The spin only magnetic moment of $\left[\mathrm{MnBr}_4\right]^{\mathrm{x}-}$ is $5.9 \mathrm{BM}$. The…
The spin only magnetic moment of $\left[\mathrm{MnBr}_4\right]^{\mathrm{x}-}$ is $5.9 \mathrm{BM}$. The geometry of the complex and $\mathrm{x}$ respectively are
tetrahedral, 1
square planar, 1
square planar, 2
tetrahedral, 2
Solution
$\mu_{\mathrm{s}}=\sqrt{\mathrm{n}(\mathrm{n}+2)}=5.9$ B. M.
$\Rightarrow \mu^2=\mathrm{n}(\mathrm{n}+2)=34.81 \Rightarrow \mathrm{n}=5$
Thus, $\mathrm{Mn}$ will have five unpaired electrons so $\mathrm{Mn}$ will be in +2 state with $\mathrm{Br}^{-}$being weak- field ligands.
Thus, $x=2$ for $\left[\mathrm{Mn} \mathrm{Br}_4\right]^{2-}$
