The correct order of the spin –only magnetic moment of metal ions in the following low-spin complexes,…

1. ${\mathrm{ }\mathrm{V}}^2+>\mathrm{C}{\mathrm{r}}^{2+}>\mathrm{R}{\mathrm{u}}^{3+}>\mathrm{F}{\mathrm{e}}^{2+}$
2. $\mathrm{C}{\mathrm{r}}^{2+}>{\mathrm{V}}^2+>\mathrm{R}{\mathrm{u}}^{3+}>\mathrm{F}{\mathrm{e}}^{2+}$
3. $\mathrm{C}{\mathrm{r}}^{2+}>\mathrm{R}{\mathrm{u}}^{3+}>\mathrm{F}{\mathrm{e}}^{2+}>{\mathrm{V}}^{2+}$
4. ${\mathrm{V}}^{2+}>\mathrm{R}{\mathrm{u}}^{3+}>\mathrm{C}{\mathrm{r}}^{2+}>\mathrm{F}{\mathrm{e}}^{2+}$

Solution

(i) $\mathrm{ }{\mathrm{V}}^{+2}\left(23\right)⟹\left[\mathrm{A}\mathrm{r}\right]3{\mathrm{d}}^{3}4{\mathrm{d}}^0⟹\mathrm{ }$ $\left(\mathrm{n}=3\right)$ $\Rightarrow \mathrm{\mu }\sqrt{3\left(3+2\right)}=3.87$
(ii) $\mathrm{C}{\mathrm{r}}^{+2}\left(24\right)⟹\left[\mathrm{A}\mathrm{r}\right]3{\mathrm{d}}^4$ $\left(\mathrm{n}=2\right)$ $\Rightarrow \mathrm{\mu }\sqrt{2\left(2+2\right)}=2.82$
(iii) $\mathrm{ }\mathrm{ }\mathrm{R}{\mathrm{u}}^{+3}\left(44\right)⟹\left[\mathrm{k}\mathrm{r}\right]4{\mathrm{d}}^5$   $\mathrm{ }(\mathrm{n}=1)$ $\Rightarrow \mathrm{\mu }\sqrt{1\left(1+2\right)}=3$
(iv) $\mathrm{ }\mathrm{ }\mathrm{F}{\mathrm{e}}^{+2}\left(26\right)\mathrm{ }⟹\left[\mathrm{A}\mathrm{r}\right]3{\mathrm{d}}^6\mathrm{ }$ $(\mathrm{n}=0)$ $\Rightarrow \mathrm{\mu }=0$
Order of magnetic moment will be $\propto$ no. of unpaired ${\mathrm{e}}^{-}$
${\mathrm{V}}^{+2}>\mathrm{ }\mathrm{C}{\mathrm{r}}^{+2}>\mathrm{R}{\mathrm{u}}^{+3}\mathrm{ }>\mathrm{F}{\mathrm{e}}^{+2}$
$\mathrm{\mu }=\sqrt{\mathrm{n}(\mathrm{n}+2)}$ B.M. Hence more the value of $\mathrm{n}$ more will be magnitude moment.

Asked in: JEE Main 2019 (08 Apr Shift 1)

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