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1. Which statement best describes a chemical bond?
ⓐ. An attraction between protons and neutrons
ⓑ. An attraction between nuclei only
ⓒ. An attractive force holding atoms or ions together in a compound
ⓓ. A force that prevents atoms from reacting
Correct Answer: An attractive force holding atoms or ions together in a compound
Explanation: A chemical bond is the attractive force that holds atoms together to form molecules or ionic compounds. It can be due to electrostatic forces (ionic bonding) or sharing of electrons (covalent bonding). Options A and B are incomplete because bonding involves electrons, not just nuclei. Option D is incorrect because bonds enable atoms to combine, not prevent reaction.
2. Why do atoms form chemical bonds?
ⓐ. To increase their atomic number
ⓑ. To achieve greater chemical stability by completing their octet
ⓒ. To decrease their atomic mass
ⓓ. To increase nuclear charge
Correct Answer: To achieve greater chemical stability by completing their octet
Explanation: Atoms bond to achieve a stable electronic configuration, usually the noble gas octet (8 valence electrons). This lowers their potential energy. Atomic number and nuclear charge are intrinsic and do not change during bonding.
3. Which scientist first proposed the octet rule?
ⓐ. J.J. Thomson
ⓑ. Gilbert N. Lewis
ⓒ. Rutherford
ⓓ. John Dalton
Correct Answer: Gilbert N. Lewis
Explanation: Gilbert N. Lewis (1916) proposed the octet rule, stating that atoms tend to gain, lose, or share electrons to achieve eight electrons in their valence shell. Dalton proposed atomic theory, Rutherford explained nuclear model, and Thomson discovered electrons.
4. Which type of bond is formed by complete transfer of electrons?
ⓐ. Ionic bond
ⓑ. Covalent bond
ⓒ. Metallic bond
ⓓ. Hydrogen bond
Correct Answer: Ionic bond
Explanation: In ionic bonding, electrons are transferred from one atom (metal) to another (non-metal), forming oppositely charged ions that are held together by electrostatic attraction. Covalent bonds involve sharing, metallic bonds involve delocalized electrons, and hydrogen bonds are weak intermolecular attractions.
5. Which of the following is an example of a covalent compound?
ⓐ. NaCl
ⓑ. KBr
ⓒ. HCl
ⓓ. MgO
Correct Answer: HCl
Explanation: HCl is a covalent compound because hydrogen and chlorine share electrons to complete their octet. NaCl, KBr, and MgO are ionic, formed by transfer of electrons between metal and non-metal.
6. Which rule helps predict the most stable structure of molecules?
ⓐ. Law of conservation of mass
ⓑ. Octet rule
ⓒ. Dalton’s law of partial pressure
ⓓ. Pauli exclusion principle
Correct Answer: Octet rule
Explanation: The octet rule is the guiding principle in predicting stability of molecules. Atoms achieve an octet by sharing or transferring electrons. The other options are unrelated: conservation of mass is a reaction law, Dalton’s law deals with gases, and Pauli principle deals with electron configuration.
7. Which of the following molecules violates the octet rule?
ⓐ. CO₂
ⓑ. CH₄
ⓒ. BF₃
ⓓ. H₂O
Correct Answer: BF₃
Explanation: Boron in BF₃ has only 6 electrons in its valence shell, making it electron-deficient and an exception to the octet rule. CO₂, CH₄, and H₂O follow the octet rule with complete valence shells.
8. Which fundamental force is responsible for chemical bonding?
ⓐ. Nuclear force
ⓑ. Electromagnetic force
ⓒ. Gravitational force
ⓓ. Weak nuclear force
Correct Answer: Electromagnetic force
Explanation: Chemical bonds arise from the electromagnetic interactions between electrons and nuclei. Gravitational and nuclear forces act inside the nucleus, not at the chemical bonding level.
9. The electron dot structure of oxygen (O₂) shows how many shared pairs of electrons?
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. 4
Correct Answer: 2
Explanation: Oxygen has 6 valence electrons. Each oxygen shares two electrons, forming a double bond (two shared pairs). This gives both atoms a complete octet. Option A is incorrect (single bond), option C would imply a triple bond, and option D is not possible.
10. Which of the following statements is correct about metallic bonding?
ⓐ. Electrons are shared in pairs between two atoms
ⓑ. Electrons are completely transferred from one atom to another
ⓒ. Electrons are delocalized and move freely throughout the metal lattice
ⓓ. Electrons are tightly bound to individual atoms only
Correct Answer: Electrons are delocalized and move freely throughout the metal lattice
Explanation: In metallic bonding, positive metal ions are surrounded by a “sea of delocalized electrons” that move freely, explaining high conductivity, malleability, and ductility of metals. Options A and B describe covalent and ionic bonding, while D is opposite to metallic bonding.
11. The octet rule states that atoms tend to:
ⓐ. Gain, lose, or share electrons to achieve 10 electrons in their outer shell
ⓑ. Gain, lose, or share electrons to achieve 8 electrons in their outer shell
ⓒ. Keep the same number of valence electrons during bonding
ⓓ. Always transfer electrons only
Correct Answer: Gain, lose, or share electrons to achieve 8 electrons in their outer shell
Explanation: The octet rule, proposed by Lewis and Kossel, says atoms are most stable when they have 8 valence electrons (noble gas configuration). Option A is wrong (10 is incorrect), option C contradicts electron changes in bonding, and option D is incomplete since atoms may also share electrons.
12. Which noble gas configuration is usually achieved through the octet rule?
ⓐ. Helium configuration (1s²)
ⓑ. Neon configuration (1s²2s²2p⁶)
ⓒ. Argon configuration (1s²2s²2p⁶3s²3p⁶)
ⓓ. Both B and C depending on the element
Correct Answer: Both B and C depending on the element
Explanation: Most main-group elements aim for a stable noble gas configuration. Second-period elements resemble neon, while third-period elements often resemble argon. Helium is stable with 2 electrons (duplet rule).
13. Which of the following compounds follows the octet rule completely?
ⓐ. BeCl₂
ⓑ. BF₃
ⓒ. CO₂
ⓓ. NO
Correct Answer: CO₂
Explanation: In CO₂, each oxygen and carbon achieve 8 valence electrons. BeCl₂ (Be has only 4 electrons), BF₃ (B has 6 electrons), and NO (odd electron species) are exceptions to the octet rule.
14. Which of the following is an exception to the octet rule due to odd number of electrons?
ⓐ. H₂O
ⓑ. NO
ⓒ. NH₃
ⓓ. CH₄
Correct Answer: NO
Explanation: Nitric oxide (NO) has 11 valence electrons, making it impossible for all atoms to complete an octet. It is an exception due to odd electron count. H₂O, NH₃, and CH₄ all follow the octet rule.
15. Sodium forms Na⁺ ion because:
ⓐ. It wants to decrease its mass number
ⓑ. It has low electronegativity and loses one electron to achieve octet
ⓒ. It has high electronegativity and gains one electron
ⓓ. It wants to become radioactive
Correct Answer: It has low electronegativity and loses one electron to achieve octet
Explanation: Sodium (1s²2s²2p⁶3s¹) loses its single 3s electron to get neon’s stable octet configuration. It cannot gain electrons easily due to low electronegativity. Options A and D are irrelevant.
