The magnetic moments for Ti^{2+},V^{3+},and N | vedclass

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(B) 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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$2.83 \, BM, 2.83 \, BM, 2.83 \, BM$

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(B) 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 $Ti^{2+}$ $([Ar] 3d^2)$: $n = 2$,so $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \, BM$.$2$. For $V^{3+}$ $([Ar] 3d^2)$: $n = 2$,so $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \, BM$.$3$. For $Ni^{2+}$ $([Ar] 3d^8)$: $n = 2$,so $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \, BM$.Thus,the magnetic moments for $Ti^{2+}$,$V^{3+}$,and $Ni^{2+}$ are $2.83 \, BM, 2.83 \, BM, 2.83 \, BM$ respectively.

Aspects	Metal
$a$. The metal which reveals a maximum number of oxidation states	$i$. Scandium
$b$. The metal although placed in $3d$ block is considered not as a transition element	$ii$. Copper
$c$. The metal which does not exhibit variable oxidation states	$iii$. Manganese
$d$. The metal which in $+1$ oxidation state in aqueous solution undergoes disproportionation	$iv$. Zinc

The magnetic moments for $Ti^{2+}$,$V^{3+}$,and $Ni^{2+}$ are respectively:

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