Question
Determine the hybridization of the central atom in: (a) BF₃, (b) NH₃, (c) H₂O, (d) PCl₅, and (e) SF₆. Show the method clearly.
(CBSE Class 11 / JEE Main / NEET pattern)
Solution — Step by Step
Steric number = number of atoms bonded to the central atom + number of lone pairs on the central atom.
This single number tells you the hybridization directly:
| Steric Number | Hybridization | Geometry (no lone pairs) |
|---|---|---|
| 2 | sp | Linear |
| 3 | sp² | Trigonal planar |
| 4 | sp³ | Tetrahedral |
| 5 | sp³d | Trigonal bipyramidal |
| 6 | sp³d² | Octahedral |
flowchart TD
A["Count bonded atoms\n+ lone pairs on\ncentral atom"] --> B["Steric Number"]
B -->|"2"| C["sp\n(Linear)"]
B -->|"3"| D["sp²\n(Trigonal planar)"]
B -->|"4"| E["sp³\n(Tetrahedral)"]
B -->|"5"| F["sp³d\n(Trigonal bipyramidal)"]
B -->|"6"| G["sp³d²\n(Octahedral)"](a) BF₃: B has 3 bonds, 0 lone pairs → Steric number = 3 → sp² (trigonal planar)
(b) NH₃: N has 3 bonds, 1 lone pair → Steric number = 4 → sp³ (tetrahedral electron geometry, but pyramidal shape)
(c) H₂O: O has 2 bonds, 2 lone pairs → Steric number = 4 → sp³ (tetrahedral electron geometry, but bent/V-shape)
(d) PCl₅: P has 5 bonds, 0 lone pairs → Steric number = 5 → sp³d (trigonal bipyramidal)
(e) SF₆: S has 6 bonds, 0 lone pairs → Steric number = 6 → sp³d² (octahedral)
Use this shortcut: Lone pairs =
For NH₃: N has 5 valence electrons, uses 3 for bonding → lone pair. For H₂O: O has 6 valence electrons, uses 2 for bonding → lone pairs.
Why This Works
Hybridization is the central atom’s way of arranging its bonding and non-bonding electron pairs as far apart as possible (VSEPR theory). Two pairs need a linear arrangement (sp), three need trigonal planar (sp²), four need tetrahedral (sp³), and so on. The steric number simply counts how many “electron pair slots” the central atom needs, and the hybridization provides exactly that many equivalent orbitals.
Lone pairs count because they occupy space too — they just do not contribute to the molecular shape we see. That is why NH₃ is sp³ hybridized (4 pairs) but has a pyramidal shape (3 visible bonds).
Alternative Method — Formula for Quick Calculation
For a molecule (A = central atom, B = bonded atoms, E = lone pairs):
where = valence electrons of central atom, = number of monovalent atoms, = charge on cation, = charge on anion.
For JEE, you will encounter ions too. For : N has 5 valence electrons, but the positive charge means one less → effective = 4. Four bonds, 0 lone pairs → sp³. For : S has 6 + 2 (from charge) = 8 effective electrons, 4 bonds, 0 lone pairs → sp³.
Common Mistake
The biggest error: forgetting to count lone pairs. Students see BF₃ (3 bonds → sp²) and then write NH₃ as sp² too (3 bonds). But NH₃ has one lone pair, making the steric number 4, not 3. The hybridization is sp³, not sp². Always count lone pairs — they determine the hybridization just as much as bonds do.