Question

The correct order of the spin only magnetic moment of metal ions in the following low spin complexes, $\left[V\right(CN{)}_6{]}^{4-}$ , $\left[Fe\right(CN{)}_6{]}^{4-}$, $\left[Ru\right(N{H}_3{)}_6{]}^{3+}$ and $\left[Cr\right(N{H}_3{)}_6{]}^{2+}$ is:

• A. $C{r}^{2+}>R{u}^{3+}>F{e}^{2+}>V^{2+}$
• B. $V^{2+}>C{r}^{2+}>R{u}^{3+}>F{e}^{2+}$
• C. $C{r}^{2+}>V^{2+}>R{u}^{3+}>F{e}^{2+}$
• D. $V^{2+}>R{u}^{3+}>C{r}^{2+}>F{e}^{2+}$

Answer 1

The correct option is B $V^{2+}>C{r}^{2+}>R{u}^{3+}>F{e}^{2+}$

We know, spin only magnetic moment is directly proportional to the number of unpaired electrons present.
Both $C{N}^{-}andN{H}_3$ are strong field ligands, so a low spin complex will be obtained.
The electronic configuration of $V^{2+},F{e}^{2+},R{u}^{3+}$ and $C{r}^{2+}$ are $t_{2g}^3{e}_g^0,t_{2g}^6{e}_g^0,t_{2g}^5{e}_g^{0}and{t}_{2g}^4{e}_g^0$ respectively. So number of unpaired electrons present in $V^{2+},F{e}^{2+},R{u}^{3+}andC{r}^{2+}$ are 3, 0, 1 and 2 respectively. So the correct order of the spin only magnetic moment of metal ions is
$V^{2+}>C{r}^{2+}>R{u}^{3+}>F{e}^{2+}$