Hybridisation: sp, sp², sp³, sp³d and sp³d²

What is hybridisation?

In plain words: an atom’s s and p orbitals point in awkward directions (the p orbitals are at 90°, the s orbital has no direction at all). We can describe the same orbital space using directional combinations, each with a large lobe pointing one way. This is not a separate physical event that happens before bonding. Mix one s with three p and you get four sp³ orbitals aimed at the corners of a tetrahedron, which is exactly the shape of methane.

Why hybridisation is needed

Simple overlap of pure orbitals puts the bonds made by p orbitals at 90° to one another in CH₄, NH₃ and H₂O (and leaves the direction of methane’s fourth, 2s-based bond undecided), because the three p orbitals are at right angles; the real angles are 109.5°, 107° and 104.5° (NCERT Section 4.5.3, pages 119–120). To explain these characteristic shapes, Pauling proposed that the atomic orbitals combine into a new set of equivalent orbitals, the hybrid orbitals, and that these, unlike pure orbitals, are used in bond formation (NCERT Section 4.6, page 120). For example, when one 2s and three 2p orbitals of carbon hybridise, four new sp³ hybrid orbitals form. The shapes of the s and p orbitals being mixed are shown on the page on shapes of atomic orbitals.

Features and conditions

Salient features (NCERT pages 120–121):

1. The number of hybrid orbitals equals the number of atomic orbitals that are hybridised.
2. Ideal sp, sp^² and sp^³ sets are equivalent; actual bond and lone-pair orbitals need not be identical, and axial/equatorial positions in a trigonal bipyramid differ.
3. Directional hybrids provide a useful localised picture of effective overlap. Changing the mathematical orbital basis does not itself lower a molecule’s energy.
4. They point in preferred directions that minimise repulsion between electron pairs, so the type of hybridisation indicates the geometry of the molecule.

Conditions (NCERT page 121):

• only orbitals of the valence shell are hybridised;
• the orbitals being mixed must have almost equal energy;
• promotion of an electron is not an essential condition before hybridisation;
• not only half-filled orbitals take part: in some cases even filled orbitals of the valence shell are hybridised (as the lone-pair orbitals of N in NH₃ and O in H₂O are).

Types of hybridisation: sp, sp², sp³

sp hybridisation (NCERT Section 4.6.1, page 121): one s and one p orbital mix to give two equivalent sp orbitals, each with 50% s and 50% p character, pointing in opposite directions along one axis (180°). It is also called diagonal hybridisation. Each sp hybrid has a large positive lobe and a very small negative lobe, which gives more effective overlap and stronger bonds. Example, BeCl₂: Be is 1s² 2s². In the excited state one 2s electron is promoted to a vacant 2p orbital; the 2s and that 2p orbital hybridise to two sp orbitals at 180°, and each overlaps axially with a 3p orbital of chlorine to form two Be–Cl σ bonds. Isolated gaseous BeCl₂ is linear; its condensed-phase structures are more complex.

sp² hybridisation (pages 121–122): one s and two p orbitals give three equivalent sp² orbitals, each with 33.3% s and 66.7% p character, in a plane at 120°. Example, BCl₃: B is 1s² 2s² 2p¹. Promoting one 2s electron leaves three unpaired electrons; the 2s and two 2p orbitals hybridise to three sp² orbitals, which overlap with 3p orbitals of chlorine to form three B–Cl bonds. BCl₃ is trigonal planar, Cl–B–Cl 120°. (NCERT’s text on pages 121–122 says 2p here, a slip: chlorine bonds through its half-filled 3p orbital; its 2p subshell is full and lies inside.)

sp³ hybridisation (page 122): one s and three p orbitals give four equivalent sp³ orbitals, each with 25% s and 75% p character, directed to the corners of a tetrahedron at 109.5°.

• CH₄: the four sp³ orbitals of carbon each overlap with an H 1s orbital: four C–H σ bonds, tetrahedral, 109.5°.
• NH₃: nitrogen (2s² 2p³) forms four sp³ orbitals; three hold one electron each and form N–H σ bonds, the fourth holds the lone pair. Lone pair–bond pair repulsion is greater than bond pair–bond pair repulsion, so the angle shrinks from 109.5° to 107°: pyramidal.
• H₂O: oxygen’s four sp³ orbitals hold two single electrons (for the two O–H bonds) and two lone pairs. Two corners of the tetrahedron are taken by H atoms and two by lone pairs; the angle falls to 104.5°: V-shaped (angular).

Hybridisation of carbon in ethane, ethene and ethyne

In ethane molecule both the carbon atoms assume sp^³ hybrid state.

NCERT Chemistry Part I (Class 11), Section 4.6.2, page 123

Ethane, C₂H₆ (sp³): one sp³ orbital of each carbon overlaps axially with the other’s to form the sp³–sp³ C–C σ bond; the other three on each carbon form sp³–s σ bonds with H. C–C is 154 pm and C–H 109 pm (NCERT page 123).

