Why molecular orbital theory is the reliable half
VSEPR and hybridisation questions require you to construct the answer - count electron pairs, decide the geometry, subtract the lone pairs. Molecular orbital questions on the common diatomics require you to recall a short list: N₂ has bond order 3 and is diamagnetic, O₂ has bond order 2 and is paramagnetic, and the oxygen ions shift that bond order by half a unit each way.
That list is perhaps ten facts and it answers a disproportionate share of the chapter's questions. It is the highest-return thing to memorise in Inorganic Chemistry, and unlike much of the subject it is derivable rather than arbitrary.
How JEE actually asks Chemical Bonding
NTA does not publish chapter-wise weightage, so figures online are coaching estimates from past papers; check the official NTA information bulletin for your session's syllabus and pattern.
The recurring types are: predict a molecular shape and bond angle from VSEPR, determine hybridisation from the steric number, compute a bond order and magnetic behaviour from molecular orbital theory, decide whether a molecule is polar, and explain an anomalous boiling point through hydrogen bonding. All five are conceptual and quick when prepared.
Key concepts, compressed
- ✓Ionic bonding involves electron transfer between atoms of very different electronegativity; covalent bonding involves sharing.
- ✓VSEPR predicts geometry by arranging electron pairs to minimise repulsion, with lone pairs repelling more strongly than bonding pairs.
- ✓Hybridisation follows from the steric number - sigma bonds plus lone pairs on the central atom.
- ✓Molecular orbital theory combines atomic orbitals into bonding and antibonding molecular orbitals across the whole molecule.
- ✓A molecule's polarity depends on both bond polarity and molecular symmetry.
- ✓Hydrogen bonding is a strong dipole interaction requiring H bonded to N, O or F, and it explains most boiling-point anomalies.
Relations you need before attempting the questions
| Quantity | Rule | Note |
|---|---|---|
| Formal charge | V − N − B/2 | valence, non-bonding, bonding electrons |
| Steric number | sigma bonds + lone pairs | pi bonds not counted |
| SN 2 / 3 / 4 | sp / sp² / sp³ | 180° / 120° / 109.5° |
| SN 5 / 6 | sp³d / sp³d² | trigonal bipyramidal / octahedral |
| Repulsion order | lp-lp > lp-bp > bp-bp | lone pairs compress angles |
| Bond order (MOT) | (Nb − Na)/2 | |
| N₂ | bond order 3, diamagnetic | 14 electrons |
| O₂ | bond order 2, paramagnetic | 16 electrons, 2 unpaired |
| O₂⁺ / O₂⁻ / O₂²⁻ | bond order 2.5 / 1.5 / 1 | |
| He₂ | bond order 0 | does not exist |
| Dipole moment | μ = q × d | vector sum over bonds |
| Hydrogen bonding | H bonded to N, O or F |
The five mistakes that cost the most marks
- ✓Naming the shape from the hybridisation. Water is sp³ but bent; ammonia is sp³ but pyramidal. The lone pairs occupy positions without appearing in the shape's name.
- ✓Assuming polar bonds mean a polar molecule. Carbon dioxide has two polar bonds and zero net dipole because they cancel.
- ✓Counting pi bonds in the steric number. Only sigma bonds and lone pairs contribute to hybridisation.
- ✓Giving oxygen a diamagnetic ground state. O₂ has two unpaired electrons, and this is the standard demonstration of molecular orbital theory's advantage.
- ✓Treating hydrogen bonding as available to any hydrogen. It requires nitrogen, oxygen or fluorine specifically.
Practice set 1: ionic and covalent bonding
1. How do ionic and covalent bonds form?
An ionic bond forms by transfer of electrons between atoms of very different electronegativity, producing oppositely charged ions held by electrostatic attraction. A covalent bond forms by sharing electron pairs between atoms of similar electronegativity. Most real bonds lie somewhere between the two extremes.
2. What is lattice energy and what does it depend on?
The energy released when gaseous ions combine to form one mole of a solid ionic crystal. It increases with higher ionic charges and decreases with larger ionic radii, so magnesium oxide has a much larger lattice energy than sodium chloride - doubled charges on both ions. It is the main determinant of an ionic compound's melting point.
3. What do Fajans' rules predict?
The degree of covalent character in an ionic bond. Covalent character increases with a small, highly charged cation, a large, highly charged anion, and a cation with a pseudo noble-gas configuration. This is why aluminium chloride is substantially covalent while sodium chloride is not.
4. What is the octet rule and which species break it?
Atoms tend to acquire eight valence electrons. Exceptions fall into three groups: incomplete octets such as BF₃ with six, expanded octets such as SF₆ with twelve - possible only from period three onward where d orbitals are accessible - and odd-electron species such as NO. Questions frequently pick an exception deliberately.
