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(D) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 3.86 \

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$23$

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(D) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 3.86 \ BM$,we have $\sqrt{n(n+2)} = 3.86$.Squaring both sides,$n(n+2) \approx 15$,which gives $n = 3$.For a metal ion in the $+2$ oxidation state to have $3$ unpaired electrons,the electronic configuration must be $d^3$.Checking the options:${}_{22} Ti^{+2} = [Ar] 3d^2$ $(n=2)$${}_{23} V^{+2} = [Ar] 3d^3$ $(n=3)$${}_{25} Mn^{+2} = [Ar] 3d^5$ $(n=5)$${}_{26} Fe^{+2} = [Ar] 3d^6$ $(n=4)$Thus,the metal is Vanadium $(V)$ with atomic number $23$.

List-$I$ Species	List-$II$ Electronic distribution
$A$. $Cr^{2+}$	$I$. $3d^8$
$B$. $Mn^{+}$	$II$. $3d^3 4s^1$
$C$. $Ni^{2+}$	$III$. $3d^4$
$D$. $V^{+}$	$IV$. $3d^5 4s^1$

$A$ first row transition metal in its $+2$ oxidation state has a spin-only magnetic moment value of $3.86 \ BM$. The atomic number of the metal is

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Amongst the following,the lowest degree of paramagnetism per mole of the compound at $298 \ K$ will be shown by:

Match List-$I$ with List-$II$

List-$I$ Species List-$II$ Electronic distribution

$A$. $Cr^{2+}$ $I$. $3d^8$

$B$. $Mn^{+}$ $II$. $3d^3 4s^1$

$C$. $Ni^{2+}$ $III$. $3d^4$

$D$. $V^{+}$ $IV$. $3d^5 4s^1$

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