Chemistry - Carbonyl compounds Question with Solution | TestHub

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