Chemical Bonding and Molecular Structure · Lesson 8 of 12
Hybridisation and the Shapes of Covalent Molecules
“Connect hybrid orbitals with molecular geometry and build the sigma–pi bonding descriptions of common molecules.”
• Explain hybridisation, its conditions, and the relationship between the number of input and output orbitals. • Describe sp, sp², and sp³ hybrids using their composition, direction, and examples. • Use hybrid and unhybridised orbitals to explain bonding in ethane, ethene, and ethyne. • Relate conventional d-orbital hybridisation schemes to the geometries discussed in the chapter. • Explain axial and equatorial bonds in PCl₅ and the octahedral arrangement of SF₆.
Equivalent bonds need an appropriate orbital description
Methane’s four C–H bonds are equivalent and point towards tetrahedral corners. Treating one bond as a pure s overlap and three as pure p overlaps does not describe that pattern well. Hybridisation expresses orbitals on the same atom as combinations directed towards suitable bonding regions.
The mixing, as an orbital model, of atomic orbitals on the same atom to form a new set of directed hybrid orbitals suited to describing bonds and lone pairs.
The chapter describes mixing orbitals of suitable, reasonably comparable energies. The number of hybrids equals the number of input orbitals. Hybridisation does not add electrons, turn one atom into several atoms, or describe overlap between different atoms. Interatomic overlap occurs when a hybrid orbital forms a bond with an orbital on a neighbouring atom.
A set of equivalent hybrids is equivalent in its ideal orbital construction. The complete molecular environment can make different positions or bonds unequal, as in axial and equatorial PCl₅ bonds. Hybrid direction and molecular shape must also be distinguished when some hybrids contain lone pairs.
Conditions and electron occupancy
The orbitals selected for the model must belong to the same central atom and have suitable energies and symmetry. Orbitals need not all be half-filled: a filled orbital can supply a lone pair in the hybrid set. Electron promotion is not an obligatory separate step for every hybridisation example.
For carbon, the familiar promoted valence configuration supplies four singly occupied orbitals in the simple valence bond construction. Its energy cost is part of the overall bonding energy balance. In ammonia and water, filled and singly occupied orbitals are combined without requiring the same promotion picture.
Linear and trigonal planar hybrids
Mixing one s orbital and one p orbital produces two sp hybrids. Their large lobes point in opposite directions, giving a linear arrangement with a 180° angle. Each hybrid contains 50% s and 50% p character, and two p orbitals remain unhybridised if we started with a full s-plus-three-p valence set.
Problem
Describe the simple orbital model of gaseous BeCl₂.
- 1.Beryllium’s valence configuration is . In the chapter’s valence bond construction, promotion gives one electron in 2s and one in 2p.
- 2.Mixing these two orbitals gives two singly occupied sp hybrids directed 180° apart.
- 3.Each hybrid overlaps head-on with a singly occupied orbital on chlorine, forming a sigma bond. The gaseous molecule is therefore described as linear in this simple molecular model.
Mixing one s and two p orbitals produces three sp² hybrids. Their large lobes lie in one plane, 120° apart. Each has one-third s character and two-thirds p character. The remaining p orbital points perpendicular to that plane and can participate in pi bonding.
Problem
Why does the sp² model fit BCl₃?
- 1.Boron has valence configuration . The chapter uses a promoted arrangement with three unpaired electrons.
- 2.One s and two p orbitals mix into three singly occupied sp² hybrids.
- 3.Head-on overlaps with chlorine orbitals form three sigma bonds in a trigonal plane. Boron counts six surrounding bonding electrons in the Lewis model, consistent with its incomplete-octet example.
Tetrahedral hybrids, bonds, and lone pairs
One s and three p orbitals give four sp³ hybrids, with 25% s character and 75% p character in each. Their large lobes point towards tetrahedral corners at ideal angles of about 109.5°. These directed lobes provide effective overlap with neighbouring orbitals.
Problem
Use four hybrids to describe three different molecular shapes.
- 1.In CH₄, four singly occupied carbon sp³ hybrids overlap with four H 1s orbitals, giving four equivalent C–H sigma bonds and a tetrahedral molecule.
