CBSE Class 11 Chemical Bonding — The Chapter Everything Depends On
If you leave one Class 11 chapter half-learnt, do not let it be this one. Organic chemistry, coordination compounds, the p-block and the d-block all quietly assume you already know it: a student who cannot draw a Lewis structure will struggle with resonance in Class 12, and one who cannot compute a bond order will find molecular orbital questions impossible. This guide covers the five things the chapter actually asks of you — ionic versus covalent character, Lewis structures, formal charge, bond parameters, and basic molecular orbital theory for diatomic molecules.
1. Ionic and covalent are two ends of one scale
A purely ionic bond would need complete transfer of an electron; a purely covalent bond would need perfectly equal sharing. Almost every real bond sits in between, and the position is decided by the electronegativity difference between the two atoms.
Small difference → polar covalent (HCl, H₂O)
Zero difference → non-polar covalent (H₂, Cl₂, O₂)
An ionic solid is held together by lattice enthalpy — the energy released when gaseous ions come together to form one mole of the solid. It rises with higher ionic charges and smaller ions, which is why MgO melts far higher than NaCl.
Fajans' rules tell you when an "ionic" compound is really significantly covalent. Covalent character increases with a smaller cation (LiCl > NaCl > KCl), a larger anion (AlI₃ > AlF₃), a higher charge on either ion (AlCl₃ > MgCl₂ > NaCl), and a cation with a pseudo noble-gas (18-electron) configuration — which is why AgCl is far more covalent than NaCl even though Ag⁺ and Na⁺ are similar in size.
2. Lewis structures — a fixed five-step method
Step 2: Choose the central atom — usually the least electronegative. Hydrogen is never central.
Step 3: Join every atom to the centre with a single bond (2 electrons each).
Step 4: Give lone pairs to the outer atoms until each has an octet (hydrogen needs 2).
Step 5: Any electrons left go on the central atom. If the centre is still short of an octet, convert an outer lone pair into a double or triple bond.
Worked example 1 — the carbonate ion, CO₃²⁻.
Step 1. Carbon gives 4 valence electrons; three oxygens give 3 × 6 = 18. Total 4 + 18 = 22, plus 2 for the 2− charge = 24 electrons = 12 pairs.
Steps 2–3. Carbon is less electronegative, so it is central. Three C–O single bonds use 3 × 2 = 6 electrons, leaving 24 − 6 = 18.
Step 4. Each oxygen needs three lone pairs (6 electrons) to complete its octet: 3 × 6 = 18 — exactly what is left. Remaining: 0.
Step 5. Carbon now has only 6 electrons, so move one lone pair from an oxygen into the bond to make a C=O double bond.
Result: one C=O and two C–O bonds, with the 2− charge on the two singly-bonded oxygens. The double bond could have gone to any of the three oxygens, so CO₃²⁻ has three resonance structures.
3. Formal charge — the test that tells you which structure is best
When several Lewis structures are possible, formal charge decides which contributes most. It is bookkeeping, not a real charge.
V = valence electrons in the free atom · L = non-bonding (lone-pair) electrons on that atom · B = electrons shared in bonds at that atom
Two checks apply every time: the formal charges must add up to the overall charge, and the best structure is normally the one with the smallest formal charges, with any negative charge on the more electronegative atom.
Worked example 2 — formal charges in CO₃²⁻.
Carbon (V = 4): no lone pairs, L = 0; four bonds (one double + two single) = 8
bonding electrons, B = 8.
FC = 4 − 0 − (8 ÷ 2) = 0
Double-bonded oxygen (V = 6): two lone pairs, L = 4; one double bond, B = 4.
FC = 6 − 4 − (4 ÷ 2) = 0
Each single-bonded oxygen (V = 6): three lone pairs, L = 6; one single bond, B = 2.
FC = 6 − 6 − (2 ÷ 2) = −1
Check: 0 + 0 + (−1) + (−1) = −2, matching the charge on the carbonate ion. The structure is correct.
Worked example 3 — ozone, O₃.
Valence electrons: 3 × 6 = 18 = 9 pairs. The accepted structure is bent, with one O=O double bond and one O–O single bond.
Central O: V = 6, L = 2 (one lone pair), B = 6 (three bonds) →
FC = 6 − 2 − 3 = +1
Double-bonded terminal O: V = 6, L = 4, B = 4 → FC = 6 − 4 − 2 = 0
Single-bonded terminal O: V = 6, L = 6, B = 2 → FC = 6 − 6 − 1 = −1
Check: (+1) + 0 + (−1) = 0, correct for a neutral molecule. This is why ozone carries formal charges although it is uncharged overall — and why its two O–O bonds are found to be equal in length: the real molecule is the resonance hybrid, not either drawing.
4. Bond parameters — length, angle, enthalpy, order
Four measurable quantities describe a bond, and they move together predictably.
| Bond | Bond order | Approximate bond length |
|---|---|---|
| C–C | 1 | 154 pm |
| C=C | 2 | 134 pm |
| C≡C | 3 | 120 pm |
Bond enthalpy behaves the same way: NCERT quotes about 435.8 kJ mol⁻¹ for H–H, 498 kJ mol⁻¹ for O=O and 946 kJ mol⁻¹ for N≡N. That very large N≡N value is why nitrogen gas is so unreactive and why ammonia synthesis needs a catalyst and high temperature.
Where resonance is involved, bond order is fractional. In CO₃²⁻ four bonds are shared over three C–O positions, giving 4 ÷ 3 ≈ 1.33 — and all three C–O bonds are found to be identical in length, which no single Lewis structure predicts.
Bond angle comes from VSEPR: electron pairs arrange themselves as far apart as possible, and lone pairs repel more strongly than bonding pairs. That is why CH₄ is 109.5°, NH₃ close to 107° and H₂O close to 104.5° — one tetrahedral arrangement squeezed progressively by one and then two lone pairs.
5. Molecular orbital theory for diatomic molecules
Lewis structures fail on one famous molecule: they predict that O₂ has all its electrons paired, yet liquid oxygen is clearly attracted by a magnet. Molecular orbital (MO) theory gets it right, which is why the chapter ends here. In MO theory the atomic orbitals of both atoms combine into bonding molecular orbitals (lower in energy) and antibonding ones marked with a star (higher in energy), and electrons fill them by the same Aufbau, Pauli and Hund rules you already use for atoms.
Nb = electrons in bonding MOs · Na = electrons in antibonding MOs
Bond order 0 → the molecule does not exist. Any unpaired electron → paramagnetic.
One ordering detail decides many answers: for diatomics up to and including N₂ the two π2p orbitals lie below σ2pz; from O₂ onwards the order flips and σ2pz lies below the π2p pair.
| Species | Electrons | Nb | Na | Bond order | Magnetic behaviour |
|---|---|---|---|---|---|
| H₂ | 2 | 2 | 0 | (2 − 0)/2 = 1 | Diamagnetic |
| He₂ | 4 | 2 | 2 | (2 − 2)/2 = 0 | Does not exist |
| B₂ | 10 | 6 | 4 | (6 − 4)/2 = 1 | Paramagnetic (2 unpaired) |
| C₂ | 12 | 8 | 4 | (8 − 4)/2 = 2 | Diamagnetic |
| N₂ | 14 | 10 | 4 | (10 − 4)/2 = 3 | Diamagnetic |
| O₂ | 16 | 10 | 6 | (10 − 6)/2 = 2 | Paramagnetic (2 unpaired) |
| F₂ | 18 | 10 | 8 | (10 − 8)/2 = 1 | Diamagnetic |
Worked example 4 — why O₂ is paramagnetic, and the O₂ family.
