Kossel - Lewis approach to chemical bond formation, the concept of ionic and covalent bonds. Ionic Bonding – Formation of ionic bonds: NEET MCQs with Answers

15 free questions on Kossel - Lewis approach to chemical bond formation, the concept of ionic and covalent bonds. Ionic Bonding – Formation of ionic bonds for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Formation of ionic bonds, factors affecting the formation of ionic bonds; calculation of lattice enthalpy.

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Question 1

Previous-year question
An element X, whose ground-state electronic configuration is 1s² 2s² 2p³, combines with magnesium to give an ionic compound. What is the simplest formula of this compound? [NEET 2018]
  1. Mg₂X₃
  2. MgX₂
  3. Mg₂X
  4. Mg₃X₂
Show answer
Answer: (D) Mg₃X₂
Explanation: X has the configuration 1s² 2s² 2p³, i.e. it is nitrogen, which gains three electrons to form X³⁻ (N³⁻). Magnesium forms Mg²⁺. Balancing charges, 3 Mg²⁺ (+6) combine with 2 X³⁻ (−6), giving Mg₃X₂ (Mg₃N₂). The other formulae do not give an electrically neutral compound.

Question 2

Previous-year question
What is the correct order of ionic character among CaH₂, BeH₂ and BaH₂? [NEET 2018]
  1. BeH₂ < CaH₂ < BaH₂
  2. CaH₂ < BeH₂ < BaH₂
  3. BeH₂ < BaH₂ < CaH₂
  4. BaH₂ < BeH₂ < CaH₂
Show answer
Answer: (A) BeH₂ < CaH₂ < BaH₂
Explanation: Down group 2 the cation size increases (Be²⁺ < Ca²⁺ < Ba²⁺), so its polarising power decreases and, by Fajans' rules, covalent character falls. BeH₂ is largely covalent (polymeric), while BaH₂ is the most ionic: BeH₂ < CaH₂ < BaH₂.

Question 3

Which of the following statements is correct?
  1. The average formal charge on each oxygen atom of the sulphate ion is −1/4
  2. The average formal charge on each oxygen atom of the carbonate ion is −2/3
  3. The C–O bond order in carbon dioxide is 3
  4. The average N–O bond order in the nitrate ion is 1.5
Show answer
Answer: (B) The average formal charge on each oxygen atom of the carbonate ion is −2/3
Explanation: The carbonate ion carries a charge of −2; with carbon at zero formal charge, this is spread over three equivalent oxygens, so each has an average formal charge of −2/3. In sulphate the −2 charge is shared by four oxygens, so the average on each oxygen cannot be −1/4 (it is −1/2 when sulphur is given zero formal charge). Nitrate has an average N–O bond order of 4/3, and each C–O bond in O=C=O is a double bond.

Question 4

In which pair do the central atoms of both molecules have fewer than eight electrons in their valence shells?
  1. NF₃ and PCl₅
  2. BeCl₂ and BCl₃
  3. CCl₄ and BCl₃
  4. SF₆ and BeCl₂
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Answer: (B) BeCl₂ and BCl₃
Explanation: In monomeric BeCl₂, beryllium has only four valence electrons, and in BCl₃ boron has six. Carbon in CCl₄ and nitrogen in NF₃ have eight, while sulphur in SF₆ and phosphorus in PCl₅ have more than eight.

Question 5

Which statement correctly describes a covalent bond formed between two atoms?
  1. Electrons are completely transferred from one atom to the other
  2. Each atom contributes electrons and the electron pair is shared between the two atoms
  3. The bonding electron pair is held by only one of the two atoms
  4. In every case one atom alone supplies both bonding electrons
Show answer
Answer: (B) Each atom contributes electrons and the electron pair is shared between the two atoms
Explanation: In a covalent bond the two atoms share a pair of electrons, each usually contributing one. Complete transfer of electrons gives an ionic bond. A bond in which one atom supplies both electrons is a special coordinate case, not the general rule.

Question 6

In which of the following molecules does the central atom have fewer than eight electrons in its valence shell?
  1. CCl₄
  2. PCl₃
  3. BCl₃
  4. SiCl₄
Show answer
Answer: (C) BCl₃
Explanation: Boron has three valence electrons and forms three B–Cl bonds, so it has only six electrons around it, an incomplete octet. Carbon and silicon in their tetrachlorides, and phosphorus in PCl₃ with its lone pair, each have eight.

Question 7

In which of the following molecules does the central atom NOT obey the octet rule?
  1. CCl₄
  2. H₂O
  3. BrF₃
  4. NCl₃
Show answer
Answer: (C) BrF₃
Explanation: In BrF₃, bromine forms three bonds and carries two lone pairs, so it has 10 electrons around it, an expanded octet. Carbon in CCl₄, nitrogen in NCl₃ and oxygen in H₂O each have exactly eight valence electrons.
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Question 8

Consider the molecules (I) SF₄, (II) NF₃, (III) PCl₅ and (IV) CH₄. In which of them does the central atom have more than eight electrons in its valence shell?
  1. II and IV
  2. I, II and III
  3. III and IV
  4. I and III
Show answer
Answer: (D) I and III
Explanation: In SF₄, sulphur has four bond pairs and one lone pair, that is 10 electrons. In PCl₅, phosphorus has five bond pairs, also 10 electrons. Nitrogen in NF₃ and carbon in CH₄ have exactly eight electrons, as second-period atoms cannot expand the octet.

