VSEPR, Hybridization & Molecular Shape
VSEPR Theory
The shapes of molecules are crucial because many physical and chemical properties depend on the three-dimensional arrangement of their atoms. The Lewis model explains how atoms bond by sharing or transferring electrons, but it does not explain why molecules adopt particular geometries. To address this, Sidgwick and Powell (1940) proposed that molecular shapes depend on how electron pairs arrange around the central atom. This idea was developed into a full theory by Nyholm and Gillespie, known as the Valence Shell Electron Pair Repulsion theory. VSEPR theory allows us to predict molecular geometry by counting bond pairs and lone pairs around the central atom and understanding how they repel each other.
Basic Assumption: All valence electron pairs around a central atom arrange themselves at maximum distance apart to minimise repulsions between them, giving the most stable geometry.
Scope: VSEPR applies to non-transition elements and successfully predicts the shapes of simple molecules by considering the arrangement of electron pairs in the valence shell of the central atom.
Limitation: VSEPR predicts shapes and bond angles but does not explain why bonds form or the nature of bonding — that requires hybridization theory and orbital overlap concepts.
VSEPR theory rests on four key postulates that determine how electron pairs behave around a central atom and how this behaviour translates into observable molecular shapes.
$$\text{lp-lp} > \text{lp-bp} > \text{bp-bp}$$
Order of electron pair repulsion strength: lone pair–lone pair repulsion is strongest, followed by lone pair–bond pair, then bond pair–bond pair which is weakest.
$\text{lp-lp}$=Repulsion between two lone pairs on the central atom(relative repulsive strength (highest))
$\text{lp-bp}$=Repulsion between a lone pair and a bond pair(relative repulsive strength (intermediate))
$\text{bp-bp}$=Repulsion between two bond pairs(relative repulsive strength (lowest))
multiple lone pairs present
→bond angles decrease progressively as more lone pairs compress bonding pairs
Electron Pair Participation: Both lone pairs and bond pairs participate in determining the geometry of the molecule — geometry is not decided by bonded atoms alone.
Maximum Separation: Electron pairs arrange around the central atom so as to remain at maximum distance apart, minimising repulsive interactions between them.
Lone Pair Dominance: Lone pairs occupy more space than bond pairs. A lone pair is attracted by only one nucleus while a bonding pair is attracted by two nuclei, so the lone pair charge cloud spreads out more in space. This means lone pairs exert greater repulsive forces on bonding pairs and compress the bond angles.
Multiple Bond Effect: Double bonds (two electron pairs) and triple bonds (three electron pairs) contain higher charge density and occupy more space than a single bond, but they behave like a single unit in determining the overall geometry of the molecule.
A bonding electron pair is attracted by both nuclei of the bonded atoms, which keeps the electron density concentrated between them. A non-bonding lone pair, however, is attracted by only one nucleus. Because it experiences less nuclear attraction, the lone pair charge cloud spreads out over a larger volume of space. As a result, lone pairs exert stronger repulsive forces on neighbouring electron pairs than bonding pairs do. This asymmetry is why molecules like NH₃ and H₂O have bond angles smaller than the ideal tetrahedral angle of 109.5°.
Predicting Molecular Shapes Using VSEPR
To predict molecular shape using VSEPR, identify the central atom, count the total number of electron pairs (bonding + lone), and determine the arrangement that minimises repulsion. The shape is then identified based on which corners of the electron pair geometry are occupied by atoms versus lone pairs.
VSEPR Shape Prediction Steps
1
Draw the Lewis structure to identify the central atom
2
Count total bond pairs around the central atom
3
Count total lone pairs on the central atom
4
Determine the electron pair geometry from total pairs
5
Identify the molecular shape from atom positions (excluding lone pairs)
When a central atom has two bond pairs and zero lone pairs (AB₂ type), the electron pairs arrange at maximum distance apart — an angle of 180° — producing a linear geometry. The two bonding pairs are directed in opposite directions along a straight line.
$$\theta = 180°$$
The bond angle in linear molecules is always 180° because two electron pairs are diametrically opposite.
$\theta$=Bond angle between the two bonded atoms and the central atom(degrees (°))
AB₂ with no lone pairs
→Always linear — examples include BeCl₂, CO₂, HgCl₂, MgCl₂
Examples: BeCl₂, CO₂, HgCl₂, MgCl₂, CaCl₂, SrCl₂, CdCl₂ — all have central atoms with two valence electrons forming two bonds.
Carbon Dioxide: Although formed from heteroatoms, CO₂ is linear because the O=C=O arrangement balances polar character on both sides of the carbon atom.
Three electron pairs around the central atom adopt a trigonal planar arrangement with bond angles of 120°. The molecular shape depends on whether all three pairs are bonding or one is a lone pair.
$$\theta = 120° \text{ (ideal trigonal planar)}$$
Three electron pairs around the central atom arrange at 120° to each other in a flat triangular geometry.