Oxygen has 8 electrons per atom, so O₂ has 16. Filling in the O₂ order gives
σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² π2py²
π*2px¹ π*2py¹
The last two electrons enter the two π* orbitals singly, by Hund's rule, so O₂ has two unpaired electrons — paramagnetic, exactly as experiment shows.
Nb = 2 + 2 + 2 + 2 + 2 = 10; Na = 2 + 2 + 1 + 1 = 6.
Bond order = (10 − 6) ÷ 2 = 2, agreeing with the O=O double bond.
Adding or removing electrons only changes the π* count:
O₂⁺ (15 e⁻): Na = 5 → BO = (10 − 5)/2 = 2.5
O₂ (16 e⁻): BO = 2
O₂⁻ (17 e⁻): Na = 7 → BO = (10 − 7)/2 = 1.5
O₂²⁻ (18 e⁻): Na = 8 → BO = (10 − 8)/2 = 1
Higher bond order means a shorter bond, so bond length increases as O₂⁺ < O₂ < O₂⁻ < O₂²⁻. This exact comparison is asked very often.
Octet rule exceptions, and a note on editions
The octet rule is a guide, not a law, and it has three failure modes: incomplete octet (BeCl₂ with 4 electrons on Be, BF₃ with 6 on B); expanded octet (PCl₅ with 10, SF₆ with 12 — only elements from period 3 onwards can do this); and odd-electron species such as NO and NO₂, where complete pairing is impossible.
For the expanded-octet molecules, NCERT explains the bonding through d-orbital participation (sp³d for PCl₅, sp³d² for SF₆). At university level this is refined to a three-centre four-electron description in which d-orbital involvement is small. Both give the same shapes; in a CBSE answer, use the NCERT hybridisation explanation.
Mistakes that cost marks
- Getting the charge adjustment backwards. Anion → add electrons; cation → subtract. Reversing it wrecks the Lewis structure from step 1.
- Putting hydrogen in the middle. Hydrogen forms only one bond, so it can never be a central atom.
- Confusing formal charge with oxidation number. Formal charge assumes bonding electrons are shared equally; oxidation number gives them entirely to the more electronegative atom. In CO₃²⁻ carbon's formal charge is 0 but its oxidation number is +4.
- Saying the molecule "keeps changing" between resonance structures. It does not. There is one real structure — the hybrid — and the drawings are our imperfect ways of describing it.
- Using the O₂ molecular-orbital order for N₂, which makes you call N₂ paramagnetic. Remember the switch that happens at O₂.
- Forgetting the ½ in the bond-order formula, or subtracting bonding from antibonding by mistake, which gives a negative answer.
- Applying Fajans' rules in the wrong direction. A smaller cation and a larger anion give more covalent character, not less.
Where this appears in your exam
| Question type | What is being tested |
|---|---|
| "Draw the Lewis structure of …" | The five-step method and correct electron count |
| "Calculate the formal charge on each atom" | V − L − ½B, plus the sum check |
| "Compare the bond lengths of …" | Bond order reasoning, often through MO theory |
| "Why is O₂ paramagnetic?" | The two singly occupied π* orbitals |
| "Which is more covalent, AgCl or NaCl?" | Fajans' rules, especially the 18-electron cation |
| "Predict the shape and bond angle of …" | VSEPR and lone-pair repulsion |
Bonding feeds directly into Class 12 — coordination compounds, the p-block and organic mechanisms all lean on it. Confirm the current chapter list and syllabus from the official CBSE notification and the latest NCERT textbook before planning your revision.
Every bonding question starts with valence electrons. The Interactive Periodic Table gives you the group, block, electron configuration and electronegativity of any element in one click — so you can count valence electrons and judge polarity before you draw anything.
Open the Interactive Periodic Table →Bonding is the chapter worth getting right the first time. ABC Chemistry runs Class 11–12 chemistry coaching from its Gurugram centre and online classes across India — details at abcchemistry.in. Families in Delhi, Noida or Gurgaon who prefer one-to-one teaching at home can also arrange home tuition.