Why atoms bond: octet rule, Lewis structures and formal charge
Link to Std 9–10 : You learnt that atoms gain, lose or share electrons to get a stable noble-gas arrangement; now we draw this with dots and lines and count charges atom by atom.
Two friends share one umbrella in the rain; both stay drier together. Atoms also share or hand over electrons so that each one ends up more stable.
The octet rule says that atoms tend to have 8 electrons in the outermost shell (2 for hydrogen). A Lewis structure shows valence electrons as dots and shared pairs as lines. Formal charge is the charge an atom would have if every shared pair were split equally.
In the figure : Lewis structures of H₂O (two single bonds, two lone pairs on O), CO₂ (two double bonds) and N₂ (one triple bond).
Rule 1: Count the valence electrons — Add the valence electrons of all atoms; add one for each negative charge and subtract one for each positive charge.
- Easy : H₂O has 2 × 1 + 6 = 8 valence electrons.
- Medium : CO₃²⁻ has 4 + 3 × 6 + 2 = 24 valence electrons.
- Exam-type : NH₄⁺ has 5 + 4 × 1 − 1 = 8 valence electrons, all of them used in four N–H bonds.
Rule 2: Draw the skeleton, then complete octets — Put the least electronegative atom (not H) in the centre, join atoms by single bonds, place the rest as lone pairs, and turn lone pairs into multiple bonds if the centre lacks an octet.
- Easy : In CO₂ the C centre needs two double bonds to get 8 electrons.
- Medium : N₂ needs a triple bond: each N then has 6 shared + 2 lone = 8 electrons.
- Exam-type : For the nitrate ion (24 electrons) one N=O double bond and two N–O⁻ single bonds are drawn, and the double bond can sit on any of the three O atoms.
Rule 3: Formal charge — Formal charge = (valence electrons) − (non-bonding electrons) − ½ (bonding electrons). Prefer the structure with charges closest to zero, and with negative charge on the more electronegative atom.
- Easy : N in NH₄⁺: 5 − 0 − 4 = +1.
- Medium : In CO₃²⁻ the double-bonded O has 6 − 4 − 2 = 0 and each single-bonded O has 6 − 6 − 1 = −1; the C has 0, so the total is −2.
- Exam-type : In O₃ the central O has 6 − 2 − 3 = +1, the single-bonded end O has −1 and the double-bonded end O has 0; the total is 0.
Rule 4: Exceptions to the octet rule — Incomplete octet (BeCl₂ has 4 electrons, BF₃ has 6), odd-electron molecules (NO has 11, NO₂ has 17) and expanded octet in period 3 and beyond (PCl₅ has 10, SF₆ has 12).
- Easy : BF₃ has only 6 electrons around boron.
- Medium : NO has 5 + 6 = 11 valence electrons, an odd number, so one electron is unpaired.
- Exam-type : SF₆ has 12 electrons around S; this is possible because S has vacant d-orbitals available in the third shell.
A lone pair belongs fully to one atom; a shared pair counts half for each atom when finding formal charge.
Forgetting the charge of an ion when counting electrons. Add electrons for a negative ion and remove them for a positive ion.
Memory clue: "V minus L minus half B" (Valence − Lone − half Bonding) gives formal charge.
Spot the difference
| Point | Formal charge | Real charge |
|---|---|---|
| Meaning | Bookkeeping by equal sharing | Actual charge distribution |
| Used for | Choosing the best Lewis structure | Predicting polarity |
| Total | Equals the ion's charge | Equals the ion's charge |
Practice questions
- 1State the octet rule.
- 2How many valence electrons are in NH₃?
- 3How many lone pairs does the O atom have in H₂O?
- 4Draw the Lewis structure of HCN and name the bond between C and N.
- 5Find the formal charge on N in NO₃⁻ if N forms one double and two single bonds and has no lone pair.
- 6Why is NO₂ called an odd-electron molecule?
- 7In CO (C≡O with one lone pair on each atom) find the formal charge on C and on O.
- 8Explain why PCl₅ exists but NCl₅ does not.
Show answers (try them yourself first)
- 1Atoms tend to have 8 electrons in the outer shell by gaining, losing or sharing electrons.
- 25 + 3 × 1 = 8 electrons.
- 3Two lone pairs.
- 4H–C≡N with a lone pair on N; the C–N bond is a triple bond (10 valence electrons in total: 1 + 4 + 5).
- 5Formal charge = 5 − 0 − 4 = +1.
- 6It has 5 + 2 × 6 = 17 valence electrons, an odd number, so one electron stays unpaired.
- 7C: 4 − 2 − 3 = −1; O: 6 − 2 − 3 = +1.
- 8P is in period 3 and can use vacant d-orbitals to hold 10 electrons; N is in period 2 with no d-orbitals, so it cannot exceed 8.
Ionic bond and lattice enthalpy
Link to Std 9–10 : You know that sodium gives an electron to chlorine to form NaCl; now we count the energy changes that make this possible.
Table salt is hard, brittle and melts only at a very high temperature. The strong pull between thousands of oppositely charged ions explains this.
An ionic bond is the electrostatic attraction between cations and anions formed by transfer of electrons. Lattice enthalpy is the energy needed to separate one mole of a solid ionic compound into gaseous ions (in this book's sign convention it is positive).
In the figure : Born–Haber cycle for NaCl: sublimation (+108), half bond dissociation (+122), ionisation (+496), electron gain (−349) and lattice formation (−788) add up to the enthalpy of formation (−411 kJ mol⁻¹).
Rule 1: Favourable conditions — Ionic bonds form easily when the metal has low ionisation enthalpy, the non-metal has a large negative electron gain enthalpy, and the lattice enthalpy is high.
- Easy : Na (low ionisation enthalpy) and Cl (high electron affinity) form an ionic bond easily.
- Medium : Electron gain by Cl is exothermic (−349 kJ mol⁻¹) but ionisation of Na costs +496 kJ mol⁻¹; the compound forms because of the large lattice energy.
- Exam-type : Ionic compounds are formed even when the first steps absorb energy, because the final release of lattice energy pays for them.
Rule 2: Born–Haber cycle — By Hess's law, formation enthalpy = sublimation + ionisation + ½ bond dissociation + electron gain enthalpy − lattice enthalpy.
- Easy : For NaCl: 108 + 496 + 122 − 349 = +377 kJ mol⁻¹ before the lattice step.
- Medium : Lattice enthalpy of NaCl = 377 − (−411) = 788 kJ mol⁻¹.
- Exam-type : A compound has sublimation 90, ionisation 420, ½ bond dissociation 120, electron gain −350 and formation enthalpy −430 kJ mol⁻¹. Lattice enthalpy = 90 + 420 + 120 − 350 + 430 = 710 kJ mol⁻¹.
Rule 3: Factors for lattice enthalpy — It rises with higher ionic charges and falls with larger ionic sizes.
- Easy : NaF has a higher lattice enthalpy than NaCl because F⁻ is smaller than Cl⁻.
- Medium : MgO has a much higher lattice enthalpy than NaCl because both ions carry a charge of 2.
- Exam-type : Arrange NaCl, MgO, NaF in increasing lattice enthalpy: NaCl < NaF < MgO.
Rule 4: Properties — High melting and boiling points, hard and brittle, conduct electricity when molten or dissolved (not as solids), and many dissolve in water.
- Easy : Solid NaCl does not conduct because its ions are locked in the lattice.
- Medium : Molten NaCl conducts because ions are free to move.
- Exam-type : Ionic solids are brittle because a small shift brings like charges face to face and the layers repel and split.
Higher lattice enthalpy means a harder, higher-melting ionic solid.
Treating lattice enthalpy as the energy of forming one ion pair. It is for one mole of the whole solid.
Memory clue: "Big charge, small size, big lattice".
