The spin-only magnetic moments of $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ and $\left[\mathrm{MnBr}_{4}\right]^{2-}$ in Bohr Magnetons, respectively, are

• A

  $5.92$ and $5.92$
• B

  $4.89$ and $1.73$
• Correct answer
  C

  $1.73$ and $5.92$
• D

  $1.73$ and $1.73$

Solution

$\begin{array}{l} {\left[\mathrm{Mn}^{+2}(\mathrm{CN})_{6}\right]^{-4}} \\ \mathrm{Mn}^{+2} \rightarrow 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0} 4 \mathrm{p} \end{array}$ $\mathrm{CN}^{-}$is strong ligands so creates back paring effect of $(\mathrm{n}-1) \mathrm{d}$ orbitals configuration
So, unpaired $\mathrm{e}^{-}=1$ $\begin{array}{l} \mu=\sqrt{\mathrm{n}(\mathrm{n}+2)} \mathrm{B}, \mathrm{M} \\ \mu=1.73 \mathrm{~B} . \mathrm{M} \end{array}$ And in $\left[\mathrm{MnBr}_{4}\right]^{-2}$ Br is a weak ligands so no back pairing effect on $(n-1)$ d orbital so, unpaired $\mathrm{e}^{-}$is $=5$ $\mu=\sqrt{5(5+2)}=\sqrt{35}=5.92 \mathrm{~B} . \mathrm{M}$

Hence, the correct answer is option: 'C'.