1. Draw the Lewis Structure
* Count valence electrons: Chlorine (Cl) has 7 valence electrons, and Oxygen (O) has 6. With two chlorine atoms, the total is 7 + 7 + 6 = 20 valence electrons.
* Connect the atoms: Place the oxygen atom in the center, as it's less electronegative. Connect it to the two chlorine atoms with single bonds.
* Complete octets: Each chlorine atom needs 6 more electrons, and the oxygen atom needs 4 more. This requires two lone pairs on each chlorine atom and two lone pairs on the oxygen atom.
2. Determine the Steric Number
* Steric number: This is the number of electron groups around the central atom. Electron groups include bonds (single, double, or triple) and lone pairs.
* In Cl₂O: The oxygen atom has two bonds (to the chlorine atoms) and two lone pairs, giving it a steric number of 4.
3. Relate Steric Number to Hybridization
* Steric Number | Hybridization
* 2 | sp
* 3 | sp²
* 4 | sp³
* 5 | sp³d
* 6 | sp³d²
4. Conclusion
Since the central oxygen atom in Cl₂O has a steric number of 4, its hybridization is sp³.