5. How is formal charge calculated?
Formal charge = V − N − B/2, where V is the valence electrons of the free atom, N the non-bonding electrons, and B the bonding electrons on that atom. It is used to choose between competing Lewis structures - the preferred structure minimises formal charges and places any negative formal charge on the most electronegative atom.
6. What is the procedure for drawing a Lewis structure?
Count the total valence electrons, place the least electronegative atom at the centre, connect the atoms with single bonds, distribute the remaining electrons as lone pairs to complete octets on the outer atoms first, and then form multiple bonds if the central atom is short. Finally check formal charges to confirm the best structure.
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Practice set 2: VSEPR and molecular shape
7. What principle underlies VSEPR theory?
Electron pairs around a central atom arrange themselves as far apart as possible to minimise repulsion. Both bonding pairs and lone pairs count in determining the arrangement, but only the positions of the atoms determine the shape's name - which is the distinction the next three questions turn on.
8. How does the steric number determine the electron geometry?
Steric number 2 gives linear, 3 trigonal planar, 4 tetrahedral, 5 trigonal bipyramidal and 6 octahedral. The steric number is the count of sigma bonds plus lone pairs on the central atom. This gives the arrangement of electron pairs; the molecular shape then follows by ignoring the lone pairs.
9. What is the shape and bond angle of methane?
Tetrahedral with bond angles of 109.5°. The carbon has four bonding pairs and no lone pairs, so the steric number is 4 and the electron geometry is tetrahedral. Because there are no lone pairs, the molecular shape is the same as the electron geometry - which is precisely what changes in the next two questions.
10. What is the shape and approximate bond angle of ammonia?
Trigonal pyramidal with bond angles of about 107°. Nitrogen has three bonding pairs and one lone pair, giving steric number 4 and a tetrahedral electron geometry - but the lone pair is not part of the shape, so the molecule is pyramidal. The lone pair also repels more strongly than bonding pairs, compressing the angle from 109.5° to about 107°.
11. What is the shape and approximate bond angle of water?
Bent, with a bond angle of about 104.5°. Oxygen has two bonding pairs and two lone pairs, so the steric number is still 4 and the electron geometry is still tetrahedral - but two lone pairs compress the angle further than ammonia's one. The progression 109.5°, 107°, 104.5° across methane, ammonia and water is worth remembering as a set.
12. What is the order of repulsion between electron pairs?
Lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair. Lone pairs are held by only one nucleus, so they spread out more and repel more strongly. This ordering is what explains the decreasing bond angles in question 11.
13. What is the shape of sulphur hexafluoride?
Octahedral, with all bond angles 90°. Sulphur has six bonding pairs and no lone pairs, giving a steric number of 6 and sp³d² hybridisation. This is an expanded octet with twelve electrons around sulphur, which is possible because sulphur is in period three and has accessible d orbitals.
Practice set 3: hybridisation
14. How do you determine hybridisation quickly?
Count sigma bonds plus lone pairs on the central atom. Two gives sp, three sp², four sp³, five sp³d and six sp³d². Pi bonds are not counted, which is why ethene's carbon is sp² despite having a double bond - the double bond contributes one sigma and one pi.
15. What bond angles correspond to sp, sp² and sp³ hybridisation?
180° for sp, 120° for sp², and 109.5° for sp³. These are the ideal angles for the corresponding electron geometries. Actual angles deviate when lone pairs are present, as in ammonia and water, but the ideal value is what the hybridisation itself predicts.
16. What is the hybridisation of each carbon in ethene?
sp². Each carbon has three sigma bonds - two to hydrogen and one to the other carbon - and no lone pairs, giving a steric number of 3. The remaining p orbital on each carbon forms the pi bond. The molecule is planar with bond angles of about 120°, which is exactly what sp² predicts.
17. What is the hybridisation of each carbon in ethyne?
sp. Each carbon has two sigma bonds - one to hydrogen and one to the other carbon - and no lone pairs, so the steric number is 2. The two remaining p orbitals on each carbon form the two pi bonds of the triple bond. The molecule is linear, and the higher s character makes the C-H bond more acidic.
18. Why are pi bonds excluded from the steric number?
Because hybrid orbitals form only sigma bonds. Pi bonds are formed by the sideways overlap of unhybridised p orbitals, which remain outside the hybridisation scheme. Counting them would give the wrong answer for every multiply-bonded species, so the exclusion is essential rather than a technicality.
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Practice set 4: molecular orbital theory
19. What are bonding and antibonding molecular orbitals?
Bonding orbitals form by constructive overlap of atomic orbitals and are lower in energy than the atomic orbitals; antibonding orbitals form by destructive overlap and are higher in energy, with a node between the nuclei. Electrons in bonding orbitals hold the molecule together; electrons in antibonding orbitals work against it.