- 2.Nitrogen in NH₃ has five valence electrons. In the sp³ description, one hybrid contains a lone pair and three contain one electron each. The three singly occupied hybrids form N–H sigma bonds.
- 3.The lone pair is not an atom, so NH₃ is trigonal pyramidal. Lone-pair repulsion helps compress the angle to about 107°.
- 4.Oxygen in H₂O has six valence electrons: two hybrids contain lone pairs and two are available for O–H sigma bonds. The resulting molecular shape is bent, with angle about 104.5°.
| Hybrid set | Input orbitals | Number of hybrids | Ideal directions | s character |
|---|---|---|---|---|
| sp | One s + one p | 2 | Linear, 180° | 50% |
| sp² | One s + two p | 3 | Trigonal planar, 120° | 33⅓% |
| sp³ | One s + three p | 4 | Tetrahedral, about 109.5° | 25% |
The “1s + 2p” wording in this table counts one s-type orbital and two p-type orbitals; it does not mean mixing a first-shell 1s orbital with second-shell 2p orbitals. The principal shell is specified separately in actual electron configurations, such as carbon’s 2s and 2p orbitals.
Ethane: the single-bond framework
In ethane, both carbon atoms are described as sp³ hybridised. One hybrid on each carbon overlaps head-on with the corresponding hybrid on the other carbon to make the C–C sigma bond. The remaining three hybrids on each carbon overlap with H 1s orbitals.
Ethane therefore contains one C–C sigma bond and six C–H sigma bonds, seven sigma bonds in total and no pi bonds. Its C–C distance is about 154 pm and the chapter gives C–H distances near 109 pm. The geometry around each carbon is approximately tetrahedral, even though the whole molecule contains more than one central atom.
Ethene and ethyne: keep some p orbitals unmixed
In ethene, each carbon uses sp² hybrids for its three sigma directions: two towards hydrogen and one towards the other carbon. Each carbon also retains one unhybridised p orbital perpendicular to this sigma-bond plane. The two parallel p orbitals overlap sideways to form one pi bond.
The C=C bond therefore contains one sp²–sp² sigma bond and one p–p pi bond. Each C–H bond is sp²–1s sigma overlap. The chapter gives a C=C distance near 134 pm and C–H near 108 pm; the angles are close to trigonal planar, with measured H–C–H and H–C–C angles differing slightly from 120°.
In ethyne, each carbon uses sp hybrids along the straight H–C–C–H axis. One sp–sp overlap forms the C–C sigma bond and the other hybrid on each carbon forms a C–H sigma bond. Each carbon retains two mutually perpendicular p orbitals. Matching p orbitals overlap sideways in two perpendicular planes, creating two pi bonds.
Problem
Count sigma and pi bonds in ethene and ethyne, including the C–H bonds.
- 1.Ethene has four C–H single bonds and one C=C double bond. Its sigma total is , and it has one pi bond.
- 2.Ethyne has two C–H single bonds and one C≡C triple bond. Its sigma total is , and it has two pi bonds.
- 3.Do not count the double bond as two sigma bonds or the triple bond as three. There is one internuclear sigma framework for each linked carbon pair in this localised description.
If a p orbital has been included in the hybrid set, it is not also an independent unhybridised p orbital. An sp² carbon retains one p orbital for pi bonding; an sp carbon retains two. Count the original orbitals and the resulting directions consistently.
Conventional schemes involving d orbitals
The chapter extends its hybridisation language to geometries with five or six domains and to some square planar coordination examples. In the conventional construction, one s, three p, and one d orbital give an sp³d set; adding a second d orbital gives an sp³d² set. Other labels specify a different selected orbital set.
| Conventional label | Orbital count | Geometry or arrangement | Source examples |
|---|---|---|---|
| dsp² | 1d + 1s + 2p = 4 | Square planar | , |
| sp³d | 1s + 3p + 1d = 5 | Trigonal bipyramidal | , |
| sp³d² | 1s + 3p + 2d = 6 | Octahedral electron arrangement; square pyramidal if one site is a lone pair | |
| sp³d² | 1s + 3p + 2d = 6 | Octahedral | , |
| d²sp³ | 2d + 1s + 3p = 6 | Octahedral |
The source contrasts sets such as 3s, 3p, 3d and 3d, 4s, 4p when discussing compatible orbital energies. These conventional labels organise geometrical models. A label alone is not experimental proof that substantial d-orbital participation is the physical cause of hypervalent bonding; the electron distribution requires a more complete quantum description.
