Oxidation numbers of P in PO_4^{3-},of S in S | vedclass

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(D) For $PO_4^{3-}$: Let the oxidation number of $P$ be $x$. Then $x + 4(-2) = -3$,which gives $x - 8 = -3$,so $x = +5$.For $SO_4^{2-}$: Let the oxida

Vedclass · Accepted Answer

$+5, +6$ and $+6$

Vedclass · Answer

(D) For $PO_4^{3-}$: Let the oxidation number of $P$ be $x$. Then $x + 4(-2) = -3$,which gives $x - 8 = -3$,so $x = +5$.For $SO_4^{2-}$: Let the oxidation number of $S$ be $x$. Then $x + 4(-2) = -2$,which gives $x - 8 = -2$,so $x = +6$.For $Cr_2O_7^{2-}$: Let the oxidation number of $Cr$ be $x$. Then $2x + 7(-2) = -2$,which gives $2x - 14 = -2$,so $2x = 12$,and $x = +6$.Thus,the oxidation numbers are $+5, +6, +6$.

List $I$ (Compound)	List $II$ (Oxidation state of $N$)
$(A)$ $NO_2$	$(1)$ $+5$
$(B)$ $HNO$	$(2)$ $-3$
$(C)$ $NH_3$	$(3)$ $+4$
$(D)$ $N_2O_5$	$(4)$ $+1$

Oxidation numbers of $P$ in $PO_4^{3-}$,of $S$ in $SO_4^{2-}$ and that of $Cr$ in $Cr_2O_7^{2-}$ are respectively

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List $I$ (Compound) List $II$ (Oxidation state of $N$)
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