Which of the following orders is correct?

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(D) $1$. $Cr^{2+}$ is a stronger reducing agent than $Fe^{2+}$ because $Cr^{2+}$ $(d^4)$ changes to $Cr^{3+}$ ($d^3$,stable $t_{2g}^3$ configuration)

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All of these

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(D) $1$. $Cr^{2+}$ is a stronger reducing agent than $Fe^{2+}$ because $Cr^{2+}$ $(d^4)$ changes to $Cr^{3+}$ ($d^3$,stable $t_{2g}^3$ configuration) while $Fe^{2+}$ $(d^6)$ changes to $Fe^{3+}$ ($d^5$,stable half-filled configuration). However,the $E^o$ value for $Cr^{3+}/Cr^{2+}$ is $-0.41 \ V$ and $Fe^{3+}/Fe^{2+}$ is $+0.77 \ V$,making $Cr^{2+}$ a better reducing agent.$2$. $Mn^{3+}$ is a stronger oxidising agent than $Cr^{3+}$ because $Mn^{3+}$ $(d^4)$ easily gains an electron to form $Mn^{2+}$ ($d^5$,stable half-filled configuration).$3$. In group $6$,stability of the highest oxidation state increases down the group. Thus,$WO_3$ is more stable than $CrO_3$.$4$. Since all statements are correct,the correct option is $D$.

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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