Consider the following compounds:underline{K} | vedclass

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(A) $1$. For $\underline{K}O_2$: Potassium $(K)$ is an alkali metal and always exhibits an oxidation state of $+1$ in its compounds.$2$. For $H_2\unde

Vedclass · Accepted Answer

$+1, -1$ and $+6$

Vedclass · Answer

(A) $1$. For $\underline{K}O_2$: Potassium $(K)$ is an alkali metal and always exhibits an oxidation state of $+1$ in its compounds.$2$. For $H_2\underline{O}_2$: Let the oxidation state of oxygen be $x$. Since hydrogen is $+1$,we have $2(+1) + 2x = 0$,which gives $2x = -2$,so $x = -1$.$3$. For $H_2\underline{S}O_4$: Let the oxidation state of sulfur be $x$. Hydrogen is $+1$ and oxygen is $-2$. Thus,$2(+1) + x + 4(-2) = 0$,which simplifies to $2 + x - 8 = 0$,so $x = +6$.Therefore,the oxidation states are $+1, -1$ and $+6$.

Column-$I$	Column-$II$
$A$. $\underline{N}H_4 Cl$	$P$. $+1$
$B$. $H\underline{N}O_3$	$Q$. $+5$
$C$. $\underline{N}_2 O$	$R$. $-3$
$D$. $\underline{N}H_2 OH$	$S$. $-1$

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

Consider the following compounds:
$\underline{K}O_2$,$H_2\underline{O}_2$ and $H_2\underline{S}O_4$.
The oxidation states of the underlined elements in them are,respectively,

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Match the following Column-$I$ to Column-$II$ based on the oxidation state of the underlined nitrogen atom:
Column-$I$ Column-$II$
$A$. $\underline{N}H_4 Cl$ $P$. $+1$
$B$. $H\underline{N}O_3$ $Q$. $+5$
$C$. $\underline{N}_2 O$ $R$. $-3$
$D$. $\underline{N}H_2 OH$ $S$. $-1$

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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Consider the following compounds:$\underline{K}O_2$,$H_2\underline{O}_2$ and $H_2\underline{S}O_4$.The oxidation states of the underlined elements in them are,respectively,