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Chemical Bonding and Molecular Structure · TOPIC LESSON

Molecular Geometry and Hybridization

Build the electron-domain arrangement, then name the shape made by the atoms.

JEE Main · JEE Advanced · NEET UG38 min study guide9 indexed sections

BY THE END, YOU CAN

  • Distinguish electron-domain geometry from molecular geometry.
  • Predict shapes for common two- through six-domain main-group species.
  • Place lone pairs appropriately in trigonal-bipyramidal and octahedral arrangements.
  • Relate sp, sp² and sp³ descriptions to local sigma bonding and remaining p orbitals.
  • Use geometry to reason about bond angles and dipole cancellation.
  • Recognize the limits of VSEPR and conventional hypervalent hybridization labels.

BEFORE YOU START

  • Lewis structures, valence counts and formal charges.
  • Sigma and pi bonds.
  • Basic three-dimensional geometry.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

Available in this lesson: JEE Advanced 2016. This is a limited lesson sample, not a complete exam-question bank.

1 sourced sample question available across the listed exams and years.

Matched topics: Molecular Geometry and Hybridization.

JEE Advanced · 2016 · Paper 1 · Code 1 · Chemistry · Q27

Paraphrased exam question

Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.

Which have exactly two lone pairs on the central atom: BrF₅, ClF₃, XeF₄, SF₄?

Reveal answer & worked solution

Answer: ClF₃ and XeF₄ (independently derived; archived B,C).

  1. For these neutral fluorides, central lone pairs = (central valence electrons − number of single bonds)/2.
  2. BrF₅: (7−5)/2=1; ClF₃: (7−3)/2=2.
  3. XeF₄: (8−4)/2=2; SF₄: (6−4)/2=1. Thus only ClF₃ and XeF₄ qualify.
  4. VSEPR then gives ClF₃: AX₃E₂, T-shaped; XeF₄: AX₄E₂, square planar.

Source: JEE (Advanced) 2016, Paper 1, Q27; independently derived; key finality unverified. Solution written for this lesson.

01 / UNDERSTAND & APPLY

1. A flat formula is not a molecular shape

A structural formula printed on a page is a connectivity map. It does not imply that all atoms lie in the page. Molecular geometry describes the positions of nuclei in three-dimensional space; electron-domain geometry describes the arrangement of bonding directions and lone-pair regions around a chosen central atom. Keeping those two descriptions separate prevents the most common naming error in this chapter.

In VSEPR, regions of valence electron density adopt arrangements that reduce unfavorable crowding. The model is a qualitative structural guide, not a complete calculation of all molecular forces. A bond to one neighboring atom counts as one domain whether it is single, double or triple. A lone pair on the central atom counts as another domain. Lone pairs on terminal atoms are not added to the central atom’s domain count.

Use AX_mE_n notation when helpful: A is the central atom, X represents bonded atoms and E represents lone pairs on A. The domain count is m+n. The notation does not say that all ligands have identical chemical identities; if they differ, angles and dipole cancellation can differ from the highly symmetric ideal example.

Before assigning a shape, construct a valid Lewis picture and check total charge. Forgetting an electron in an anion can remove a lone pair and produce the wrong entire geometry. Multiple bonds should be represented correctly for valence accounting even though they count as one domain for the initial geometric arrangement.

02 / UNDERSTAND & APPLY

2. Two and three domains: linear and planar foundations

Two domains lie on opposite sides of a central atom in the ideal arrangement, giving a linear electron-domain geometry and a 180° separation. CO₂ has two bonding domains around carbon and no carbon lone pairs, so its molecular geometry is linear. A triple bond also counts as one domain: carbon in H–C≡N has two directions, toward H and toward N.

Three domains occupy a trigonal plane with ideal 120° separations. In BF₃, three bonds and no boron lone pair give trigonal-planar molecular geometry. The electron deficiency of boron does not prevent this domain classification. In a three-domain AX₂E species, one position is a lone pair, so the atomic shape is bent rather than trigonal planar.

SO₂ illustrates the distinction: two S–O directions and one sulfur lone pair give three domains in the usual VSEPR description. Its atoms form a bent shape even though the electron-domain arrangement is based on a trigonal plane. Resonance-equivalent bonding should not be interpreted as the molecule alternating between two shapes.

Ideal angles describe a symmetric reference. A multiple-bond domain often occupies more angular space than a single-bond domain, and a lone pair can compress the angle between bonded atoms. VSEPR gives useful trends, but precise angles require experimental or computational information. Writing exactly 120° for every three-domain molecule overstates the model.

