What hybridisation means
The shapes of polyatomic molecules such as $\ce{CH4}$, $\ce{NH3}$ and $\ce{H2O}$ cannot be read off directly from the pure $s$ and $p$ orbitals of the central atom. To account for them, Pauling introduced the idea that the atomic orbitals of an atom combine to form a new set of equivalent orbitals, the hybrid orbitals. Unlike the pure orbitals, it is these hybrid orbitals that are actually used in bond formation.
Formally, hybridisation is the process of intermixing the orbitals of slightly different energies belonging to the same atom so as to redistribute their energies, resulting in a new set of orbitals of equivalent energy and shape. When one $2s$ and three $2p$ orbitals of carbon hybridise, for instance, four new $sp^3$ hybrid orbitals are produced. The number of hybrid orbitals always equals the number of atomic orbitals that entered the mix.
Each hybridisation directs its hybrid orbitals towards fixed corners so that electron-pair repulsion is minimised. The geometries below are the skeletons every example in this note hangs on.
Salient features and conditions
The power of the model lies in a small set of rules. The hybrid orbitals are equivalent in energy and shape, they form stronger bonds than the pure orbitals, and they orient themselves so as to keep electron-pair repulsion minimal — which is precisely why the type of hybridisation fixes the geometry of the molecule.
| Salient feature | What it tells you |
|---|---|
| Number conservation | The number of hybrid orbitals equals the number of atomic orbitals that get hybridised. |
| Equivalence | The hybridised orbitals are always equivalent in energy and shape. |
| Bonding strength | Hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals. |
| Directionality | Hybrid orbitals point in preferred directions to minimise electron-pair repulsion, fixing the geometry. |
Four conditions govern when hybridisation occurs. Only the orbitals present in the valence shell are hybridised, and they should be of nearly equal energy. The promotion of an electron is not an essential prerequisite, and it is not necessary that only half-filled orbitals participate — in some cases even filled valence-shell orbitals take part. This last point matters because it allows lone pairs, which sit in filled or singly filled orbitals, to occupy hybrid orbitals.
sp hybridisation (linear)
sp hybridisation mixes one $s$ and one $p$ orbital to give two equivalent $sp$ hybrid orbitals, each with 50% $s$-character and 50% $p$-character. The two hybrids point in opposite directions along one axis, giving an angle of $180°$ and a linear geometry; this is also called diagonal hybridisation.
The standard example is $\ce{BeCl2}$. The ground-state configuration of beryllium is $\ce{1s^2 2s^2}$; in the excited state one $2s$ electron is promoted to a vacant $2p$ orbital to account for the divalency. One $2s$ and one $2p$ orbital then hybridise into two $sp$ orbitals at $180°$, each overlapping axially with a $2p$ orbital of chlorine to form two $\ce{Be-Cl}$ sigma bonds.
sp² hybridisation (trigonal planar)
sp² hybridisation involves one $s$ and two $p$ orbitals, producing three equivalent $sp^2$ hybrid orbitals arranged in a trigonal planar geometry at $120°$ to one another. The classic case is $\ce{BCl3}$: ground-state boron is $\ce{1s^2 2s^2 2p^1}$, and after promotion of one $2s$ electron the three orbitals (one $2s$, two $2p$) hybridise to three $sp^2$ orbitals that overlap with chlorine to give three $\ce{B-Cl}$ bonds with a $\ce{Cl-B-Cl}$ angle of $120°$.
$\ce{BCl3}$ is flat with the boron in the same plane as the three chlorines; $\ce{CH4}$ pushes its four bonds into three dimensions to reach the tetrahedral angle.
sp³ hybridisation (tetrahedral)
sp³ hybridisation mixes one $s$ orbital and three $p$ orbitals to form four $sp^3$ hybrid orbitals, each with 25% $s$-character and 75% $p$-character, directed towards the four corners of a tetrahedron at $109.5°$. In $\ce{CH4}$ the four hybrids overlap with the $1s$ orbitals of four hydrogen atoms, giving a perfectly tetrahedral molecule.
