Chemical bonding and molecular structure for NEET

Chemical bonding is the chapter the rest of chemistry leans on. Shapes explain dipole moments, hybridisation explains organic reactivity, and molecular orbital theory explains why oxygen sticks to a magnet. In inorganic chemistry especially, "why" questions end up here.

NEET asks this chapter in a small number of forms: the shape or hybridisation of a molecule or ion, bond order and magnetic behaviour from MO theory, which molecule has zero dipole moment, the order of bond angles, and counting sigma and pi bonds. Each of these has a method, and the methods are worth drilling until they are quick.

Lewis structures and formal charge

The octet rule says atoms tend to gain, lose or share electrons to reach eight in their valence shell. It explains many compounds, but it has three kinds of exception:

• Incomplete octet: the central atom has fewer than eight (BeCl₂ has 4, BF₃ has 6).
• Odd-electron molecules: NO and NO₂ cannot have every electron paired.
• Expanded octet: elements from period 3 onwards can use more than eight (PCl₅ has 10, SF₆ has 12).

When several Lewis structures are possible, formal charge helps pick the best one:

Formal charge = V − L − ½B, where V is the number of valence electrons in the free atom, L the non-bonding (lone pair) electrons and B the bonding electrons around it.

The most stable structure has the smallest formal charges.

Resonance: when one Lewis structure cannot describe a molecule, the real structure is a hybrid of several. In O₃, both O–O bonds are equal, in between a single and a double bond. In CO₃²⁻, each C–O bond has a bond order of 4/3 ≈ 1.33.

Ionic bonds and Fajans' rules

An ionic bond forms most easily between a metal with a low ionisation enthalpy and a non-metal with a strongly negative electron gain enthalpy. What makes the solid stable overall is the large lattice enthalpy released when the ions pack together.

No bond is perfectly ionic. A small, highly charged cation distorts (polarises) the electron cloud of a large anion, giving the bond some covalent character. By Fajans' rules, covalent character increases with:

• a smaller cation, or one with higher charge
• a larger anion, or one with higher charge
• a cation with a d-electron (pseudo-noble-gas) configuration

So LiCl is more covalent than NaCl, and AlCl₃ is more covalent than NaCl.

Polarity and dipole moment

Dipole moment μ = charge × distance, measured in debye (D). It is a vector, so the molecule's dipole moment is the vector sum of its bond dipoles.

• Zero dipole moment (symmetric shapes cancel): CO₂, BF₃, CH₄, CCl₄, PCl₅, SF₆, XeF₄, BeCl₂.
• Non-zero: H₂O, NH₃, SO₂, CHCl₃, CH₂Cl₂, NF₃.

A favourite comparison: NH₃ has a much larger dipole moment than NF₃, even though N–F bonds are more polar. In NH₃, the lone pair's moment adds to the N–H bond moments. In NF₃, it opposes the N–F bond moments.

VSEPR theory and shapes

The shape of a molecule depends on the total number of electron pairs around the central atom and how many of them are lone pairs. Pairs repel in the order lone pair–lone pair > lone pair–bond pair > bond pair–bond pair, which is why lone pairs squeeze bond angles.

Counting the pairs quickly: electron pairs on the central atom = ½(V + M − C + A), where V is the central atom's valence electrons, M the number of monovalent atoms attached (H or halogens), C the positive charge of a cation and A the negative charge of an anion. Oxygen atoms count as 0 in M. Then lone pairs = electron pairs − atoms bonded.

For NH₄⁺: ½(5 + 4 − 1) = 4 pairs, 4 atoms bonded, no lone pair: sp³, tetrahedral.

Sigma and pi bonds

• A σ bond forms by head-on overlap along the internuclear axis. It is stronger, and rotation about it is free.
• A π bond forms by sideways overlap of parallel p orbitals. It is weaker, and it prevents rotation.
• A single bond is 1σ. A double bond is 1σ + 1π. A triple bond is 1σ + 2π.

Hybrid orbitals form only σ bonds. The π bonds use unhybridised p orbitals.

Molecular orbital theory

When atomic orbitals combine, they form bonding MOs (lower energy) and antibonding MOs (higher energy, marked with ∗).

Bond order = ½(bonding electrons − antibonding electrons). A higher bond order means a shorter, stronger bond. A bond order of zero means the molecule does not exist. Any unpaired electron makes the species paramagnetic.