ORIGINAL WORKED EXAMPLE

Count directions, not lines

Predict the geometry around carbon in H–C≡N and around sulfur in SO₂ using domain counting.

  1. HCN carbon has two bonded neighbors. The C≡N triple bond is one domain, and C has no lone pair: AX₂, linear.
  2. SO₂ sulfur has two bond directions and one lone pair in the usual description: AX₂E.
  3. Its three domains have a trigonal-planar reference arrangement, but the two oxygen nuclei form a bent molecular shape.

Answer HCN is linear around carbon; SO₂ is bent around sulfur.

03 / UNDERSTAND & APPLY

3. Four domains do not always mean a tetrahedral molecule

Four equivalent directions form a tetrahedron, with ideal angle arccos(−1/3) ≈ 109.5°. CH₄ is AX₄, so both electron-domain and molecular geometry are tetrahedral. A tetrahedron has four vertices, not four atoms arranged in a square around the center. A square planar drawing is a different spatial arrangement with different angles.

NH₃ has three bonds and one lone pair: AX₃E. Its electron-domain arrangement is approximately tetrahedral, while the hydrogen positions define a trigonal pyramid. H₂O has two bonds and two lone pairs: AX₂E₂. Its electron-domain arrangement is likewise approximately tetrahedral, but its molecular shape is bent.

For the familiar CH₄, NH₃ and H₂O comparison, bond angles decrease from about 109.5° to 107° to 104.5°. The qualitative lone-pair crowding model rationalizes that progression. Do not use those exact values as universal angles for every AX₃E or AX₂E₂ species: ligand electronegativity, central-atom size and orbital character also matter.

NH₄⁺ is tetrahedral because it has four N–H bonds and no nitrogen lone pair. Protonation of ammonia uses the lone pair to form a bond, changing the atomic geometry classification from trigonal pyramidal to tetrahedral. The positive charge alone does not dictate the geometry; the resulting electron count does.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

NH₃ and H₂O: domains versus atomsWedge and hashed-wedge projections distinguish NH3 with three hydrogen atoms and one lone pair from H2O with two hydrogen atoms and two lone pairs. Both have tetrahedral electron-domain references; their molecular shapes are pyramidal and bent.Electron domains and atom shapesNH₃ · AX₃Elone pairHHHNDomain reference: tetrahedralAtoms: trigonal pyramidalH−N−H ≈ 107°H₂O · AX₂E₂two lone pairsHHODomain reference: tetrahedralAtoms: bentH−O−H ≈ 104.5°Solid wedge: toward. Hashed wedge: away.Schematic projection; lone-pair lobes are not atoms.
Schematic projection; lone-pair lobes are electron regions, not atoms. Approximate measured bond angles are labeled separately.

Turn the molecule. Keep the geometry.

The static diagram above remains available. This optional model lets you inspect atom arrangements from different viewpoints.

Interactive rotation loads when this section is in view. With JavaScript disabled, use the labeled static diagram above.

Idealized molecular geometry; the static figure distinguishes electron domains from atom positions.

Drag with a mouse or touch. Focus the model and use arrow keys, or use the direction buttons. Rotation is manual; there is no automatic motion.

ORIGINAL WORKED EXAMPLE

Compare an isoelectronic family

Classify CH₄, NH₃, H₂O and NH₄⁺ by domains and molecular shape.

  1. CH₄: four bonds, no lone pairs, AX₄; tetrahedral.
  2. NH₃: three bonds, one lone pair, AX₃E; trigonal pyramidal.
  3. H₂O: two bonds, two lone pairs, AX₂E₂; bent.
  4. NH₄⁺: four bonds, no lone pairs, AX₄; tetrahedral.
  5. All have four central electron domains in the elementary picture, but their atomic shapes need not match.

Answer Electron-domain arrangement: tetrahedral for all four; molecular shapes differ as listed.

04 / UNDERSTAND & APPLY

4. Five domains have unequal axial and equatorial sites

Five domains adopt a trigonal-bipyramidal reference arrangement: three equatorial positions form a plane at 120° to one another, while two axial positions lie above and below that plane. An axial direction is 90° from all three equatorial directions; an equatorial direction has only two 90° interactions, with the axial pair. The two types of site are therefore not geometrically equivalent.

A lone pair normally prefers an equatorial position in this simple VSEPR setting because it has fewer 90° interactions. For AX₄E, exemplified by SF₄, the remaining atom positions form a seesaw. For AX₃E₂, exemplified by ClF₃, two equatorial lone pairs leave a T-shaped arrangement of atoms. The angles are distorted from the ideal reference by unequal interactions.

