds and thus it has $$s{p^3}$$ hybridisation.
TIPS/Formulae:
$$4\sigma \,{\text{bonds}} - s{p^3}\,{\text{hybridisation}}$$
$$2\sigma \,{\text{and}}\,2\pi \,{\text{bonds}} - s{p^2}\,{\text{hybridisation}}\,$$
$$1\sigma \,{\text{and}}\,3\pi \,{\text{bonds}} - sp\,\,\,{\text{hybridisation}}$$
[For hybridization only $$\sigma {\text{ - bonds}}$$ are considered]
$$\left( {\text{a}} \right)3\sigma ,1\pi \,\,\left( {\text{b}} \right)3\sigma ,1\pi \,\,\left( {\text{c}} \right)4\sigma \,\,\left( {\text{d}} \right)3\sigma ,1\pi $$
$$\therefore {\left( {C{H_3}} \right)_3}COH$$ has $$4\sigma $$ bonds and thus it has $$s{p^3}$$ hybridisation.