Question

The hybridization of $Cr$ in $\left[Cr\right(H_{2}O{)}_6{]}^{3+}$ ion and $\left[Cr\right(CN{)}_6{]}^{3-}$ ion are respectively :

• A. $d^{2}s{p}^3,s{p}^3{d}^3$
• B. $s{p}^3{d}^2,d^{2}s{p}^3$
• C. $d^{2}s{p}^3,d^{2}s{p}^3$
• D. $s{p}^3{d}^2,s{p}^3{d}^2$

Answer 1

Answer: (C)

$d^{2}s{p}^3,d^{2}s{p}^3$

$\left(d^{2}s{p}^{3}hybridization\right)$
Six empty orbitals (two d, one s and three p) are available, so whether the ligand is weak $\left(H_{2}O\right)$ or strong $\left(C{N}^{-}\right)$ the six orbitals hybridize to give six equivalent $d^{2}s{p}^3$ hybrid orbitals. Hence, in both $\left[Cr\right(H_{2}O{)}_6{]}^{3+}$ and $\left[Cr\right(CN{)}_6{]}^{3-}$ ions chromium is in a state of $d^{2}s{p}^3$ hybridization.