$\theta$=Ideal bond angle for three electron pairs with no lone pairs(degrees (°))
one lone pair present (AB₂E)
→bond angle reduces below 120°, giving a bent or angular molecular shape
AB₃ — Trigonal Planar: When all three pairs are bonding (no lone pairs), the molecular shape is trigonal planar with 120° bond angles. Examples: BH₃, BF₃, AlCl₃, AlH₃, GaH₃, InH₃, TlH₃.
AB₂E — Bent (Angular): When one pair is a lone pair, the remaining two bond pairs form a bent shape with bond angle less than 120°. Example: SnCl₂ in vapour phase has a lone pair occupying one corner of the triangle, giving a distorted structure.
SO₂: One corner occupied by a lone pair, two corners by S=O double bonds. The double bonds have higher charge density but still behave as single units for geometry, giving a bent shape with bond angle less than 120°.
SO₃: All three corners occupied by S=O double bonds — perfectly trigonal planar since no lone pairs are present.
Four electron pairs around the central atom adopt a tetrahedral arrangement. The ideal bond angle is 109.5°. The molecular shape depends on how many of the four pairs are lone pairs versus bond pairs.
$$\theta = 109.5° \text{ (ideal tetrahedral)}$$
Four electron pairs around the central atom adopt a regular tetrahedral arrangement with bond angles of 109.5° to minimise repulsion.
$\theta$=Ideal tetrahedral bond angle(degrees (°))
1 lone pair (AB₃E)
→angle reduces to ~107.5° (e.g., NH₃)
2 lone pairs (AB₂E₂)
→angle reduces to ~104.5° (e.g., H₂O)
AB₄ — Tetrahedral: All four pairs are bonding — the molecular shape matches the electron pair geometry exactly. Examples: CH₄, SiH₄, GeH₄, CCl₄, SiCl₄, BF₄⁻, NH₄⁺, SO₄²⁻. The tetrahedron has four corners, four faces, six edges and six bond angles.
AB₃E — Trigonal Pyramidal: Three bonding pairs and one lone pair. The lone pair occupies one corner of the tetrahedron, pushing the three bonding pairs slightly closer together. Bond angle reduces from 109.5° to ~107.5°. Examples: NH₃, NF₃, PH₃, AsH₃, SbH₃.
AB₂E₂ — Bent (Angular): Two bonding pairs and two lone pairs. Two corners of the tetrahedron are occupied by lone pairs, and the remaining two by bonding pairs. The strong lone pair–lone pair repulsion plus lone pair–bond pair repulsion compresses the bond angle further to ~104.5°. Examples: H₂O, H₂S, H₂Se, H₂Te.
Effect of Lone Pairs on Bond Angles in AB₄ Systems
1
AB₄ (0 lone pairs): 109.5° — CH₄, CCl₄, SiCl₄
2
AB₃E (1 lone pair): 107.5° — NH₃, PH₃
3
AB₂E₂ (2 lone pairs): 104.5° — H₂O, H₂S
4
Each additional lone pair compresses bond angles further due to stronger repulsion
NF₃ and BF₃ both have formula XF₃ but have different shapes. In BF₃, boron has only three bond pairs and no lone pairs, giving a trigonal planar geometry with 120° bond angles. In NF₃, nitrogen has three bond pairs and one lone pair, giving a trigonal pyramidal geometry with bond angle of 102° (even smaller than NH₃'s 107.5°). The substitution of hydrogen with electronegative fluorine atoms further reduces the angle because the polarity of the N–F bond pulls the lone pair closer to the nitrogen nucleus, increasing its repulsion on bonding pairs.
BF₃ Geometry: Boron has 3 valence electrons → 3 bonding pairs, 0 lone pairs → trigonal planar (120°).
NF₃ Geometry: Nitrogen has 5 valence electrons → 3 bonding pairs, 1 lone pair → trigonal pyramidal (102°).
Why NF₃ angle < NH₃ angle: The electronegative F atoms pull N–F bonding electrons closer to F, making the bond pairs farther from N. This reduces bp-bp repulsion relative to lp-bp repulsion, so the lone pair compresses the angle more aggressively in NF₃.
The series NH₂⁻, NH₃, NH₄⁺ illustrates how changing the number of lone pairs affects bond angles while the central nitrogen remains sp³ hybridised in all three species.
NH₂⁻ (2 bond pairs, 2 lone pairs): Bond angle ≈ 105° — two lone pairs exert maximum compression on bond pairs.
NH₃ (3 bond pairs, 1 lone pair): Bond angle = 107.5° — one lone pair compresses bond angles moderately.
NH₄⁺ (4 bond pairs, 0 lone pairs): Bond angle = 109.5° — perfect tetrahedral angle since no lone pairs are present to distort geometry.
Hybridization and Bond Formation
VSEPR theory predicts molecular shapes but does not explain why bonds form. Hybridization theory addresses this by explaining how atomic orbitals mix to form new orbitals that determine both bonding and molecular geometry. According to this concept, atomic orbitals of slightly different energies intermix to form new hybrid orbitals that differ from the parent orbitals in shape and possess specific geometry. The energy required for exciting electrons to higher orbitals is compensated by the energy released during hybridization and subsequent bond formation.
