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(B) . $2O_2^{-} \rightarrow O_2 + O_2^{2-}$ is a disproportionation reaction (oxidation state of $O$ changes from $-0.5$ to $0$ and $-1$) and also a r

Vedclass · Accepted Answer

$A-p, s; B-r; C-p, q; D-p$

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(B) . $2O_2^{-} \rightarrow O_2 + O_2^{2-}$ is a disproportionation reaction (oxidation state of $O$ changes from $-0.5$ to $0$ and $-1$) and also a redox reaction. Thus,$A-p, s$.$B$. $2CrO_4^{2-} + 2H^{+} \rightarrow Cr_2O_7^{2-} + H_2O$. $Cr_2O_7^{2-}$ is a dimeric bridged tetrahedral metal ion. Thus,$B-r$.$C$. $2MnO_4^{-} + 5NO_2^{-} + 6H^{+} \rightarrow 2Mn^{2+} + 5NO_3^{-} + 3H_2O$. This is a redox reaction. $NO_3^{-}$ has a trigonal planar structure. Thus,$C-p, q$.$D$. $NO_3^{-} + 4H^{+} + 3Fe^{2+} \rightarrow NO + 3Fe^{3+} + 2H_2O$. This is a redox reaction. Thus,$D-p$.Matching: $A-p, s; B-r; C-p, q; D-p$.

Column $I$	Column $II$
$A$. $O_2^{-} \rightarrow O_2 + O_2^{2-}$	$p$. redox reaction
$B$. $CrO_4^{2-} + H^{+} \rightarrow$	$q$. one of the products has trigonal planar structure
$C$. $MnO_4^{-} + NO_2^{-} + H^{+} \rightarrow$	$r$. dimeric bridged tetrahedral metal ion
$D$. $NO_3^{-} + H_2SO_4 + Fe^{2+} \rightarrow$	$s$. disproportionation

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