Hybridisation

Carbon in methane forms four bonds that are all identical — same length, same strength, same $109.5°$ angle between any pair. But a carbon atom's ground-state electron arrangement is $2s^2\,2p_x^1\,2p_y^1$ : two paired $s$ electrons and only two unpaired $p$ electrons pointing at $90°$ to each other. Neither the count (two bonds, not four) nor the geometry ( $90°$ , not $109.5°$ ) matches what we actually observe. Hybridisation is valence bond theory's fix: atomic orbitals of similar energy on one atom mix mathematically into a new, equal-sized set of hybrid orbitals, spaced to minimise repulsion and match the shape VSEPR already predicts. This article builds the five hybridisation schemes on the ISC syllabus — $sp$ , $sp^2$ , $sp^3$ , $sp^3d$ , $sp^3d^2$ — around a single counting rule, works through ammonia's compressed bond angle and ethyne's triple bond, and flags the traps examiners reward.

The core insight

Counting the σ-bonded regions plus lone pairs on a central atom (its steric number) tells you exactly how many atomic orbitals mixed to form hybrid orbitals, and that number fixes both the hybridisation type and the molecular geometry.

The core idea

Valence bond theory builds covalent bonds from orbital overlap, but an atom's unhybridised $s$ and $p$ orbitals rarely point where a molecule's observed shape needs them to. Hybridisation resolves this by mixing atomic orbitals of comparable energy on the same atom into a new set of equivalent hybrid orbitals — same total count in as out, new shapes and directions, chosen to keep electron pairs as far apart as possible.

The five schemes on this syllabus, each named for the orbitals mixed:

$sp$ (1 $s$ + 1 $p$ → 2 orbitals): linear, $180°$ . Example: $\text{BeCl}_2$ .
$sp^2$ (1 $s$ + 2 $p$ → 3 orbitals): trigonal planar, $120°$ . Example: $\text{BF}_3$ .
$sp^3$ (1 $s$ + 3 $p$ → 4 orbitals): tetrahedral, $109.5°$ ideal. Examples: $\text{CH}_4$ ( $109.5°$ ), $\text{NH}_3$ ( $\approx107°$ ), $\text{H}_2\text{O}$ ( $\approx104.5°$ ) — all $sp^3$ , differing only in how many of the four hybrid orbitals hold lone pairs.
$sp^3d$ (1 $s$ + 3 $p$ + 1 $d$ → 5 orbitals): trigonal bipyramidal. Example: $\text{PCl}_5$ .
$sp^3d^2$ (1 $s$ + 3 $p$ + 2 $d$ → 6 orbitals): octahedral, all adjacent angles $90°$ . Example: $\text{SF}_6$ .

Every scheme obeys the same bookkeeping: the number of hybrid orbitals produced always equals the number of atomic orbitals mixed, and that number is fixed by a single quantity — the steric number, the count of σ-bonded regions plus lone pairs on the central atom:

N = (\text{number of }\sigma\text{-bonded regions}) + (\text{number of lone pairs}). \qquad (1)

where: $N$ is the steric number (a whole number), a σ-bonded region is one atom directly attached to the central atom regardless of whether that bond is single, double, or triple (a double or triple bond still counts as one region, since only its σ component uses a hybrid orbital), and a lone pair is an unshared electron pair on the central atom. Matching $N$ to a scheme is direct: $N=2\to sp$ , $N=3\to sp^2$ , $N=4\to sp^3$ , $N=5\to sp^3d$ , $N=6\to sp^3d^2$ . Any $\pi$ bonds are extra and do not add to $N$ — they form separately from unhybridised $p$ orbitals overlapping sideways, never from hybrid orbitals.

