Chapter 3 – Classification of Elements and Periodicity of Properties (2 Marks Questions with Answers)
Q1. Why does atomic radius decrease across a period from left to right in the periodic table?
Answer:
As we move across a period, the number of protons increases, leading to a higher nuclear charge. Electrons are added to the same shell, so the attraction between nucleus and electrons increases, causing the atomic radius to decrease.
Q2. Why is the ionization enthalpy of oxygen lower than that of nitrogen, despite oxygen having a higher nuclear charge?
Answer:
Oxygen has a paired electron in one of its p-orbitals, resulting in increased electron–electron repulsion. This makes it easier to remove an electron from oxygen than nitrogen, so its ionization enthalpy is lower.
Q3. Write the electronic configuration of an element with atomic number 35. Identify its group and period.
Answer:
Atomic number 35 = Bromine (Br)
Electronic configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁵
It belongs to Group 17 and Period 4.
Q4. Compare the atomic sizes of Na, Mg, and Al. Explain the trend.
Answer:
Atomic size decreases from Na → Mg → Al because nuclear charge increases while electrons are added to the same shell. Hence, the attraction between nucleus and electrons increases.
Q5. Explain why the electron affinity of chlorine is greater than that of fluorine.
Answer:
Fluorine’s small atomic size causes strong inter-electronic repulsion in its compact 2p shell. Chlorine’s larger size allows easier addition of an electron, so chlorine has greater electron affinity than fluorine.
Q6. What is a diagonal relationship? Give one example.
Answer:
Elements of the second period show similarities with diagonally placed elements of the third period due to similar charge/radius ratio.
Example: Li and Mg, Be and Al.
Q7. Arrange the following in increasing metallic character: Li, K, Na, Rb. Explain.
Answer:
Order: Li < Na < K < Rb
Metallic character increases down a group because ionization energy decreases as atomic size increases.
Q8. State the reason why noble gases are placed in Group 18, although their electronic configuration differs from other groups.
Answer:
Noble gases have completely filled valence shells (ns² np⁶). They are chemically inert and show similar properties, hence placed together in Group 18.
Q9. Why does electronegativity increase across a period but decrease down a group?
Answer:
Across a period, atomic size decreases and nuclear charge increases, causing stronger attraction for bonding electrons.
Down a group, atomic size increases, reducing the attraction for shared electrons.
Q10. Predict the type of oxide formed by an element with configuration 1s² 2s² 2p⁶ 3s² 3p⁵.
Answer:
The configuration belongs to Chlorine (Cl), a nonmetal.
It forms acidic oxides like Cl₂O₇ and ClO₂.
Q11. Explain why alkali metals are stored under kerosene oil.
Answer:
Alkali metals are highly reactive and react vigorously with air and water. They are stored under kerosene oil to prevent oxidation and moisture contact.
Q12. Compare the first ionization enthalpies of lithium and sodium. Give reasons.
Answer:
Sodium has lower ionization enthalpy than lithium because its atomic size is larger, so the outer electron is farther from the nucleus and easier to remove.
Q13. Why is van der Waals radius greater than covalent radius for noble gases?
Answer:
Noble gases exist as monoatomic molecules and do not form covalent bonds. The van der Waals forces are weak, keeping atoms farther apart; thus, van der Waals radius > covalent radius.
Q14. Arrange F, Cl, Br, I in order of increasing electron affinity and justify.
Answer:
Order: I < Br < F < Cl
Due to small size, fluorine experiences electron repulsion, making Cl have the highest electron affinity.
Q15. Identify the block of the element with configuration [Xe] 4f¹⁴ 5d¹ 6s².
Answer:
The element has electrons in d-orbital, so it belongs to the d-block (transition elements).
Q16. Explain Mendeleev’s periodic law and give one limitation of his periodic table.
Answer:
Mendeleev’s Law: Properties of elements are periodic functions of their atomic weights.
Limitation: Could not explain position of isotopes and hydrogen properly.
Q17. State the Modern Periodic Law and explain how it differs from Mendeleev’s law.
Answer:
Modern Law: Properties of elements are periodic functions of their atomic numbers.
Difference: Mendeleev used atomic weight, while Moseley used atomic number, which resolved many anomalies.
Q18. Compare Dobereiner’s Triads and Newlands’ Octaves with examples.
Answer:
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Dobereiner’s Triads: Group of 3 elements with the middle one having atomic mass equal to the average of the other two.
Example: Li, Na, K. -
Newlands’ Octaves: Every 8th element shows repetition of properties.
Example: Na and K.
Q19. Explain the trends in atomic radius across a period and down a group with examples.
Answer:
Across a period, atomic radius decreases (e.g., Na > Mg > Al).
Down a group, it increases due to addition of new shells (e.g., Li < Na < K).