Spot the difference
| Point | Ionic compound | Molecular compound |
|---|---|---|
| Units | Ions in a lattice | Separate molecules |
| Melting point | High | Usually low |
| Conducts when molten | Yes | No |
Practice questions
- 1Define ionic bond.
- 2Why does solid NaCl not conduct electricity?
- 3Which has the higher lattice enthalpy, NaF or NaCl?
- 4Name the five steps of the Born–Haber cycle for NaCl.
- 5Compare the lattice enthalpy of MgO and NaCl and give a reason.
- 6For a compound the steps before the lattice step total +377 kJ mol⁻¹ and its formation enthalpy is −411 kJ mol⁻¹. Find the lattice enthalpy.
- 7A compound has sublimation 90, ionisation 420, ½ bond dissociation 120 and electron gain −350 kJ mol⁻¹, with formation enthalpy −430 kJ mol⁻¹. Find the lattice enthalpy.
- 8Why do ionic compounds form even though ionisation of the metal needs energy?
Show answers (try them yourself first)
- 1The electrostatic attraction between oppositely charged ions formed by electron transfer.
- 2The ions are fixed in the crystal and cannot move.
- 3NaF.
- 4Sublimation of Na, ionisation of Na, dissociation of Cl₂, electron gain by Cl, formation of the lattice.
- 5MgO is higher because both ions carry a charge of 2 and are small.
- 6377 + 411 = 788 kJ mol⁻¹.
- 790 + 420 + 120 − 350 + 430 = 710 kJ mol⁻¹.
- 8The energy released when the lattice forms is larger than the energy absorbed in the earlier steps.
Covalent bond: bond parameters, polarity and dipole moment
Link to Std 9–10 : You know that non-metals share electrons; here we measure the bond with length, angle and energy, and see why sharing is sometimes unequal.
In a tug of war between a strong and a weak team, the rope sits closer to the strong team. In HCl the shared pair sits closer to the more electronegative Cl.
Bond length is the distance between the nuclei of two bonded atoms. Bond enthalpy is the energy to break one mole of that bond in gaseous molecules. Dipole moment (μ) = charge (Q) × distance (r) between the charges; it is a vector pointing towards the negative end.
In the figure : Left: bond lengths C–C 154 pm, C=C 134 pm, C≡C 120 pm. Right: HCl with δ+ on H, δ− on Cl and the dipole arrow.
Rule 1: Bond order, length and strength — Bond order is the number of bonds between two atoms. Higher bond order means shorter bond and higher bond enthalpy.
- Easy : The bond order of N≡N is 3.
- Medium : C≡C (120 pm) is shorter than C–C (154 pm), so C≡C is stronger.
- Exam-type : Arrange C–C, C=C and C≡C by bond enthalpy: C–C < C=C < C≡C.
Rule 2: Polar and non-polar bonds — A bond between identical atoms (H–H, Cl–Cl) is non-polar. A bond between atoms of different electronegativity is polar, with partial charges δ+ and δ−.
- Easy : H–H is non-polar; H–Cl is polar.
- Medium : In H–F the shared pair is pulled towards F, so F is δ− and H is δ+.
- Exam-type : A larger electronegativity difference gives a more polar bond, so H–F is more polar than H–Cl.
Rule 3: Dipole moment — μ = Q × r, measured in coulomb metre (C m); 1 debye (D) = 3.336 × 10⁻³⁰ C m. The dipole of a molecule is the vector sum of its bond dipoles.
- Easy : CO₂ is linear and the two C=O dipoles cancel, so μ = 0.
- Medium : A bond with charges ±2.4 × 10⁻²⁰ C at a distance of 150 pm has μ = 2.4 × 10⁻²⁰ × 150 × 10⁻¹² = 3.6 × 10⁻³⁰ C m = 1.08 D.
- Exam-type : HCl has μ = 1.03 D and bond length 127 pm. If it were fully ionic, μ = 4.803 × 1.27 = 6.10 D, so the ionic character is 1.03 ÷ 6.10 ≈ 17%.
