Match the following:List-IList-IIA. Oxidation | vedclass

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(A-V, B-II, C-VI, D-IV) Oxidation state of $V$ in $VOCl_2$:Let oxidation state of $V = x$.$x + (-2) + 2(-1) = 0$ $\Rightarrow x - 4 = 0$ $\Rightarrow

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(A-V, B-II, C-VI, D-IV) Oxidation state of $V$ in $VOCl_2$:Let oxidation state of $V = x$.$x + (-2) + 2(-1) = 0$ $\Rightarrow x - 4 = 0$ $\Rightarrow x = +4$.Thus,$(A)$ matches with $(V)$.$(B)$ Oxidation state of $Mn$ in $MnO_4^{2-}$:Let oxidation state of $Mn = x$.$x + 4(-2) = -2$ $\Rightarrow x - 8 = -2$ $\Rightarrow x = +6$.Electronic configuration of $Mn^{6+}$ is $[Ar] 3d^1 4s^0$. It has $1$ unpaired electron.Thus,$(B)$ matches with $(II)$.$(C)$ Oxidation state of $Ni$ in $[NiCl_4]^{2-}$:Let oxidation state of $Ni = x$.$x + 4(-1) = -2 \Rightarrow x = +2$.Electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8 4s^0$. It has $2$ unpaired electrons.Thus,$(C)$ matches with $(VI)$.$(D)$ All lanthanide ions exhibit an oxidation state of $+3$.Thus,$(D)$ matches with $(IV)$.

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

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