Given below are two statements:Statement I: T | vedclass

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(C) Statement $I$: Ions with $d^0$ or $d^{10}$ configuration are colourless. $Sc^{3+}$ $(d^0)$, $Ti^{4+}$ $(d^0)$, and $Zn^{2+}$ $(d^{10})$ are colour

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Both Statement $I$ and Statement $II$ are false

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(C) Statement $I$: Ions with $d^0$ or $d^{10}$ configuration are colourless. $Sc^{3+}$ $(d^0)$, $Ti^{4+}$ $(d^0)$, and $Zn^{2+}$ $(d^{10})$ are colourless. - $[Sc^{3+}, Ti^{3+}]$: $Sc^{3+}$ is colourless, $Ti^{3+}$ is coloured. - $[Mn^{2+}, Cr^{2+}]$: Both are coloured ($Mn^{2+}$ is $d^5$, $Cr^{2+}$ is $d^4$). - $[Cu^{2+}, Zn^{2+}]$: $Cu^{2+}$ is coloured $(d^9)$, $Zn^{2+}$ is colourless. - $[Ni^{2+}, Ti^{4+}]$: $Ni^{2+}$ is coloured $(d^8)$, $Ti^{4+}$ is colourless. Only $1$ pair $([Mn^{2+}, Cr^{2+}])$ has both ions coloured. Thus, Statement $I$ is false. Statement $II$: $Th^{4+}$ is the most stable oxidation state of Thorium and cannot be oxidized further, so it cannot act as a reducing agent. $Eu^{2+}$ is a strong reducing agent as it tends to change to $Eu^{3+}$. Thus, Statement $II$ is false.

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	$IV$. $3$
	$V$. $4$
	$VI$. $2$

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