Consider the hydrated ions of Ti^{2+},V^{2+}, | vedclass

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

Vedclass · Accepted Answer

$Sc^{3+} < Ti^{3+} < Ti^{2+} < V^{2+}$

Vedclass · Answer

(B) The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.$1. Sc^{3+} (Z=21): [Ar] 3d^0$. Number of unpaired electrons $(n) = 0$. $\mu = 0 \ B.M.$$2. Ti^{3+} (Z=22): [Ar] 3d^1$. Number of unpaired electrons $(n) = 1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \ B.M. \approx 1.73 \ B.M.$$3. Ti^{2+} (Z=22): [Ar] 3d^2$. Number of unpaired electrons $(n) = 2$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \ B.M. \approx 2.83 \ B.M.$$4. V^{2+} (Z=23): [Ar] 3d^3$. Number of unpaired electrons $(n) = 3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \ B.M. \approx 3.87 \ B.M.$Comparing the values: $0 Therefore,the correct order is $Sc^{3+} < Ti^{3+} < Ti^{2+} < V^{2+}$.

List-$I$ (Element)	List-$II$ (Block)
$A$. $Cd$	$I$. $f$-block
$B$. $Eu$	$II$. $s$-block
$C$. $Se$	$III$. $d$-block
$D$. $Ba$	$IV$. $p$-block

Consider the hydrated ions of $Ti^{2+}$,$V^{2+}$,$Ti^{3+}$,and $Sc^{3+}$. The correct order of their spin-only magnetic moments is

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