The formal charge and P-O bond order in PO₄³⁻ respectively are

The correct answer is -0.75, 1.25.

Let's understand how to determine both the formal charge on each oxygen atom and the P-O bond order in the phosphate ion (${\mathrm{PO}}_4^{3-}$) step-by-step.

1. Finding the Formal Charge on Oxygen:
The phosphate ion (${\mathrm{PO}}_4^{3-}$) has a total net charge of -3 distributed over its four oxygen atoms because of resonance.
Since all four P-O bonds are equivalent due to resonance, the total negative charge is shared equally among the four oxygen atoms.
The formal charge ($F_c$) on each oxygen atom is calculated as:

$F_c=\frac{\text{Total Charge on Ion}}{\text{Number of Oxygen Atoms}}$

$F_c=\frac{-3}{4}=-0.75$

2. Finding the P-O Bond Order:
To find the bond order, we look at the Lewis structure of the phosphate ion. Phosphorus (P) is the central atom, bonded to four oxygen (O) atoms. To satisfy the valency of phosphorus and minimize formal charges, the stable Lewis structure consists of:
- Three P-O single bonds ($P-O^{-}$)
- One P=O double bond ($P=O$)
This gives a total of 5 bonding pairs (1 double bond + 3 single bonds) shared across 4 bond positions.
The bond order (B.O.) is calculated using the formula:

$\text{Bond Order}=\frac{\text{Total Number of Bonding Pairs}}{\text{Number of Resonating Positions}}$

$\text{Bond Order}=\frac{5}{4}=1.25$

Therefore, the formal charge on each oxygen atom is -0.75 and the P-O bond order is 1.25.