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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
A
+3, +6 and +6
B
+5, +6 and +6
C
+3, +6 and +5
D
+5, +3 and +6
Detailed Solution
Take the oxidation number of O as −2; the sum of oxidation numbers equals the charge on the ion.
$PO_4^{3-}$: $x + 4(-2) = -3$, so $x = +5$
$SO_4^{2-}$: $x + 4(-2) = -2$, so $x = +6$
$Cr_2O_7^{2-}$: $2x + 7(-2) = -2$, so $2x = +12$ and $x = +6$
The oxidation numbers are +5, +6 and +6.
$PO_4^{3-}$: $x + 4(-2) = -3$, so $x = +5$
$SO_4^{2-}$: $x + 4(-2) = -2$, so $x = +6$
$Cr_2O_7^{2-}$: $2x + 7(-2) = -2$, so $2x = +12$ and $x = +6$
The oxidation numbers are +5, +6 and +6.