The hybridization of the orbitals of Cr atom in $\mathrm{Cr}(\mathrm{CO})_{6}$ is\

Question

Professor PC Thomas Classes · Answer

$\mathrm{d}^{2} \mathrm{sp}^{3}$ Cr $4 s^{1} 3 d^{5}$. The lone electron of $4 s$ and the five electrons of 3d get paired in there 3d orbitals. Then 4 s orbials hybridized with two vacant 3 d orbitals and three 4 p orbitals\