Match List-I with List-II:List-I (text{Transi | vedclass

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

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$A-III, B-I, C-IV, D-II$

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(B) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$, where $n$ is the number of unpaired electrons.$A$. $Ti^{3+}$ $(3d^1)$: $n = 1$, $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ B.M.$ $(III)$$B$. $V^{2+}$ $(3d^3)$: $n = 3$, $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ B.M.$ $(I)$$C$. $Ni^{2+}$ $(3d^8)$: $n = 2$, $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84 \ B.M.$ $(IV)$$D$. $Sc^{3+}$ $(3d^0)$: $n = 0$, $\mu = \sqrt{0(0+2)} = 0.00 \ B.M.$ $(II)$Therefore, the correct matching is $A-III, B-I, C-IV, D-II$.

List-$I$ ($\text{Transition metal ion}$)	List-$II$ ($\text{Spin only magnetic moment (B.M.)}$)
$A$. $Ti^{3+}$	$I$. $3.87$
$B$. $V^{2+}$	$II$. $0.00$
$C$. $Ni^{2+}$	$III$. $1.73$
$D$. $Sc^{3+}$	$IV$. $2.84$

Match List-$I$ with List-$II$:
List-$I$ ($\text{Transition metal ion}$) List-$II$ ($\text{Spin only magnetic moment (B.M.)}$)
$A$. $Ti^{3+}$ $I$. $3.87$
$B$. $V^{2+}$ $II$. $0.00$
$C$. $Ni^{2+}$ $III$. $1.73$
$D$. $Sc^{3+}$ $IV$. $2.84$

Choose the correct answer from the options given below:

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Match List-$I$ with List-$II$:List-$I$ ($\text{Transition metal ion}$)List-$II$ ($\text{Spin only magnetic moment (B.M.)}$)$A$. $Ti^{3+}$$I$. $3.87$$B$. $V^{2+}$$II$. $0.00$$C$. $Ni^{2+}$$III$. $1.73$$D$. $Sc^{3+}$$IV$. $2.84$Choose the correct answer from the options given below: