Answer
a) In $CrO_{4}^{2-}$ ion, oxidation state of $Cr = +6, O = -2$
b) In $Cr_{2}O_{7}^{2-}$ ion, oxidation state of $Cr = +6, O = -2$
c) In $PO_{4}^{3-}$ ion, oxidation state of $ P = +5, O = -2$
d) In $MnO_{4}^{-}$ ion, oxidation state of $Mn = +7, O = -2$
Work Step by Step
Algebraic sum of oxidation state for a Polyatomic ion will be equal to the charge of the ion.
a) In $CrO_{4}^{2-}$, oxidation state of O =-2. Let's say oxidation state of Cr = $x$
Hence, $x + (-2)\times4 = -2$, by solving we get $ x = +6$
So, oxidation state of $ Cr = +6$.
b) In $Cr_{2}O_{7}^{2-}$, oxidation state of O =-2. Let's say oxidation state of Cr = $x$
Hence, $x\times2+ (-2)\times7 = -2$, by solving we get $ x = +6$
So, oxidation state of $ Cr = +6$.
c) In $PO_{4}^{3-}$, oxidation state of O = -2, Let's say oxidation state of P = $x$
Hence, $x+ (-2)\times4 = -3$, by solving we get $ x = +5$
So, oxidation state of $ P= +5$.
d) In $MnO_{4}^{-}$, oxidation state of O = -2, Let's say oxidation state of Mn = $x$
Hence, $x+ (-2)\times4 = -1$, by solving we get $ x = +7$
So, oxidation state of $ Mn = +7$.
