Which of the following ions has the lowest ma | vedclass

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(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. For $Ti^

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$Ti^{3+}$

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(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. For $Ti^{3+}$ $(3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$2$. For $Co^{3+}$ $(3d^6)$: In the presence of weak field ligands,$n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.$3$. For $Cr^{3+}$ $(3d^3)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.$4$. For $Fe^{3+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.Comparing the values,$Ti^{3+}$ has the lowest number of unpaired electrons $(n=1)$,therefore it has the lowest magnetic moment.

List-$I$ (Coordination compounds)	List-$II$ (Number of unpaired electrons)
$A$. $[Cr(CN)_6]^{3-}$	$I$. $0$
$B$. $[Fe(H_2O)_6]^{2+}$	$II$. $3$
$C$. $[Co(NH_3)_6]^{3+}$	$III$. $2$
$D$. $[Ni(NH_3)_6]^{2+}$	$IV$. $4$

Which of the following ions has the lowest magnetic moment?

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