20. How is bond order calculated in molecular orbital theory?
Bond order = (number of bonding electrons − number of antibonding electrons)/2. A bond order of zero means the species does not exist as a stable molecule. Higher bond order means a shorter and stronger bond, which is how the theory predicts physical properties rather than just describing them.
21. What are the bond order and magnetic behaviour of O₂?
Bond order 2, and paramagnetic with two unpaired electrons. Oxygen has 16 electrons, giving 10 bonding and 6 antibonding, so the bond order is (10 − 6)/2 = 2. The two unpaired electrons occupy degenerate antibonding pi orbitals singly by Hund's rule. This is the single most-examined result in the chapter.
22. What are the bond order and magnetic behaviour of N₂?
Bond order 3, and diamagnetic. Nitrogen has 14 electrons, giving 10 bonding and 4 antibonding, so the bond order is (10 − 4)/2 = 3 - a triple bond. All electrons are paired, so it is diamagnetic. The high bond order explains nitrogen's exceptional stability and inertness.
23. Why does He₂ not exist?
Because its bond order is zero. Helium contributes four electrons in total, filling both the bonding and antibonding sigma orbitals equally, so (2 − 2)/2 = 0. The stabilisation from the bonding pair is exactly cancelled by the destabilisation from the antibonding pair, leaving no net bond.
24. What are the bond orders of O₂⁺, O₂⁻ and O₂²⁻?
2.5, 1.5 and 1 respectively. Starting from O₂ with 16 electrons and bond order 2: removing an electron removes an antibonding one, raising the order to 2.5; adding one lowers it to 1.5; adding two lowers it to 1. Bond length increases in the same sequence, since higher bond order means a shorter bond.
25. Why is molecular orbital theory needed when valence bond theory already exists?
Because valence bond theory cannot explain oxygen's paramagnetism. Its Lewis structure shows all electrons paired, predicting diamagnetism, which contradicts experiment. Molecular orbital theory places two electrons singly in degenerate antibonding orbitals and predicts the observed behaviour correctly. It also explains fractional bond orders and why species like He₂ do not exist.
Practice set 5: polarity and intermolecular forces
26. Why is carbon dioxide non-polar while water is polar?
Because of symmetry. Carbon dioxide is linear, so its two bond dipoles point in exactly opposite directions and cancel to zero. Water is bent, so its two bond dipoles have a resultant. Both molecules contain polar bonds - the difference is entirely geometric, which is why shape must be determined before polarity.
27. How does electronegativity difference relate to bond type?
A small difference gives a non-polar covalent bond, a moderate difference a polar covalent bond, and a large difference an ionic bond. The boundaries are gradual rather than sharp - the commonly quoted figure of about 1.7 is a rough guide, not a rule, and many compounds sit ambiguously between the categories.
28. What is required for hydrogen bonding?
Hydrogen must be covalently bonded to nitrogen, oxygen or fluorine, and there must be a lone pair on an electronegative atom to accept it. These three elements are small and highly electronegative enough to create the necessary charge concentration. Hydrogen bonded to carbon or chlorine does not qualify, however polar the bond appears.
29. Why does water have an anomalously high boiling point?
Because of extensive hydrogen bonding. Comparing hydrides down group 16, water should boil around −80 °C by extrapolation from H₂S, H₂Se and H₂Te, and it boils at 100 °C instead. Each water molecule can form up to four hydrogen bonds, which requires far more energy to overcome. Ammonia and hydrogen fluoride show the same anomaly.
30. What are van der Waals forces and what do they depend on?
Weak intermolecular attractions arising from instantaneous and induced dipoles, present between all molecules. They strengthen with increasing molecular size and surface area, which is why boiling points rise down a homologous series and why branched isomers boil lower than straight-chain ones - branching reduces the contact area.
How to study this chapter efficiently
- ✓Learn the molecular orbital results for the common diatomics as a fixed list. It is the highest-return memorisation in Inorganic Chemistry.
- ✓For every shape question, write the steric number first, then the electron geometry, then remove the lone pairs to name the shape. Three steps, always in that order.
- ✓Determine geometry before deciding polarity - symmetry is what cancels bond dipoles.
- ✓Remember the methane-ammonia-water angle progression as a single fact: 109.5°, 107°, 104.5°.
- ✓Check that any hydrogen bond you invoke involves N, O or F.
- ✓Study this chapter with Atomic Structure, since the orbital concepts underpin both.
Turn this into active practice
This chapter's questions are recognition-based, which means they are fast when prepared and impossible when not - there is no way to derive a bond order under exam pressure if the electron count is unfamiliar. That makes it unusually well suited to repeated short testing.
The JEE Chemical Bonding quiz on QUFF generates fresh questions across VSEPR, hybridisation, molecular orbital theory, polarity and intermolecular forces, marks them instantly and explains each answer. Do short frequent sets rather than one long session - recognition consolidates with spacing better than with volume.
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