Phosphorus pentachloride and sulfur hexafluoride
The chapter’s PCl₅ construction starts from phosphorus with valence arrangement and uses a promoted, five-unpaired-electron arrangement to build five directed bonding orbitals. The conventional sp³d description points to the five corners of a trigonal bipyramid.
Three P–Cl bonds are equatorial and 120° apart. Two are axial, perpendicular to the equatorial plane and 180° apart from each other. Each axial bonding domain has three close 90° interactions with equatorial domains, while an equatorial domain has two such interactions with axial domains. The axial bonds are consequently longer and weaker in the chapter’s repulsion explanation.
For SF₆, sulfur has valence configuration . The conventional promoted-orbital construction supplies six singly occupied orbitals, combined as sp³d². Six equivalent S–F directions point to octahedral vertices: neighbouring directions are at 90° and opposite directions at 180°. Unlike PCl₅, the ideal octahedral molecule has one equivalent set of S–F bonds.
Problem
Compare Al in AlCl₃ and in the product of .
- 1.In a simple isolated-molecule description, AlCl₃ has three bonding directions and is trigonal planar, described as sp².
- 2.The product has four Al–Cl bonding directions and a tetrahedral arrangement around Al, described as sp³.
- 3.The change concerns the orbital description of the bonded atom. It is not an alteration of aluminium’s identity or atomic number.
Problem
Explain the bonding change when BF₃ accepts a lone pair from NH₃.
- 1.In the separate molecules, boron has three bonding directions and is described as sp², while nitrogen has three bonds plus a lone pair in an sp³ description.
- 2.Nitrogen supplies both electrons of its lone pair to a new B–N shared pair. No new electron is created: the original lone pair becomes a bonding pair.
- 3.Boron now has four bonded directions and is described as sp³. Nitrogen retains four domains but now all four correspond to bonds, so it remains sp³ in this model.
- 4.The example shows why it is not a universal requirement that each atom contribute exactly one electron to every covalent bond. Electron-pair donation can produce a covalent shared pair as well.
Check your understanding
Track three counts independently: the input orbitals, the hybrid directions, and the electrons or lone pairs occupying them. Then identify the sigma and pi overlaps formed between atoms.
Quiz
Mixing one s and two p orbitals produces:
How many unhybridised p orbitals remain on each sp carbon in ethyne?
How many sigma and pi bonds are in ethene?
Why is ammonia pyramidal in the sp³ description?
Why are axial P–Cl bonds longer in the chapter’s PCl₅ model?
Practice Problems
- Describe the input and output orbitals for sp, sp², and sp³ hybridisation, including ideal directions and s character.
- Use BeCl₂, BCl₃, and CH₄ to explain the link between the hybrid set and molecular geometry.
- Explain the differences between the electron-domain arrangements and molecular shapes of CH₄, NH₃, and H₂O.
- Draw sigma frameworks and the remaining p orbitals for ethene and ethyne. Count all sigma and pi bonds.
- Identify the hybridisation of each carbon in ethane, propene, ethanol, ethanal, and acetic acid using the local bonding directions.
- Explain the conventional hybridisation changes when BF₃ and NH₃ form an F₃B–NH₃ adduct.
- Compare axial and equatorial positions in PCl₅ and explain why SF₆ does not have two such inequivalent bond sets.
- Relate the conventional labels dsp², sp³d, sp³d², and d²sp³ to the geometries listed in the chapter.
Key Takeaways
• Hybridisation mixes orbital descriptions on one atom; bonding overlap connects different atoms. • sp, sp², and sp³ sets provide two, three, and four directed hybrid orbitals. • Lone pairs can occupy hybrid orbitals and change molecular shape. • Unhybridised p orbitals provide the pi interactions in ethene and ethyne. • Conventional d-orbital schemes organise additional geometries; PCl₅ has unequal axial and equatorial bonds, while SF₆ is ideally octahedral.