Definition: Atomic orbitals differing slightly in energy intermix to form new orbitals called hybrid atomic orbitals, which have different shapes and specific spatial arrangements.
Simultaneous Process: Promotion of electrons to excited states and hybridization occur simultaneously — the energy cost of excitation is recovered during bond formation.
Purpose: Hybridization explains both the valency of elements and the observed bond angles and geometries of molecules that pure atomic orbitals cannot account for.
In sp³ hybridization, one s orbital and three p orbitals of the central atom intermix to form four equivalent sp³ hybrid orbitals. These four hybrid orbitals are directed towards the corners of a regular tetrahedron with bond angles of 109.5°. The sp³ model explains the bonding and geometry of methane, ammonia, and water.
$$1s + 3p \rightarrow 4\,sp^3$$
One s and three p atomic orbitals combine to form four equivalent sp³ hybrid orbitals arranged tetrahedrally.
$1s$=One spherically symmetric s orbital(atomic orbital)
$3p$=Three p orbitals (px, py, pz) oriented along x, y, z axes(atomic orbital)
$4\,sp^3$=Four equivalent hybrid orbitals with tetrahedral geometry(hybrid orbital)
all four sp³ orbitals form bonds
→tetrahedral molecular geometry (CH₄)
three form bonds, one holds lone pair
→trigonal pyramidal (NH₃)
two form bonds, two hold lone pairs
→bent/angular (H₂O)
Carbon in CH₄: Ground state carbon ($1s^2 2s^2 2p_x^1 2p_y^1 2p_z^0$) promotes one electron from 2s to 2p, giving four unpaired electrons ($1s^2 2s^1 2p_x^1 2p_y^1 2p_z^1$). These four orbitals then hybridize to form four equivalent sp³ orbitals, each overlapping with a 1s orbital of hydrogen to form four sigma bonds. The result is a tetrahedral geometry with bond angle 109.5°.
Nitrogen in NH₃: Nitrogen ($1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$) hybridizes its valence orbitals to form four sp³ orbitals. One sp³ orbital holds a lone pair (completely filled), and the other three overlap with hydrogen 1s orbitals. The lone pair repulsion compresses the H–N–H angle from 109.5° to 107.5°, giving a trigonal pyramidal shape.
Oxygen in H₂O: Oxygen ($1s^2 2s^2 2p_x^2 2p_y^1 2p_z^1$) forms four sp³ orbitals. Two hold lone pairs and two overlap with hydrogen 1s orbitals. The two lone pairs exert strong repulsion on each other and on the bond pairs, reducing the H–O–H angle from 109.5° to 104.5°, giving a bent molecular shape.
In sp² hybridization, one s orbital and two p orbitals intermix to form three equivalent sp² hybrid orbitals arranged in a plane at 120° to each other. One unhybridized p orbital remains perpendicular to the plane. This explains trigonal planar geometries and the formation of double bonds containing one sigma bond and one pi bond.
$$1s + 2p \rightarrow 3\,sp^2 + 1\,p$$
One s and two p atomic orbitals combine to form three equivalent sp² hybrid orbitals in a plane, leaving one unhybridized p orbital perpendicular to that plane.
$3\,sp^2$=Three equivalent hybrid orbitals at 120° in one plane(hybrid orbital)
$1\,p$=One remaining unhybridized p orbital (perpendicular to the sp² plane)(atomic orbital)
unhybridized p orbitals of two atoms overlap sideways
→a pi (π) bond forms, creating a double bond with the sigma bond
BF₃: Boron ($1s^2 2s^2 2p_x^1$) promotes to excited state ($1s^2 2s^1 2p_x^1 2p_y^1$), then hybridizes to three sp² orbitals. These overlap with p orbitals of three fluorine atoms to form three sigma bonds in a trigonal planar arrangement with 120° bond angles.
Ethene (C₂H₄): Each carbon undergoes sp² hybridization, forming three sp² orbitals. Two sp² orbitals on each carbon form sigma bonds with hydrogen atoms (sp²–s overlaps). The remaining sp² orbital on each carbon overlaps with the other carbon's sp² orbital (sp²–sp² overlap) to form a C–C sigma bond. The unhybridized p orbitals on each carbon overlap sideways to form a pi bond. The result is a planar molecule with one C=C double bond (one sigma + one pi).
In sp hybridization, one s orbital and one p orbital intermix to form two equivalent sp hybrid orbitals oriented at 180° to each other in a linear geometry. Two unhybridized p orbitals remain perpendicular to the sp axis and to each other. This explains linear geometries and the formation of triple bonds containing one sigma bond and two pi bonds.
$$1s + 1p \rightarrow 2\,sp + 2\,p$$
One s and one p atomic orbital combine to form two equivalent sp hybrid orbitals at 180°, leaving two unhybridized p orbitals perpendicular to the sp axis.
