Question

The hybridization of $Cr$ in the complex ${\left[Cr{\left({NO}_2\right)}_4{\left({NH}_3\right)}_2\right]}^{-}$ is:

• A. ${sp}^3{d}^2$
• B. ${sp}^{3}d$
• C. $d^2{sp}^3$
• D. ${sp}^3$

Answer 1

The correct option is C $d^2{sp}^3$

${\left[Cr{\left({NO}_2\right)}_4{\left({NH}_3\right)}_2\right]}^{-}$

$x+4\times (-1)+2\times 0=-1$

$x=-1+4=+3$

Oxidation state of $Cr=+3.$

See the given figure.

Hybridization $=d^{2}s{p}^3$

No. of unpaired $e^{-}=3$

Magnetic behavior $=$ Paramagnetic.

Option $C$ is the correct answer.