Chemistry - Coordination Chemistry Question with Solution | TestHub

Negative value of electrode potential for $\raisebox{1ex}{$M^{3+}$}\!\left/ \!\raisebox{-1ex}{$M^{2+}$}\right.$ is for $\raisebox{1ex}{${\mathrm{Cr}}^{+3}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{Cr}}^{+2}$}\right.$

${\mathrm{E}}_{\raisebox{1ex}{${\mathrm{Cr}}^{+3}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{Cr}}^{+2}$}\right.}^0=-0.41\mathrm{V}$

${\mathrm{Cr}}^{+2}$ acts as a reducing agent readily

Electronic configuration of ${\mathrm{Cr}}^{+3} \mathrm{ion}$ is

$\left[\mathrm{Ar}\right]4{\mathrm{s}}^{0}3{\mathrm{d}}^3$

Number of unpaired electrons,$\mathrm{n}=$3

Magnetic moment,

$$\begin{gathered} \mathrm{\mu }=\sqrt{\mathrm{n}\left(\mathrm{n}+2\right)}\mathrm{B}.\mathrm{M} \\ =\sqrt{3\left(3+2\right)}\mathrm{B}.\mathrm{M} \end{gathered}$$

$=\sqrt{15} \mathrm{B}.\mathrm{M}\simeq 4 \mathrm{B}.\mathrm{M}$