Rule 4: Shape decides polarity — Molecules with polar bonds can still be non-polar if the shape is symmetric (CO₂, BF₃, CCl₄). Bent or pyramidal molecules are usually polar.
- Easy : H₂O (bent) is polar with μ = 1.85 D.
- Medium : NH₃ (pyramidal, 1.47 D) is polar, but NF₃ has a much smaller dipole (0.24 D) because the N–F dipoles oppose the lone-pair dipole.
- Exam-type : BF₃ is non-polar because three equal B–F dipoles at 120° cancel.
A molecule is polar when its bond dipoles do not cancel; symmetry gives non-polar molecules.
Saying that polar bonds always mean a polar molecule. CO₂ has polar bonds but is non-polar.
Memory clue: "Symmetric means zero". If the molecule looks the same from opposite sides, the dipole cancels.
Spot the difference
| Point | Polar covalent bond | Non-polar covalent bond |
|---|---|---|
| Atoms | Different | Identical |
| Shared pair | Closer to one atom | Exactly in the middle |
| Example | H–Cl | Cl–Cl |
Practice questions
- 1Define bond length.
- 2What is the bond order in O=O?
- 3Is Cl–Cl polar or non-polar?
- 4Why is the dipole moment of CO₂ zero?
- 5Calculate μ in D when charges ±2.4 × 10⁻²⁰ C are 150 pm apart.
- 6Arrange C–C, C=C, C≡C in increasing bond length.
- 7HCl has μ = 1.03 D and bond length 127 pm. Find the percentage ionic character.
- 8Why is NF₃ much less polar than NH₃ although F is more electronegative than H?
Show answers (try them yourself first)
- 1The distance between the nuclei of two bonded atoms.
- 2Two.
- 3Non-polar.
- 4The two equal C=O dipoles point in opposite directions along a straight line and cancel.
- 5μ = 2.4 × 10⁻²⁰ × 150 × 10⁻¹² = 3.6 × 10⁻³⁰ C m, which is 1.08 D.
- 6C≡C < C=C < C–C.
- 7μ(ionic) = 4.803 × 1.27 = 6.10 D; ionic character = 1.03 ÷ 6.10 ≈ 17%.
- 8In NF₃ the bond dipoles point towards F, opposite to the lone-pair dipole, so they partly cancel; in NH₃ they point the same way as the lone pair and add up.
Resonance and VSEPR shapes of molecules
Link to Std 9–10 : You drew simple molecules like water and methane; now we predict their three-dimensional shapes from electron-pair repulsion.
Tie balloons together at the knot: they spread out and push each other away. Electron pairs around an atom do exactly the same and decide the shape.
VSEPR theory says that electron pairs around the central atom arrange themselves as far apart as possible to reduce repulsion. Resonance is the use of two or more Lewis structures to describe one molecule whose real structure is the hybrid of them.
Resonance and VSEPR shapes of molecules
In the app you can move this figure with your finger.
Open in the appIn the figure : Basic shapes for 2 to 6 electron pairs, and the way lone pairs squeeze the bond angle (109.5° > 107° > 104.5°). The live 3D model below lets you turn the shapes.
Rule 1: Shape from electron pairs — Count bond pairs (BP) and lone pairs (LP) on the central atom. The electron-pair geometry follows the total; the molecular shape follows the atoms only.
- Easy : BeCl₂ (2 BP) is linear, 180°.
- Medium : CH₄ (4 BP) is tetrahedral, NH₃ (3 BP + 1 LP) is pyramidal and H₂O (2 BP + 2 LP) is bent.
- Exam-type : SF₄ (4 BP + 1 LP) is see-saw, ClF₃ (3 BP + 2 LP) is T-shaped and XeF₂ (2 BP + 3 LP) is linear.
Rule 2: Order of repulsion — LP–LP > LP–BP > BP–BP. Lone pairs take more room, so bond angles shrink.
- Easy : The H–O–H angle in water is 104.5°, smaller than the tetrahedral angle 109.5°.
- Medium : The bond angle falls CH₄ (109.5°) > NH₃ (107°) > H₂O (104.5°) as lone pairs increase from 0 to 1 to 2.