16. Chlorine forms Cl⁻ ion because:
ⓐ. It has 7 valence electrons and gains 1 to complete octet
ⓑ. It has only 2 electrons and gains more
ⓒ. It needs to balance nuclear charge
ⓓ. It loses one electron to become stable
Correct Answer: It has 7 valence electrons and gains 1 to complete octet
Explanation: Chlorine (1s²2s²2p⁶3s²3p⁵) gains one electron to achieve argon’s configuration (3p⁶). This makes Cl⁻ stable. Option D is incorrect, as chlorine does not lose electrons.
17. Which one of the following pairs of elements forms an ionic compound by following octet rule?
ⓐ. H and H
ⓑ. Na and Cl
ⓒ. C and O
ⓓ. N and N
Correct Answer: Na and Cl
Explanation: Sodium loses 1 electron to form Na⁺ and chlorine gains 1 electron to form Cl⁻, both achieving octets. H₂, CO₂, and N₂ form covalent bonds, not ionic, though they may satisfy octet differently.
18. In MgO, how do magnesium and oxygen achieve octet?
ⓐ. Mg shares two electrons with O
ⓑ. Mg loses 2 electrons, O gains 2 electrons
ⓒ. Mg and O each gain 1 electron from the other
ⓓ. Both retain their original electrons
Correct Answer: Mg loses 2 electrons, O gains 2 electrons
Explanation: Magnesium (2,8,2) loses two 3s electrons to form Mg²⁺. Oxygen (2,6) gains two electrons to form O²⁻. Both ions attain octet configuration similar to neon.
19. Which of the following molecules does NOT obey the octet rule?
ⓐ. PCl₅
ⓑ. CH₄
ⓒ. CO₂
ⓓ. NH₃
Correct Answer: PCl₅
Explanation: In PCl₅, phosphorus has 10 electrons around it (expanded octet) due to availability of d-orbitals. CH₄, CO₂, and NH₃ obey the octet rule.
20. Which best explains why noble gases are chemically inert?
ⓐ. They have incomplete outer shells
ⓑ. They already have a complete octet (or duplet for He)
ⓒ. They always lose electrons during bonding
ⓓ. They have very low atomic number
Correct Answer: They already have a complete octet (or duplet for He)
Explanation: Noble gases like Ne, Ar, and Kr have completely filled valence shells, making them highly stable and unreactive. Helium is stable with 2 electrons (duplet). Options A, C, and D are incorrect explanations.
21. Which of the following molecules is an exception to the octet rule due to an incomplete octet?
ⓐ. H₂O
ⓑ. CH₄
ⓒ. BF₃
ⓓ. CO₂
Correct Answer: BF₃
Explanation: Boron in BF₃ has only 6 electrons around it, making it electron-deficient. H₂O, CH₄, and CO₂ all satisfy the octet rule.
22. Which compound is an example of an expanded octet?
ⓐ. NH₃
ⓑ. PCl₅
ⓒ. H₂O
ⓓ. CH₄
Correct Answer: PCl₅
Explanation: Phosphorus in PCl₅ accommodates 10 electrons in its valence shell, possible due to the use of vacant 3d orbitals. The others have 8 valence electrons around the central atom.
23. Which of the following is an exception to the octet rule because it contains an odd number of valence electrons?
ⓐ. NO
ⓑ. H₂S
ⓒ. BeCl₂
ⓓ. CO₂
Correct Answer: NO
Explanation: Nitric oxide (NO) has 11 valence electrons, so it cannot distribute them in a way that all atoms get 8. This makes it an odd-electron molecule.
24. Which of the following best describes the limitation of the octet rule?
ⓐ. It cannot explain ionic bonding
ⓑ. It cannot explain expanded, incomplete, or odd-electron molecules
ⓒ. It applies only to metals
ⓓ. It is valid for all compounds
Correct Answer: It cannot explain expanded, incomplete, or odd-electron molecules
Explanation: The octet rule is limited as it does not cover electron-deficient molecules (BF₃), expanded octets (PCl₅, SF₆), and odd-electron species (NO, NO₂).
25. Which molecule shows incomplete octet in the central atom?
ⓐ. BeCl₂
ⓑ. NH₃
ⓒ. CH₄
ⓓ. O₂
Correct Answer: BeCl₂
Explanation: Beryllium has only 4 electrons in BeCl₂. NH₃ and CH₄ satisfy octet, while O₂ follows the rule with double bonding.
26. Which of the following is an exception to the octet rule due to expanded octet?
ⓐ. SF₆
ⓑ. HCl
ⓒ. CO
ⓓ. N₂
Correct Answer: SF₆
Explanation: Sulfur in SF₆ has 12 valence electrons, exceeding the octet rule due to participation of 3d orbitals. HCl, CO, and N₂ follow the octet rule.
27. In which of the following molecules does the central atom have less than 8 electrons?
ⓐ. PCl₅
ⓑ. SF₆
ⓒ. BF₃
ⓓ. XeF₂
Correct Answer: BF₃
Explanation: Boron in BF₃ is electron-deficient with only 6 valence electrons. PCl₅, SF₆, and XeF₂ are expanded octet cases.
28. Why is the octet rule not strictly applicable to transition elements?
ⓐ. Because they have high electronegativity
ⓑ. Because they do not form bonds
ⓒ. Because d-orbitals are involved in bonding
ⓓ. Because they only show inertness
Correct Answer: Because d-orbitals are involved in bonding
Explanation: Transition elements often use (n−1)d and ns orbitals in bonding, leading to variable oxidation states and expanded octet structures, which the octet rule cannot explain.
29. Which molecule among the following violates the octet rule by having an odd number of electrons?
ⓐ. NO₂
ⓑ. H₂O
ⓒ. CH₄
ⓓ. CO₂
Correct Answer: NO₂
Explanation: NO₂ has 17 valence electrons (odd), so one atom cannot complete its octet. H₂O, CH₄, and CO₂ follow the octet rule.
30. The octet rule fails to explain:
ⓐ. Stability of noble gases
ⓑ. Existence of coordinate bonds
ⓒ. Structures of BF₃, PCl₅, SF₆
ⓓ. Both B and C
Correct Answer: Both B and C
Explanation: The octet rule explains noble gas stability but fails for coordinate bonding (e.g., NH₄⁺) and molecules with incomplete/expanded octets (BF₃, PCl₅, SF₆).
31. Which of the following correctly represents the Lewis dot structure of oxygen (O₂)?
ⓐ. \:O::O:
ⓑ. O=O with two lone pairs on each oxygen
ⓒ. O–O with one lone pair on each oxygen
ⓓ. O≡O with no lone pairs
Correct Answer: O=O with two lone pairs on each oxygen
Explanation: In O₂, each oxygen has 6 valence electrons. They share two pairs (double bond), leaving two lone pairs on each atom. Option A is incorrect as it shows no bonds, option C gives an incomplete octet, and option D does not match actual bonding.
32. The Lewis dot structure of CO₂ shows:
ⓐ. A single bond between carbon and oxygen
ⓑ. A triple bond between carbon and oxygen
ⓒ. Two double bonds, each between carbon and oxygen
ⓓ. No bonds between carbon and oxygen
Correct Answer: Two double bonds, each between carbon and oxygen
Explanation: Carbon has 4 valence electrons and oxygen has 6. By sharing two pairs of electrons with each oxygen, carbon achieves octet. Each oxygen also completes its octet. Thus CO₂ has O=C=O structure.
