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BITSAT 2019 Slot 2 Solved Paper

Section: Chemistry

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Question : 71 of 150

Marks: +1, -0

Spin only magnetic moment of

\mathrm{Mn}^{4+}

ion is nearly

3 BM
6 BM
4 BM
5 BM

Solution:

Electronic configuration of

\mathrm{Mn}

is

(Z=25)=[\mathrm{Ar}] 3 d^{5} 4 s^{2}

\therefore

Electronic configuration of

\mathrm{Mn}^{4+}

is

\mathrm{Mn}^{4+}=[\mathrm{Ar}] 3 d^{3}

Thus,

\mathrm{Mn}^{4+}

has 3 -unpaired electron and its spin only magnetic moment

(\mu)

is,

\mu=\sqrt{n(n+2)}\,\mathrm{BM}

where,

n=

number of unpaired electrons.

\mu=\sqrt{3(3+2)}\,\cdot\,\mathrm{BM}

\Rightarrow\mu=\sqrt{15}=3.87\,\mathrm{BM}\approx 4\,\mathrm{BM}

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