Identify the order in which the spin-only mag | vedclass

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

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$III, II, IV, I$

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(D) The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For the given ions:$(I) Fe^{2+} ([Ar] 3d^6)$: Number of unpaired electrons $(n)$ = $4$.$(II) Ti^{2+} ([Ar] 3d^2)$: Number of unpaired electrons $(n)$ = $2$.$(III) Cu^{2+} ([Ar] 3d^9)$: Number of unpaired electrons $(n)$ = $1$.$(IV) V^{2+} ([Ar] 3d^3)$: Number of unpaired electrons $(n)$ = $3$.Comparing the number of unpaired electrons: $1 (III) Therefore,the increasing order of magnetic moment is $III < II < IV < I$.

Identify the order in which the spin-only magnetic moment (in $BM$) increases for the following four ions:
$(I) Fe^{2+}$
$(II) Ti^{2+}$
$(III) Cu^{2+}$
$(IV) V^{2+}$

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Identify the order in which the spin-only magnetic moment (in $BM$) increases for the following four ions:$(I) Fe^{2+}$$(II) Ti^{2+}$$(III) Cu^{2+}$$(IV) V^{2+}$