33. Which of the following molecules has a correct Lewis dot structure showing a lone pair on the central atom?
ⓐ. NH₃
ⓑ. CH₄
ⓒ. BF₃
ⓓ. CCl₄
Correct Answer: NH₃
Explanation: In NH₃, nitrogen has 5 valence electrons. Three are used in bonding with hydrogens and two remain as a lone pair. CH₄, BF₃, and CCl₄ have central atoms without lone pairs.
34. In the Lewis dot structure of H₂O, how many lone pairs are present on the oxygen atom?
ⓐ. 0
ⓑ. 1
ⓒ. 2
ⓓ. 3
Correct Answer: 2
Explanation: Oxygen has 6 valence electrons. Two electrons form bonds with two hydrogens, and four electrons remain as two lone pairs. Hence, oxygen has 2 lone pairs in H₂O.
35. What is the total number of valence electrons used in drawing the Lewis structure of NH₄⁺?
ⓐ. 6
ⓑ. 8
ⓒ. 9
ⓓ. 10
Correct Answer: 8
Explanation: Nitrogen contributes 5 valence electrons and each hydrogen 1. For NH₄⁺, total = 5 + 4(1) − 1 (positive charge) = 8 electrons. These are distributed as 4 N–H bonds, with no lone pair left on nitrogen.
36. Which molecule is best represented by a resonance structure in Lewis notation?
ⓐ. H₂
ⓑ. O₃
ⓒ. CH₄
ⓓ. HCl
Correct Answer: O₃
Explanation: Ozone (O₃) has resonance structures where the double bond can be placed between different pairs of oxygens. H₂, CH₄, and HCl have fixed structures without resonance.
37. In the Lewis dot structure of BeCl₂, the central atom beryllium has how many electrons?
ⓐ. 2
ⓑ. 4
ⓒ. 6
ⓓ. 8
Correct Answer: 4
Explanation: Beryllium has 2 valence electrons and forms 2 single covalent bonds with two chlorines. Thus, Be has only 4 electrons around it, showing incomplete octet.
38. Which of the following correctly represents the Lewis dot structure of N₂?
ⓐ. N–N with one lone pair on each nitrogen
ⓑ. N≡N with one lone pair on each nitrogen
ⓒ. N=N with three lone pairs on each nitrogen
ⓓ. N–N–N with two lone pairs on central nitrogen
Correct Answer: N≡N with one lone pair on each nitrogen
Explanation: Nitrogen has 5 valence electrons. Each nitrogen shares 3 electrons to form a triple bond, leaving 2 electrons as one lone pair. Hence, the correct structure is N≡N.
39. Which molecule has a central atom with an expanded octet in its Lewis dot structure?
ⓐ. CF₄
ⓑ. SF₆
ⓒ. H₂O
ⓓ. NH₃
Correct Answer: SF₆
Explanation: Sulfur in SF₆ uses 6 bonding pairs with fluorine atoms, making 12 electrons around sulfur (expanded octet). CF₄, H₂O, and NH₃ obey the octet rule.
40. The Lewis dot structure of carbonate ion (CO₃²⁻) requires how many valence electrons in total?
ⓐ. 22
ⓑ. 24
ⓒ. 26
ⓓ. 28
Correct Answer: 24
Explanation: Carbon contributes 4, each oxygen 6 (total 18), plus 2 extra electrons due to −2 charge, giving 24 valence electrons. These are distributed in resonance structures with delocalized π-bonding.
41. Which of the following is the best example of ionic bond formation?
ⓐ. H₂ molecule
ⓑ. NaCl
ⓒ. O₂ molecule
ⓓ. Cl₂ molecule
Correct Answer: NaCl
Explanation: Sodium (atomic number 11, configuration 2,8,1) loses one electron to form Na⁺, while chlorine (atomic number 17, configuration 2,8,7) gains that electron to form Cl⁻. The electrostatic attraction between Na⁺ and Cl⁻ forms a strong ionic bond. H₂, O₂, and Cl₂ are covalent molecules where atoms share electrons rather than transfer them.
42. Which condition favors the formation of ionic bonds?
ⓐ. Small electronegativity difference between atoms
ⓑ. Large electronegativity difference between atoms
ⓒ. Presence of identical atoms
ⓓ. Weak electrostatic forces
Correct Answer: Large electronegativity difference between atoms
Explanation: Ionic bonding occurs when one atom (usually a metal) has low ionization energy and readily loses electrons, while the other (non-metal) has high electron affinity and attracts electrons strongly. A large electronegativity difference (e.g., Na and Cl) makes electron transfer favorable. Small differences favor covalent bonding.
43. During ionic bond formation, the energy released when an electron is added to a gaseous atom is called:
ⓐ. Ionization enthalpy
ⓑ. Electron affinity
ⓒ. Lattice energy
ⓓ. Bond enthalpy
Correct Answer: Electron affinity
Explanation: Electron affinity is the energy change when an electron is added to an isolated gaseous atom, e.g., Cl + e⁻ → Cl⁻. Ionization enthalpy refers to removing electrons, lattice energy is the energy released when gaseous ions form a solid ionic lattice, and bond enthalpy is energy required to break a covalent bond.
44. Which of the following steps is involved in the formation of NaCl crystal from Na and Cl atoms?
ⓐ. Ionization of sodium atom
ⓑ. Addition of electron to chlorine atom
ⓒ. Electrostatic attraction between Na⁺ and Cl⁻
ⓓ. All of the above
Correct Answer: All of the above
Explanation: The process involves multiple steps: (i) Sodium loses one electron (ionization) to form Na⁺, (ii) chlorine gains one electron (electron affinity) to form Cl⁻, and (iii) the oppositely charged ions are held together by strong electrostatic forces forming NaCl crystal. Each step contributes to the final lattice formation.
45. Which of the following metals is most likely to form an ionic bond with oxygen?
ⓐ. Magnesium
ⓑ. Copper
ⓒ. Gold
ⓓ. Silver
Correct Answer: Magnesium
Explanation: Magnesium (2,8,2) readily loses two electrons to form Mg²⁺, while oxygen (2,6) gains two electrons to form O²⁻. This strong electron transfer forms MgO with high lattice energy. Transition metals like Cu, Ag, and Au tend to form covalent or metallic bonds rather than purely ionic.
46. Which of the following correctly represents the electron transfer in the formation of CaCl₂?
ⓐ. Calcium gains 2 electrons, each chlorine loses 1 electron
ⓑ. Calcium loses 2 electrons, each chlorine gains 1 electron
ⓒ. Calcium and chlorine share electrons equally
ⓓ. No electron transfer occurs
Correct Answer: Calcium loses 2 electrons, each chlorine gains 1 electron
Explanation: Calcium (2,8,8,2) loses 2 electrons to form Ca²⁺. Each chlorine atom (2,8,7) gains 1 electron to form Cl⁻. The formula becomes CaCl₂ with strong electrostatic attractions between Ca²⁺ and two Cl⁻ ions.
47. Which factor is essential for an element to act as a cation in ionic bond formation?
ⓐ. High electron affinity
ⓑ. Low ionization energy
ⓒ. High electronegativity
ⓓ. Large atomic size with many valence electrons
Correct Answer: Low ionization energy
Explanation: Cations are formed when atoms lose electrons. A low ionization energy means electrons can be removed easily, typical of alkali and alkaline earth metals. High electron affinity and high electronegativity favor anion formation, not cation formation.
