This MCQ module is based on: Ionic Covalent Bonding
TOPIC 14 OF 28
Ionic Covalent Bonding
🎓 Class 11
Chemistry
CBSE
Theory
Ch 4 – Chemical Bonding and Molecular Structure
⏱ ~14 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📝 Worksheet / Assessment
▲
This assessment will be based on: Ionic Covalent Bonding
Upload images, PDFs, or Word documents to include their content in assessment generation.
Ionic and Covalent Bonding
4.1 Kössel–Lewis Approach to Chemical Bonding
Matter is made up of one or more different kinds of elements. Under normal conditions no other element exists as an independent atom in nature, except noble gases. A group of atoms is found to exist together as one species having characteristic properties. Such a group of atoms is called a molecule. The attractive force which holds the constituents (atoms, ions, etc.) together in a molecule is called a chemical bond.
Several theories have been proposed to explain bond formation. Two early but important theories are:
- Kössel–Lewis approach (1916) — based on the octet rule and electron transfer/sharing.
- Valence Shell Electron Pair Repulsion (VSEPR) theory, Valence Bond Theory (VBT) and Molecular Orbital Theory (MOT) — modern approaches.
In 1916, Kössel and Lewis independently provided a logical explanation of the electronic theory of valence based on the inertness of noble gases.
4.1.1 Octet Rule
Lewis pictured the atom as a positively charged kernel (nucleus + inner electrons) and an outer shell of electrons. He suggested that the outer shell can hold a maximum of eight electrons. He further proposed that atoms combine either by transfer or sharing of valence electrons in such a way that they attain the noble gas configuration of having eight electrons in their outermost shell — known as the octet rule.
4.1.2 Significance of Lewis Symbols
The number of dots around the symbol gives the number of valence electrons. This number helps to calculate the common or group valence of the element: it equals 8 minus the number of dots (for non-metals) or equals the number of dots (for metals).
4.2 Ionic or Electrovalent Bond
From the Kössel–Lewis treatment of ionic bonding, the following key facts emerge: (i) highly electropositive metals (low IE) and highly electronegative non-metals (high EA) form ionic bonds; (ii) electrons are transferred from metal to non-metal; (iii) the resulting cation and anion are bound by electrostatic (Coulombic) attraction.
Ionic (electrovalent) bond: The electrostatic force of attraction that holds together the oppositely charged ions formed by complete transfer of one or more electrons from one atom to another.
Consider the formation of NaCl:
Na (2,8,1) → Na⁺ (2,8) + e⁻
Cl (2,8,7) + e⁻ → Cl⁻ (2,8,8)
Na⁺ + Cl⁻ → Na⁺Cl⁻ (NaCl)
Both ions now possess a noble-gas electronic configuration (Ne and Ar respectively). In sodium chloride, each Na⁺ is surrounded by six Cl⁻ ions and vice-versa, forming a 3-D crystal lattice — there are no discrete NaCl molecules!
4.2.1 Lattice Enthalpy
Lattice enthalpy of an ionic solid is the energy required to completely separate one mole of the solid ionic compound into gaseous ions:
NaCl(s) → Na⁺(g) + Cl⁻(g); ΔLH = +788 kJ mol⁻¹
NaCl(s) → Na⁺(g) + Cl⁻(g); ΔLH = +788 kJ mol⁻¹
The greater the lattice enthalpy, the more stable the ionic crystal. Lattice enthalpies cannot be measured directly — they are calculated using the Born–Haber cycle, which combines several measurable thermochemical steps.
Interactive: Ionic vs Covalent Predictor
Pick two elements and the simulator predicts the bond type using the electronegativity (EN) difference. Rule of thumb: |ΔEN| ≥ 1.7 → predominantly ionic; 0.4–1.7 → polar covalent; < 0.4 → non-polar covalent.
|ΔEN| = 2.23
Predicted bond: Ionic
Difference is large — the more electronegative atom strips an electron from the other.
4.3 Bond Parameters
4.3.1 Bond Length
Bond length is the equilibrium distance between the nuclei of two bonded atoms in a molecule. It is measured by spectroscopic, X-ray and electron-diffraction techniques.
| Bond | Length (pm) | Bond | Length (pm) |
|---|---|---|---|
| H–H (in H₂) | 74 | O=O (in O₂) | 121 |
| F–F (in F₂) | 144 | N≡N (in N₂) | 110 |
| Cl–Cl (in Cl₂) | 199 | C–C (single) | 154 |
| O–H (in H₂O) | 96 | C=C (double) | 134 |
| C–H (in CH₄) | 109 | C≡C (triple) | 120 |
4.3.2 Bond Angle
Bond angle is the angle between the orbitals containing bonding electron pairs around the central atom in a molecule/complex ion. Examples: H–O–H angle in water = 104.5°, H–N–H in NH₃ = 107°, H–C–H in CH₄ = 109.5°.