For AX₂E₃, the three lone pairs occupy the equatorial plane, leaving the two bonded atoms axial and opposite. XeF₂ is therefore linear despite having five domains around xenon. It should not be called “trigonal bipyramidal” when the question asks for the molecular shape rather than the electron-domain reference.

PCl₅ is a standard gaseous molecular AX₅ example with trigonal-bipyramidal geometry. The axial and equatorial P–Cl bonds occupy different environments. The solid-state structure can be ionic rather than simply an assembly of identical gas-phase PCl₅ molecules, so specify the phase when the question makes it relevant. A molecular-shape rule cannot be transferred blindly across phase-dependent structural changes.

ORIGINAL WORKED EXAMPLE

Locate the lone pairs in XeF₂

Use valence counting to explain the linear molecular shape of XeF₂.

  1. Total valence electrons = 8 + 2(7) = 22.
  2. Two Xe–F single bonds use four electrons; completing the two F octets uses twelve more, leaving six electrons on Xe.
  3. Those six form three Xe lone pairs. Xe is AX₂E₃ with five domains.
  4. The lone pairs occupy equatorial positions in the trigonal-bipyramidal reference, leaving opposite axial F atoms.

Answer XeF₂ is linear, with three equatorial lone pairs in the VSEPR reference arrangement.

05 / UNDERSTAND & APPLY

5. Six domains give octahedral, pyramidal or square-planar shapes

Six equivalent directions form an octahedral arrangement. Four can be pictured in a square plane with one above and one below, all centered on A. Neighboring directions are 90° apart and opposite directions 180°. SF₆ is AX₆ and has octahedral molecular geometry in the simple isolated-molecule model.

With one lone pair, AX₅E leaves five atoms in a square pyramid, as in BrF₅. With two lone pairs, the preferred simple arrangement puts them opposite, reducing their mutual crowding. The four remaining atoms form a square plane, as in XeF₄. The molecular geometry is square planar while the six-domain reference is octahedral.

Valence counting makes the difference between XeF₂ and XeF₄ clear. XeF₄ has four bonds and two lone pairs on xenon, rather than two bonds and three lone pairs. Both species can be nonpolar by symmetry with identical ligands, but their geometry and domain counts are different. Counting just the number of fluorines without accounting for the remaining electrons misses the reason.

Square-planar transition-metal complexes should not automatically be assigned this main-group AX₄E₂ explanation. Their geometry depends on d-electron configuration, ligand-field effects and bonding. The same geometric label can arise from different electronic models. Use the coordination-compound model for those species rather than force central lone-pair counting to explain them.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

SF₄ and XeF₄: different domain frameworksSF4 has a trigonal-bipyramidal domain reference with an equatorial lone pair and a seesaw atom arrangement. XeF4 has an octahedral reference with opposite lone pairs and four fluorine atoms in a square plane.Lone-pair positions and atom shapesSF₄ · AX₄EXeF₄ · AX₄E₂FFFFSEquatorial lone pair → seesawVertical F positions are axialFFFFXeOpposite lone pairs → square planarFour F atoms lie in the shaded planeSchematic projection, not measured angles.Lobes with dots represent lone-pair domains.
Schematic, not a measured-angle drawing. Lone-pair dots are not additional bonded atoms.

ORIGINAL WORKED EXAMPLE

Count XeF₄ completely

Determine the number of xenon lone pairs and the molecular geometry of XeF₄.

  1. Total valence electrons = 8 + 4(7) = 36.
  2. Four Xe–F bonds use eight electrons; four F octets require 24 more nonbonding electrons on F.
  3. Four electrons remain on Xe, giving two lone pairs. The central pattern is AX₄E₂.
  4. Six domains use an octahedral reference; opposite lone pairs leave a square-planar arrangement of F atoms.

Answer Two xenon lone pairs; square-planar molecular geometry.

06 / UNDERSTAND & APPLY

6. What sp, sp² and sp³ mean in a localized model

Hybridization is a mathematical combination of atomic orbitals on the same atom to produce a useful set of directional orbitals. Combining one s and one p orbital gives two sp hybrids; one s and two p orbitals give three sp² hybrids; one s and three p orbitals give four sp³ hybrids. The number of orbitals is conserved. Hybridization does not create additional electrons or require an atom to physically execute a timed mixing step before bonding.

In the idealized model, sp hybrids are linear with 50% s character; sp² hybrids are trigonal planar with one-third s character; sp³ hybrids are tetrahedral with one-quarter s character. Unused p orbitals remain available for π bonding: two in an sp description, one in sp² and none in sp³. These counts help connect local sigma geometry with multiple bonds.