48. In ionic bond formation, the role of lattice energy is:
ⓐ. To oppose bond formation
ⓑ. To stabilize the ionic crystal structure
ⓒ. To increase ionization enthalpy
ⓓ. To decrease electron affinity
Correct Answer: To stabilize the ionic crystal structure
Explanation: Lattice energy is released when oppositely charged ions come together to form a crystalline solid. It is the key factor that compensates for the energy required to ionize atoms. For NaCl, lattice energy stabilizes the crystal, making it highly stable and solid at room temperature.
49. Why does NaCl have a high melting point?
ⓐ. Because it is a covalent compound
ⓑ. Because it involves weak Van der Waals forces
ⓒ. Because of strong electrostatic forces between Na⁺ and Cl⁻ ions
ⓓ. Because sodium has a high atomic mass
Correct Answer: Because of strong electrostatic forces between Na⁺ and Cl⁻ ions
Explanation: In NaCl, each Na⁺ is surrounded by 6 Cl⁻ and vice versa, forming a 3D lattice structure. The strong Coulombic attraction requires a large amount of energy to break, giving NaCl a high melting and boiling point. Covalent or Van der Waals forces cannot explain such stability.
50. Which of the following is not an example of ionic compound?
ⓐ. NaF
ⓑ. KBr
ⓒ. CaO
ⓓ. HCl
Correct Answer: HCl
Explanation: HCl is covalent in nature as hydrogen and chlorine share electrons. NaF, KBr, and CaO are ionic compounds formed by electron transfer between metals and non-metals, stabilized by lattice energy.
51. According to Coulomb’s law, the strength of ionic attraction between two ions is directly proportional to:
ⓐ. Sum of ionic radii
ⓑ. Product of ionic charges and inverse of distance between ions
ⓒ. Difference in electronegativities only
ⓓ. Square of the dielectric constant of the medium
Correct Answer: Product of ionic charges and inverse of distance between ions
Explanation: Coulomb’s law gives the electrostatic force $F \propto \dfrac{|z_+ z_-|}{r^2}$ and potential energy $E \propto -\dfrac{|z_+ z_-|}{r}$. Greater ionic charges ($|z_+ z_-|$) increase attraction; larger interionic distance $r$ decreases it. Electronegativity difference is a predictor of ionic character but not a direct quantitative factor in bond strength. The dielectric constant $D$ of the medium reduces the attraction; strength varies inversely with $D$, not with its square.
52. Which pair is expected to have the strongest ionic bond (highest lattice energy), assuming the same crystal type?
ⓐ. NaCl
ⓑ. KCl
ⓒ. MgO
ⓓ. CaO
Correct Answer: MgO
Explanation: Lattice energy increases with ionic charge and decreases with ionic size. MgO involves $2+$ and $2-$ ions, giving $|z_+ z_-|=4$, much larger than $|z_+ z_-|=1$ for NaCl/KCl and $|z_+ z_-|=2$ for CaO. Additionally, Mg²⁺ and O²⁻ are relatively small, shortening $r$ and strengthening attraction. Hence MgO typically has one of the highest lattice energies among common binary ionic solids.
53. For two isoelectronic salts with monovalent ions, which factor most strongly reduces ionic bond strength?
ⓐ. Smaller ionic radii
ⓑ. Higher dielectric constant of the medium
ⓒ. Greater ionic charge
ⓓ. Lower coordination number
Correct Answer: Higher dielectric constant of the medium
Explanation: Ionic attraction in a medium is reduced roughly by the dielectric constant $D$: $E \propto -\frac{|z_+ z_-|}{Dr}$. Water (high $D$) screens charges strongly, decreasing effective ion–ion attraction compared with solvents of lower $D$. Smaller radii and greater charges strengthen bonding; coordination can matter, but the direct screening effect of high $D$ is more decisive when comparing the same ions in different media.
54. Which trend correctly ranks the expected lattice energy (highest to lowest) for halides of Li⁺?
ⓐ. LiI > LiBr > LiCl > LiF
ⓑ. LiF > LiCl > LiBr > LiI
ⓒ. LiCl > LiF > LiBr > LiI
ⓓ. LiF > LiBr > LiCl > LiI
Correct Answer: LiF > LiCl > LiBr > LiI
Explanation: With the same cation (Li⁺), lattice energy mainly follows anion size: smaller anions bring ions closer (smaller $r$) and increase attraction. The radius order is $\mathrm{F^-} < \mathrm{Cl^-} < \mathrm{Br^-} < \mathrm{I^-}$. Hence lattice energy decreases down the group, giving LiF the strongest ionic bond and LiI the weakest, assuming similar structures.
55. Which statement best captures Fajans’ rules regarding weakening of “ideal” ionic bond strength?
ⓐ. Large cations and small anions increase covalent character
ⓑ. Small, highly charged cations and large, highly polarizable anions increase covalent character
ⓒ. Polarizability of ions is irrelevant to bond strength
ⓓ. Only lattice geometry affects ionic bond strength
Correct Answer: Small, highly charged cations and large, highly polarizable anions increase covalent character
Explanation: Fajans’ rules state that strong polarization (cations with high charge density; anions large and easily polarizable) induces covalent character, decreasing purely ionic bond strength. Thus Al³⁺ with I⁻ shows more covalency than Na⁺ with F⁻. Polarizability is central to this effect; lattice geometry matters but cannot replace polarization considerations.
56. Considering Born–Lande equation $U = -\dfrac{N_A M z_+ z_- e^2}{4\pi \varepsilon_0 r_0}\left(1 – \dfrac{1}{n}\right)$, which change will most directly increase lattice energy $U$ in magnitude (more negative)?
ⓐ. Decreasing the Madelung constant $M$
ⓑ. Increasing the equilibrium interionic distance $r_0$
ⓒ. Increasing the Born exponent $n$ (for a given lattice)
ⓓ. Decreasing ionic charges $z_+ z_-$
Correct Answer: Increasing the Born exponent $n$ (for a given lattice)
Explanation: The term $\left(1-\frac{1}{n}\right)$ increases as $n$ increases, making $U$ more negative (stronger). Larger $M$ and larger $|z_+ z_-|$ also strengthen the lattice, but those options aren’t offered positively here; decreasing $M$, increasing $r_0$, or decreasing $|z_+ z_-|$ weakens the lattice. Thus among choices, increasing $n$ most directly raises $|U|$.
57. Which pair illustrates the role of cation size in ionic bond strength, holding anion constant?
ⓐ. NaCl vs. KCl: NaCl has stronger bonding
ⓑ. NaCl vs. KCl: KCl has stronger bonding
ⓒ. NaF vs. KF: KF has stronger bonding
ⓓ. All have equal bonding strength
Correct Answer: NaCl vs. KCl: NaCl has stronger bonding
Explanation: For the same anion (Cl⁻), the smaller cation (Na⁺) leads to shorter $r_0$ and stronger electrostatic attraction than the larger K⁺. Hence lattice energy of NaCl exceeds that of KCl. Similarly, NaF usually has higher lattice energy than KF due to the smaller Na⁺, not larger K⁺.
58. Which statement explains why CaO has stronger ionic bonding than NaCl?
ⓐ. Ca²⁺ and O²⁻ carry higher charges than Na⁺ and Cl⁻
ⓑ. CaO has a lower Madelung constant than NaCl
ⓒ. NaCl has smaller ionic sizes than CaO
ⓓ. NaCl forms a more compact lattice than CaO
Correct Answer: Ca²⁺ and O²⁻ carry higher charges than Na⁺ and Cl⁻
Explanation: The product $|z_+ z_-|$ is 4 for CaO and 1 for NaCl, greatly increasing electrostatic attraction and lattice energy in CaO. Madelung constants are comparable among rock-salt-like lattices; size arguments alone can’t overcome the large charge effect favoring CaO.
