Hybridization prediction — count lone pairs and bond pairs to determine shape

Question

How do we predict the hybridization and shape of a molecule? Determine the hybridization of the central atom in $\text{NH}_3$ , $\text{H}_2\text{O}$ , $\text{BF}_3$ , and $\text{SF}_6$ .

(CBSE 11 + JEE Main + NEET)

Solution — Step by Step

Steric number = number of atoms bonded to central atom + number of lone pairs on central atom.

This is the key number that determines everything:

Steric Number	Hybridization	Geometry (no lone pairs)
2	$sp$	Linear (180°)
3	$sp^2$	Trigonal planar (120°)
4	$sp^3$	Tetrahedral (109.5°)
5	$sp^3d$	Trigonal bipyramidal
6	$sp^3d^2$	Octahedral (90°)

N has 5 valence electrons. It bonds with 3 H atoms and has 1 lone pair.

Steric number = $3 + 1 = 4$ → $sp^3$ hybridization

Shape: Trigonal pyramidal (not tetrahedral — lone pair pushes bonds closer).

Bond angle: $\approx 107°$ (less than 109.5° due to lone pair repulsion).

O has 6 valence electrons. It bonds with 2 H atoms and has 2 lone pairs.

Steric number = $2 + 2 = 4$ → $sp^3$ hybridization

Shape: Bent/V-shaped.

Bond angle: $\approx 104.5°$ (even smaller — two lone pairs compress more).

$\text{BF}_3$ : B has 3 valence electrons, bonds with 3 F, no lone pairs. Steric number = 3 → $sp^2$ , trigonal planar.

$\text{SF}_6$ : S has 6 valence electrons, bonds with 6 F, no lone pairs. Steric number = 6 → $sp^3d^2$ , octahedral.

flowchart TD
    A["Find hybridization"] --> B["Count bonds to central atom"]
    B --> C["Count lone pairs on central atom"]
    C --> D["Steric Number = bonds + lone pairs"]
    D --> E{"Steric Number?"}
    E -- 2 --> F["sp — Linear"]
    E -- 3 --> G["sp² — Trigonal planar"]
    E -- 4 --> H["sp³ — Tetrahedral base"]
    E -- 5 --> I["sp³d — Trigonal bipyramidal"]
    E -- 6 --> J["sp³d² — Octahedral"]
    H --> K{"Lone pairs?"}
    K -- 0 --> L["Tetrahedral"]
    K -- 1 --> M["Trigonal pyramidal"]
    K -- 2 --> N["Bent / V-shape"]

Why This Works

Hybridization is a model that explains observed molecular shapes. When an atom forms bonds, its atomic orbitals mix (“hybridize”) to create new orbitals that point in directions that minimize electron pair repulsion (VSEPR theory). The number of hybrid orbitals equals the steric number.

Lone pairs occupy hybrid orbitals too — they are “invisible bonds” that take up space. That is why $\text{NH}_3$ is pyramidal (not flat) and $\text{H}_2\text{O}$ is bent (not linear), even though both have $sp^3$ hybridization.

Alternative Method

Quick formula for steric number:

\text{Steric Number} = \frac{V + M - C + A}{2}

where $V$ = valence electrons of central atom, $M$ = monovalent atoms bonded, $C$ = positive charge, $A$ = negative charge.

For $\text{NH}_4^+$ : $V = 5$ , $M = 4$ , $C = 1$ , $A = 0$ → SN = $(5 + 4 - 1)/2 = 4$ → $sp^3$ .

For NEET MCQs, memorise these common molecules: $\text{BeCl}_2$ ( $sp$ , linear), $\text{BF}_3$ ( $sp^2$ , trigonal planar), $\text{CH}_4$ ( $sp^3$ , tetrahedral), $\text{PCl}_5$ ( $sp^3d$ , TBP), $\text{SF}_6$ ( $sp^3d^2$ , octahedral). Then use lone pair count to modify the shape.

Common Mistake

Students confuse electron geometry with molecular shape. The hybridization and electron geometry of $\text{NH}_3$ is tetrahedral ( $sp^3$ ), but the molecular shape is trigonal pyramidal. The shape describes where the ATOMS are, not where the lone pairs are. Always report the molecular shape in exams, not the electron geometry, unless specifically asked.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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