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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
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The correct answer is:
$1.73$ and $5.92$
$\begin{array}{l}
{\left[\mathrm{Mn}^{+2}(\mathrm{CN})_{6}\right]^{-4}} \\
\mathrm{Mn}^{+2} \rightarrow 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0} 4 \mathrm{p}
\end{array}$
$\mathrm{CN}^{-}$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 $=5$ $\mu=\sqrt{5(5+2)}=\sqrt{35}=5.92 \mathrm{~B} . \mathrm{M}$
{\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 $=5$ $\mu=\sqrt{5(5+2)}=\sqrt{35}=5.92 \mathrm{~B} . \mathrm{M}$
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