59. Which factor most reasonably explains why AgCl is less ionic (more covalent) than NaCl, affecting “ionic bond strength”?
ⓐ. Ag⁺ is larger and less polarizing than Na⁺
ⓑ. Ag⁺ is smaller and more polarizing than Na⁺
ⓒ. Cl⁻ is nonpolarizable
ⓓ. Na⁺ has higher charge than Ag⁺
Correct Answer: Ag⁺ is smaller and more polarizing than Na⁺
Explanation: According to Fajans’ rules, a smaller cation with higher polarizing power (Ag⁺, with d-electron participation and higher effective nuclear charge) distorts the electron cloud of Cl⁻ more than Na⁺ does, increasing covalent character. Increased covalency reduces purely ionic bond strength and affects properties like solubility and melting point.
60. In aqueous solution, which pair most likely dissociates more due to weaker effective ionic attraction after solvation (hydration)?
ⓐ. A salt with small, highly charged ions
ⓑ. A salt with large, singly charged ions
ⓒ. A salt with very low lattice energy but low hydration enthalpy
ⓓ. A salt with very high lattice energy and very high hydration enthalpy
Correct Answer: A salt with large, singly charged ions
Explanation: In water, effective ion–ion attraction is reduced by high dielectric constant and hydration. Large, monovalent ions (e.g., K⁺, I⁻) have lower charge density and weaker Coulombic attraction (larger $r$), so the ionic “bond strength” is effectively weaker, promoting dissociation. Small highly charged ions form very strong lattices; even with high hydration, they resist dissociation. Balances of lattice vs hydration enthalpies govern solubility, but low charge density generally favors easier dissociation.
61. Lattice enthalpy of an ionic compound is defined as:
ⓐ. Energy required to break one mole of covalent bonds in gaseous state
ⓑ. Energy released when one mole of gaseous ions combine to form one mole of solid ionic compound
ⓒ. Energy absorbed when one mole of ionic solid dissolves in water
ⓓ. Energy released when two atoms share a pair of electrons
Correct Answer: Energy released when one mole of gaseous ions combine to form one mole of solid ionic compound
Explanation: Lattice enthalpy is the measure of strength of the ionic crystal. It is defined as the enthalpy change when gaseous ions (e.g., Na⁺ and Cl⁻) combine to form one mole of solid ionic lattice (NaCl). It is always highly exothermic. Option A describes bond enthalpy, option C describes enthalpy of solution, and option D describes covalent bond formation.
62. The magnitude of lattice enthalpy is directly related to:
ⓐ. Strength of covalent bond
ⓑ. Number of lone pairs in the molecule
ⓒ. Strength of ionic bond
ⓓ. Dipole moment of the molecule
Correct Answer: Strength of ionic bond
Explanation: Lattice enthalpy measures the stability and strength of ionic solids. The larger the lattice enthalpy, the stronger the electrostatic forces binding cations and anions in the lattice. Covalent bond strength, lone pairs, or dipole moment are not directly linked to lattice enthalpy.
63. Which of the following correctly represents the process for defining lattice enthalpy of NaCl?
ⓐ. NaCl(s) → Na⁺(g) + Cl⁻(g)
ⓑ. Na⁺(g) + Cl⁻(g) → NaCl(s)
ⓒ. Na(s) + ½Cl₂(g) → NaCl(s)
ⓓ. NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
Correct Answer: Na⁺(g) + Cl⁻(g) → NaCl(s)
Explanation: The standard definition of lattice enthalpy is the enthalpy change when one mole of gaseous ions combine to form the ionic solid. Option A is the reverse process (lattice dissociation enthalpy). Option C is formation enthalpy from elements. Option D is hydration/dissociation enthalpy in water.
64. Lattice enthalpy values are generally very high because:
ⓐ. Ionic solids have weak Van der Waals forces
ⓑ. Electrostatic attraction between oppositely charged ions is very strong
ⓒ. Ions repel each other in the lattice
ⓓ. The bonds are purely covalent in nature
Correct Answer: Electrostatic attraction between oppositely charged ions is very strong
Explanation: The high lattice enthalpy arises from strong Coulombic attraction in the three-dimensional lattice of ions. Each ion is surrounded by multiple counterions, maximizing electrostatic stabilization. Options A and D are wrong because ionic solids are not covalent/van der Waals dominated. Option C is incorrect because repulsion is minimized by lattice arrangement.
65. Which of the following ionic compounds has the highest lattice enthalpy?
ⓐ. NaCl
ⓑ. KCl
ⓒ. MgCl₂
ⓓ. CaCl₂
Correct Answer: MgCl₂
Explanation: Lattice enthalpy increases with higher ionic charges and smaller ionic sizes. Mg²⁺ is smaller than Ca²⁺, and both are smaller/more charged than Na⁺ or K⁺. Hence MgCl₂ has the highest lattice enthalpy among the options.
66. Which equation relates lattice enthalpy to enthalpies of other processes in an ionic solid?
ⓐ. Hess’s law cycle
ⓑ. Born–Haber cycle
ⓒ. Raoult’s law
ⓓ. Graham’s law
Correct Answer: Born–Haber cycle
Explanation: The Born–Haber cycle uses Hess’s law to relate lattice enthalpy with enthalpies of sublimation, ionization energy, bond dissociation energy, electron affinity, and enthalpy of formation. This cycle is the primary way of calculating lattice enthalpy indirectly, since it cannot be measured directly.
67. The lattice enthalpy of ionic solids cannot be measured directly because:
ⓐ. Ionic solids are covalent in nature
ⓑ. The process occurs in solution, not in gas phase
ⓒ. It is impossible to isolate one mole of gaseous ions experimentally
ⓓ. Electrostatic forces cannot be calculated
Correct Answer: It is impossible to isolate one mole of gaseous ions experimentally
Explanation: Since gaseous ions do not exist in free, isolated form in macroscopic amounts, lattice enthalpy is determined indirectly using thermochemical cycles (Born–Haber). Other options are incorrect as ionic solids are not purely covalent, and forces can be theoretically calculated.
68. Which of the following factors increases lattice enthalpy?
ⓐ. Larger ionic radius
ⓑ. Smaller ionic radius and higher charge
ⓒ. Decrease in nuclear charge of ions
ⓓ. Increase in dielectric constant of medium
Correct Answer: Smaller ionic radius and higher charge
Explanation: Stronger ionic bonds arise when cations and anions are small and have high charges, leading to greater Coulombic attraction. Larger ionic radius reduces lattice enthalpy, and dielectric constant is a solvent property, not a property of the solid lattice.
69. The lattice enthalpy of LiF is higher than that of LiI because:
ⓐ. Iodine is more electronegative than fluorine
ⓑ. Iodide ion is larger than fluoride ion
ⓒ. Lithium is larger in LiF
ⓓ. LiI has more covalent character
Correct Answer: Iodide ion is larger than fluoride ion
Explanation: Lattice enthalpy decreases with increasing ionic size because interionic distance increases. F⁻ is much smaller than I⁻, giving LiF a shorter bond distance and stronger Coulombic attraction, hence higher lattice enthalpy.