$2\,sp$=Two equivalent hybrid orbitals oriented at 180° (linear)(hybrid orbital)
$2\,p$=Two remaining unhybridized p orbitals (perpendicular to each other and to the sp axis)(atomic orbital)
both unhybridized p orbitals of two atoms overlap sideways
→two pi bonds form, creating a triple bond with the sigma bond
BeCl₂: Beryllium ($1s^2 2s^2$) promotes one electron ($1s^2 2s^1 2p_x^1$) and hybridizes to two sp orbitals at 180°. Each sp orbital overlaps with a chlorine p orbital to form a linear Cl–Be–Cl molecule.
Ethyne (C₂H₂): Each carbon undergoes sp hybridization, forming two sp orbitals and leaving two unhybridized p orbitals. Each carbon forms one sp–s sigma bond with hydrogen and one sp–sp sigma bond with the other carbon. The two pairs of unhybridized p orbitals (py–py and pz–pz) overlap sideways to form two pi bonds. The result is a linear molecule with a C≡C triple bond (one sigma + two pi). The two pi bonds' electron clouds surround the sigma bond in a cylindrical drum shape.
Sigma and Pi Bonds in Hybridization
A sigma bond is formed when two partially filled atomic orbitals overlap along the internuclear axis (end-to-end or head-on overlap). The electron density is concentrated along the line joining the two nuclei. A sigma bond is formed in all types of orbital overlap — s-s, s-p, p-p (head-on), and hybrid orbital overlaps. Every covalent bond contains at least one sigma bond.
Formation: Sigma bonds result from direct overlap along the line connecting the two nuclei — whether the orbitals are s, p, or hybrid (sp, sp², sp³).
Characteristics: The electron density is symmetric around the internuclear axis. Sigma bonds are the strongest type of covalent bond because of maximum orbital overlap.
Free Rotation: Atoms connected by a single sigma bond can freely rotate about the bond axis because the electron density is cylindrically symmetric.
A pi bond is formed by the sideways overlap of two half-filled p orbitals that are parallel to each other and perpendicular to the internuclear axis. The electron density is concentrated above and below (or in front of and behind) the internuclear axis, with a nodal plane (zero electron density) along the axis itself.
Formation Requirement: A pi bond can only form between two atoms that are already bonded by a sigma bond — pi bonds are always additional to a sigma bond.
Characteristics: Pi bonds have lower electron density between the nuclei than sigma bonds (because the overlap is sideways, not direct), making them weaker and more reactive than sigma bonds.
No Free Rotation: The parallel alignment required for pi bond overlap prevents free rotation around a double bond, which is why molecules like ethene have restricted rotation around the C=C bond.
Multiple Pi Bonds: When two p orbitals on each atom remain unhybridized (as in sp hybridization), two separate pi bonds can form (as in the C≡C triple bond of ethyne).
Bond Composition Summary
•
Single bond (C–C): 1 sigma bond (sp³–sp³ overlap)
•
Double bond (C=C): 1 sigma + 1 pi (sp²–sp² sigma + p–p pi)
•
Triple bond (C≡C): 1 sigma + 2 pi (sp–sp sigma + two p–p pi bonds)
Connecting Hybridization to Molecular Shape
The type of hybridization a central atom undergoes directly determines its electron pair geometry and hence the molecular shape. By counting bond pairs and lone pairs around the central atom, you can identify the hybridization and predict the shape.
Quick Method: To determine hybridization quickly, count the sigma bonds plus lone pairs on the central atom. Two regions → sp, three regions → sp², four regions → sp³.
Why Lone Pairs Count: Lone pairs occupy hybrid orbitals just like bond pairs do. Even though they are not bonded to another atom, they occupy space in the hybridization scheme and affect the geometry.
Hybridization – Geometry – Shape Reference Table
•
sp hybridization (2 electron pairs): linear geometry, 180° — BeCl₂, CO₂, C₂H₂
•
sp² hybridization (3 electron pairs): trigonal planar, 120° — BF₃, SO₃, C₂H₄
•
sp³ hybridization (4 electron pairs): tetrahedral, 109.5° — CH₄, CCl₄, SiCl₄
•
sp³ with 1 lone pair: trigonal pyramidal, 107.5° — NH₃, PH₃, NF₃
•
sp³ with 2 lone pairs: bent/angular, 104.5° — H₂O, H₂S, H₂Se
Multiple bonds (double and triple) behave as a single electron pair region when determining molecular geometry via VSEPR, but they occupy slightly more space than a single bond due to their higher charge density. This means that in molecules with mixed single and multiple bonds, the bond angles involving the multiple bond region are slightly larger than ideal.
Double Bond Space: A C=O or S=O double bond occupies more space than a C–O or S–O single bond, pushing other bonding pairs slightly closer together. However, the double bond is treated as ONE region for shape classification.
Triple Bond Space: A C≡C triple bond has even higher charge density and occupies the most space, but still counts as one region for VSEPR shape determination.
Shape Remains: Despite the space differences, the overall molecular shape is still determined by the number of electron pair regions, not the type of bond.
Application Strategy
Worked Examples
Problem Statement
Determine the molecular shape, hybridization, and bond angle of CH₄ using VSEPR theory and hybridization concepts.
VSEPRsp³ hybridizationtetrahedral geometry
Strategic Toolkit
$$\text{Steric number} = \text{bond pairs} + \text{lone pairs}$$
Step 1
Draw the Lewis structure
Carbon has 4 valence electrons and each hydrogen has 1. Carbon shares one electron with each of the four hydrogen atoms, forming four C–H single bonds. No lone pairs remain on carbon.
$$\text{C: 4e}^- + 4 \times \text{H: 1e}^- = 8\,e^- \text{ (4 bond pairs)}$$
Step 2
Count electron pairs on the central atom
Carbon has 4 bonding pairs and 0 lone pairs. The steric number is 4, which determines both the hybridization and the electron pair geometry.
$$\text{Steric number} = 4 + 0 = 4 \rightarrow sp^3$$
Step 3
Identify geometry and shape
With 4 electron pairs and no lone pairs, the electron pair geometry is tetrahedral and the molecular shape is also tetrahedral. All positions are occupied by atoms.
$$\theta_{\text{H-C-H}} = 109.5°$$
Problem Statement
Explain why NH₃ has a trigonal pyramidal shape with a bond angle of 107.5° instead of the tetrahedral angle of 109.5°.
VSEPRlone pair repulsionsp³ hybridization
Strategic Toolkit
$$\text{Steric number} = \text{bond pairs} + \text{lone pairs}$$
Step 1
Draw the Lewis structure and count pairs
Nitrogen has 5 valence electrons. It forms 3 bonds with hydrogen atoms, using 3 electrons. The remaining 2 electrons form one lone pair. Total steric number is 4.
$$\text{N: 5e}^- \rightarrow 3\,\text{bond pairs} + 1\,\text{lone pair}, \quad \text{Steric number} = 3 + 1 = 4 \rightarrow sp^3$$
Step 2
Determine electron pair geometry
Four electron pairs adopt a tetrahedral arrangement of sp³ hybrid orbitals. One corner is occupied by the lone pair and three corners by hydrogen atoms.
$$\text{Electron pair geometry: tetrahedral (109.5° ideal)}$$
Step 3
Determine molecular shape and actual angle
The lone pair occupies more space than bond pairs and exerts stronger repulsion on the three bonding pairs, compressing them closer together. The molecular shape (considering only atom positions) is trigonal pyramidal.
$$\theta_{\text{H-N-H}} = 107.5° < 109.5° \quad (\text{lone pair compression})$$
Problem Statement
Why is H₂O bent with a bond angle of 104.5° rather than linear at 180° as a simple AB₂ molecule might suggest?
VSEPRlone pair repulsionbent geometrysp³ hybridization
Strategic Toolkit
$$\text{Steric number} = \text{bond pairs} + \text{lone pairs}$$
Step 1
Count valence electrons and electron pairs
Oxygen has 6 valence electrons. It forms 2 bonds with hydrogen, using 2 electrons. The remaining 4 electrons form 2 lone pairs. The steric number is 4, not 2.
$$\text{O: 6e}^- \rightarrow 2\,\text{bond pairs} + 2\,\text{lone pairs}, \quad \text{Steric number} = 2 + 2 = 4 \rightarrow sp^3$$
Step 2
Identify electron pair geometry
Four electron pairs adopt a tetrahedral electron pair geometry with sp³ hybridization. Two corners hold lone pairs and two corners hold hydrogen atoms.
$$\text{Electron pair geometry: tetrahedral}$$
Step 3
Explain the bent shape and compressed angle
The two lone pairs exert strong repulsion on each other (lp–lp) and on the bond pairs (lp–bp). This triple compression reduces the H–O–H bond angle from the ideal 109.5° to 104.5°, giving a bent molecular shape.
$$\theta_{\text{H-O-H}} = 104.5° < 107.5° < 109.5°$$
Problem Statement
Determine the molecular shape, hybridization, and bond angle of boron trifluoride (BF₃).
VSEPRsp² hybridizationtrigonal planar geometry
Strategic Toolkit
$$\text{Steric number} = \text{bond pairs} + \text{lone pairs}$$
Step 1
Count electron pairs on boron
Boron has 3 valence electrons. It forms 3 bonds with fluorine atoms, using all 3 electrons. No lone pairs remain on boron. Steric number is 3.
$$\text{B: 3e}^- \rightarrow 3\,\text{bond pairs} + 0\,\text{lone pairs}, \quad \text{Steric number} = 3 + 0 = 3 \rightarrow sp^2$$
Step 2
Determine geometry and bond angle
Three electron pairs with no lone pairs arrange at 120° in a plane. The molecular shape is trigonal planar, and boron is sp² hybridized with one unhybridized p orbital.
$$\theta_{\text{F-B-F}} = 120°, \quad \text{Shape: trigonal planar}$$
Problem Statement
Explain why BeCl₂ is a linear molecule and identify the hybridization of beryllium.
