How to Draw the Lewis Structure for PO₄³⁻ and Determine Its Electron and Molecular Geometries

To draw the Lewis structure for PO₄³⁻ (phosphate ion), follow these steps:

1. Count the total number of valence electrons: Phosphorus (P) has 5 valence electrons, and each oxygen (O) has 6 valence electrons. Since there are 4 oxygen atoms, the total number of valence electrons from oxygen is 4 × 6 = 24. Additionally, the ion has a 3- charge, which means we need to add 3 more electrons. So, the total number of valence electrons is 5 (from P) + 24 (from O) + 3 (from the charge) = 32.
2. Place the least electronegative atom in the center: Phosphorus is less electronegative than oxygen, so it will be the central atom.
3. Connect the outer atoms to the central atom with single bonds: Draw single bonds between phosphorus and each of the four oxygen atoms. This uses up 4 × 2 = 8 electrons, leaving us with 32 – 8 = 24 electrons.
4. Distribute the remaining electrons as lone pairs: Place the remaining 24 electrons as lone pairs on the oxygen atoms. Each oxygen atom will have 6 electrons (3 lone pairs) after forming a single bond with phosphorus.
5. Check for octet rule: Phosphorus has 8 electrons around it (4 from the single bonds), and each oxygen atom also has 8 electrons (6 from lone pairs and 2 from the single bond).
6. Assign formal charges: The formal charge on phosphorus is 5 – (0 + 8/2) = 0, and the formal charge on each oxygen is 6 – (6 + 2/2) = -1. The overall charge of the ion is 3-, which matches the given charge.

Electron Geometry: The electron geometry of PO₄³⁻ is tetrahedral because there are four regions of electron density around the central phosphorus atom.

Molecular Geometry: The molecular geometry of PO₄³⁻ is also tetrahedral because all four regions of electron density are bonding pairs, and there are no lone pairs on the central atom.