VSEPR Shapes of Molecules, Bond Angles, Electronegativity & Hydrogen Bonding
Valence Shell Electron Pair Repulsion (VSEPR) theory predicts the 3D geometry and bond angles of any covalent molecule or polyatomic ion from the number of bonding pairs and lone pairs of electrons surrounding the central atom. Because non-bonding lone pairs are held closer to the central nucleus than shared bonding pairs, lone-pair repulsion compresses standard tetrahedral, trigonal bipyramidal, or octahedral bond angles by approximately 2.5° per lone pair. Combining molecular 3D symmetry with Pauling electronegativity differences determines whether bond dipoles cancel out or create a net permanent molecular dipole. Examiners test your ability to deduce shapes of complex ions (like ICl₄⁻ or SF₆) and compare boiling points across van der Waals (induced dipole-dipole), permanent dipole-dipole, and hydrogen bonding interactions.
- VSEPR Core Rules & The 2.5° Lone-Pair Compression Rule
- Electron pairs in the outer shell of a central atom negatively repel one another and arrange themselves in 3D space as far apart as possible to minimise repulsion. Lone pairs are held closer to the central nucleus and occupy more spatial volume than shared bonding pairs, so strength of repulsion follows: Lone Pair-Lone Pair (LP-LP) > Lone Pair-Bonding Pair (LP-BP) > Bonding Pair-Bonding Pair (BP-BP). Each lone pair reduces the bond angle from the parent geometry by ~2.5°.
- Standard geometries without lone pairs (2 to 6 bonding pairs)
- 2 bonding pairs, 0 lone pairs (e.g. BeCl₂, CO₂): Linear, 180°. 3 bonding pairs, 0 lone pairs (e.g. BF₃, AlCl₃, CO₃²⁻): Trigonal planar, 120°. 4 bonding pairs, 0 lone pairs (e.g. CH₄, NH₄⁺, SiCl₄): Tetrahedral, 109.5°. 5 bonding pairs, 0 lone pairs (e.g. PCl₅): Trigonal bipyramidal, 90° (axial-equatorial) and 120° (equatorial). 6 bonding pairs, 0 lone pairs (e.g. SF₆): Octahedral, 90°.
- Geometries with lone pairs (Pyramidal, Bent, Seesaw, square planar)
- 3 BP + 1 LP (e.g. NH₃, H₃O⁺, PCl₃): Trigonal pyramidal, 107°. 2 BP + 2 LP (e.g. H₂O, H₂S): Bent / V-shaped / Non-linear, 104.5°. 4 BP + 1 LP (e.g. SF₄): Seesaw / Distorted tetrahedral, ~87° and ~102°. 3 BP + 2 LP (e.g. ClF₃): T-shaped, ~87.5°. 5 BP + 1 LP (e.g. BrF₅, IF₅): Square pyramidal, ~81.9°. 4 BP + 2 LP (e.g. XeF₄, ICl₄⁻): Square planar, 90° (lone pairs sit at 180° opposite each other so their repulsions cancel symmetrically).
VSEPR Shapes of Molecules, Bond Angles, Electronegativity & Hydrogen Bonding: 3D Molecular, Lattice, DNA & Vector Explorer
Drag directly on the 3D model to rotate it through any angle, or switch tabs to inspect VSEPR bond angles, crystal lattices, DNA base pairing, and 3D space diagonals.
CH₄ / NH₄⁺ — Bond Angle: 109.5°
- Bonding Pairs:
- 4
- Lone Pairs on Central Atom:
- 0
- Molecule Polarity:
- Non-polar
4 bonding pairs and 0 lone pairs repel equally in 3D space to the vertices of a regular tetrahedron (109.5°).
Key Concepts & Exam Definitions (6)
- VSEPR Core Rules & The 2.5° Lone-Pair Compression Rule
- Electron pairs in the outer shell of a central atom negatively repel one another and arrange themselves in 3D space as far apart as possible to minimise repulsion. Lone pairs are held closer to the central nucleus and occupy more spatial volume than shared bonding pairs, so strength of repulsion follows: Lone Pair-Lone Pair (LP-LP) > Lone Pair-Bonding Pair (LP-BP) > Bonding Pair-Bonding Pair (BP-BP). Each lone pair reduces the bond angle from the parent geometry by ~2.5°.
- Standard geometries without lone pairs (2 to 6 bonding pairs)
- 2 bonding pairs, 0 lone pairs (e.g. BeCl₂, CO₂): Linear, 180°. 3 bonding pairs, 0 lone pairs (e.g. BF₃, AlCl₃, CO₃²⁻): Trigonal planar, 120°. 4 bonding pairs, 0 lone pairs (e.g. CH₄, NH₄⁺, SiCl₄): Tetrahedral, 109.5°. 5 bonding pairs, 0 lone pairs (e.g. PCl₅): Trigonal bipyramidal, 90° (axial-equatorial) and 120° (equatorial). 6 bonding pairs, 0 lone pairs (e.g. SF₆): Octahedral, 90°.
