Match List-I with List-II List-I Species | vedclass

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(B) The electronic configurations of the given species are as follows:$1$. $Cr$ $(Z=24)$: $[Ar] 3d^5 4s^1$. Therefore,$Cr^{2+}$ is $[Ar] 3d^4$ (Matche

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

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(B) The electronic configurations of the given species are as follows:$1$. $Cr$ $(Z=24)$: $[Ar] 3d^5 4s^1$. Therefore,$Cr^{2+}$ is $[Ar] 3d^4$ (Matches $III$).$2$. $Mn$ $(Z=25)$: $[Ar] 3d^5 4s^2$. Therefore,$Mn^{+}$ is $[Ar] 3d^5 4s^1$ (Matches $IV$).$3$. $Ni$ $(Z=28)$: $[Ar] 3d^8 4s^2$. Therefore,$Ni^{2+}$ is $[Ar] 3d^8$ (Matches $I$).$4$. $V$ $(Z=23)$: $[Ar] 3d^3 4s^2$. Therefore,$V^{+}$ is $[Ar] 3d^3 4s^1$ (Matches $II$).Thus,the correct matching is $A-III, B-IV, C-I, D-II$.

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

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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Match List-$I$ with List-$II$ List-$I$ Species List-$II$ Electronic distribution $A$. $Cr^{2+}$ $I$. $3d^8$ $B$. $Mn^{+}$ $II$. $3d^3 4s^1$ $C$. $Ni^{2+}$ $III$. $3d^4$ $D$. $V^{+}$ $IV$. $3d^5 4s^1$ Choose the correct answer from the options given below: