In `PO_(4)^(3-)` ion, the formal charge on the oxygen
atom of P-O bond is

Options

A. `+1`
B. -1
C. `-0.75`
D. `+ 0.75`

Explanation

To determine the formal charge on the oxygen atom in the phosphate ion \( PO_4^{3-} \), we can follow these steps:

### Step 1: Determine the Lewis Structure of \( PO_4^{3-} \)
- The phosphate ion consists of one phosphorus atom (P) and four oxygen atoms (O).
- Phosphorus is in group 15 and has 5 valence electrons. Each oxygen atom is in group 16 and has 6 valence electrons.
- The total number of valence electrons for \( PO_4^{3-} \) is calculated as follows:
- Phosphorus: 5 electrons
- Oxygen: 4 atoms × 6 electrons = 24 electrons
- Total = 5 + 24 + 3 (for the 3- charge) = 32 electrons

### Step 2: Draw the Lewis Structure
- Place phosphorus in the center and surround it with the four oxygen atoms.
- Form single bonds between phosphorus and each oxygen atom. This uses 8 electrons (4 bonds).
- Distribute the remaining 24 electrons to satisfy the octet rule for the oxygen atoms. Each oxygen will receive 6 additional electrons, making them each have 8 electrons (octet).
- One of the oxygen atoms can form a double bond with phosphorus to reduce the formal charge on the oxygen atoms.

### Step 3: Calculate the Formal Charge
The formal charge (FC) can be calculated using the formula:
\[ \text{FC} = V - (N + \frac{B}{2}) \]
Where:
- \( V \) = number of valence electrons in the free atom
- \( N \) = number of non-bonding electrons (lone pairs)
- \( B \) = number of bonding electrons (shared in bonds)

For the oxygen atom involved in the P-O bond:
- \( V \) = 6 (valence electrons for oxygen)
- \( N \) = 4 (assuming it has 2 lone pairs)
- \( B \) = 2 (since it is involved in one single bond with phosphorus)

Substituting these values into the formula:
\[ \text{FC} = 6 - (4 + \frac{2}{2}) = 6 - (4 + 1) = 6 - 5 = 1 \]
Thus, the formal charge on the oxygen atom in the P-O bond is +1.

### Final Answer
The formal charge on the oxygen atom of the P-O bond in the \( PO_4^{3-} \) ion is +1.

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