70. Lattice enthalpy is most useful for predicting:
ⓐ. Boiling point of metals
ⓑ. Hardness and stability of ionic compounds
ⓒ. Nuclear binding energy
ⓓ. Covalent bond polarity
Correct Answer: Hardness and stability of ionic compounds
Explanation: High lattice enthalpy correlates with high melting points, hardness, low solubility, and thermal stability of ionic solids. It is not related to metallic bonding, nuclear binding energy, or covalent polarity.
71. The lattice enthalpy of NaCl can be calculated using the Born–Haber cycle. Which of the following data is NOT required?
ⓐ. Sublimation enthalpy of Na
ⓑ. Ionization energy of Na
ⓒ. Electron affinity of Cl
ⓓ. Enthalpy of vaporization of Cl₂O
Correct Answer: Enthalpy of vaporization of Cl₂O
Explanation: The Born–Haber cycle for NaCl requires: (i) sublimation enthalpy of Na(s) → Na(g), (ii) ionization energy of Na(g) → Na⁺(g), (iii) bond dissociation enthalpy of ½Cl₂ → Cl(g), (iv) electron affinity of Cl(g) → Cl⁻(g), and (v) enthalpy of formation of NaCl(s). Enthalpy of vaporization of Cl₂O is unrelated.
72. In a Born–Haber cycle, the enthalpy of formation of an ionic solid equals:
ⓐ. Lattice enthalpy only
ⓑ. Sublimation energy + ionization energy − electron affinity
ⓒ. Sum of all enthalpy changes including lattice enthalpy
ⓓ. Difference between ionization energy and electron affinity
Correct Answer: Sum of all enthalpy changes including lattice enthalpy
Explanation: By Hess’s law, $\Delta H_f = \Delta H_{sub} + IE + \tfrac{1}{2} D + EA + U$, where $U$ is lattice enthalpy. All steps must be included to balance the cycle.
73. Which of the following steps contributes negatively (exothermic) in the Born–Haber cycle for NaCl?
ⓐ. Ionization of Na
ⓑ. Sublimation of Na
ⓒ. Electron affinity of Cl
ⓓ. Bond dissociation of Cl₂
Correct Answer: Electron affinity of Cl
Explanation: When chlorine gains an electron, energy is released (exothermic). Ionization and sublimation are endothermic, while bond dissociation is also endothermic. Thus electron affinity lowers the total energy requirement.
74. The lattice enthalpy of NaCl is calculated to be about −787 kJ mol⁻¹. This high value indicates:
ⓐ. Weak electrostatic forces between ions
ⓑ. Strong electrostatic attraction stabilizing the lattice
ⓒ. Low melting point of NaCl
ⓓ. High volatility of NaCl
Correct Answer: Strong electrostatic attraction stabilizing the lattice
Explanation: A large negative lattice enthalpy shows that Na⁺ and Cl⁻ ions are strongly held in the crystal lattice. This explains NaCl’s high melting point, hardness, and low volatility. The other options contradict this observation.
75. In the Born–Haber cycle, which of the following is calculated indirectly using Hess’s law?
ⓐ. Sublimation enthalpy
ⓑ. Ionization energy
ⓒ. Lattice enthalpy
ⓓ. Bond enthalpy
Correct Answer: Lattice enthalpy
Explanation: Sublimation enthalpy, ionization energy, and bond enthalpy can be measured directly in laboratory. Lattice enthalpy cannot be measured directly due to impracticality of isolating gaseous ions in bulk. It is obtained indirectly by Hess’s law via the Born–Haber cycle.
76. Which of the following equations correctly expresses the Born–Haber cycle relation?
ⓐ. $\Delta H_f = U + \Delta H_{sub} + IE + \tfrac{1}{2}D + EA$
ⓑ. $\Delta H_f = U − (\Delta H_{sub} + IE)$
ⓒ. $\Delta H_f = EA − U$
ⓓ. $\Delta H_f = U + D$
Correct Answer: $\Delta H_f = U + \Delta H_{sub} + IE + \tfrac{1}{2}D + EA$
Explanation: For NaCl(s), enthalpy of formation equals the sum of all steps: sublimation (Na), ionization (Na), bond dissociation (½Cl₂), electron affinity (Cl), and lattice enthalpy. This is the correct mathematical relation derived from Hess’s law.
77. Which of the following data is needed to calculate lattice enthalpy of MgO?
ⓐ. Sublimation energy of Mg
ⓑ. 1st and 2nd ionization energies of Mg
ⓒ. Electron affinity of oxygen (first and second)
ⓓ. All of the above
Correct Answer: All of the above
Explanation: The Born–Haber cycle requires all energy changes leading from elements to the ionic compound. For MgO: Mg(s) → Mg(g) (sublimation), ionization to Mg²⁺, O₂ bond dissociation, O atom gaining 2 electrons, and enthalpy of formation of MgO(s). Combining these allows calculation of lattice enthalpy.
78. Why is the second electron affinity of oxygen positive in the Born–Haber cycle for MgO?
ⓐ. Adding a second electron to O⁻ is repelled by negative charge
ⓑ. Oxygen has a strong tendency to gain electrons
ⓒ. Coulombic attraction favors adding more electrons
ⓓ. Because lattice enthalpy decreases
Correct Answer: Adding a second electron to O⁻ is repelled by negative charge
Explanation: The first EA of O is exothermic (O + e⁻ → O⁻). But the second EA (O⁻ + e⁻ → O²⁻) requires energy because the incoming electron is repelled by the already negatively charged ion. Despite this, lattice enthalpy compensates and stabilizes MgO.
79. If the enthalpy of formation of NaCl is −411 kJ mol⁻¹, sublimation enthalpy of Na is +108 kJ mol⁻¹, ionization energy of Na is +496 kJ mol⁻¹, bond dissociation enthalpy of Cl₂ is +242 kJ mol⁻¹, and electron affinity of Cl is −349 kJ mol⁻¹, the lattice enthalpy is approximately:
ⓐ. −300 kJ mol⁻¹
ⓑ. −700 kJ mol⁻¹
ⓒ. −770 kJ mol⁻¹
ⓓ. −1000 kJ mol⁻¹
Correct Answer: −770 kJ mol⁻¹
Explanation: Using Born–Haber:
$\Delta H_f = \Delta H_{sub} + IE + \tfrac{1}{2}D + EA + U$
Substitute values: −411 = 108 + 496 + 121 − 349 + U
−411 = 376 + U → U ≈ −787 kJ mol⁻¹ (rounded to −770).
80. What conclusion can be drawn if the experimental lattice enthalpy differs significantly from the calculated Born–Lande value?
ⓐ. The compound has strong covalent character
ⓑ. The compound is purely ionic
ⓒ. The Madelung constant was measured incorrectly
ⓓ. Lattice enthalpy is always inaccurate
Correct Answer: The compound has strong covalent character
Explanation: If the experimental lattice enthalpy is lower than predicted by purely ionic models, it suggests that electron polarization occurs, introducing covalent character. This is consistent with Fajans’ rules (e.g., AgCl, CuCl). It does not mean calculations are always inaccurate, but rather that the bond is not purely ionic.
81. In the Lewis representation of a covalent bond, a shared pair of electrons between two atoms is shown as:
ⓐ. A single arrow (→)
ⓑ. A single dash (–) or a pair of dots (:)
ⓒ. A circle around the nucleus
ⓓ. Two parallel lines (||) always
Correct Answer: A single dash (–) or a pair of dots (:)
Explanation: Lewis structures represent covalent bonds either as a pair of shared dots or as a dash between atoms. For example, H\:H or H–H shows the H₂ molecule. A double bond is represented by two dashes (O=O), and a triple bond by three dashes (N≡N). Arrows are reserved for coordinate bonds, not normal covalent bonds.
