Question
Given any molecule or ion, how do we determine the hybridization of the central atom using the steric number method?
Solution — Step by Step
The steric number (SN) is the total count of electron groups around the central atom:
Each bond — whether single, double, or triple — counts as one electron group. A double bond between C and O is still one group.
| Steric Number | Hybridization | Geometry (no lone pairs) |
|---|---|---|
| 2 | sp | Linear |
| 3 | sp | Trigonal planar |
| 4 | sp | Tetrahedral |
| 5 | spd | Trigonal bipyramidal |
| 6 | spd | Octahedral |
This mapping works universally — memorise it once, use it forever.
Example 1: HO
- Central atom: O
- Bonded atoms: 2 (two H)
- Lone pairs on O: 2
- SN = 2 + 2 = 4 → sp
Example 2: BF
- Central atom: B
- Bonded atoms: 3 (three F)
- Lone pairs on B: 0
- SN = 3 + 0 = 3 → sp
Example 3: SF
- Central atom: S
- Bonded atoms: 6
- Lone pairs: 0
- SN = 6 → spd
For molecules where counting lone pairs is tricky, use:
where V = valence electrons of central atom, M = number of monovalent atoms bonded, C = charge for cation, A = charge for anion.
Example: ICl
- V = 7 (iodine), M = 4 (four Cl), charge = -1 (anion, A = 1)
- SN = (7 + 4 + 1)/2 = 6 → spd
flowchart TD
A["Given: molecule or ion"] --> B["Count bonded atoms around central atom"]
B --> C["Count lone pairs on central atom"]
C --> D["SN = bonded atoms + lone pairs"]
D --> E{"SN value?"}
E -->|"2"| F["sp — Linear"]
E -->|"3"| G["sp2 — Trigonal planar"]
E -->|"4"| H["sp3 — Tetrahedral"]
E -->|"5"| I["sp3d — Trigonal bipyramidal"]
E -->|"6"| J["sp3d2 — Octahedral"]Why This Works
Hybridization is the mixing of atomic orbitals to form new orbitals that minimise repulsion between electron groups. The number of hybrid orbitals formed equals the number of electron groups (steric number). Two groups need two hybrid orbitals (one s + one p = sp), three need three (s + 2p = sp), and so on.
Lone pairs count because they also occupy space and contribute to repulsion — even though they do not form bonds.
Alternative Method
For organic molecules specifically, there is an even faster shortcut: count the number of sigma bonds + lone pairs on the atom. A carbon with 4 sigma bonds and 0 lone pairs is sp. A carbon in C=O (2 sigma bonds from C, including the one in C=O, plus potentially one more) — just count all sigma connections.
Common Mistake
The number one error: counting a double bond as two groups. A double bond (one sigma + one pi) counts as one electron group for steric number purposes. CO has SN = 2 (not 4), making it sp hybridized and linear. If you counted each bond separately, you would wrongly predict sp and tetrahedral — which is completely wrong. This appears in nearly every JEE and NEET paper.