Ethene, C₂H₄ (sp²): one sp² orbital of each carbon forms the C–C σ bond; the other two form sp²–s σ bonds with H. The unhybridised 2p orbital on each carbon, perpendicular to the plane of the molecule, overlaps sideways with the other to form a weak π bond, with two equal electron clouds above and below the plane. So the C=C bond is one sp²–sp² σ bond plus one π bond, 134 pm long; C–H is 108 pm; H–C–H is 117.6° and H–C–C 121° (NCERT page 123).

Ethyne, C₂H₂ (sp): each carbon is sp hybridised and keeps two unhybridised p orbitals, 2p_x and 2p_y. One sp orbital of each carbon forms the C–C σ bond and the other forms a σ bond with H, so the molecule is linear. The two pairs of unhybridised p orbitals overlap sideways to form two π bonds: the triple bond is one σ and two π bonds (NCERT page 124). The C=C of every alkene and the C≡C of every alkyne are built the same way as in ethene and ethyne.

More s character holds the electrons closer to the nucleus, so bonds from sp² and sp carbons are shorter: C–H is 109 pm from sp³ carbon and 108 pm from sp² carbon (NCERT page 123); beyond the NCERT text, the C–H bond in ethyne (sp, 50% s) is shorter still, about 106 pm. The C–C bond also shortens, from 154 to 134 to about 120 pm, but mainly because of the extra π bonds.

Hybridisation involving d orbitals

The following is NCERT’s traditional d-hybridisation model for third-period molecules such as PCl₅ and SF₆. Modern bonding calculations do not require significant 3d participation: these labels are useful school-level geometry labels, not literal proof of electron promotion into 3d orbitals. The 3d orbitals are also comparable to 4s and 4p, which allows 3d, 4s, 4p hybridisation in transition-metal ions such as [Ni(CN)₄]²⁻ (NCERT Section 4.6.3, page 124). The important schemes:

Gaseous PCl₅, traditional sp³d description: phosphorus (Z = 15) is 3s² 3p³; in the excited state one 3s electron moves to a 3d orbital, leaving five half-filled orbitals (one s, three p, one d). These hybridise to five sp³d orbitals pointing to the corners of a trigonal bipyramid, and each overlaps with a singly occupied 3p orbital of a chlorine atom to form a P–Cl σ bond (NCERT pages 124–125). The bond angles are not all equal: three equatorial bonds lie in one plane at 120° to each other, and two axial bonds, one above and one below, are at 90° to that plane. The axial bond pairs suffer more repulsion from the equatorial bond pairs, so the axial bonds are slightly longer and weaker, which makes PCl₅ more reactive (page 125). Beyond the NCERT text: the rule that hybrids are equivalent holds for sp, sp², sp³ and sp³d², but not for sp³d. Its three equatorial and two axial orbitals differ, which is one more reason the axial and equatorial bonds are not the same.

NCERT’s orbital-box diagram for PCl₅ (page 124) is captioned as sp³d orbitals filled by electron pairs donated by the five Cl atoms. That would describe coordinate bonds, and it is a slip: each P–Cl bond is an ordinary covalent bond, with one electron from a half-filled sp³d orbital of phosphorus and one from the singly occupied orbital of chlorine, as NCERT’s own text on page 125 says.

SF₆, traditional sp³d² description: sulphur is 3s² 3p⁴; in the excited state one s, three p and two d orbitals are singly occupied. They hybridise to six sp³d² orbitals towards the corners of a regular octahedron, which overlap with singly occupied orbitals of six fluorine atoms: six S–F σ bonds, octahedral (page 125).

Beyond the NCERT text: modern calculations show that d orbitals contribute little to the bonding in molecules such as PCl₅ and SF₆. The sp³d and sp³d² descriptions are still the ones used in NCERT, JEE and NEET, and they give the right shapes.

Finding the hybridisation quickly

A common shortcut (beyond the NCERT text) for an atom in a molecule or ion:

A steric number of 2, 3, 4, 5 or 6 means sp, sp², sp³, sp³d or sp³d². Count every single, double or triple bond as one σ bond; π bonds use unhybridised p orbitals and do not count. Find lone pairs from a complete Lewis structure. For an individual atom, nonbonding electrons = its valence count − its formal charge − the sum of its bond orders; divide by two for lone pairs. Do not assign the overall ion charge to the central atom unless that is its formal charge; for example, S in SO₂ has 6 − 4 = 2 electrons left, one lone pair, so it is sp². For hypervalent main-group species, 5, 6 and 7 domains are traditionally labelled sp³d, sp³d² and sp³d³ (IF₇). Conjugated lone pairs are another limitation: an amide nitrogen is approximately planar/sp^² because its lone pair occupies a p orbital, despite a naive count of three bonds plus one lone pair. The shortcut does not decide between dsp² and sp³, or d²sp³ and sp³d², in complex ions; that needs the d-electron count, covered with coordination compounds in Class 12.