VSEPRsp hybridizationlinear geometry
Strategic Toolkit
$$\text{Steric number} = \text{bond pairs} + \text{lone pairs}$$
Step 1
Count electron pairs on beryllium
Beryllium has 2 valence electrons. It forms 2 bonds with chlorine atoms, using both electrons. No lone pairs remain. Steric number is 2.
$$\text{Be: 2e}^- \rightarrow 2\,\text{bond pairs} + 0\,\text{lone pairs}, \quad \text{Steric number} = 2 + 0 = 2 \rightarrow sp$$
Step 2
Determine geometry and hybridization
Two electron pairs with no lone pairs arrange at 180° on opposite sides of beryllium. Beryllium is sp hybridized with two unhybridized p orbitals. The molecular shape is linear.
$$\theta_{\text{Cl-Be-Cl}} = 180°, \quad \text{Shape: linear}$$
Problem Statement
Describe the types of bonds formed between the carbon atoms in ethene, including the hybridization involved and the nature of sigma and pi bonds.
sp² hybridizationsigma bondpi bonddouble bond
Strategic Toolkit
$$1s + 2p \rightarrow 3\,sp^2 + 1\,p$$
Step 1
Identify carbon hybridization
Each carbon atom in ethene is bonded to two hydrogen atoms and one other carbon atom (three regions). The steric number is 3, so each carbon is sp² hybridized with one unhybridized p orbital remaining.
$$\text{Steric number} = 3 \rightarrow sp^2\,\text{hybridization per carbon}$$
Step 2
Identify the sigma bonds
Each carbon forms two C–H sigma bonds (sp²–s overlap) and one C–C sigma bond (sp²–sp² overlap). That gives a total of 5 sigma bonds in ethene.
$$\text{Total sigma bonds: 4 (C-H)} + 1 (C-C) = 5$$
Step 3
Identify the pi bond
The unhybridized p orbitals on each carbon are parallel and overlap sideways to form one pi bond. Combined with the C–C sigma bond, this gives a C=C double bond. The molecule is planar with all atoms in one plane.
$$\text{C=C} = 1\,\sigma + 1\,\pi$$
Problem Statement
Describe the hybridization of each carbon in ethyne and explain the composition of the C≡C triple bond.
sp hybridizationsigma bondpi bondtriple bond
Strategic Toolkit
$$1s + 1p \rightarrow 2\,sp + 2\,p$$
Step 1
Identify carbon hybridization
Each carbon in ethyne is bonded to one hydrogen and one other carbon (two regions). Steric number is 2, so each carbon is sp hybridized with two unhybridized p orbitals.
$$\text{Steric number} = 2 \rightarrow sp\,\text{hybridization per carbon}$$
Step 2
Identify sigma bonds
Each carbon forms one C–H sigma bond (sp–s overlap) and one C–C sigma bond (sp–sp overlap). That gives 2 sigma bonds in the carbon–carbon framework plus 2 C–H sigma bonds.
$$\text{Total sigma bonds: 2 (C-H)} + 1 (C-C)} = 3$$
Step 3
Identify the two pi bonds
The two unhybridized p orbitals on each carbon are perpendicular to each other. One pair of p orbitals (py–py) and another pair (pz–pz) overlap sideways to form two separate pi bonds. Combined with the sigma bond, this gives a C≡C triple bond.
$$\text{C≡C} = 1\,\sigma + 2\,\pi$$
Memorize formulas
Electron Pair Repulsion Order
Lone pair–lone pair repulsion is strongest, lone pair–bond pair is intermediate, and bond pair–bond pair is weakest. This hierarchy determines how much bond angles are compressed by lone pairs.
Formula
$$\text{lp-lp} > \text{lp-bp} > \text{bp-bp}$$
Steric Number to Hybridization
Steric number = sigma bonds + lone pairs on central atom. Determines hybridization type and ideal bond angle: 2 → sp (180°), 3 → sp² (120°), 4 → sp³ (109.5°).
Formula
$$\text{Steric number} = n_{\sigma} + n_{\text{lp}}$$
Ideal Tetrahedral Angle
The bond angle in a perfect tetrahedral arrangement with 4 sp³ hybrid orbitals. Lone pairs reduce this: 1 lp → 107.5° (NH₃), 2 lp → 104.5° (H₂O).
Formula
$$\theta_{\text{tetrahedral}} = \arccos\left(-\frac{1}{3}\right) \approx 109.5°$$
sp² Hybrid Orbitals Formed
One s + two p orbitals form three sp² hybrid orbitals at 120° in a plane, leaving one unhybridized p orbital for pi bond formation.
