(D) For $PO_4^{3-}$: $P + 4(-2) = -3 \implies P - 8 = -3 \implies P = +5$. For $SO_4^{2-}$: $S + 4(-2) = -2 \implies S - 8 = -2 \implies S = +6$. For

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(D) For $PO_4^{3-}$: $P + 4(-2) = -3 \implies P - 8 = -3 \implies P = +5$. For $SO_4^{2-}$: $S + 4(-2) = -2 \implies S - 8 = -2 \implies S = +6$. For $Cr_2O_7^{2-}$: $2Cr + 7(-2) = -2 \implies 2Cr - 14 = -2 \implies 2Cr = +12 \implies Cr = +6$. Thus,the oxidation states are $+5, +6, +6$.

Class 11 Chemistry Redox Reactions Oxidation number and Oxidation state

Medium

The oxidation states of $P$ in $PO_4^{3-}$,$S$ in $SO_4^{2-}$,and $Cr$ in $Cr_2O_7^{2-}$ are respectively:

A
$+3, +6, +5$
B
$+5, +3, +6$
C
$-3, +6, +6$
D
$+5, +6, +6$

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Class 11 Questions

Class 11

Chemistry Questions

Chapter

Redox Reactions

Subtopic

Oxidation number and Oxidation state

Select the compound in which chlorine is assigned the oxidation number $+5$.

Easy

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Chlorine is in $+1$ oxidation state in

Medium

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Match the items in Column-$I$ with relevant items in Column-$II$.

Column-$I$ Column-$II$

$A$. Positively charged ion $1$. $+7$

$B$. Oxidation number of all neutral atoms $2$. $-1$

$C$. Oxidation number of hydrogen $(H^+)$ $3$. $+1$

$D$. Oxidation number of fluorine in $NaF$ $4$. $0$

$E$. Negatively charged ion $5$. Cation

$6$. Anion

Column-$I$	Column-$II$
$A$. Positively charged ion	$1$. $+7$
$B$. Oxidation number of all neutral atoms	$2$. $-1$
$C$. Oxidation number of hydrogen $(H^+)$	$3$. $+1$
$D$. Oxidation number of fluorine in $NaF$	$4$. $0$
$E$. Negatively charged ion	$5$. Cation
	$6$. Anion