4.3.3 Bond Enthalpy
Bond enthalpy is the amount of energy required to break one mole of bonds of a particular type between two atoms in the gaseous state. Examples: H–H = 435.8 kJ mol⁻¹, O=O = 498 kJ mol⁻¹, N≡N = 946.0 kJ mol⁻¹.
For polyatomic molecules, the average bond enthalpy is taken (mean over identical bonds). For H₂O the two O–H bonds break at 502 and 427 kJ mol⁻¹ respectively, giving an average value of 464.5 kJ mol⁻¹.
4.3.4 Bond Order
According to the Lewis description, the bond order is the number of bonds between two atoms in a molecule:
- H₂ (H–H): bond order 1
- O₂ (O=O): bond order 2
- N₂ (N≡N): bond order 3
Iso-electronic species have the same bond order: N₂, CO and NO⁺ all have 14 electrons and bond order 3.
4.3.5 Resonance & 4.3.6 Polarity
(Detailed in Part 2.)
4.4 Covalent Bond
Langmuir (1919) refined the Lewis postulates by introducing the term covalent bond for the shared-pair bond. Each combining atom contributes at least one electron to the shared pair and both shared electrons belong to both atoms.
4.4.1 Lewis Representation of Simple Molecules
Steps to draw a Lewis dot structure:
- Compute the total number of valence electrons of all atoms (add 1 e⁻ per −ve charge, subtract 1 per +ve charge).
- Identify the central atom (usually the least electronegative); place terminal atoms around it.
- Distribute electrons as bonding pairs first, then complete octets of terminal atoms with lone pairs.
- Place remaining electrons on the central atom; if octet is incomplete, form double or triple bonds.
Worked Example 4.1 — Lewis Structure of CO₂
Step 1: Valence electrons = 4 (C) + 2 × 6 (O) = 16.
Step 2: C is central; arrangement O–C–O.
Step 3: Distribute the 16 electrons. To satisfy octet on every atom we need two C=O double bonds.
Final: Ö=C=Ö (each O has 2 lone pairs).
Step 2: C is central; arrangement O–C–O.
Step 3: Distribute the 16 electrons. To satisfy octet on every atom we need two C=O double bonds.
Final: Ö=C=Ö (each O has 2 lone pairs).
Worked Example 4.2 — Lewis Structure of Nitrite Ion (NO₂⁻)
Total valence electrons = 5 + 2 × 6 + 1 = 18.
Skeleton: O–N–O. Place lone pairs to fill octets — one N=O double bond and one N–O single bond plus a lone pair on N. The negative charge is delocalised, hence two equivalent resonance structures (covered in Part 2).
Skeleton: O–N–O. Place lone pairs to fill octets — one N=O double bond and one N–O single bond plus a lone pair on N. The negative charge is delocalised, hence two equivalent resonance structures (covered in Part 2).
4.4.2 Formal Charge
Formal charge on an atom in a Lewis structure = (Total valence electrons in the free atom) − (Total non-bonding electrons) − ½ × (Total bonding electrons).
For ozone (O₃), all three oxygens have 6 valence electrons in the free state. In the Lewis structure with the central O double-bonded to one terminal O and single-bonded to the other (with a positive formal charge on central O and a negative on one terminal O), the formal charges sum to zero, matching the neutral molecule.
4.4.3 Limitations of the Octet Rule
(a) Incomplete octet of central atom: Li, Be, B form compounds with fewer than 8 e⁻ (LiCl, BeH₂, BCl₃).
(b) Odd-electron molecules: NO, NO₂ have an odd number of electrons; not all atoms achieve an octet.
(c) Expanded octet: Elements of period 3 and beyond (using vacant d-orbitals) can hold more than 8 e⁻ — e.g., PF₅ (10), SF₆ (12), H₂SO₄ (12 around S), [PF₆]⁻ (12).
