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Question

(A) The magnetic moment of the metal ions is determined by the number of unpaired electrons $(n)$ using the formula $\mu_{spin} = \sqrt{n(n+2)} \ BM$.

Vedclass · Accepted Answer

$Mn^{2+}$

Vedclass · Answer

(A) The magnetic moment of the metal ions is determined by the number of unpaired electrons $(n)$ using the formula $\mu_{spin} = \sqrt{n(n+2)} \ BM$.$1$. For $Mn^{2+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \ BM \approx 5.92 \ BM$.$2$. For $Cu^{2+}$ $(3d^9)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \ BM \approx 1.73 \ BM$.$3$. For $Sc^{2+}$ $(3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \ BM \approx 1.73 \ BM$.$4$. For $Cu^{+}$ $(3d^{10})$: $n = 0$,$\mu = 0 \ BM$.Since $Mn^{2+}$ has the highest number of unpaired electrons $(n=5)$,it exhibits the maximum paramagnetic behaviour.

Atom/Ion	Electronic Configuration
$(1) \ Cu$	$(A) 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}$
$(2) \ Cu^{2+}$	$(B) 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2$
$(3) \ Zn^{2+}$	$(C) 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^1$
$(4) \ Cr^{3+}$	$(D) 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^9$
	$(E) 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^3$

The metal ion having maximum paramagnetic behaviour among the first transition series elements is,

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