How to Use This Page
This page provides complete, step-by-step NCERT solutions for Class 11 Chemistry Chapter 3 — Classification of Elements and Periodicity in Properties. Every intext question and chapter-end exercise question is answered with detailed explanations. Use the section headings to navigate directly to the question you need.
Key Concepts (Quick Recap)
- Mendeleev's periodic table: Arranged 63 elements by increasing atomic mass; left gaps for undiscovered elements; limitations — position of hydrogen, anomalous pairs (Ar/K, Co/Ni), position of isotopes
- Modern periodic law: Properties of elements are a periodic function of their atomic number (Moseley)
- Groups: 18 vertical columns; elements in same group have same valence electron configuration
- Periods: 7 horizontal rows; period number = number of electron shells
- s-block: Groups 1 (alkali metals) and 2 (alkaline earth metals); valence config: ns¹–ns²
- p-block: Groups 13–18; valence config: ns²np¹–ns²np⁶
- d-block: Groups 3–12 (transition metals); valence config: (n−1)d¹–¹⁰ns⁰–²
- f-block: Lanthanides (4f) and actinides (5f); placed separately at bottom
- Atomic radius: Increases down a group (new shells added); decreases across a period (increased nuclear charge)
- Ionization enthalpy (IE): Energy to remove outermost electron; increases across a period, decreases down a group
- Electron gain enthalpy (EGE): Energy released when electron added; generally more negative across a period (exceptions: noble gases, N, Be, Mg)
- Electronegativity: Tendency to attract shared electrons; increases across a period, decreases down a group
- Metallic character: Decreases across a period, increases down a group
- Diagonal relationships: Li–Mg, Be–Al, B–Si show similar properties due to similar charge/size ratios
Intext Questions
Page 52–53 (Need for Classification, Mendeleev's Table)
Q3.1: What was the need for classification of elements?
AnswerWith the discovery of more elements, it became essential to organise them systematically so that their properties could be studied and predicted efficiently. Classification helps: (1) study properties of elements collectively rather than individually, (2) predict properties of undiscovered elements, (3) understand relationships between different elements, and (4) remember the vast information in a manageable way.
Q3.2: State the merits of Mendeleev's periodic table.
Answer(1) Systematic study of elements — elements with similar properties were grouped together. (2) Prediction of properties of undiscovered elements — Mendeleev predicted properties of eka-silicon (Ge), eka-aluminium (Ga), and eka-boron (Sc). (3) Correction of doubtful atomic masses — Be, In, Te were placed correctly. (4) Placement of noble gases — accommodated in a separate group.
Q3.3: What are the limitations of Mendeleev's periodic table?
Answer(1) Position of hydrogen — resembles both alkali metals and halogens, could not be placed definitively. (2) Anomalous pairs — Ar (39.9) before K (39.0), Co (58.9) before Ni (58.7) violated the increasing mass rule. (3) Position of isotopes — no place for isotopes with different atomic masses. (4) Position of metals and non-metals — some similar elements were placed in different groups. (5) No distinction between lanthanides and actinides.
Page 60–62 (Modern Periodic Table, Groups, Periods)
Q3.4: How does the modern periodic table differ from Mendeleev's table?
AnswerThe modern periodic table is based on atomic number (not atomic mass), which eliminates the anomalous pair problem. It has 18 groups (instead of 8) and 7 periods. It clearly separates s, p, d, and f blocks. The position of hydrogen is still debated. Isotopes occupy the same position since they have the same atomic number.
Q3.5: Write the electronic configuration of: (i) group 1 elements, (ii) group 17 elements, (iii) group 18 elements.
Answer
- (i) Group 1 (alkali metals): ns¹ (e.g., Li: 2s¹, Na: 3s¹, K: 4s¹)
- (ii) Group 17 (halogens): ns²np⁵ (e.g., F: 2s²2p⁵, Cl: 3s²3p⁵)
- (iii) Group 18 (noble gases): ns²np⁶ except He (1s²) (e.g., Ne: 2s²2p⁶, Ar: 3s²3p⁶)
Q3.6: How many elements are present in each period?