- Geometries with lone pairs (Pyramidal, Bent, Seesaw, square planar)
- 3 BP + 1 LP (e.g. NH₃, H₃O⁺, PCl₃): Trigonal pyramidal, 107°. 2 BP + 2 LP (e.g. H₂O, H₂S): Bent / V-shaped / Non-linear, 104.5°. 4 BP + 1 LP (e.g. SF₄): Seesaw / Distorted tetrahedral, ~87° and ~102°. 3 BP + 2 LP (e.g. ClF₃): T-shaped, ~87.5°. 5 BP + 1 LP (e.g. BrF₅, IF₅): Square pyramidal, ~81.9°. 4 BP + 2 LP (e.g. XeF₄, ICl₄⁻): Square planar, 90° (lone pairs sit at 180° opposite each other so their repulsions cancel symmetrically).
- Electronegativity, Bond Polarity & Molecular Symmetry
- Electronegativity is the power of an atom to attract the shared pair of electrons in a covalent bond (increases across a period as nuclear charge rises and radius falls; decreases down a group as radius and shielding increase; F > O > N > Cl). A difference in electronegativity creates a polar covalent bond (δ⁺ - δ⁻). However, if a molecule is completely symmetrical (e.g. linear CO₂, trigonal planar BF₃, tetrahedral CCl₄, octahedral SF₆), the individual bond dipoles cancel out in 3D space so the overall molecule is non-polar.
- Van der Waals (Induced Dipole-Dipole) & Permanent Dipole Forces
- 1. Induced dipole-dipole (London dispersion / van der Waals) forces exist between all atoms and molecules: random movement of electrons creates an instantaneous dipole in one molecule, which induces a dipole in a neighbouring molecule. Strength increases with total number of electrons (Mr) and straight-chain surface area contact. 2. Permanent dipole-dipole forces act between asymmetrical polar molecules (e.g. HCl, CHCl₃, propanone) in addition to van der Waals forces.
- Hydrogen Bonding & Anomalous Properties of Water/Ice
- The strongest intermolecular force (~10% the strength of a covalent bond), occurring only when Hydrogen is covalently bonded to a very small, highly electronegative atom with a lone pair: Nitrogen (N), Oxygen (O), or Fluorine (F). The strongly δ⁺ H atom is attracted to a lone pair on an N, O, or F atom of a neighbouring molecule (shown with a dashed line along a 180° bond angle). Gives HF, H₂O, and NH₃ anomalously high boiling points compared to Group 15/16/17 hydrides, and causes ice to be less dense than liquid water because each H₂O molecule forms 4 hydrogen bonds in an open 3D tetrahedral lattice with empty spaces.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
How do you calculate the number of bonding pairs and lone pairs around a central atom X in a molecule or ion?
State the shapes and exact bond angles of CH₄, NH₃, and H₂O.
What is the shape and bond angle of XeF₄ (or ICl₄⁻), and why?
Why is CCl₄ a non-polar molecule even though each C-Cl bond is strongly polar?
Why does H₂O have a higher boiling point (100°C) than HF (19.5°C) and NH₃ (-33°C), even though Fluorine is more electronegative than Oxygen?
Why does pentane have a higher boiling point than 2,2-dimethylpropane, even though both have the molecular formula C₅H₁₂?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Predict and explain the shapes and bond angles of the boron trifluoride molecule (BF₃), the ammonia molecule (NH₃), and the ammonium ion (NH₄⁺). (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. In BF₃, boron is in Group 13 (3 valence electrons) and forms 3 covalent bonds with fluorine, giving 3 bonding pairs and 0 lone pairs [1 mark]. 2. The 3 bonding pairs repel equally to get as far apart as possible, giving BF₃ a trigonal planar shape with a bond angle of 120° [1 mark]. 3. In NH₃, nitrogen is in Group 15 (5 valence electrons) and forms 3 covalent bonds with hydrogen, giving 3 bonding pairs and 1 lone pair [1 mark]. 4. Because lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion, the bond angle is compressed from 109.5° to 107°, giving NH₃ a trigonal pyramidal shape [1 mark]. 5. In NH₄⁺, the lone pair on nitrogen forms a dative covalent (coordinate) bond with an H⁺ ion, resulting in 4 bonding pairs and 0 lone pairs [1 mark]. 6. All 4 bonding pairs repel equally, giving NH₄⁺ a symmetrical tetrahedral shape with a bond angle of 109.5° [1 mark].
