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An ion ${Mn}^{a +}$ bas the magnetic moment equal to 4.9 BM. The value of a is : (atomic no. of Mn = 25)

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detailed solution

Correct option is A

The fact is,

$Magnetic moment = \sqrt{n (n + 1)}$ ;

n is the number of unpaired electrons in this instance.

$4.9 = \sqrt{n (n + 2)}$

This ion possesses 4 unpaired electrons, or $n = 4$ .

electronically configured for $Mn = 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{5} 4 s^{2}$ have 5 unpaired electrons, hence in order to produce an ion with 4 unpaired electrons, it must let go of 3 electrons.

$\begin{array}{l} _{25} {Mn}^{3 +} : 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{4} \\ ∴ a = 3 \end{array}$

Therefore, choice A is the best one.

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An ion Mna+ bas the magnetic moment equal to 4.9 BM. The value of a is : (atomic no. of Mn = 25)