- Exam-type : XeF₄ (4 BP + 2 LP) is square planar because the two lone pairs sit opposite each other, as far apart as possible.
Rule 3: Resonance and bond order — Resonance spreads a double bond over several bonds. Bond order in the hybrid = (total bonds) ÷ (number of equal positions).
- Easy : Ozone O₃ has two resonance structures; each O–O bond has order 1.5.
- Medium : CO₃²⁻ has 4 bonds shared over 3 positions, so each C–O bond has order 4 ÷ 3 ≈ 1.33.
- Exam-type : All three N–O bonds in NO₃⁻ are equal in length, with bond order 1.33, which proves the real structure is a hybrid, not the separate drawings.
Rule 4: Limits of VSEPR — It predicts shape well but cannot explain why bonds form or give their energies; that needs valence bond and molecular orbital theories.
- Easy : VSEPR cannot tell the bond energy of H–H.
- Medium : It cannot say why the bond angle in H₂S (92°) is smaller than in H₂O.
- Exam-type : VSEPR is weak for transition metal compounds where d-orbitals matter.
Shape name comes from the position of atoms, not lone pairs.
Calling H₂O "tetrahedral". Its electron pairs point to a tetrahedron, but the molecule is bent.
Memory clue: "BP + LP gives the family, LP removes corners": 4 pairs = tetrahedron; remove one corner = pyramid; remove two = bent.
Spot the difference
| Point | Electron-pair geometry | Molecular shape |
|---|---|---|
| Counts | Bond pairs and lone pairs | Only atoms |
| H₂O | Tetrahedral | Bent |
| NH₃ | Tetrahedral | Pyramidal |
Practice questions
- 1State the main idea of VSEPR theory.
- 2What is the shape of CH₄?
- 3What is the bond order of each O–O bond in O₃?
- 4Predict the shape of SO₂.
- 5Predict the shape of NH₄⁺ and its bond angle.
- 6Find the C–O bond order in CO₃²⁻.
- 7Predict the shapes of ClF₃ and XeF₄.
- 8Why is the H–O–H angle smaller than H–N–H in NH₃?
Show answers (try them yourself first)
- 1Electron pairs around the central atom stay as far apart as possible, which decides the shape.
- 2Tetrahedral, bond angle 109.5°.
- 31.5.
- 42 BP + 1 LP, so bent (V-shaped).
- 54 BP and no lone pair, so tetrahedral, 109.5°.
- 64 ÷ 3 ≈ 1.33.
- 7ClF₃ has 3 BP + 2 LP: T-shaped. XeF₄ has 4 BP + 2 LP: square planar.
- 8Water has two lone pairs that repel the bond pairs more than the single lone pair in NH₃ does, so its angle is smaller (104.5° against 107°).
Valence bond theory and hybridisation
Link to Std 9–10 : You know carbon forms four bonds; now we see why all four C–H bonds in methane are identical although carbon has one s and three p electrons.
Mixing different paint colours gives new, identical shades. Mixing orbitals of different kinds gives new, equal hybrid orbitals.
In valence bond theory a bond forms by overlap of half-filled atomic orbitals. Hybridisation is the mixing of atomic orbitals of the same atom into the same number of equivalent hybrid orbitals.
Valence bond theory and hybridisation
In the app you can move this figure with your finger.
Open in the appIn the figure : sp (linear, 180°), sp² (trigonal planar, 120°) and sp³ (tetrahedral, 109.5°) with the s-character 50%, 33% and 25%. The live model below shows the same shapes in 3D.
Rule 1: Sigma and pi bonds — A σ bond forms by head-on overlap and is stronger; a π bond forms by sideways overlap of p-orbitals above and below the axis. A single bond is 1 σ; a double bond is 1 σ + 1 π; a triple bond is 1 σ + 2 π.
- Easy : Ethene (C₂H₄) has 5 σ bonds and 1 π bond.