Each carbon in ethene is approximately sp²: three sigma directions lie in a plane and an unhybridized p orbital participates in the π system. Each carbon in ethyne is approximately sp: two sigma directions are linear and two perpendicular p orbitals form two π components. The methyl carbon in ethane is ordinarily described as sp³ with four sigma directions.

A hybridization label is a model, not a direct synonym for molecular geometry. A lone pair can occupy a directional orbital in the localized description, leaving fewer atom positions than total orbitals. Thus ammonia is commonly described with approximately sp³ nitrogen even though its molecular shape is pyramidal, not tetrahedral. Real orbitals need not contain the exact ideal s:p ratio assumed by a simple label.

Conjugation can favor an approximately planar arrangement and alter the simple isolated-atom domain picture. An amide nitrogen is approximately planar because its lone pair participates in conjugation with the carbonyl, often described using a p orbital and sp² sigma framework. Blindly counting three bonds plus one lone pair and assigning a strongly pyramidal sp³ geometry would miss this electronic effect.

ORIGINAL WORKED EXAMPLE

Assign local carbon descriptions

Give elementary hybridization descriptions for the carbon atoms in CH₃–CH=CH₂ and state the unhybridized p-orbital count at each.

  1. The CH₃ carbon has four sigma directions, so use sp³ and no unhybridized p orbital in the simple model.
  2. Each alkene carbon has three sigma directions and one pi participation, so use sp².
  3. Each sp² carbon retains one unhybridized p orbital, and these overlap to form the π system.
  4. The molecule contains different local geometries; there is no single hybridization assigned to the whole molecule.

Answer CH₃ carbon: sp³, zero unhybridized p. Each alkene carbon: sp², one unhybridized p.

07 / UNDERSTAND & APPLY

7. Treat conventional d-hybrid labels as limited bookkeeping

Some introductory and examination conventions label five-domain geometries sp³d and six-domain geometries sp³d². If a question explicitly asks for that conventional hybridization assignment, trigonal-bipyramidal and octahedral reference arrangements correspond to those labels in that scheme. Recognizing the convention is useful, but it should not be confused with a modern demonstration of strong d-orbital participation.

For main-group hypervalent molecules such as PF₅, SF₆ or XeF₂, modern bonding descriptions can explain the structure through molecular orbitals and multicenter interactions without requiring the central atom’s d orbitals to form the literal equivalent hybrids of the older picture. The VSEPR shape prediction and the claimed orbital explanation are separate statements; accepting the former does not force the latter.

VSEPR also has limits for transition-metal complexes, some radicals, strongly delocalized systems and stereochemically inactive lone pairs. The domain method is most useful for the common main-group closed-shell cases taught here. If a problem gives measured geometry or a more specific electronic model, use that information rather than override it with an overgeneralized steric-number rule.

Bond angle is influenced by more than lone-pair count. Differences in central atom, ligand electronegativity, multiple bonding, steric crowding and conjugation can all matter. Use broad lone-pair repulsion trends within comparable examples, and avoid claiming an exact numerical angle from domain count alone.

08 / UNDERSTAND & APPLY

8. Symmetry decides whether bond dipoles cancel

After finding geometry, compare the bond-dipole vectors. Equal vectors arranged symmetrically can cancel: linear CO₂, trigonal-planar BF₃, tetrahedral CF₄, trigonal-bipyramidal PF₅ and octahedral SF₆ have symmetry-enforced cancellation in their ideal isolated structures. The existence of polar bonds alone does not make the molecule polar.

Lone pairs often reduce atomic symmetry and prevent complete cancellation, as in bent H₂O or pyramidal NH₃, but “has a lone pair” is not a universal polarity test. Linear XeF₂ and square-planar XeF₄ have central lone pairs while their equivalent bond dipoles cancel by symmetry. Always inspect the actual atom arrangement rather than stop at the lone-pair count.

For an AX₄ tetrahedron with unequal substituents, such as CH₃Cl, the four contributions are not equivalent, so tetrahedral geometry alone does not imply zero dipole. Likewise, a linear heteronuclear triatomic molecule need not be nonpolar: cancellation requires equal opposing contributions. Symmetry is a sufficient reason in the standard examples; accidental numerical cancellation in unsymmetrical cases cannot be assumed.

For two equal bond dipoles of magnitude μ_b separated by angle θ, the vector-sum magnitude is 2μ_b cos(θ/2). It vanishes at 180° but not at a typical bent angle. This simple vector model provides intuition, although an actual molecular dipole includes the full electron distribution and need not be exactly the sum of transferable isolated bond-dipole values.