The same hybridisation explains $\ce{NH3}$ and $\ce{H2O}$, but here lone pairs enter the picture. In ammonia, three of the $sp^3$ orbitals hold bond pairs while the fourth holds a lone pair; because lone-pair repulsion exceeds bond-pair repulsion, the molecule is distorted and the angle falls to $107°$, giving a pyramidal shape. In water, two $sp^3$ orbitals hold lone pairs, compressing the angle further to $104.5°$ and producing the angular or V-shape.
Lone pairs count towards hybridisation, not towards the visible shape
A frequent error is to count only the bonded atoms. The hybridisation is set by the steric number — bond pairs plus lone pairs. $\ce{NH3}$ has three bonds but is $sp^3$, not $sp^2$, because the lone pair occupies a fourth hybrid orbital. Likewise $\ce{H2O}$ is $sp^3$ with two lone pairs. The lone pairs are invisible in the reported shape (pyramidal, bent) yet they govern both the hybridisation and the squeezed bond angle.
$\ce{CH4}$, $\ce{NH3}$ and $\ce{H2O}$ are all $sp^3$; the angle order $109.5° > 107° > 104.5°$ tracks the rising number of lone pairs.
sp³d hybridisation (trigonal bipyramidal)
From the third period onward the central atom has $d$ orbitals whose energies are comparable to the $s$ and $p$ orbitals of the same shell, allowing the octet to expand. sp³d hybridisation mixes one $s$, three $p$ and one $d$ orbital to give five hybrid orbitals directed towards the corners of a trigonal bipyramid.
The benchmark molecule is $\ce{PCl5}$. Phosphorus ($Z=15$) promotes an electron so that five orbitals (one $s$, three $p$, one $d$) become available, yielding five $sp^3d$ orbitals. Crucially, the bond angles are not all equal: three equatorial $\ce{P-Cl}$ bonds lie in one plane at $120°$, while two axial bonds lie at $90°$ to that plane.
Axial bond pairs face more repulsion from the equatorial bonds, so the axial $\ce{P-Cl}$ bonds are slightly longer and weaker — the structural reason $\ce{PCl5}$ is reactive.
sp³d² hybridisation (octahedral)
sp³d² hybridisation mixes one $s$, three $p$ and two $d$ orbitals to give six hybrid orbitals pointing to the corners of a regular octahedron, all bond angles $90°$. In $\ce{SF6}$ the sulphur ground state is $\ce{3s^2 3p^4}$; in the excited state six orbitals (one $s$, three $p$, two $d$) become singly occupied, hybridise to six $sp^3d^2$ orbitals and overlap with six fluorine atoms to give six equivalent $\ce{S-F}$ sigma bonds.
The same scheme also produces square pyramidal species such as $\ce{BrF5}$, where one of the six hybrid orbitals carries a lone pair instead of a bond pair. Coordination complexes extend this further: $\ce{[CrF6]^3-}$ uses $sp^3d^2$ while $\ce{[Co(NH3)6]^3+}$ uses the inner-orbital $d^2sp^3$ set, both octahedral.
Master table of hybridisation
The five schemes can be compressed into a single reference. Memorise the steric number against the hybridisation and the geometry follows automatically.
