The difference in the 'spin-only' magnetic mo | vedclass

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(B) In $KMnO_4$,the oxidation state of $Mn$ is $+7$. The electronic configuration is $[Ar] 3d^0$. The number of unpaired electrons $n = 0$. The spin-o

Vedclass · Accepted Answer

$6$

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(B) In $KMnO_4$,the oxidation state of $Mn$ is $+7$. The electronic configuration is $[Ar] 3d^0$. The number of unpaired electrons $n = 0$. The spin-only magnetic moment $\mu = \sqrt{n(n+2)} = 0 \ BM$.During the titration of $KMnO_4$ with oxalic acid in an acidic medium,$Mn^{7+}$ is reduced to $Mn^{2+}$.The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$. The number of unpaired electrons $n = 5$.The spin-only magnetic moment $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.The difference in magnetic moments is $|5.92 - 0| = 5.92 \ BM$.The nearest integer is $6$.

List-$I$	List-$II$
$A$. Oxidation state of $V$ in $VOCl_2$	$I$. $0$
$B$. Number of unpaired electrons in $MnO_4^{2-}$ ion	$II$. $1$
$C$. Number of unpaired electrons in $[NiCl_4]^{2-}$ ion	$III$. $5$
$D$. This oxidation state is exhibited by all lanthanide ions	$IV$. $3$
	$V$. $4$
	$VI$. $2$

Column-$I$	Column-$II$
$(1)$ General electronic configuration of $d$-block elements	$(A)$ Transition elements
$(2)$ $d$-block elements	$(B)$ $(n-1)d^{1-10}ns^{1-2}$
$(3)$ Groups $3$ to $12$	$(C)$ $Z = 58$ to $71$
$(4)$ Lanthanoid elements	$(D)$ Last electron enters $d$-orbital

The difference in the 'spin-only' magnetic moment values of $KMnO_4$ and the manganese product formed during titration of $KMnO_4$ against oxalic acid in acidic medium is .................... $BM$. (nearest integer)

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