For the tetrahedral complex [MnBr_4]^{2-}, th | vedclass

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(B) The central metal ion is $Mn^{2+}$.The atomic number of $Mn$ is $25$, so the electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$.In the tetrahedr

Vedclass · Accepted Answer

$5.9$

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(B) The central metal ion is $Mn^{2+}$.The atomic number of $Mn$ is $25$, so the electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$.In the tetrahedral complex $[MnBr_4]^{2-}$, $Br^-$ is a weak field ligand, so the electrons remain unpaired.Number of unpaired electrons $(n)$ = $5$.The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.

For the tetrahedral complex $[MnBr_4]^{2-}$, the spin-only magnetic moment value is $....$ (At. no. of $Mn = 25$) (in $BM$)

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