Other limitations: Octet rule cannot explain the relative stability of molecules in terms of bond energy; it is silent about geometry and ignores noble gas chemistry (XeF₂, XeF₄, XeF₆ are well-known compounds).
Activity 4.1 — Lewis Builder
Setup: Take three molecules: H₂CO (methanal), HCN (hydrogen cyanide), and SO₂ (sulphur dioxide).
Predict: For each molecule (i) total valence electrons, (ii) the central atom, (iii) the Lewis structure including any multiple bonds and lone pairs.
H₂CO: 1+1+4+6 = 12 e⁻. Central C has two H and one O. Structure: H–C(=O)–H with one C=O double bond and a lone pair on O. C has no lone pair; O has 2 lone pairs.
HCN: 1+4+5 = 10 e⁻. Linear H–C≡N: triple bond between C and N, lone pair on N.
SO₂: 6+12 = 18 e⁻. Central S with one S=O double and one S–O single bond plus a lone pair on S. Resonance averages the two structures (Part 2).
Worked Example 4.3 — Formal charges in CO₃²⁻
Calculate the formal charge on each atom in the Lewis structure of carbonate ion CO₃²⁻ (one C=O double + two C–O single bonds).
Total valence e⁻ = 4 + 18 + 2 = 24.
Central C: 4 − 0 − ½(8) = 0.
Doubly bonded O: 6 − 4 − ½(4) = 0.
Each singly bonded O: 6 − 6 − ½(2) = −1.
Sum = 0 + 0 + (−1) + (−1) = −2 ✓ matches the charge of the ion.
Central C: 4 − 0 − ½(8) = 0.
Doubly bonded O: 6 − 4 − ½(4) = 0.
Each singly bonded O: 6 − 6 − ½(2) = −1.
Sum = 0 + 0 + (−1) + (−1) = −2 ✓ matches the charge of the ion.
Competency-Based Questions
Q1. The octet rule is most strictly obeyed by:L1 Remember
Answer: (d) CH₄. C and each H satisfy the octet/duplet rule. PCl₅ (10) and SF₆ (12) expand octet; NO₂ has an odd electron.
Q2. Which of the following bonds is the shortest?L2 Understand
Answer: (c) C≡C (120 pm). Higher bond order → shorter bond.
Q3. Calculate the formal charge on the chlorine atom of ClO₄⁻ (perchlorate) where Cl is doubly bonded to two oxygens and singly bonded to two oxygens.L3 Apply
Cl valence = 7. Non-bonding e⁻ on Cl = 0. Bonding e⁻ around Cl = 12 (two doubles + two singles). Formal charge = 7 − 0 − 6 = +1. Each singly bonded O carries −1, doubly bonded O = 0. Sum = +1 + 2(−1) + 2(0) = −1 ✓.
Q4. Why does sodium chloride exist as a giant 3-D crystal lattice rather than as discrete NaCl molecules?L4 Analyse
The Coulombic attraction between Na⁺ and Cl⁻ is non-directional (radial). Each Na⁺ attracts as many Cl⁻ neighbours as can geometrically fit around it (six in NaCl). This 6 : 6 octahedral packing extends throughout the crystal, making the entire lattice a single "molecule." The lattice enthalpy (788 kJ mol⁻¹) reflects this enormous cumulative stabilisation.
Q5. Predict whether MgF₂ or AlCl₃ has greater ionic character. Justify using EN values: Mg = 1.31, Al = 1.61, F = 3.98, Cl = 3.16.L5 Evaluate
|ΔEN| MgF₂ = 3.98 − 1.31 = 2.67 (highly ionic). |ΔEN| AlCl₃ = 3.16 − 1.61 = 1.55 (polar covalent, ~25% ionic by Pauling). MgF₂ is far more ionic. Also, Al³⁺ has high charge density and polarises Cl⁻ heavily (Fajans' rules), giving AlCl₃ significant covalent character.
Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: Bond enthalpy of N₂ (946 kJ mol⁻¹) is greater than that of O₂ (498 kJ mol⁻¹).
R: N₂ has a triple bond whereas O₂ has only a double bond.
Answer: (A). Both true; R correctly explains A. Higher bond order → shorter, stronger bond.
A: SF₆ exists but SH₆ does not.
R: Sulphur has vacant 3d orbitals that can accept additional electron pairs from highly electronegative atoms like F.
Answer: (A). Both true; R explains A. Six small, electronegative F atoms can pack around S and stabilise the expanded octet; six bulky, less electronegative H atoms cannot.