Formula
$$1s + 2p \rightarrow 3\,sp^2 + 1\,p$$
sp Hybrid Orbitals Formed
One s + one p orbital form two sp hybrid orbitals at 180° (linear), leaving two unhybridized p orbitals for two pi bonds.
Formula
$$1s + 1p \rightarrow 2\,sp + 2\,p$$
Review
1
What is the fundamental principle of Valence Shell Electron Pair Repulsion (VSEPR) theory?
Ans
Electron pairs around a central atom repel each other and position themselves to minimize this repulsion
Review
Tap to reveal
Identify the core principle
VSEPR theory is based on electrostatic repulsion between electron pairs in the valence shell of the central atom.
$$\text{Electron pairs repel} \rightarrow \text{adopt positions of minimum repulsion}$$
Determine molecular shape
The arrangement of electron pairs dictates the geometry, which is why molecules with the same number of electron pairs share similar basic geometries.
$$\text{Geometry depends on number of electron pairs}$$
2
Which sequence correctly ranks electron pair repulsions from strongest to weakest?
Ans
Lone pair–lone pair > lone pair–bond pair > bond pair–bond pair
Review
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Recall the repulsion order
The repulsion order is a fundamental VSEPR concept that governs how electron pair interactions affect molecular geometry.
$$\text{lp-lp} > \text{lp-bp} > \text{bp-bp}$$
Understand the reasoning
Lone pairs are localized on one atom and spread more into space, so they repel more strongly than bond pairs which are constrained between two nuclei.
$$\text{More localized charge} \rightarrow \text{stronger repulsion}$$
3
What is the steric number of the carbon atom in CH₄?
Ans
4
Review
Tap to reveal
Count sigma bonds
Carbon in CH₄ forms four single bonds to four hydrogen atoms, each of which is a sigma bond.
$$n_{\sigma} = 4$$
Count lone pairs
Carbon has no lone pairs on its valence shell after forming four bonds.
$$n_{\text{lp}} = 0$$
Calculate steric number
The steric number is the sum of sigma bonds and lone pairs.
$$\text{Steric number} = 4 + 0 = 4 \rightarrow sp^3$$
4
What type of hybridization does the nitrogen atom undergo in NH₃?
Ans
sp³
Review
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Count electron groups on nitrogen
Nitrogen forms 3 sigma bonds with H atoms and has 1 lone pair.
$$\text{Bond pairs} = 3, \quad \text{Lone pairs} = 1$$
Calculate steric number
Steric number = number of sigma bonds + number of lone pairs.
$$\text{Steric number} = 3 + 1 = 4$$
Determine hybridization
A steric number of 4 corresponds to sp³ hybridization.
$$\text{sp}^3 \rightarrow \text{trigonal pyramidal geometry}$$
5
Which molecular geometry does BF₃ exhibit?
Ans
Trigonal planar
Review
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Count electron groups on boron
Boron has 3 bond pairs (to three F atoms) and zero lone pairs.
$$\text{Steric number} = 3 + 0 = 3$$
Identify hybridization and shape
A steric number of 3 gives sp² hybridization with trigonal planar geometry.
$$sp^2 \rightarrow \text{trigonal planar}, \quad \theta = 120°$$
6
Both NH₃ and H₂O have sp³ hybridized central atoms with bond angles less than the ideal tetrahedral angle of 109.5°. The bond angle in NH₃ is about 107° while in H₂O it is about 104.5°. Why is the angle smaller in H₂O?
Ans
H₂O has two lone pairs that repel the bonding pairs more strongly than the single lone pair in NH₃
Review
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Compare electron pair arrangements
NH₃ has 3 bond pairs + 1 lone pair. H₂O has 2 bond pairs + 2 lone pairs.
$$\text{NH}_3: 3\text{bp} + 1\text{lp}, \quad \text{H}_2\text{O}: 2\text{bp} + 2\text{lp}$$
Apply repulsion hierarchy
The extra lone pair–lone pair repulsion in H₂O compresses the bond angle more than in NH₃.
$$\text{lp-lp} > \text{lp-bp} > \text{bp-bp}$$
Compare resulting bond angles
More lone pairs means more compression of bond angles away from 109.5°.
$$107° \text{ (NH}_3\text{)} > 104.5° \text{ (H}_2\text{O)}$$
7
How many sigma (σ) bonds are present in a molecule of ethene (C₂H₄)?
Ans
5
Review
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Count the C=C sigma bond
The double bond between the two carbons consists of one sigma bond and one pi bond.
$$\text{C–C } \sigma \text{ bonds} = 1$$
Count C–H sigma bonds
Each carbon is bonded to 2 hydrogen atoms, giving 4 C–H sigma bonds total.
$$\text{C–H } \sigma \text{ bonds} = 2 + 2 = 4$$
Total sigma bonds
Add all sigma bonds together.
$$\text{Total } \sigma \text{ bonds} = 1 + 4 = 5$$
8
The ideal tetrahedral bond angle is approximately 109.5°. In NH₃, the observed bond angle is about 107°. What best explains this deviation?