| Steric no. | Hybridisation | Orbitals mixed | Geometry | Bond angle | Examples |
|---|---|---|---|---|---|
| 2 | sp | s + p | Linear | 180° | $\ce{BeCl2}$, $\ce{C2H2}$ |
| 3 | sp² | s + p(2) | Trigonal planar | 120° | $\ce{BCl3}$, $\ce{BF3}$, $\ce{C2H4}$ |
| 4 | sp³ | s + p(3) | Tetrahedral | 109.5° | $\ce{CH4}$, $\ce{NH3}$, $\ce{H2O}$ |
| 5 | sp³d | s + p(3) + d | Trigonal bipyramidal | 120° & 90° | $\ce{PCl5}$, $\ce{PF5}$ |
| 6 | sp³d² | s + p(3) + d(2) | Octahedral | 90° | $\ce{SF6}$, $\ce{[CrF6]^3-}$ |
Two further $d$-orbital sets appear in coordination chemistry but follow the same logic: $\ce{dsp^2}$ gives square planar species such as $\ce{[Ni(CN)4]^2-}$ and $\ce{[PtCl4]^2-}$, while $\ce{d^2sp^3}$ gives octahedral inner-orbital complexes such as $\ce{[Co(NH3)6]^3+}$. The geometry depends only on the number of hybrid orbitals, not on whether the $d$ orbitals are mixed in before or after the $s$ and $p$ orbitals.
The steric-number method
For NEET speed, do not draw the molecule — compute the steric number. It is the number of sigma bonds on the central atom plus the number of lone pairs, and it maps directly onto the hybridisation in the master table. For many main-group species the steric number is found without a Lewis structure using
$$H = \tfrac{1}{2}\,(V + M - c + a)$$where $V$ = valence electrons of the central atom, $M$ = number of monovalent surrounding atoms (e.g. H, F, Cl), $c$ = charge if the species is a cation, and $a$ = charge if it is an anion. $H = 2$ is $sp$, $3$ is $sp^2$, $4$ is $sp^3$, $5$ is $sp^3d$, $6$ is $sp^3d^2$.
Find the hybridisation of the central atom in $\ce{SF6}$.
Sulphur has $V = 6$ valence electrons. There are $M = 6$ monovalent fluorine atoms, and the molecule is neutral so $c = a = 0$.
$H = \tfrac{1}{2}(6 + 6 - 0 + 0) = \tfrac{1}{2}(12) = 6.$
A steric number of $6$ means $sp^3d^2$ — octahedral, consistent with the six $\ce{S-F}$ sigma bonds and no lone pair.
Find the hybridisation of nitrogen in $\ce{NH3}$ and verify the lone pair.
Nitrogen has $V = 5$. There are $M = 3$ hydrogen atoms; the species is neutral so $c = a = 0$.
$H = \tfrac{1}{2}(5 + 3 - 0 + 0) = \tfrac{1}{2}(8) = 4.$
$H = 4$ means $sp^3$. Three of the four hybrid orbitals form $\ce{N-H}$ bonds and the fourth holds a lone pair — exactly why the geometry is pyramidal at $107°$ rather than the ideal $109.5°$.
The $\tfrac{1}{2}(V+M-c+a)$ formula sign on charge
Add the magnitude of the charge for an anion ($+a$) and subtract it for a cation ($-c$). Reversing these gives a wrong steric number and a wrong hybridisation. Also, $M$ counts only monovalent terminal atoms; oxygen bonded by a double bond does not add to $M$. When in doubt for polyatomic ions, fall back to counting sigma bonds plus lone pairs directly.
Anion: $+a$. Cation: $-c$. Only monovalent atoms enter $M$.
Hybridisation in one screen
- Hybridisation intermixes valence orbitals of nearly equal energy into equivalent hybrid orbitals; their number equals the orbitals mixed.
- $sp \to$ linear $180°$ ($\ce{BeCl2}$); $sp^2 \to$ trigonal planar $120°$ ($\ce{BCl3}$); $sp^3 \to$ tetrahedral $109.5°$ ($\ce{CH4}$).
- $sp^3d \to$ trigonal bipyramidal ($\ce{PCl5}$, axial $90°$, equatorial $120°$); $sp^3d^2 \to$ octahedral $90°$ ($\ce{SF6}$).
- Lone pairs occupy hybrid orbitals: $\ce{NH3}$ and $\ce{H2O}$ are $sp^3$ with angles squeezed to $107°$ and $104.5°$.
- Compute the steric number with $H=\tfrac12(V+M-c+a)$: $2,3,4,5,6 \to sp, sp^2, sp^3, sp^3d, sp^3d^2$.