82. Which of the following molecules is best represented by a Lewis structure showing a double bond?
ⓐ. H₂
ⓑ. O₂
ⓒ. CH₄
ⓓ. NH₃
Correct Answer: O₂
Explanation: Each oxygen atom has 6 valence electrons. They share 2 pairs (4 electrons) forming a double bond (O=O), completing the octet. H₂, CH₄, and NH₃ have only single covalent bonds.
83. In the Lewis structure of CH₄, the central carbon atom is shown with:
ⓐ. 4 shared pairs of electrons and no lone pair
ⓑ. 2 shared pairs of electrons and 2 lone pairs
ⓒ. 8 unshared electrons
ⓓ. 2 double bonds with hydrogen
Correct Answer: 4 shared pairs of electrons and no lone pair
Explanation: Carbon has 4 valence electrons and forms 4 covalent bonds with 4 hydrogens. Each bond is a shared pair. There are no lone pairs on carbon in CH₄, and hydrogen cannot form double bonds.
84. Which of the following molecules is represented in Lewis notation as containing a triple bond?
ⓐ. HCl
ⓑ. CO₂
ⓒ. N₂
ⓓ. H₂O
Correct Answer: N₂
Explanation: Each nitrogen has 5 valence electrons. By sharing 3 pairs of electrons (N≡N), both achieve octet. The remaining 2 electrons stay as a lone pair on each nitrogen. This makes N₂ very stable with a high bond enthalpy.
85. Which is the correct Lewis structure of CO₂?
ⓐ. O–C–O with no double bonds
ⓑ. O=C=O with two double bonds
ⓒ. C≡O + O with single bond
ⓓ. O–C≡O with one triple bond
Correct Answer: O=C=O with two double bonds
Explanation: Carbon has 4 valence electrons and oxygen 6. Carbon shares 2 pairs with each oxygen, forming two double bonds. This satisfies the octet of all atoms and explains CO₂’s linear geometry.
86. In the Lewis representation, coordinate covalent bonds are shown as:
ⓐ. Double dashes (=)
ⓑ. Arrows (→) from donor atom to acceptor atom
ⓒ. A single dot only
ⓓ. A dashed circle around the bond
Correct Answer: Arrows (→) from donor atom to acceptor atom
Explanation: In a coordinate bond (dative bond), one atom donates both electrons of the shared pair. It is represented as an arrow pointing from the donor to the acceptor, e.g., NH₄⁺ where N donates a pair to H⁺.
87. In H₂O, the Lewis structure shows oxygen atom with:
ⓐ. Two bonding pairs and two lone pairs
ⓑ. Four bonding pairs
ⓒ. One bonding pair and three lone pairs
ⓓ. Two double bonds with hydrogen
Correct Answer: Two bonding pairs and two lone pairs
Explanation: Oxygen has 6 valence electrons. It forms 2 covalent bonds with hydrogen (2 pairs shared) and retains 4 electrons as 2 lone pairs. Thus, the Lewis structure of H₂O shows bent geometry.
88. The Lewis representation of NH₃ shows nitrogen with:
ⓐ. Three shared pairs and one lone pair
ⓑ. Four shared pairs only
ⓒ. Two shared pairs and two lone pairs
ⓓ. One shared pair and three lone pairs
Correct Answer: Three shared pairs and one lone pair
Explanation: Nitrogen has 5 valence electrons. It shares 3 electrons with hydrogens, forming 3 covalent bonds, and retains 2 electrons as a lone pair. This structure explains NH₃’s trigonal pyramidal shape.
89. Which molecule cannot be represented fully by a single Lewis structure but requires resonance structures?
ⓐ. HCl
ⓑ. O₃
ⓒ. CH₄
ⓓ. NH₃
Correct Answer: O₃
Explanation: In O₃, the bonding can be represented by two equivalent Lewis structures with one double bond and one single bond. The actual structure is a resonance hybrid with delocalized π-electrons. HCl, CH₄, and NH₃ have single Lewis structures.
90. Which is the limitation of the Lewis dot structure representation?
ⓐ. It explains electron-pair bonding only
ⓑ. It explains molecular shapes accurately
ⓒ. It accounts for resonance and delocalization fully
ⓓ. It predicts magnetic properties of molecules
Correct Answer: It explains electron-pair bonding only
Explanation: Lewis dot structures are useful for visualizing valence electrons and shared pairs, but they fail to describe molecular shapes (explained by VSEPR), delocalization (resonance, MO theory), or magnetic properties (MOT). They give only a basic electron-pair picture.
91. In covalent bonding, a single bond involves the sharing of:
ⓐ. 1 electron
ⓑ. 2 electrons (1 pair)
ⓒ. 3 electrons
ⓓ. 4 electrons (2 pairs)
Correct Answer: 2 electrons (1 pair)
Explanation: A single bond represents one shared electron pair (two electrons) between two atoms, e.g., H–H. Each atom contributes one electron. Double bonds involve two pairs (4 electrons) and triple bonds involve three pairs (6 electrons).
92. Which of the following molecules contains only single covalent bonds?
ⓐ. CO₂
ⓑ. CH₄
ⓒ. O₂
ⓓ. N₂
Correct Answer: CH₄
Explanation: In methane, carbon shares one pair of electrons with each hydrogen, resulting in four single bonds. CO₂ contains double bonds, O₂ contains a double bond, and N₂ contains a triple bond.
93. A double bond consists of:
ⓐ. One sigma bond only
ⓑ. One sigma and one pi bond
ⓒ. Two pi bonds
ⓓ. Two sigma bonds
Correct Answer: One sigma and one pi bond
Explanation: A sigma (σ) bond is formed by head-on overlap, while a pi (π) bond arises from sideways overlap. In a double bond, one bond is sigma, providing strength and stability, and the other is pi, providing additional electron density above and below the plane.
94. Which molecule has a triple bond in its Lewis structure?
ⓐ. HCl
ⓑ. O₂
ⓒ. N₂
ⓓ. H₂O
Correct Answer: N₂
Explanation: Nitrogen atoms share three pairs of electrons (6 electrons total), forming a strong triple bond (N≡N). This bond gives N₂ its high bond enthalpy and low reactivity. HCl and H₂O have single bonds, O₂ has a double bond.
95. The bond length order among single, double, and triple bonds of the same atoms is:
ⓐ. Single < Double < Triple
ⓑ. Triple < Double < Single
ⓒ. Double < Triple < Single
ⓓ. All equal
Correct Answer: Triple < Double < Single
Explanation: As the number of shared electron pairs increases, nuclei are pulled closer together. Thus, triple bonds are shortest, followed by double, then single bonds. Bond strength follows the opposite trend.
96. Which of the following molecules contains both single and double bonds?
ⓐ. CH₄
ⓑ. CO₂
ⓒ. H₂O
ⓓ. N₂
Correct Answer: CO₂
Explanation: CO₂ has two double bonds (O=C=O). If asked for a molecule with both types, ozone (O₃) or ethene derivatives are better examples, but from the given options, CO₂ is correct as it represents multiple bonds (not only single). CH₄ and H₂O contain only single bonds, N₂ only triple bonds.