- Q2
State the shapes and bond angles of SF₆ and ClF₄⁻. Show how you deduced the number of bonding pairs and lone pairs in each species. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. In SF₆, sulfur (Group 16) contributes 6 valence electrons and the 6 fluorine atoms contribute 6 electrons, giving 12 electrons = 6 bonding pairs and 0 lone pairs [1 mark]. 2. Equal repulsion between 6 bonding pairs gives SF₆ an octahedral shape with a bond angle of 90° [1 mark]. 3. In ClF₄⁻, chlorine (Group 17) contributes 7 valence electrons, 4 fluorine atoms contribute 4 electrons, and the -1 charge adds 1 electron, giving 12 electrons = 6 pairs (4 bonding pairs and 2 lone pairs) [1 mark]. 4. The 2 lone pairs arrange themselves at 180° opposite each other to minimise lone pair-lone pair repulsion, giving ClF₄⁻ a square planar shape with a bond angle of 90° [1 mark].
- Q3
Explain why the boiling points of the Group 16 hydrides increase from H₂S (-60°C) to H₂Se (-41°C) to H₂Te (-2°C), whereas H₂O has a much higher boiling point (+100°C). (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. H₂O molecules have hydrogen atoms covalently bonded directly to a small, highly electronegative oxygen atom with lone pairs, so H₂O molecules attract one another via strong intermolecular hydrogen bonds [1 mark]. 2. Considerably more thermal energy is required to overcome hydrogen bonds in H₂O than the weaker intermolecular forces in H₂S, H₂Se, and H₂Te [1 mark]. 3. From H₂S to H₂Se to H₂Te, the total number of electrons per molecule (and molecular size/Mr) increases down the group [1 mark]. 4. More electrons create larger temporary and induced dipoles, increasing the strength of the van der Waals (induced dipole-dipole) forces between molecules and requiring more heat energy to boil [1 mark].
- Q4
Draw or describe a diagram showing how two molecules of ethanol (CH₃CH₂OH) interact via a hydrogen bond, and explain why ethanol is completely miscible with water whereas chloroethane (CH₃CH₂Cl) is almost insoluble in water. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Diagram description: Show the O-H bond of both ethanol molecules with partial charges (δ⁻ on O, δ⁺ on H), at least one lone pair on each oxygen atom, and a dashed line representing the hydrogen bond from the δ⁺ H of one ethanol molecule to a lone pair on the δ⁻ O of the second molecule, with a linear 180° O - H---:O angle [2 marks]. 2. Ethanol can form intermolecular hydrogen bonds with water molecules (between the -OH group of ethanol and H₂O), releasing sufficient solvation energy to overcome the hydrogen bonds between water molecules [1 mark]. 3. Chloroethane lacks an H atom bonded to N, O, or F and can only form weaker permanent dipole-dipole and van der Waals forces, which cannot disrupt the strong hydrogen-bonded network between water molecules [1 mark].
Common Exam Questions (4)
How do double bonds (like C=O in CO₂ or S=O in SO₂) affect VSEPR shape predictions?
In VSEPR theory, a double bond (4 electrons) or triple bond (6 electrons) is treated as a single 'bonding region' (electron domain) when determining the molecular geometry. For example, SO₂ has two S=O double-bond regions and one lone pair on the sulfur atom (3 electron regions total: 2 bonding regions + 1 lone pair), giving SO₂ a bent (non-linear) shape with a bond angle of ~119°.
What is a dative covalent (coordinate) bond, and does it behave differently from a normal covalent bond once formed?
A dative covalent bond is a covalent bond in which both shared electrons come from the same atom (a donor atom with a lone pair, like N in NH₃ donating to an electron-deficient H⁺ ion or B in BF₃). Once formed, a dative covalent bond is identical in length, strength, and repulsion to any ordinary covalent bond.
Why doesn't chlorine (Cl) form hydrogen bonds in HCl, even though chlorine has the same Pauling electronegativity (3.0) as nitrogen (3.0)?
A chlorine atom is in Period 3 and has a much larger atomic radius than Period 2 nitrogen, oxygen, or fluorine. Because chlorine's lone pairs are spread over a much larger, more diffuse 3p volume, the charge density on Cl is too low to form a true hydrogen bond.
Why does iodine (I₂) sublime at a relatively low temperature whereas diamond (C) has a melting point above 3500°C?
Iodine has a simple molecular crystal lattice where I₂ molecules are held together by weak intermolecular van der Waals forces that require very little energy to break (the strong I-I covalent bonds are not broken on melting/subliming). Diamond is a giant covalent (macromolecular) lattice where every carbon atom is tetrahedrally bonded to four others by strong covalent bonds throughout the entire crystal.
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