For two equal bond vectors: μ_resultant = 2μ_b cos(θ/2)

A geometrical model for equal contributions separated by angle θ.

ORIGINAL WORKED EXAMPLE

Lone pairs do not force polarity

Compare ideal isolated SF₄ and XeF₄ for geometry and dipole cancellation with identical F ligands.

  1. SF₄ is AX₄E with a seesaw shape from a five-domain reference. Its bond vectors do not cancel by the high symmetry of a square plane or tetrahedron.
  2. XeF₄ is AX₄E₂ with opposite lone pairs and square-planar F positions.
  3. The Xe–F vectors cancel in opposite pairs.
  4. The different domain arrangements matter more than the fact that both species have central lone pairs.

Answer SF₄ is polar; ideal square-planar XeF₄ is nonpolar.

09 / UNDERSTAND & APPLY

9. Original shape and orbital practice

These original problems refer to isolated molecular species in the stated elementary models unless a phase is specified. Count valence electrons first for an unfamiliar ion or noble-gas compound. Then report electron-domain geometry and molecular geometry separately when both are requested.

For a multiple-center molecule, assign geometry locally. For a polarity question, add a symmetry argument after the shape. For a hybridization question, state whether you are using a conventional introductory label or a more detailed bonding interpretation. This prevents a correct geometric result from carrying an unjustified orbital claim.

ORIGINAL PRACTICE · 1

What is the molecular geometry of NO₂⁻ in a standard VSEPR description?

Reveal answer & reasoning

Bent, with a trigonal-planar three-domain reference.

Nitrite has two N–O directions and one nitrogen lone pair. Resonance does not create an extra domain or remove the lone pair.

ORIGINAL PRACTICE · 2

Classify BF₃ and NH₃. Why are both not simply “trigonal”?

Reveal answer & reasoning

BF₃ is trigonal planar; NH₃ is trigonal pyramidal.

BF₃ has three bond domains and no central lone pair. NH₃ has three bonds plus one lone pair, giving four domains and a pyramidal atom arrangement.

ORIGINAL PRACTICE · 3

Where does the lone pair go in the usual SF₄ VSEPR model?

Reveal answer & reasoning

An equatorial site.

The five-domain trigonal-bipyramidal reference gives an equatorial lone pair two 90° interactions rather than the three of an axial site, leaving a seesaw atom arrangement.

ORIGINAL PRACTICE · 4

What molecular shape follows from AX₃E₂ with five domains?

Reveal answer & reasoning

T-shaped.

The two lone pairs occupy equatorial sites. Two axial bonded atoms and the remaining equatorial bonded atom make the T arrangement.

ORIGINAL PRACTICE · 5

Predict BrF₅ geometry from its central bond and lone-pair count.

Reveal answer & reasoning

Square pyramidal.

Five Br–F bonds and one central lone pair give AX₅E. Removing one atom position from an octahedral reference leaves a square pyramid.

ORIGINAL PRACTICE · 6

How many hybrid orbitals and unhybridized p orbitals occur in an sp² model?

Reveal answer & reasoning

Three sp² hybrids and one unhybridized p orbital.

One s and two p orbitals combine to form three hybrids. The third p orbital remains available for π overlap.

ORIGINAL PRACTICE · 7

Must a molecule with a central lone pair have a permanent dipole?

Reveal answer & reasoning

No.

XeF₂ is linear with three central lone pairs and XeF₄ is square planar with two; identical bond contributions cancel by symmetry in both ideal structures.

ORIGINAL PRACTICE · 8

A student calls an amide nitrogen necessarily pyramidal because it has three bonds and a lone pair. What is missing?

Reveal answer & reasoning

Conjugation of the nitrogen lone pair with the carbonyl.

The delocalized interaction favors a roughly planar nitrogen environment and a p-orbital role for the lone pair. Simple isolated-domain counting is not enough for this case.

KEEP THE BIG IDEAS

Quick revision

  • A multiple bond is one electron domain, while a central lone pair adds one domain.
  • Name electron-domain geometry and molecular shape separately.
  • Five-domain axial and equatorial sites differ; lone pairs usually prefer equatorial positions.
  • Two lone pairs in an octahedral reference commonly occupy opposite sites, giving AX₄E₂ square-planar shape.
  • Hybridization is a local orbital model with a conserved number of orbitals.
  • Conjugation and hypervalent bonding require care beyond a mechanical steric-number label.
  • Polarity requires geometry and equivalent vector contributions, not just polar bonds or lone pairs.
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