In which of the following, the oxidation stat | vedclass

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Question

(B) To find the oxidation state of nitrogen $(N)$ in each compound, we assign the oxidation states of other elements (Hydrogen = $+1$, Oxygen = $-2$,

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$NH_4Cl$

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(B) To find the oxidation state of nitrogen $(N)$ in each compound, we assign the oxidation states of other elements (Hydrogen = $+1$, Oxygen = $-2$, Chlorine = $-1$):$1$. In $NH_2OH$: $x + 2(+1) + (-2) + (+1) = 0 \implies x + 1 = 0 \implies x = -1$.$2$. In $NH_4Cl$: $x + 4(+1) + (-1) = 0 \implies x + 3 = 0 \implies x = -3$.$3$. In $N_2H_4$: $2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$.$4$. In $HNO_2$: $(+1) + x + 2(-2) = 0 \implies x - 3 = 0 \implies x = +3$.Comparing the values: $-1, -3, -2, +3$. The lowest oxidation state is $-3$ in $NH_4Cl$.

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, the oxidation state of nitrogen is lowest?

Similar Questions

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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