Central atom of which of the following is associated with ${\mathrm{sp}}^3$ hybridization?

$$\begin{gathered} \left(\mathrm{A}\right) {\mathrm{CCl}}_4 \\ \left(\mathrm{B}\right) {\mathrm{ClO}}_2 \\ \left(\mathrm{C}\right) {\mathrm{NCl}}_3 \end{gathered}$$

CCl₄ has 4 + 4(7) = 32 valence electrons. CCl₄ has a tetrahedral arrangement of the electron pairs about the carbon atom that requires $S{P}^3$ hybridization. The four $S{P}^3$ hybrid orbitals from carbon are used to form the four bonds to chlorine.

In ClO₂ , Cl shows $S{P}^3$ hybridization & the molecule contains an odd electron. Large O=Cl=O. bond angle (111°). VSEPR theory predicts that the bond angle should be less than 109° due to lone-pair repulsion.

Hence, the correct option is option (A) A and B.