Q20. Explain the variation of ionization energy in a period and a group.
Answer:
Across a period, ionization energy increases because of higher nuclear charge.
Down a group, it decreases due to increased atomic size and shielding effect.
Chapter 4 – Chemical Bonding (2 Marks Questions with Answers)
Q1. Explain why ionic compounds conduct electricity in molten state but not in solid state.
Answer:
In solid state, ions are fixed in lattice positions and cannot move.
In molten or aqueous state, ions are free to move, allowing conduction of electricity.
Q2. Differentiate between sigma (σ) and pi (π) bonds.
Answer:
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Sigma bond (σ): Formed by end-to-end overlap of orbitals; strong bond; allows free rotation.
-
Pi bond (π): Formed by sideways overlap of p-orbitals; weaker; restricts rotation.
Q3. Write the shapes of the following molecules using VSEPR theory: BeCl₂ and H₂O.
Answer:
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BeCl₂: Linear shape, bond angle 180°.
-
H₂O: Bent (V-shaped), bond angle 104.5° due to lone pair repulsion.
Q4. Explain why NH₃ is polar but BF₃ is nonpolar.
Answer:
NH₃ has trigonal pyramidal shape and non-symmetric distribution of charge → polar.
BF₃ is trigonal planar and symmetric → dipoles cancel out → nonpolar.
Q5. Compare the bond angles of CH₄, NH₃, and H₂O. Give reasons.
Answer:
Bond angles: CH₄ = 109.5°, NH₃ = 107°, H₂O = 104.5°.
Lone pairs on N and O repel more strongly, reducing bond angles progressively.
Q6. Why is the bond length of a double bond shorter than a single bond?
Answer:
A double bond has both sigma and pi components, resulting in greater electron density between atoms → stronger attraction → shorter bond length.
Q7. Explain why resonance increases the stability of benzene.
Answer:
Resonance allows delocalization of electrons across the ring, reducing energy and increasing stability of benzene.
Q8. Describe the hybridization of carbon in ethylene (C₂H₄).
Answer:
Each carbon is sp² hybridized, forming three sigma bonds (two C–H and one C–C). The unhybridized p orbitals overlap to form a π bond.
Q9. Explain the concept of bond order. Calculate bond order of O₂.
Answer:
Bond order = (Nb – Na)/2
For O₂, Nb = 10, Na = 6 → (10–6)/2 = 2
Hence, O₂ has a double bond.
Q10. Why does H₂O have a bent shape while CO₂ is linear?
Answer:
H₂O has two lone pairs on oxygen → electron pair repulsion → bent shape.
CO₂ has no lone pairs on central atom → linear shape.
Q11. Write one example each of molecules with sp, sp², and sp³ hybridization.
Answer:
-
sp: BeCl₂
-
sp²: BF₃
-
sp³: CH₄
Q12. Why is HF a liquid at room temperature while HCl is a gas?
Answer:
HF molecules form strong hydrogen bonds, increasing intermolecular attraction.
HCl has weak van der Waals forces → remains gaseous.
Q13. Explain metallic bonding with the help of the electron-sea model.
Answer:
Metal atoms release valence electrons forming a “sea of electrons” around positive ions.
This results in conductivity, malleability, and metallic lustre.
Q14. Differentiate between ionic and covalent bonds with examples.
Answer:
| Property | Ionic Bond | Covalent Bond |
|---|---|---|
| Formation | Electron transfer | Electron sharing |
| Type | Metal + Nonmetal | Nonmetal + Nonmetal |
| Example | NaCl | H₂O |
Q15. Explain why ionic compounds generally have high melting and boiling points.
Answer:
Ionic compounds have strong electrostatic forces between oppositely charged ions, requiring high energy to break.
Q16. Draw Lewis structure of H₂O / CO₂ / NH₃.
Answer:
-
H₂O: O connected to 2 H atoms, two lone pairs on O.
-
CO₂: O=C=O (no lone pairs on central C).
-
NH₃: N connected to 3 H atoms, one lone pair on N.
Q17. Predict the shape of BF₃ / CH₄ / NH₄⁺.
Answer:
-
BF₃: Trigonal planar (120°)
-
CH₄: Tetrahedral (109.5°)
-
NH₄⁺: Tetrahedral (109.5°)
Q18. Calculate formal charge of N in NH₄⁺.
Answer:
Formal charge = Valence e⁻ – (Non-bonding e⁻ + ½ Bonding e⁻)
= 5 – (0 + ½×8) = +1
Hence, nitrogen carries +1 charge.
Q19. Explain resonance with an example.
Answer:
Resonance occurs when a molecule can be represented by two or more valid Lewis structures.
Example: CO₃²⁻ has three equivalent structures with delocalized π electrons.
