An ion with a definite magnetic moment (spin | vedclass

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(D) The magnetic moment (spin only) is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$Sc^{3+}$ $([Ar]

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Both $(b)$ and $(c)$

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(D) The magnetic moment (spin only) is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$Sc^{3+}$ $([Ar] 3d^0)$: $n = 0$,so $\mu = 0 \ BM$.$Ti^{3+}$ $([Ar] 3d^1)$: $n = 1$,so $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$Cu^{2+}$ $([Ar] 3d^9)$: $n = 1$,so $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.Since both $Ti^{3+}$ and $Cu^{2+}$ have $n=1$,they possess a definite non-zero magnetic moment.

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

An ion with a definite magnetic moment (spin only) is:

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