The pair of electron in the given carbanion, $$C{H_3}C \equiv {C^ - },$$   is present in which orbitals?

Solution :
$$\eqalign{ & {\text{Hybridisation}} \cr & = \frac{{{\text{Number of }}\sigma {\text{ - }}\,{\text{electrons}}}}{2} \cr & = \frac{{2 + 2{\text{(negative ion)}}}}{2} \cr & = 2 \cr & = sp \cr} $$
Hence, in the carbanion, $$C{H_3}C \equiv {C^\Theta },$$    pair of electron as $$\left( - \right)ve$$   charge is present in $$sp$$ - hybridised - orbital.