The formal charge and bond order of every oxygen in ${\mathrm{PO}}_4^{3-}$ ion is

It is given to find out the formal charge and bond order of every oxygen in ${\mathrm{PO}}_4^{3-}$ ion. 
Explanation 

Formal Charge =[ Number of valence electrons on atom ]-[ non-bonded electrons + number of bonds]

Formal Charge =$\left[6\right]-[4+2]=0.$

O atom is doubly bonded to the P atom. 

In a given resonance structure, the formal charge of $\mathrm{O}$ is 0, and the remaining O atoms have a formal charge of -1. 

In the resonance hybrid, a total of -3 charge is distributed over O atoms. 

 The O atom's formal charge is $\frac{-3}{4}=-0.75$.

$\text{ Bond order }=\frac{\text{ Number of bonds }}{\text{ Number of Resonating structures }}=\frac{5}{4}=1.25$

So, option C is correct.