Ans
The lone pair on nitrogen occupies more space than a bonding pair and compresses the H–N–H angle
Review
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Identify the electron groups on nitrogen
Nitrogen has 3 bonding pairs and 1 lone pair in a tetrahedral electron geometry.
$$3\text{bp} + 1\text{lp} \rightarrow \text{steric number} = 4$$
Analyze the lone pair effect
The lone pair occupies more space and repels bonding pairs more than they repel each other.
$$\text{lp-bp} > \text{bp-bp} \rightarrow \text{compressed angles}$$
Determine the result
The bonding pairs are pushed together, reducing the bond angle below 109.5°.
$$109.5° \rightarrow 107°$$
9
A central atom has 3 bond pairs and 1 lone pair. What is its steric number and corresponding hybridization?
Ans
Steric number = 4, sp³ hybridization
Review
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Calculate steric number
Steric number includes both bond pairs and lone pairs.
$$\text{Steric number} = 3\text{bp} + 1\text{lp} = 4$$
Determine hybridization
A steric number of 4 always gives sp³ hybridization.
$$\text{SN} = 4 \rightarrow sp^3$$
Predict molecular shape
With 3 bonded atoms and 1 lone pair, the shape is trigonal pyramidal.
$$\text{Geometry: trigonal pyramidal}$$
10
What is the molecular shape of BeCl₂?
Ans
Linear
Review
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Count electron groups on beryllium
Beryllium forms 2 sigma bonds with chlorine atoms and has zero lone pairs.
$$\text{Steric number} = 2 + 0 = 2$$
Identify hybridization and shape
A steric number of 2 gives sp hybridization and linear geometry.
$$sp \rightarrow \text{linear}, \quad \theta = 180°$$
11
How many pi (π) bonds are present in a molecule of ethyne (C₂H₂)?
Ans
2
Review
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Analyze the carbon–carbon bond
The C≡C triple bond contains 1 sigma bond and 2 pi bonds.
$$\text{C≡C: } 1\sigma + 2\pi$$
Count remaining bonds
The two C–H bonds are single bonds (sigma only).
$$\text{C–H bonds: } 2\sigma + 0\pi$$
Total pi bonds
Only the triple bond contributes pi bonds.
$$\text{Total } \pi \text{ bonds} = 2$$
12
A molecule has a central atom with sp³ hybridization, 2 bond pairs, and 2 lone pairs. What is its molecular shape and approximate bond angle?
Ans
Bent (angular), 104.5°
Review
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Verify steric number
Steric number = 2 bond pairs + 2 lone pairs = 4.
$$\text{SN} = 4 \rightarrow sp^3 \text{ electron geometry (tetrahedral)}$$
Determine molecular shape
With 2 bonded atoms and 2 lone pairs, only the atom positions define the molecular shape.
$$\text{2 atoms + 2 lone pairs} \rightarrow \text{bent (angular)}$$
Predict bond angle
Two lone pairs compress the bond angle more than one lone pair would.
$$\theta \approx 104.5°$$
13
Which statement correctly describes the formation of sp² hybrid orbitals?
Ans
One s and two p orbitals combine to form three sp² hybrid orbitals, leaving one unhybridized p orbital
Review
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Identify the mixing components
sp² hybridization involves 1 s orbital and 2 p orbitals.
$$1s + 2p$$
Determine the products
These combine to form 3 sp² hybrid orbitals, leaving 1 unhybridized p orbital.
$$1s + 2p \rightarrow 3\,sp^2 + 1\,p$$
Connect to molecular geometry
The three sp² orbitals arrange at 120° in a trigonal planar geometry.
$$\theta_{sp^2} = 120°$$
14
How many sigma (σ) bonds are present in a molecule of ethyne (C₂H₂)?
Ans
3
Review
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Count C–C sigma bond
The C≡C triple bond has exactly 1 sigma bond.
$$\text{C–C } \sigma = 1$$
Count C–H sigma bonds
Each carbon has 1 hydrogen, contributing 2 sigma bonds.
$$\text{C–H } \sigma = 1 + 1 = 2$$
Total sigma bonds
Sum all sigma bonds.
$$\text{Total } \sigma = 1 + 2 = 3$$
15
A molecule has a central atom with a steric number of 3 and no lone pairs. What is its molecular shape and hybridization?
Ans
Trigonal planar, sp²
Review
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Interpret the steric number
Steric number 3 with no lone pairs means 3 bond pairs.
$$3\text{bp} + 0\text{lp} = \text{SN } 3$$
Determine hybridization
Steric number 3 corresponds to sp² hybridization.
$$\text{SN } 3 \rightarrow sp^2$$
Identify molecular shape
With all 3 positions occupied by bonded atoms and no lone pairs, the shape is trigonal planar.
$$\text{Shape: trigonal planar}, \quad \theta = 120°$$
Question 1 of 15
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1
What is the fundamental principle of Valence Shell Electron Pair Repulsion (VSEPR) theory?