JAM Chemical Bonding and Molecular Structure
Bonding questions on IIT-JAM lean heavily on two skills applied to inorganic molecules rather than organic ones: working out a molecule's shape from VSEPR and hybridisation, and reasoning quantitatively about how "ionic" or "covalent" a bond really is. This article covers both, using expanded-octet examples that go beyond the organic hybridisation already covered elsewhere on this site, plus dipole moment and hydrogen bonding.
VSEPR — the geometry follows from electron-pair count
SN 2 → linear · SN 3 → trigonal planar · SN 4 → tetrahedral · SN 5 → trigonal bipyramidal · SN 6 → octahedral
Lone pairs repel more strongly than bonding pairs (lone pair–lone pair > lone pair–bond pair > bond pair–bond pair), which is why NH₃'s bond angle compresses to about 107° and H₂O's to about 104.5°, both below the ideal tetrahedral 109.5° — the lone pair(s) push the bonding pairs closer together. In a trigonal-bipyramidal (SN 5) arrangement, a lone pair preferentially occupies an equatorial position, since equatorial positions have only two close 90° neighbours while axial positions have three, so an equatorial lone pair experiences less repulsion overall.
Hybridisation with d-orbitals — inorganic examples
Hybridisation as it applies to organic acidity and resonance (sp/sp²/sp³ in carbon chains) is covered in the companion resonance article. Here the focus is on central atoms that use d orbitals to accommodate more than four electron domains — a case organic carbon never needs, since carbon has no accessible d orbitals in its valence shell.
Worked example 1 — geometry and hybridisation of ClF₃.
Chlorine has 7 valence electrons. Three are used to form σ bonds to the three fluorine atoms, leaving 4 electrons as 2 lone pairs.
Total electron domains = 3 bonding + 2 lone pair = 5 → hybridisation sp³d, electron geometry trigonal bipyramidal. Following the rule that lone pairs prefer equatorial positions, both lone pairs occupy equatorial sites, leaving the three fluorines in a T-shaped molecular arrangement.
Worked example 2 — geometry and hybridisation of XeF₄.
Xenon has 8 valence electrons. Four are used to form σ bonds to the four fluorine atoms, leaving 4 electrons as 2 lone pairs.
Total electron domains = 4 bonding + 2 lone pair = 6 → hybridisation sp³d², electron geometry octahedral. In an octahedral arrangement, two lone pairs minimise repulsion by sitting opposite each other (180° apart, on the two axial positions), leaving the four fluorines arranged in a square planar molecular shape.
Note the distinction both examples rely on: the electron geometry (trigonal bipyramidal, octahedral) accounts for lone pairs; the molecular shape (T-shaped, square planar) is the name given only to where the atoms actually sit, with lone pairs invisible in the name but not in the reasoning that placed them.
Dipole moment and percentage ionic character
% ionic character = (μ observed ÷ μ calculated for 100% ionic character) × 100
Worked example 3 — percentage ionic character of HCl. HCl has a bond length of 1.274 Å and an observed dipole moment of 1.03 D. Find its percentage ionic character.
First find the dipole moment HCl would have if the bond were 100% ionic — a full electronic charge fully separated by the bond length. In Debye, this works out to μ(100% ionic) = 4.803 × bond length in Å.
μ(100% ionic) = 4.803 × 1.274 = 6.12 D.
% ionic character = (1.03 ÷ 6.12) × 100 = 16.8%.
This is the standard way a textbook value like "HCl is about 17% ionic" is actually obtained — never assume it, calculate it from the measured bond length and dipole moment.
Resonance in an inorganic species
Resonance is not only an organic tool. Ozone, O₃, is a bent molecule described by two equivalent resonance structures, each with one O=O double bond and one O–O single bond, differing only in which terminal oxygen carries the double bond. Because the two contributors are equivalent, they contribute equally, and the two experimentally measured O–O bond lengths in ozone are identical — intermediate between a pure single and a pure double bond, exactly as equivalent resonance forms predict for organic examples such as the carbonate ion.
Molecular orbital theory — where the anomaly comes from
Simple valence bond structures predict O₂ should be diamagnetic (every electron paired in a clean double bond), yet liquid O₂ is measurably paramagnetic. This is resolved only by molecular orbital theory, which places two electrons singly in a degenerate pair of antibonding π orbitals. The full derivation of the second-row diatomic MO diagrams, the s–p mixing that reorders σ and π levels between nitrogen and oxygen, and the bond-order table for B₂ through Ne₂ are worked through completely in a dedicated article, linked below, rather than repeated here.
Hydrogen bonding and its structural consequences
Hydrogen bonding requires H attached to a small, highly electronegative atom — N, O or F — interacting with a lone pair on another such atom nearby; it is markedly stronger than an ordinary dipole-dipole interaction, which is why NH₃, H₂O and HF all show boiling points far above what their molecular weight alone would predict compared with their heavier group congeners (PH₃, H₂S, HCl).
Intramolecular hydrogen bonding can change physical properties in the opposite direction from intermolecular bonding. ortho-Nitrophenol forms a hydrogen bond between its –OH and the adjacent nitro group within the same molecule, which both closes off that –OH from bonding to other molecules and gives the compound a lower boiling point and lower water solubility than its para isomer. para-Nitrophenol has no such intramolecular option (the groups are too far apart), so its –OH is free to form extensive intermolecular hydrogen-bonded networks with neighbouring molecules, raising its boiling point well above the ortho isomer's.
Common mistakes
- Naming the molecular shape as the same as the electron-domain geometry. Lone pairs are counted for hybridisation but are excluded from the shape's name — SN 5 with 2 lone pairs is trigonal bipyramidal electron geometry but a T-shaped molecule.
- Placing lone pairs axially in a trigonal-bipyramidal arrangement. They preferentially occupy the less-crowded equatorial positions.
- Comparing raw dipole moments across different bonds to judge "more ionic". A fair comparison needs the percentage ionic character, normalised against each bond's own 100%-ionic value — a longer bond has a larger theoretical maximum, which changes the comparison.
- Assuming any O–H or N–H counts as hydrogen bonding strength. The effect is specific to N, O and F as both the donor and acceptor atom; a C–H···O contact is far weaker and not conventional hydrogen bonding.
- Forgetting intramolecular hydrogen bonding can lower, not raise, a boiling point, by removing that –OH from the pool available for intermolecular bonding.
Exam relevance
| Question style | What to check first |
|---|---|
| Predict a molecular shape | Steric number (bonds + lone pairs), then where lone pairs sit to minimise repulsion |
| Numerical on percentage ionic character | μ(100% ionic) = 4.803 × bond length in Å, then compare to the observed μ |
| Explain equal bond lengths in a molecule with more than one Lewis structure | Equivalent resonance contributors |
| Explain an anomalous magnetic property | Whether VBT/Lewis structures miss something MO theory predicts correctly |
| Compare boiling points of positional isomers with H-bonding groups | Whether intramolecular H-bonding is geometrically possible (ortho) or not (para) |
Check the scientific constants and molar masses behind a bonding calculation. The suite's constants tool and molar mass calculator both help verify a percentage-ionic-character or dipole-moment numerical quickly.
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