In which of the following pairs is the differ | vedclass

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(A) To find the difference in oxidation numbers,we calculate the oxidation state of the underlined element in each compound:$A$: In $\underline{N}O_2$

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$\underline{N}O_2$ and $\underline{N}_2H_4$

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(A) To find the difference in oxidation numbers,we calculate the oxidation state of the underlined element in each compound:$A$: In $\underline{N}O_2$,$x + 2(-2) = 0 \implies x = +4$. In $\underline{N}_2H_4$,$2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$. Difference = $|4 - (-2)| = 6$.$B$: In $\underline{P}_2O_5$,$2x + 5(-2) = 0 \implies 2x = 10 \implies x = +5$. In $\underline{P}_4O_{10}$,$4x + 10(-2) = 0 \implies 4x = 20 \implies x = +5$. Difference = $|5 - 5| = 0$.$C$: In $\underline{N}_2O$,$2x + (-2) = 0 \implies 2x = 2 \implies x = +1$. In $\underline{N}O$,$x + (-2) = 0 \implies x = +2$. Difference = $|2 - 1| = 1$.$D$: In $\underline{S}O_2$,$x + 2(-2) = 0 \implies x = +4$. In $\underline{S}O_3$,$x + 3(-2) = 0 \implies x = +6$. Difference = $|6 - 4| = 2$.Comparing the differences $(6, 0, 1, 2)$,the greatest difference is $6$ in option $A$.

Column-$I$	Column-$II$
$A$. $Cr_2O_7^{2-}$	$1$. $+3$
$B$. $MnO_4^{-}$	$2$. $+4$
$C$. $VO_3^{-}$	$3$. $+5$
$D$. $FeF_6^{3-}$	$4$. $+6$
	$5$. $+7$

In which of the following pairs is the difference in the oxidation number of the underlined elements the greatest?

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What is the oxidation number of underlined species in $\underline{P}F_6^{-}$ and $\underline{V}_2 O_7^{4-}$ ions respectively?

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Match the Column-$I$ with Column-$II$ for the oxidation states of the central atoms.

Column-$I$ Column-$II$

$A$. $Cr_2O_7^{2-}$ $1$. $+3$

$B$. $MnO_4^{-}$ $2$. $+4$

$C$. $VO_3^{-}$ $3$. $+5$

$D$. $FeF_6^{3-}$ $4$. $+6$

$5$. $+7$

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