Figure

Five hybridisation geometries in a row: linear sp at 180 degrees, trigonal planar sp2 at 120 degrees, tetrahedral sp3 at 109.5 degrees, trigonal bipyramidal sp3d, and octahedral sp3d2, each labelled with its steric number

Derivation

Derivation ASteric number fixes hybridisation and predicts the geometry, illustrated on phosphorus pentachloride

The goal is to show, starting from nothing but a Lewis structure and the definition of steric number in equation (1), that counting bonded regions plus lone pairs on a central atom is enough to name its hybridisation and predict its molecular geometry without any further assumption — then to check the prediction against the trigonal bipyramidal shape actually observed for $\text{PCl}_5$ , so the counting rule earns its keep on a molecule that needs a $d$ orbital, not just the familiar $s/p$ -only cases.

Step 1. Draw the Lewis structure of $\text{PCl}_5$ . Phosphorus, a period 3 element with 5 valence electrons, forms one single bond to each of 5 chlorine atoms. Every valence electron of phosphorus is used in a bond, so there are no lone pairs left on the central atom.

Step 2. Apply equation (1). The number of σ-bonded regions is 5 (one P–Cl σ bond to each chlorine — all single bonds, so each is trivially one region), and the number of lone pairs is 0. So $N = 5 + 0 = 5$ .

Step 3. Map $N=5$ to a hybridisation scheme. Following the $N \to$ scheme correspondence built from equation (1) ( $2\to sp$ , $3 \to sp^2$ , $4 \to sp^3$ , $5 \to sp^3d$ , $6\to sp^3d^2$ ), $N=5$ means phosphorus mixes 1 $s$ , 3 $p$ , and 1 $d$ orbital into 5 equivalent $sp^3d$ hybrid orbitals. This is only possible because phosphorus, like sulphur, is a period 3 element with an energetically accessible empty $3d$ orbital — the same reasoning that permits its expanded octet.

Step 4. Read off the geometry from the scheme. Five $sp^3d$ hybrid orbitals do not space out symmetrically the way four or six do; the lowest-repulsion arrangement is a trigonal bipyramid: three hybrid orbitals in an equatorial plane at $120°$ to each other, and two axial orbitals perpendicular to that plane, each $90°$ from every equatorial orbital and $180°$ from each other. Each of the 5 hybrid orbitals overlaps head-on with a chlorine $3p$ orbital, giving 5 equivalent σ bonds — matching the experimentally observed shape of $\text{PCl}_5$ and confirming that VSEPR (electron-pair repulsion) and hybridisation (orbital mixing) are two routes to the identical answer. Check: 3 equatorial + 2 axial = 5 hybrid orbitals = $N$ , consistent with 1 $s$ + 3 $p$ + 1 $d$ = 5 atomic orbitals mixed — the orbital count balances on both sides, as it must for any hybridisation scheme.

Worked example

Determine the hybridisation of nitrogen in ammonia, $\text{NH}_3$ , and explain why its bond angle is slightly less than the ideal tetrahedral value.

Step 1. Draw the Lewis structure. Nitrogen forms 3 single bonds to 3 hydrogen atoms and retains 1 lone pair (nitrogen has 5 valence electrons; 3 are used in the N–H bonds, leaving 2 as one lone pair).

Step 2. Apply equation (1): steric number $N = 3\ (\sigma\text{-bonded regions}) + 1\ (\text{lone pair}) = 4$ . By the $N \to$ scheme correspondence, $N=4$ means nitrogen is $sp^3$ hybridised — one $s$ orbital mixes with three $p$ orbitals to give 4 equivalent $sp^3$ hybrid orbitals arranged tetrahedrally.

Step 3. Assign the 4 hybrid orbitals: 3 hold the bonding pairs shared with hydrogen (each overlapping head-on with an H $1s$ orbital to form an N–H σ bond), and the 4th holds the lone pair.

Step 4. Explain the angle. If all four hybrid orbitals held bonding pairs, the H–N–H angle would be the ideal $sp^3$ value of $109.5°$ , exactly as in $\text{CH}_4$ . But a lone pair is held closer to the nitrogen nucleus by only one atom's attraction (unlike a bonding pair, shared between two nuclei), so it spreads out more and repels the three bonding pairs more strongly than they repel each other. This squeezes the three N–H bonds closer together. Answer: nitrogen in $\text{NH}_3$ is $sp^3$ hybridised, and the extra lone-pair repulsion compresses the bond angle from the ideal $109.5°$ to the observed $\approx107°$ .