- Medium : Ethyne (C₂H₂) has 3 σ and 2 π bonds.
- Exam-type : Benzene (C₆H₆) has 12 σ and 3 π bonds; acetaldehyde (CH₃CHO) has 6 σ and 1 π.
Rule 2: Types of hybridisation — Count σ bonds + lone pairs on the atom: 2 gives sp, 3 gives sp², 4 gives sp³, 5 gives sp³d and 6 gives sp³d².
- Easy : C in CH₄ has 4 σ bonds, so it is sp³.
- Medium : B in BF₃ has 3 σ bonds, so it is sp²; Be in BeCl₂ has 2, so it is sp.
- Exam-type : In PCl₅ (sp³d) the shape is trigonal bipyramidal; in SF₆ (sp³d²) it is octahedral.
Rule 3: Lone pairs also count — The central atom in NH₃ and H₂O is sp³ even though the shape is pyramidal or bent.
- Easy : N in NH₃ has 3 σ + 1 lone pair = 4, so it is sp³.
- Medium : O in H₂O has 2 σ + 2 lone pairs = 4, so it is sp³ and the angle is 104.5°.
- Exam-type : In CH₂=CH–C≡N the carbons are sp², sp² and sp, and the nitrogen is sp.
Rule 4: s-character — More s-character gives a shorter, stronger bond with a more electronegative carbon: sp (50%) > sp² (33%) > sp³ (25%).
- Easy : The C–H bond in ethyne is shorter than in ethane.
- Medium : sp carbon is more electronegative than sp³ carbon because s-orbitals hold electrons closer to the nucleus.
- Exam-type : Acidity of C–H follows the same trend: ethyne is more acidic than ethene, which is more acidic than ethane.
Hybrid orbitals make σ bonds and hold lone pairs; π bonds use the unhybridised p-orbitals.
Counting a π bond as a separate hybrid orbital. Only σ bonds and lone pairs are counted.
Memory clue: "2-3-4-5-6 gives sp, sp², sp³, sp³d, sp³d²" (σ bonds plus lone pairs).
Spot the difference
| Point | σ bond | π bond |
|---|---|---|
| Overlap | Head-on | Sideways |
| Strength | Stronger | Weaker |
| Free rotation | Yes | No |
Practice questions
- 1Define hybridisation.
- 2What is the hybridisation of C in CH₄?
- 3How many σ and π bonds does ethyne have?
- 4Give the hybridisation of B in BF₃ and of Be in BeCl₂.
- 5Why is the H–O–H angle 104.5° although O is sp³?
- 6Count the σ and π bonds in CH₃CHO.
- 7Give the hybridisation of each carbon and of N in CH₂=CH–C≡N.
- 8Why is ethyne more acidic than ethane?
Show answers (try them yourself first)
- 1Mixing of atomic orbitals of one atom to form equal hybrid orbitals.
- 2sp³.
- 33 σ and 2 π bonds.
- 4B is sp² and Be is sp.
- 5Two lone pairs on O repel the bond pairs more strongly and shrink the angle from 109.5°.
- 66 σ and 1 π.
- 7The first two carbons are sp² and the third carbon is sp; N is sp.
- 8The sp carbon (50% s) holds the C–H pair closer to the nucleus than sp³ (25% s), so the hydrogen leaves more easily as H⁺.
Molecular orbital theory and bond order
Link to Std 9–10 : You know that atoms share electrons; here we treat the molecule as a whole, with orbitals spread over both nuclei.
Two ripples on a pond can add up to a bigger wave or cancel each other. Atomic orbitals do the same: adding gives a bonding orbital, subtracting gives an anti-bonding one.
In molecular orbital (MO) theory, n atomic orbitals combine to give n molecular orbitals. A bonding MO has lower energy than the parent orbitals; an anti-bonding MO (marked *) has higher energy. Bond order = ½ (bonding electrons − anti-bonding electrons).
Molecular orbital theory and bond order
In the app you can move this figure with your finger.
