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D and F Block Elements question

2022 · 29 Jun · Shift 2 · Q19

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D and F Block Elements
2022 · 29 Jun · Shift 2 · Q19

D and F Block Elements question

2022 · 29 Jun · Shift 2 · Q19

JEE MainChemistryD and F Block ElementsNumerical+4 / −1

Spin only magnetic moment of

[MnBr_6]^{4-}

is

\underline{\hspace{2cm}}

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

Numerical answer

View written solutionFree

Correct answer: 6

Find the oxidation state of Mn in $[MnBr_6]^{4-}$

Let oxidation state of Mn be $x$ .

Each $Br^-$ ligand has charge $-1$ , so:

$x + 6(-1) = -4$ $x - 6 = -4$ $x = +2$

So, the metal ion is $Mn^{2+}$ .

Write the electronic configuration of $Mn^{2+}$

Atomic number of Mn = 25.

Neutral Mn:

$[Ar] \, 3d^5 4s^2$

For $Mn^{2+}$ , remove 2 electrons from $4s$ first:

$Mn^{2+} = [Ar] \, 3d^5$

Determine the number of unpaired electrons

$Br^-$ is a weak field ligand, so the complex is high spin.

For $3d^5$ high-spin configuration, the number of unpaired electrons is:

$n = 5$

Calculate spin-only magnetic moment

Formula:

$\mu = \sqrt{n(n+2)} \text{ B.M.}$

Substitute $n=5$ :

$\mu = \sqrt{5(5+2)} = \sqrt{35} \text{ B.M.}$

$\mu \approx 5.92 \text{ B.M.}$

Round off to the closest integer

$5.92 \approx 6$

Final Answer

Spin-only magnetic moment of $[MnBr_6]^{4-}$ is:

$\boxed{6 \text{ B.M.}}$

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