Question

The hybridization and magnetic nature of $\left[Mn\right(CN{)}_6{]}^{4-}$ and $\left[Fe\right(CN{)}_6{]}^{3-},$ respectively are:

• A. $d^{2}s{p}^3$ and paramagnetic
• B. $s{p}^3{d}^2$ and paramagnetic
• C. $d^{2}s{p}^3$ and diamagnetic
• D. $s{p}^3{d}^2$ and diamagnetic

Answer 1

The correct option is A $d^{2}s{p}^3$ and paramagnetic

$\left[Mn\right(CN{)}_6{]}^{4-}\to M{n}^{2+}=3d^5\left(t_{2g}^5{e}_g^0\right)\therefore \text{Number of unpaired electrons}\left(n\right)=1\mu =\sqrt{n(n+2)}\therefore \mu =\sqrt{3}\text{Hybridization}=d^{2}s{p}^3$

$\left[Fe\right(CN{)}_6{]}^{3-}\to F{e}^{3+}=3d^5\left(t_{2g}^5{e}_g^0\right)\therefore \text{Number of unpaired electrons}\left(n\right)=1\mu =\sqrt{n(n+2)}\therefore \mu =\sqrt{3}\text{Hybridization}=d^{2}s{p}^3$