Open in the appIn the figure : Energy levels for molecules up to N₂: blue lines are bonding and red lines are anti-bonding orbitals. For O₂ and F₂ the σ2p level falls below π2p.
Rule 1: Filling rules — Fill orbitals from low to high energy, with the Aufbau principle, Pauli exclusion principle and Hund's rule. Up to N₂ the order is σ1s < σ*1s < σ2s < σ*2s < π2p = π2p < σ2p < π*2p < σ*2p.
- Easy : H₂ has 2 electrons in σ1s, bond order = ½ (2 − 0) = 1.
- Medium : He₂ has 2 bonding and 2 anti-bonding electrons, bond order 0, so it does not exist.
- Exam-type : Li₂ has σ2s² as the valence pair, bond order ½ (4 − 2) = 1.
Rule 2: Bond order and stability — Higher bond order means a shorter, stronger and more stable bond; a bond order of zero means no bond.
- Easy : N₂ has bond order ½ (10 − 4) = 3.
- Medium : O₂ has bond order ½ (10 − 6) = 2.
- Exam-type : Stability of oxygen species: O₂⁺ (2.5) > O₂ (2) > O₂⁻ (1.5) > O₂²⁻ (1); bond length increases in the opposite order.
Rule 3: Magnetic behaviour — Molecules with all electrons paired are diamagnetic; with one or more unpaired electrons are paramagnetic.
- Easy : N₂ has no unpaired electron, so it is diamagnetic.
- Medium : O₂ has two unpaired electrons in π*2p, so it is paramagnetic (liquid oxygen is attracted to a magnet).
- Exam-type : B₂ (10 electrons, two unpaired in π2p) is paramagnetic with bond order 1; C₂ (12 electrons) is diamagnetic with bond order 2.
Rule 4: Compare with other theories — MO theory explains the paramagnetism of O₂, which the Lewis picture cannot.
- Easy : The Lewis structure of O₂ shows all electrons paired, which is wrong.
- Medium : N₂⁺ (13 electrons) has bond order 2.5 and is paramagnetic.
- Exam-type : Be₂ has 4 bonding and 4 anti-bonding electrons, bond order 0, so it is not formed.
Electrons in anti-bonding orbitals cancel the bonding effect of an equal number in bonding orbitals.
Using the O₂ order of energy levels for N₂ and below. The σ2p and π2p order is swapped.
Memory clue: "Bond order = (Nb − Na) ÷ 2" — count bonding, subtract anti-bonding, halve.
Spot the difference
| Point | Bonding MO | Anti-bonding MO |
|---|---|---|
| Energy | Lower than atomic orbitals | Higher than atomic orbitals |
| Electron density | Between nuclei | Away from the region between nuclei |
| Effect | Holds atoms together | Weakens the bond |
Practice questions
- 1Define bond order in MO theory.
- 2What is the bond order of H₂?
- 3Is N₂ paramagnetic or diamagnetic?
- 4Find the bond order of O₂.
- 5Why does He₂ not exist?
- 6Arrange O₂, O₂⁺ and O₂⁻ in increasing bond length.
- 7Find the bond order of O₂²⁻ and state its magnetic nature.
- 8Compare the bond order and magnetism of B₂ and C₂.
Show answers (try them yourself first)
- 1Half of the difference between the number of bonding and anti-bonding electrons.
- 2½ (2 − 0) = 1.
- 3Diamagnetic.
- 4½ (10 − 6) = 2.
- 5Bond order is ½ (2 − 2) = 0; the extra stability from bonding electrons is cancelled by anti-bonding ones.
- 6O₂⁺ < O₂ < O₂⁻ (bond orders 2.5, 2, 1.5).
- 7½ (10 − 8) = 1; all electrons are paired, so it is diamagnetic.
- 8B₂ has bond order 1 and is paramagnetic (two unpaired electrons); C₂ has bond order 2 and is diamagnetic.
Hydrogen bonding
Link to Std 9–10 : You know that water is liquid at room temperature while hydrogen sulphide is a gas; the reason is a special force between molecules.
