Draw the Lewis structure of XeCl2 and then determine the hybridization of the central atom. Draw the Lewis structure of SOF4 (with minimized formal charges) and then determine the hybridization of the central atom. Determine the hybridization of the central atom. Draw the Lewis structure of phosphate (PO43-) with minimized formal charges and then determine the hybridization of the central atom.
The Correct Answer and Explanation is:
Let’s examine the Lewis structures and determine the hybridization of the central atoms for XeCl₂, SOF₄, and PO₄³⁻, along with an explanation.
1. XeCl₂ (Xenon Dichloride)
Lewis Structure:
- Xe is the central atom with 8 valence electrons.
- Each Cl brings 7 valence electrons, totaling 14 from two Cl atoms.
- Total valence electrons = 8 (Xe) + 14 (Cl) = 22 electrons.
After forming two single bonds with Cl atoms (4 electrons used), 18 electrons remain. Place three lone pairs (6 electrons) on Xe to complete its octet and minimize formal charges.
Electron geometry: Trigonal bipyramidal
Molecular geometry: Linear
Hybridization of Xe: sp³d
Explanation: 2 bonding pairs + 3 lone pairs = 5 regions of electron density → sp³d hybridization
2. SOF₄ (Thionyl Tetrafluoride)
Lewis Structure:
- S is central: 6 valence electrons
- O contributes 6, and each F contributes 7 × 4 = 28
- Total = 6 (S) + 6 (O) + 28 (F) = 40 electrons
Make 4 single bonds with F and one double bond with O to minimize formal charges. This uses 10 electrons. The rest are used to complete octets for F and O.
Electron geometry: Trigonal bipyramidal
Hybridization of S: sp³d
Explanation: 5 regions of electron density (4 single bonds with F, 1 double bond with O) → sp³d hybridization
3. Phosphate ion (PO₄³⁻)
Lewis Structure:
- P = 5 valence electrons
- O = 6 × 4 = 24
- Extra 3 electrons for the 3⁻ charge → Total = 32 electrons
To minimize formal charges, make one P=O double bond and three P–O⁻ single bonds. Each O gets lone pairs to complete octets.
Electron geometry: Tetrahedral
Hybridization of P: sp³
Explanation: 4 bonding domains (1 double bond + 3 single bonds) = 4 regions → sp³ hybridization
Summary Table:
| Molecule/Ion | Central Atom | Lewis Structure Geometry | Hybridization |
|---|---|---|---|
| XeCl₂ | Xe | Linear (5 electron regions) | sp³d |
| SOF₄ | S | Trigonal bipyramidal | sp³d |
| PO₄³⁻ | P | Tetrahedral | sp³ |
Final Notes:
Hybridization correlates with the number of electron domains (bonding + lone pairs). This concept helps in predicting molecular geometry and chemical reactivity. Accurate Lewis structures help determine how atoms are bonded and how electrons are distributed, which is critical in understanding hybridization and molecular shape.