The filling order has two versions:

• For up to 14 electrons (B₂, C₂, N₂): σ1s, σ∗1s, σ2s, σ∗2s, π2p (x, y), then σ2p, π∗2p, σ∗2p.
• For O₂, F₂ and beyond: σ2p comes before π2p.

So the bond lengths run O₂⁺ < O₂ < O₂⁻ < O₂²⁻, and the bond strengths the other way round.

Hydrogen bonding

A hydrogen bond forms when H, bonded to a small, highly electronegative atom (F, O or N), is attracted to a lone pair on another such atom.

• Intermolecular hydrogen bonds join separate molecules. They explain the high boiling points of H₂O, HF and NH₃ compared with the hydrides below them in their groups.
• Intramolecular hydrogen bonds form within one molecule. o-Nitrophenol has one, so it cannot hydrogen-bond as well with its neighbours. It is more volatile, with a lower boiling point, than p-nitrophenol, which forms intermolecular bonds.

Worked NEET-style questions

Example 1: shape of XeF₄

• Pairs = ½(8 + 4) = 6. Atoms bonded = 4, so there are 2 lone pairs.
• Six pairs: sp³d². The two lone pairs sit opposite each other to reduce repulsion.
• Shape: square planar, and the dipole moment is zero.

Example 2: bond order of O₂⁻

• O₂⁻ has 17 electrons: 10 in bonding MOs and 7 in antibonding MOs.
• Bond order = ½(10 − 7) = 1.5.
• One electron in π∗2p is unpaired, so O₂⁻ is paramagnetic.

Example 3: counting σ and π bonds

How many σ and π bonds are there in acrylonitrile, CH₂=CH–C≡N?

• C–H bonds: 3, all σ.
• C=C: 1σ + 1π. C–C: 1σ. C≡N: 1σ + 2π.
• Total: 6 σ and 3 π.

Example 4: formal charge in ozone

Find the formal charge on the central O in the Lewis structure O=O–O.

• The central O has V = 6, one lone pair (L = 2) and three bonding pairs (B = 6).
• Formal charge = 6 − 2 − ½(6) = +1. The singly bonded end O carries −1, and the doubly bonded end O is 0.

Practice MCQs

1. The hybridisation of carbon in ethene is: (a) sp (b) sp² (c) sp³ (d) sp³d
2. Which is paramagnetic? (a) N₂ (b) C₂ (c) F₂ (d) O₂
3. The shape of SF₄ is: (a) see-saw (b) square planar (c) tetrahedral (d) trigonal pyramidal
4. The bond order of N₂⁺ is: (a) 2 (b) 2.5 (c) 3 (d) 3.5
5. The correct order of bond angle is: (a) CH₄ > NH₃ > H₂O (b) H₂O > NH₃ > CH₄ (c) NH₃ > CH₄ > H₂O (d) all equal
6. Which has zero dipole moment? (a) NH₃ (b) H₂O (c) CHCl₃ (d) BF₃
7. The shape of ClF₃ is: (a) trigonal planar (b) see-saw (c) trigonal pyramidal (d) T-shaped
8. Which has intramolecular hydrogen bonding? (a) water (b) p-nitrophenol (c) o-nitrophenol (d) ethanol

Answers

1. (b) Each carbon forms three σ bonds and one π bond.
2. (d) O₂ has two unpaired electrons in π∗2p.
3. (a) ½(6 + 4) = 5 pairs: 4 bond pairs and 1 lone pair.
4. (b) 13 electrons: ½(9 − 4) = 2.5.
5. (a) 109.5° > about 107° > about 104.5°, as lone pairs increase.
6. (d) Trigonal planar; the three B–F moments cancel.
7. (d) ½(7 + 3) = 5 pairs: 3 bond pairs and 2 lone pairs.
8. (c) The –OH and –NO₂ groups are next to each other.

What to do next

• Rebuild the VSEPR table from memory, with one example per row.
• Draw the MO diagrams for N₂ and O₂ and derive every row of the bond-order table from them.
• Solve 15 hybridisation questions with the ½(V + M − C + A) formula, including ions such as NH₄⁺, NO₃⁻ and I₃⁻.
• Review periodic trends for the electronegativity ideas behind polarity, then move on to chemical thermodynamics.