Spin only magnetic moment of $\left[\mathrm{MnBr}_{6}\right]^{4-}$ is ____ B.M. (round off to the closest integer)

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Coordination Compounds

Spin only magnetic moment of

[MnBr_{6}]^{4 -}

is ____
B.M. (round off to the closest integer)

Answer: 6

Solution

[MnBr_{6}]^{4 -}

, manganese is in the +2 oxidation state.

Mn^{2 +}

has electronic configuration

3 d^{5}

with 5 unpaired electrons in a high-spin octahedral complex (Br⁻ is a weak field ligand). The spin-only magnetic moment is given by

μ_{s} = n (n + 2)

where

n

is the number of unpaired electrons. Thus

μ_{s} = 5 (5 + 2) = 35 = 5.91 \approx 6

B.M.

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Integer answer

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