Question

The spin-only magnetic moments of $[Mg(CN{)}_6{]}^{2-}$ and $[MnB{r}_4{]}^{2-}$ in Bohr magnetons, respectively are

• A. 5.92 and 5.92
• B. 4.89 and 1.73
• C. 1.73 and 5.92
• D. 1.73 and 1.73

Answer 1

Answer: (C)

The oxidation state for $Mn$ in $Mn[(CN{)}_6{]}^{2-}$ is $+4$. The electronic
configuration for $M{n}^{4+}$ is $[Ar]3d^3$.

As $C{N}^{-}$ is a strong field ligand, pairing of electrons occur

In case of $[Mn(CN{)}_6{]}^{2-}$
$n=1$
$\therefore \mu =\sqrt{1\left(1+2\right)}=\sqrt{3}=1.73BM$
The oxidation state of $Mn$ in $[MnB{r}_4{]}^{2-}$ is $+2$ .
The electronic configuration for $M{n}^{2+}$ is $[Ar]3d^{5}4s^0$.

As $B{r}^{-}$ is a weak field ligand, therefore pairing of electrons will not occur.
In this case,
$n=5$
$\therefore \mu =\sqrt{5\left(5+2\right)}$
$=\sqrt{35}=5.92BM$