(ii) oxidation number of oxygen $=-2$

Let the oxidation state of $\mathrm{P}$ in $\mathrm{PO}_{4}^{3-}$ " is $\mathrm{x}$. $\mathrm{PO}_{4}^{3-}$

$x+4(-2)=-3$

$x-8=-3$

$x=+5$

Let the oxidation state of $\mathrm{s}$ in $\mathrm{SO}_{4}^{2-}$  is $\mathrm{y}$

$y+4(-2)=-2$

$y-8=-2$

$y=+6$

Let the oxidation state of $\mathrm{Cr}$ in $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ is $\mathrm{z}$

$2 \times z+7(-2)=-2$

$2 z-14=-2$

$z=+6$

Hence, oxidation state of $\mathrm{P}, \mathrm{S}$ and $\mathrm{Cr}$ are $+5,+6$ and $+6$