Solved examples

Common mistakes to avoid

Quick revision summary

• Hybridisation (Pauling): valence orbitals of nearly equal energy mix into an equal number of directional hybrid orbitals; ideal equivalent sets describe familiar geometries without implying a literal pre-bonding mixing process.
• sp: 2 orbitals, 180°, 50% s (BeCl₂, C₂H₂). sp²: 3, 120° (BCl₃, C₂H₄). sp³: 4, 109.5°, 25% s (CH₄, NH₃ 107°, H₂O 104.5°).
• Carbon: ethane sp³ (C–C 154 pm), ethene sp² (1σ + 1π, 134 pm), ethyne sp (1σ + 2π).
• Traditional main-group d-hybrid labels (modern bonding need not use significant d character): sp³d trigonal bipyramidal (PCl₅, axial bonds longer), sp³d² octahedral (SF₆), dsp² square planar.
• Steric number (σ bonds + lone pairs) 2 to 6 gives sp to sp³d².

Questions students ask

What is the difference between sp² and sp³ hybridisation?

In sp² hybridisation one s and two p orbitals mix into three hybrids in a plane at 120°, each with 33.3% s character, while in sp³ hybridisation one s and three p orbitals mix into four hybrids at 109.5°, pointing to the corners of a tetrahedron, each with 25% s character. An sp² atom keeps one unhybridised p orbital, perpendicular to the plane, which can form a π bond, so sp² carbon is found in double bonds (C=C in alkenes, C=O in aldehydes); an sp³ atom has no p orbital left over and forms only σ bonds, as carbon does in alkanes.

What is the hybridisation of carbon in CO₂?

The carbon atom in CO₂ is sp hybridised: it forms two σ bonds, one to each oxygen, and has no lone pair, so its steric number is 2 and the molecule is linear, O=C=O, with a bond angle of 180°. The two perpendicular sets of p orbitals form π systems delocalised over O–C–O. Assigning one p orbital exclusively to one oxygen is only a localised Lewis picture, not the full molecular-orbital description.

What is the hybridisation of XeF₄ and XeF₆?

In the traditional school model, XeF₄ is labelled sp³d² and XeF₆ sp³d³; these labels do not establish substantial d-orbital participation. In XeF₄ its 8 valence electrons form 4 Xe–F bonds and leave 2 lone pairs, a steric number of 6, so the six hybrid orbitals point to the corners of an octahedron; the lone pairs take opposite corners, and the molecule is square planar. In XeF₆, 6 bonds and 1 lone pair give a steric number of 7, and the molecule is a distorted octahedron.

How does hybridisation affect electronegativity?

The more s character a hybrid orbital has, the more electronegative the atom using it, so carbon’s electronegativity rises in the order sp³ (25% s) < sp² (33.3% s) < sp (50% s): an s orbital holds its electrons closer to the nucleus than a p orbital does. An sp carbon therefore pulls the electrons of a C–H bond towards itself more strongly, which is why the hydrogen atoms of ethyne are weakly acidic: ethyne reacts with sodium to give sodium ethynide and hydrogen, while ethene and ethane do not.

What is the difference between sp³ and dsp² hybridisation?

Both sp³ and dsp² hybridisation give four hybrid orbitals, but sp³ mixes one s and three p orbitals into a tetrahedral set (109.5°), while dsp² mixes one inner d orbital, of the shell below, with one s and two p orbitals into a square planar set (90°). The difference matters for complexes of nickel(II), studied in Class 12: [NiCl₄]²⁻ is sp³, tetrahedral and paramagnetic, with two unpaired electrons, while [Ni(CN)₄]²⁻ is dsp², square planar and diamagnetic.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp² and sp³ hybrid orbitals (NCERT Exercise 4.24, page 134).

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   Hybridisation is the intermixing of orbitals of slightly different energies on one atom to give an equal number of new orbitals of equal energy and shape. Each hybrid has one large lobe and a very small one. Two sp hybrids point in opposite directions (linear, 180°); three sp² hybrids lie in a plane at 120° (trigonal planar); four sp³ hybrids point to the corners of a tetrahedron at 109.5°.

2. Q2. Is there any change in the hybridisation of B and N atoms as a result of the reaction BF₃ + NH₃ → F₃B.NH₃ (NCERT Exercise 4.26, page 135)?

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   Boron in BF₃ has three σ bonds (sp², trigonal planar); after accepting nitrogen’s lone pair it has four σ bonds (sp³, tetrahedral). Nitrogen is sp³ in NH₃ (three σ bonds + one lone pair) and stays sp³ in the adduct (four σ bonds). So only boron changes.

3. Q3. Describe how the double bond in C₂H₄ and the triple bond in C₂H₂ are formed (based on NCERT Exercise 4.27, page 135).

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   C₂H₄: each carbon is sp²; sp²–sp² overlap gives the C–C σ bond, and the unhybridised 2p orbitals, perpendicular to the molecular plane, overlap sideways to give one π bond (clouds above and below the plane). C₂H₂: each carbon is sp; sp–sp overlap gives the σ bond, and the two unhybridised p orbitals on each carbon, at right angles, overlap sideways to give two π bonds. The figure above shows both.