Competitive stretch

Extend the same logic to ethyne, $\text{HC}{\equiv}\text{CH}$ , which has a carbon–carbon triple bond. Each carbon is bonded to 1 hydrogen and to the other carbon (1 σ-bonded region for the C–C triple bond, since only its σ component counts) — steric number $N = 1+1 = 2$ , so each carbon is $sp$ hybridised (linear, $180°$ ). The two $sp$ hybrid orbitals on each carbon form 1 C–H σ bond and 1 C–C σ bond; that leaves 2 unhybridised $p$ orbitals per carbon, mutually perpendicular, each overlapping sideways with its counterpart on the other carbon to give 2 π bonds. The C≡C triple bond is therefore 1 σ + 2 π, and the whole molecule is linear — consistent with the general rule that any multiple bond beyond the first σ is built entirely from unhybridised $p$ orbitals, never from extra hybrid orbitals.

Common traps

Counting a double or triple bond as more than one σ-bonded region in the steric number — it is always exactly 1, no matter how many π bonds ride along with it.
Forgetting that lone pairs occupy a hybrid orbital and count toward $N$ just as bonding pairs do — this is why $\text{NH}_3$ and $\text{H}_2\text{O}$ are both $sp^3$ despite having fewer than four bonds.
Assuming π bonds use hybrid orbitals — they never do; π bonds always come from sideways overlap of leftover, unhybridised $p$ orbitals.
Treating hybridisation and VSEPR as competing theories rather than two views of the same geometry — if they disagree on a structure, recheck the Lewis structure and the steric-number count, not the rule itself.

Ask the tutor: Want to try counting the steric number for a molecule of your own choosing, like $\text{SO}_2$ or $\text{XeF}_4$ , and check the hybridisation and shape you get against VSEPR?

Why does a double or triple bond only count as one region in the steric number?

Because the steric number counts σ-bonded regions, and every covalent bond — single, double, or triple — has exactly one σ bond as its backbone; any extra bonds in a double or triple bond are π bonds, built from leftover unhybridised $p$ orbitals, not from hybrid orbitals. Since hybridisation only concerns how many hybrid orbitals form and how they're arranged in space, and only σ bonds and lone pairs occupy hybrid orbitals, a triple bond contributes the same "1" to $N$ as a single bond does. This is exactly why ethyne's carbons come out $sp$ (steric number 2) even though each carbon has a triple bond plus a single bond.

If NH3 and CH4 are both sp3, why don't they look the same shape?

Hybridisation tells you the orbital geometry — how the four hybrid orbitals are arranged around the central atom — which is tetrahedral in both cases. But the molecular shape, the shape you'd draw from the positions of the atoms alone, depends on how many of those four hybrid orbitals hold bonding pairs versus lone pairs. $\text{CH}_4$ has all 4 as bonding pairs, so its molecular shape is tetrahedral. $\text{NH}_3$ has 3 bonding and 1 lone pair, so removing the "invisible" lone-pair vertex leaves a pyramidal molecular shape, even though the underlying $sp^3$ orbital arrangement is still tetrahedral.

Why can sulphur and phosphorus use d orbitals but carbon and nitrogen cannot?

Carbon and nitrogen are period 2 elements, and their outermost shell is $n=2$ , which has only $s$ and $p$ subshells — no $d$ subshell exists at that energy level for them to use. Phosphorus and sulphur are period 3, with outermost shell $n=3$ , which does have a $3d$ subshell close enough in energy to the $3s$ and $3p$ orbitals to mix in. That is why $sp^3d$ and $sp^3d^2$ hybridisation, and the expanded octets they go with (as in $\text{PCl}_5$ and $\text{SF}_6$ ), are only ever seen from period 3 onward.