Chemistry - Carbonyl compounds Question with Solution | TestHub

In the formation of${\mathrm{d}}^2{\mathrm{sp}}^3$hybrid orbitals, two$\left(\mathrm{n}-1\right)\mathrm{d}$orbitals of${\mathrm{e}}_{\mathrm{g}}$set i.e.,$\left(\mathrm{n}-1\right)\mathrm{ }{\mathrm{d}}_{{\mathrm{z}}^2}\mathrm{and}\mathrm{ }\left(\mathrm{n}-1\right){\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$orbitals, one$\mathrm{ns}\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{three}\mathrm{ }\mathrm{np}\left({\mathrm{np}}_{\mathrm{x}},\mathrm{ }{\mathrm{np}}_{\mathrm{y}}\mathrm{ }\mathrm{and}\mathrm{ }{\mathrm{np}}_{\mathrm{z}}\right)$orbitals combine together and form six${\mathrm{d}}^2{\mathrm{sp}}^3$hybrid orbitals.