Oxidation numbers of P in PO_{4}^{3-},of S in | vedclass

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(D) $(i)$ Sum of oxidation states of all atoms $=$ charge of ion.$(ii)$ Oxidation number of oxygen $= -2$.Let the oxidation state of $P$ in $PO_{4}^{3

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) $(i)$ Sum of oxidation states of all atoms $=$ charge of ion.$(ii)$ Oxidation number of oxygen $= -2$.Let the oxidation state of $P$ in $PO_{4}^{3-}$ be $x$:$x + 4(-2) = -3 \implies x - 8 = -3 \implies x = +5$.Let the oxidation state of $S$ in $SO_{4}^{2-}$ be $y$:$y + 4(-2) = -2 \implies y - 8 = -2 \implies y = +6$.Let the oxidation state of $Cr$ in $Cr_{2}O_{7}^{2-}$ be $z$:$2z + 7(-2) = -2 \implies 2z - 14 = -2 \implies 2z = 12 \implies z = +6$.Thus,the oxidation states of $P, S$ and $Cr$ are $+5, +6$ and $+6$ respectively.

Oxidation numbers of $P$ in $PO_{4}^{3-}$,of $S$ in $SO_{4}^{2-}$ and that of $Cr$ in $Cr_{2}O_{7}^{2-}$ are respectively

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