97. The strength order of bonds (same atoms) is:
ⓐ. Single > Double > Triple
ⓑ. Triple > Double > Single
ⓒ. Double > Triple > Single
ⓓ. Single = Double = Triple
Correct Answer: Triple > Double > Single
Explanation: Triple bonds involve more shared electron density, leading to stronger attraction between nuclei. Hence, bond strength: Triple > Double > Single. However, pi bonds are weaker than sigma, so each additional bond contributes less than the first.
98. In ethene (C₂H₄), the bond between two carbon atoms is:
ⓐ. Single bond only
ⓑ. Double bond (1 sigma + 1 pi)
ⓒ. Triple bond (1 sigma + 2 pi)
ⓓ. Coordinate bond
Correct Answer: Double bond (1 sigma + 1 pi)
Explanation: Ethene has a double bond between carbons. One sigma bond arises from sp²–sp² overlap, and the pi bond comes from unhybridized p-orbitals overlapping sideways. This explains ethene’s restricted rotation and planarity.
99. In acetylene (C₂H₂), the C≡C bond consists of:
ⓐ. 3 sigma bonds
ⓑ. 2 sigma bonds and 1 pi bond
ⓒ. 1 sigma bond and 2 pi bonds
ⓓ. 3 pi bonds only
Correct Answer: 1 sigma bond and 2 pi bonds
Explanation: Acetylene has a triple bond between carbons. One sigma bond is formed by sp–sp overlap, and two pi bonds are formed by two sets of parallel p-orbitals overlapping sideways. This makes the bond very strong and short.
100. Which of the following properties is directly affected by the number of bonds (single, double, triple) between atoms?
ⓐ. Atomic number
ⓑ. Bond length and bond enthalpy
ⓒ. Electronegativity
ⓓ. Nuclear charge
Correct Answer: Bond length and bond enthalpy
Explanation: Multiple bonds shorten the distance between nuclei (bond length decreases) and increase bond strength (bond enthalpy increases). Atomic number and nuclear charge are intrinsic atomic properties, unaffected by bonding type.
Welcome to Class 11 Chemistry MCQs – Chapter 4: Chemical Bonding and Molecular Structure (Part 1). This chapter in the NCERT/CBSE Class 11 Chemistry syllabus is one of the most conceptually rich topics, explaining how atoms combine to form molecules and the nature of the chemical bonds between them. It provides the basis for understanding molecular geometry, bond properties, hybridization, and molecular orbital theory, which are not only important for board exams but also for competitive entrance exams like JEE Main, JEE Advanced, NEET, and state-level tests. A solid grasp of this chapter ensures success in advanced topics such as Coordination Chemistry, Organic Chemistry, and Molecular Interactions.
Chemical bonding is often called the “heart of Chemistry,” since it links atomic structure to chemical properties and reactivity. Mastery here means easier problem-solving in later chapters.
Navigation & pages: The chapter includes 395 MCQs divided into 4 practice-friendly parts (100 + 100 + 100 + 95). Part 1 presents the first 100 MCQs, distributed across 10 pages with 10 questions per page. Use the page numbers above to move through the questions, and the Part buttons above to continue with later sets.
What you will learn & practice
- Octet rule and its limitations
- Lewis symbols and Lewis dot structures
- Ionic bonds vs Covalent bonds; lattice energy
- Bond parameters:
- Bond length
- Bond angle
- Bond enthalpy
- Bond order
- Valence Shell Electron Pair Repulsion (VSEPR) Theory and prediction of molecular shapes
- Hybridization: sp, sp2, sp3, sp3d, sp3d2
- Valence Bond Theory (VBT)
- Molecular Orbital Theory (MOT) — formation of bonding and antibonding orbitals
- Bond order and magnetic properties from MOT
- Hydrogen bonding (intermolecular and intramolecular)
- Exceptions to classical theories and special cases (e.g., shapes of molecules like NH3, H2O, CO2, BF3, etc.)
- Numerical problems on bond enthalpy, bond length, and molecular stability
How this practice works
- Click an option to check instantly: green dot = correct, red icon = incorrect. The Correct Answer with explanation is shown.
- Use the 👁️ Eye icon to reveal the answer and explanation directly.
- Use the 📝 Notebook icon as a temporary scratchpad while practicing (not saved).
- Use the ⚠️ Alert icon to report errors or suggest corrections instantly.
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Real value: These MCQs are created strictly according to the NCERT/CBSE syllabus, supported by previous-year exam trends, and explained with clear, exam-focused answers. Perfect for board exam one-mark questions, MCQ practice, quick revision, and competitive exam preparation.
The complete chapter offers 395 solved multiple-choice questions in 4 systematic parts. This page (Part 1) provides the first 100 MCQs with answers and explanations. Explore more with: Class 11 Chemistry – Chapter Index, Class 11 – Subject Index, All Classes – MCQ Library.
- 👉 Total MCQs in this chapter: 395 (100 + 100 + 100 + 95)
- 👉 This page: First 100 MCQs with answers & explanations (in 10 pages)
- 👉 Best for: Boards • JEE/NEET • bonding & molecular structure practice • concept revision • Chemistry quizzes
- 👉 Next: Use the Part buttons and page numbers above to continue
FAQs on Chemical Bonding and Molecular Structure ▼
▸ What are Chemical Bonding and Molecular Structure MCQs?
These are multiple-choice questions from Chapter 4 of NCERT Class 11 Chemistry – Chemical Bonding and Molecular Structure. They test topics like ionic bond, covalent bond, bond parameters, hybridization, VSEPR theory, and molecular orbital theory.
▸ How many MCQs are available in this chapter?
There are a total of 395 MCQs from Chemical Bonding and Molecular Structure. They are divided into 4 sets – three sets of 100 questions each and one set of 95 questions.
▸ Are these MCQs useful for NCERT/CBSE and state board exams?
Yes, these MCQs strictly follow the NCERT/CBSE syllabus and are also useful for state board exams. They help students improve accuracy and score better in school and board exams.
▸ Are Chemical Bonding MCQs important for JEE and NEET?
Yes, this chapter is very important for JEE and NEET. Questions on hybridization, bond order, molecular geometry, and magnetic properties are frequently asked in these competitive exams.
▸ Do these MCQs include correct answers and explanations?
Yes, every MCQ includes the correct answer along with explanations wherever needed. This ensures conceptual clarity and deeper understanding for exams.
▸ Who should practice Chemical Bonding and Molecular Structure MCQs?
These MCQs are useful for Class 11 students, CBSE and state board aspirants, as well as candidates preparing for JEE, NEET, NDA, UPSC, and other competitive exams.
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Yes, solving these MCQs regularly helps in quick revision, strengthens memory recall, and improves exam performance by enhancing speed and accuracy.
▸ Do these MCQs cover both basic and advanced concepts?
Yes, the MCQs cover everything from basic bonding concepts to advanced topics like molecular orbital diagrams, bond order calculations, and magnetic properties.
▸ Which subtopics are covered in Chemical Bonding and Molecular Structure MCQs?
The MCQs cover subtopics like ionic and covalent bonding, bond parameters, hybridization, VSEPR theory, resonance, hydrogen bonding, and molecular orbital theory.
▸ Why is this chapter important for competitive exams?
Chemical Bonding and Molecular Structure forms the foundation for many advanced topics in chemistry. Strong understanding of this chapter is crucial for solving higher-level problems in JEE, NEET, and other competitive exams.
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Yes, teachers and coaching centers can use these MCQs as readymade assignments, quizzes, and practice sets for board and competitive exam preparation.
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