The spin only magnetic moment of [MnBr_{4}]^{ | vedclass

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(A) The coordination number of the $Mn^{2+}$ ion in the complex ion $[MnBr_{4}]^{2-}$ is $4$. This implies it can have either tetrahedral ($sp^{3}$ hy

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Tetrahedral

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(A) The coordination number of the $Mn^{2+}$ ion in the complex ion $[MnBr_{4}]^{2-}$ is $4$. This implies it can have either tetrahedral ($sp^{3}$ hybridization) or square planar ($dsp^{2}$ hybridization) geometry. The given spin-only magnetic moment is $5.9 \ BM$. Using the formula $\mu = \sqrt{n(n+2)}$,where $n$ is the number of unpaired electrons: $5.9 \approx \sqrt{n(n+2)} \implies n \approx 5$. Since $Mn^{2+}$ has a $d^{5}$ configuration,having $5$ unpaired electrons indicates a high-spin tetrahedral complex where the $d$-orbitals are not forced to pair up. Therefore,the geometry is tetrahedral.

The spin only magnetic moment of $[MnBr_{4}]^{2-}$ is $5.9 \ BM$. The geometry of this complex ion is

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