Question 9

In which of the following species does the central atom have fewer than eight electrons in its valence shell?
  1. NH₄⁺
  2. SnCl₄
  3. SnCl₂ (gaseous molecule)
  4. BH₄⁻
Show answer
Answer: (C) SnCl₂ (gaseous molecule)
Explanation: In SnCl₂ tin forms two Sn–Cl bonds and keeps one lone pair, so it has only 6 electrons around it, an incomplete octet. In SnCl₄, BH₄⁻ and NH₄⁺ the central atom has four bond pairs, that is 8 electrons.

Question 10

Which scientist received the Nobel Prize in Chemistry for research into the nature of the chemical bond, and also introduced an electronegativity scale and the idea of orbital hybridisation?
  1. Walther Kossel
  2. Gilbert N. Lewis
  3. Niels Bohr
  4. Linus Pauling
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Answer: (D) Linus Pauling
Explanation: Linus Pauling was awarded the 1954 Nobel Prize in Chemistry for his work on the nature of the chemical bond. He developed the electronegativity scale named after him and the concept of hybridisation of atomic orbitals. Lewis and Kossel proposed the electronic theory of valence, and Bohr is known for his model of the atom.

Question 11

A K⁺ ion and a Br⁻ ion held at a fixed distance in air attract each other with a force F. If the same ion pair, at the same separation, is placed in methanol, whose dielectric constant is about 32, the force of attraction becomes about
  1. F/16
  2. 32F
  3. F/√32
  4. F/32
Show answer
Answer: (D) F/32
Explanation: In a medium, Coulomb's law becomes F = q₁q₂/(4πε₀ε_r r²), so the force is inversely proportional to the dielectric constant. Air has ε_r ≈ 1 and methanol has ε_r ≈ 32, so the force falls to F/32. This weakening of interionic attraction by solvents of high dielectric constant helps ionic solids dissolve and dissociate.

Question 12

A solid is built from positively charged and negatively charged ions that are held in a regular three-dimensional arrangement by electrostatic attraction. Such a solid is best classified as
  1. a covalent network solid
  2. an ionic compound
  3. a molecular solid held by dipole–dipole forces
  4. a metallic solid
Show answer
Answer: (B) an ionic compound
Explanation: Complete transfer of electrons from a metal to a non-metal gives cations and anions, and the strong non-directional Coulombic attraction between them packs the ions into a crystal lattice; this is an ionic compound. A metallic solid has cations in a sea of delocalised electrons, not anions. Covalent network solids share electron pairs, and dipole–dipole forces act between neutral polar molecules.

Question 13

Which of the following is NOT a drawback of the octet theory?
  1. It does not account for the shapes of molecules
  2. It cannot show how sodium and chlorine atoms reach noble gas configurations when they form NaCl
  3. It says nothing about the energy of a molecule, so it cannot explain relative stabilities
  4. It rests on the chemical inertness of noble gases, yet xenon and krypton form compounds such as XeF₂ and KrF₂
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Answer: (B) It cannot show how sodium and chlorine atoms reach noble gas configurations when they form NaCl
Explanation: The octet idea explains NaCl well: sodium loses one electron and chlorine gains one, so that each ion has a noble gas configuration. The real drawbacks are different. Noble gases, on whose inertness the rule is based, do form compounds, and the theory is silent about both molecular shape and molecular energy.

Question 14

For sodium chloride, the first ionisation enthalpy of Na(g) is +496 kJ/mol, the electron gain enthalpy of Cl(g) is −349 kJ/mol and the lattice enthalpy of NaCl(s) is +788 kJ/mol. What is the enthalpy change for Na(g) + Cl(g) → NaCl(s), and what makes the ionic solid stable?
  1. −935 kJ/mol; the ionisation of sodium and the lattice formation both release energy
  2. +641 kJ/mol; the solid forms only because its ions have noble gas configurations
  3. +147 kJ/mol; the electron gain by chlorine alone stabilises the solid
  4. −641 kJ/mol; the energy released in forming the crystal lattice outweighs the net energy needed to form the gaseous ions
Show answer
Answer: (D) −641 kJ/mol; the energy released in forming the crystal lattice outweighs the net energy needed to form the gaseous ions
Explanation: Forming Na⁺(g) and Cl⁻(g) from the gaseous atoms needs a net +496 − 349 = +147 kJ/mol, so the ions are not favoured by this step alone. Packing the ions into the crystal releases 788 kJ/mol. The overall change is +147 − 788 = −641 kJ/mol, so the lattice enthalpy is what stabilises an ionic solid.

Question 15

For KCl(s), the following values in kJ mol⁻¹ are given: enthalpy of formation of KCl(s) = −437, enthalpy of sublimation of K(s) = +89, first ionisation enthalpy of K = +419, bond dissociation enthalpy of Cl₂ = +242 and electron gain enthalpy of Cl = −349. Using the Born–Haber cycle, what is the lattice enthalpy of KCl, taken as the enthalpy needed to separate one mole of the solid into gaseous ions?
  1. +838 kJ mol⁻¹
  2. +157 kJ mol⁻¹
  3. +1415 kJ mol⁻¹
  4. +717 kJ mol⁻¹
Show answer
Answer: (D) +717 kJ mol⁻¹
Explanation: One mole of KCl needs only half a mole of Cl₂, so the chlorine step is +121 kJ. The cycle gives −437 = 89 + 419 + 121 − 349 − L, where L is the lattice enthalpy. This becomes −437 = 280 − L, so L = +717 kJ mol⁻¹.

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