A: Lattice enthalpy of CsCl is less negative than that of NaCl.
R: The cation Cs⁺ is larger than Na⁺, so the inter-ionic distance is greater and the Coulombic attraction is weaker.
Answer: (A). Both true; R correctly explains A. |U| ∝ 1/r₀.
Frequently Asked Questions — Ionic and Covalent Bonding
What is the Kossel-Lewis approach to chemical bonding?
The Kossel-Lewis approach (1916) explained chemical bonding using the octet rule: atoms combine to attain a stable noble-gas configuration of eight electrons in the valence shell. Kossel emphasised the role of electron transfer (forming ionic bonds), while Lewis emphasised electron sharing (forming covalent bonds). Lewis introduced dot symbols to represent valence electrons and Lewis structures to depict bonding. The Kossel-Lewis approach is foundational in NCERT Class 11 Chemistry Chapter 4 and explains most simple compounds, though exceptions exist for species with expanded octets (PCl₅, SF₆) or fewer than 8 electrons (BCl₃).
What is the octet rule and what are its exceptions?
The octet rule states that atoms tend to gain, lose or share electrons to attain eight electrons (an octet) in their outermost shell, like the nearest noble gas. The rule explains formation of NaCl, H₂O, CH₄ and many simple compounds. Exceptions in NCERT Class 11 Chemistry Chapter 4 include: (1) incomplete octet (LiCl, BeH₂, BCl₃ — central atom has fewer than 8 electrons); (2) odd-electron molecules (NO, NO₂); (3) expanded octet (PCl₅, SF₆, IF₇ — central atom has more than 8 electrons due to availability of d-orbitals in Period 3 onwards). The octet rule is a useful guide but not absolute.
What is lattice enthalpy and how is it calculated?
Lattice enthalpy is the energy released when one mole of an ionic crystal is formed from its gaseous ions: M⁺(g) + X⁻(g) → MX(s), ΔH = lattice enthalpy. It indicates the strength of ionic bonding — higher lattice enthalpy means stronger ionic crystal. Lattice enthalpy cannot be measured directly and is calculated indirectly using the Born-Haber cycle, which uses Hess's law with sublimation, ionisation, dissociation, electron affinity and formation enthalpies. NCERT Class 11 Chemistry uses lattice enthalpy to compare ionic compound stabilities (e.g., NaCl vs KCl) and to explain solubility patterns.
How do you draw Lewis structures of covalent molecules?
To draw Lewis structures in NCERT Class 11 Chemistry Chapter 4: (1) count total valence electrons of all atoms, (2) write skeletal structure with the least electronegative atom in the centre (except H), (3) place bonding pairs between atoms, (4) distribute remaining electrons as lone pairs starting from outer atoms to complete octets, (5) if central atom is short of octet, form double or triple bonds. Calculate formal charges to check the most plausible structure. Examples: CO₂ has two C=O double bonds; CH₄ has four C-H single bonds with no lone pairs on C; H₂O has two O-H bonds and two lone pairs on O.
What are Fajans rules and how do they predict covalent character?
Fajans rules predict the degree of covalent character in an ionic bond. Covalent character increases when: (1) cation is small (high charge density polarises anion strongly), (2) cation has high charge, (3) anion is large (electron cloud more polarisable), (4) anion has high charge, (5) cation has pseudo-noble-gas configuration (d¹⁰) such as Cu⁺, Ag⁺, which polarise more than noble-gas-configured ions of similar size. For example, AgCl is more covalent than NaCl because Ag⁺ has 4d¹⁰ configuration. NCERT Class 11 Chemistry uses Fajans rules to explain why some 'ionic' compounds dissolve in organic solvents.
What are bond length, bond angle and bond enthalpy?
Bond length is the equilibrium distance between the nuclei of two bonded atoms, measured in picometres. Bond angle is the angle between two bonds at a central atom (e.g., H–O–H angle in water is 104.5°). Bond enthalpy is the energy required to break one mole of bonds in the gaseous state. NCERT Class 11 Chemistry Chapter 4 lists average bond energies (e.g., C–H = 414 kJ/mol, C=C = 614 kJ/mol, C≡C = 839 kJ/mol). Multiple bonds are shorter and stronger than single bonds. Bond parameters are used in thermochemistry calculations and to predict molecular shape and stability.
AI Tutor
Chemistry Class 11 Part I – NCERT (2025-26)
Ready