What is the hybridization of the central atom in (a) PBr5, (b) CH2O, (c) O3, (d) NO2?
The correct answer and explanation is:
The hybridization of the central atom in each of the following molecules is:
- (a) PBr₅ – sp³d
- (b) CH₂O – sp²
- (c) O₃ – sp²
- (d) NO₂ – sp²
Explanation (300 words):
Hybridization is the mixing of atomic orbitals to form new, hybrid orbitals that explain the molecular geometry and bonding in molecules. To determine hybridization, we examine the number of electron regions (bonding and lone pairs) around the central atom.
(a) PBr₅ – sp³d
Phosphorus is the central atom, surrounded by 5 bromine atoms. There are 5 bonding pairs and no lone pairs on phosphorus. This gives 5 regions of electron density, which corresponds to sp³d hybridization, forming a trigonal bipyramidal shape.
(b) CH₂O (Formaldehyde) – sp²
Carbon is the central atom with 2 single bonds to hydrogen and 1 double bond to oxygen. That’s a total of 3 regions of electron density (each single/double bond counts as one region), corresponding to sp² hybridization, giving a trigonal planar geometry.
(c) O₃ (Ozone) – sp²
Ozone has a central oxygen atom bonded to two other oxygen atoms, with resonance between single and double bonds and a lone pair on the central atom. This results in 3 regions of electron density (2 bonding pairs and 1 lone pair), so the central oxygen is sp² hybridized. The molecular shape is bent due to lone pair repulsion.
(d) NO₂ – sp²
Nitrogen is the central atom bonded to two oxygen atoms, with one single bond (with a formal negative charge on oxygen) and one double bond. There’s also one unpaired electron (a single electron counts as a region of electron density). This totals 3 regions, so nitrogen is sp² hybridized, resulting in a bent geometry. NO₂ is an example of a free radical with an odd number of electrons.
In summary, hybridization helps explain molecular shape, and it is determined by counting the electron domains around the central atom.