Match the metal ions given in Column-I with t | vedclass

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Question

(A) The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.$

Vedclass · Accepted Answer

$(iv) \quad (v) \quad (ii) \quad (i)$

Vedclass · Answer

(A) The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.$1. \ Co^{3+} (3d^6)$: $n = 4$,so $\mu = \sqrt{4(4+2)} = \sqrt{24} \ B.M.$ (Matches $iv$)$2. \ Cr^{3+} (3d^3)$: $n = 3$,so $\mu = \sqrt{3(3+2)} = \sqrt{15} \ B.M.$ (Matches $v$)$3. \ Fe^{3+} (3d^5)$: $n = 5$,so $\mu = \sqrt{5(5+2)} = \sqrt{35} \ B.M.$ (Matches $ii$)$4. \ Ni^{2+} (3d^8)$: $n = 2$,so $\mu = \sqrt{2(2+2)} = \sqrt{8} \ B.M.$ (Matches $i$)Therefore,the correct sequence is $(a)-(iv), (b)-(v), (c)-(ii), (d)-(i)$.

Column-$I$	Column-$II$
$(a) \ Co^{3+}$	$(i) \ \sqrt{8} \ B.M.$
$(b) \ Cr^{3+}$	$(ii) \ \sqrt{35} \ B.M.$
$(c) \ Fe^{3+}$	$(iii) \ \sqrt{3} \ B.M.$
$(d) \ Ni^{2+}$	$(iv) \ \sqrt{24} \ B.M.$
	$(v) \ \sqrt{15} \ B.M.$

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