Ice floats on water. Hydrogen bonds hold water molecules in an open cage in ice, so ice takes more space and is lighter than water.
A hydrogen bond is the attractive force between a hydrogen atom bonded to a highly electronegative atom (F, O or N) and a lone pair on another such atom nearby.
In the figure : An H–F molecule joined to the next one by a hydrogen bond (dotted line), and boiling points that stay high for H₂O and HF because of it.
Rule 1: Conditions — Hydrogen must be attached to F, O or N, which are small and highly electronegative, and the other molecule must have a lone pair on such an atom.
- Easy : HF, H₂O and NH₃ form hydrogen bonds.
- Medium : HCl does not form strong hydrogen bonds because Cl is larger and less electronegative.
- Exam-type : CH₄ does not form hydrogen bonds because C–H is almost non-polar.
Rule 2: Strength — A hydrogen bond is weaker than a covalent bond (usually tens of kJ mol⁻¹ compared with hundreds) but much stronger than ordinary van der Waals forces.
- Easy : Breaking an O–H covalent bond needs far more energy than breaking a hydrogen bond between water molecules.
- Medium : Boiling a liquid breaks hydrogen bonds, not covalent bonds, so water stays unchanged as H₂O vapour.
- Exam-type : The strength order is covalent bond > hydrogen bond > weak van der Waals attraction.
Rule 3: Effect on properties — Hydrogen bonding raises melting and boiling points and increases solubility in water.
- Easy : H₂O boils at 100 °C while H₂S, with no hydrogen bond, boils near −60 °C.
- Medium : Ethanol (hydrogen bonds) boils at a higher temperature than dimethyl ether of the same formula C₂H₆O.
- Exam-type : In HF, hydrogen bonding makes the boiling point (about 20 °C) far higher than that of HCl (about −85 °C).
Rule 4: Intermolecular and intramolecular — Intermolecular H-bonds join different molecules; intramolecular H-bonds form inside one molecule.
- Easy : o-Nitrophenol has an intramolecular hydrogen bond.
- Medium : p-Nitrophenol has intermolecular hydrogen bonds, so it boils higher and is less volatile than o-nitrophenol.
- Exam-type : o-Nitrophenol is steam volatile because its internal H-bond leaves nothing to link molecules together.
A hydrogen bond is a force between molecules (or parts of molecules), not a true chemical bond inside a molecule.
Saying that the H–O bond breaks when water boils. The covalent bond stays; the hydrogen bonds are broken.
Memory clue: "FON" — hydrogen needs F, O or N next to it.
Spot the difference
| Point | Intermolecular H-bond | Intramolecular H-bond |
|---|---|---|
| Where | Between molecules | Inside one molecule |
| Effect on boiling point | Raises it | Little effect |
| Example | p-Nitrophenol | o-Nitrophenol |
Practice questions
- 1Define hydrogen bond.
- 2Which three elements are needed with hydrogen for strong H-bonding?
- 3Why does ice float on water?
- 4Why is the boiling point of H₂O higher than that of H₂S?
- 5Which is stronger, a covalent bond or a hydrogen bond?
- 6Which of ethanol and dimethyl ether boils higher, and why?
- 7Why is o-nitrophenol more volatile than p-nitrophenol?
- 8Why does CH₄ not form hydrogen bonds?
Show answers (try them yourself first)
- 1The attraction between H bonded to F, O or N and a lone pair on another such atom.
- 2Fluorine, oxygen and nitrogen.
- 3Hydrogen bonds hold water molecules in an open cage in ice, so ice is less dense than liquid water.
- 4Water molecules form hydrogen bonds; H₂S does not.
- 5The covalent bond.
- 6Ethanol, because it forms intermolecular hydrogen bonds through its O–H group.
- 7o-Nitrophenol has an intramolecular hydrogen bond, so its molecules are not joined to each other; p-Nitrophenol forms intermolecular H-bonds.
- 8C is not electronegative enough, so C–H is almost non-polar and cannot provide the partial positive hydrogen.