AnswerPeriod 1: 2 elements, Period 2: 8, Period 3: 8, Period 4: 18, Period 5: 18, Period 6: 32, Period 7: incomplete (progressively filled).
Page 68–70 (Periodic Trends, Atomic Radius, IE)
Q3.7: Why does atomic radius decrease across a period?
AnswerAcross a period, the nuclear charge increases (more protons) while electrons are added to the same principal shell. The increased effective nuclear charge pulls electrons closer to the nucleus, reducing the atomic radius.
Q3.8: Why does ionization enthalpy decrease down a group?
AnswerDown a group: (1) Atomic size increases — the outermost electron is farther from the nucleus. (2) Shielding effect increases — inner electron shells screen the nuclear charge. Both factors make it easier to remove the outermost electron, so ionization enthalpy decreases.
Q3.9: Why is the first ionization enthalpy of B less than that of Be?
AnswerBe: 1s² 2s² (fully filled 2s, stable). B: 1s² 2s² 2p¹ (the 2p¹ electron is easier to remove because: (1) 2p is farther from the nucleus than 2s, (2) 2p electron is shielded by 2s electrons, (3) a half-filled or fully filled subshell is more stable, so removing an electron from 2p¹ is easier than from 2s²).
Q3.10: Why does electronegativity increase across a period?
AnswerAcross a period, atomic size decreases and nuclear charge increases. The smaller the atom and the greater the nuclear charge, the stronger the pull on shared electrons. Hence, electronegativity increases from left to right.
Chapter-End Exercise
Q1. Explain why the first ionization enthalpy of nitrogen is higher than that of oxygen.
AnswerN (1s² 2s² 2p³) has a half-filled 2p subshell, which is extra stable. O (1s² 2s² 2p⁴) has one electron paired in 2p, creating electron-electron repulsion. The extra stability of half-filled 2p in N makes it harder to remove an electron compared to O, where the paired electron is easier to remove. Hence, IE₁(N) > IE₁(O).
Q2. Arrange the following in order of increasing ionization enthalpy: B, C, N, O.
AnswerExpected order based on increasing Z: B < C < N < O. But N has a half-filled 2p³ configuration (extra stability), so: O < N (N higher due to half-filled). Also O has lower IE than N. Actual order: B < C < O < N. Note: The general increase is B < C, and N > O, so the full order is B < C < O < N.
Q3. Which of the following has the largest atomic radius: Na, Mg, Al, Si?
AnswerAll are in period 3. Atomic radius decreases across a period (increasing nuclear charge). Na has the largest atomic radius.
Q4. How does atomic radius vary down a group and across a period?
AnswerDown a group: Atomic radius increases — new electron shells are added, and the outermost electrons are farther from the nucleus. Across a period: Atomic radius decreases — nuclear charge increases while electrons are added to the same shell, pulling them closer to the nucleus.
Q5. Consider the following electronic configurations: (a) 1s² 2s² 2p⁶ 3s¹ (b) 1s² 2s² 2p⁶ 3s² (c) 1s² 2s² 2p⁶ 3s² 3p¹ (d) 1s² 2s² 2p⁶ 3s² 3p⁴ Which has the highest first ionization enthalpy?
Answer(b) represents Mg — fully filled 3s² gives it stability. But (a) is Na (3s¹ — easy to remove), (c) is Al (3p¹ — easy to remove), (d) is S. Actually, (b) Mg has IE₁ higher than (a) Na and (c) Al. Among all four, (b) Mg has the highest IE₁ because removing an electron from a stable 3s² configuration requires more energy than from 3s¹ (Na) or 3p¹ (Al), and S has paired 3p electrons. However, Ne (noble gas) would be higher. Among these four: (b) Mg has the highest IE₁.
Q6. Explain why Cl has a higher electron gain enthalpy than F.
AnswerAlthough F is more electronegative, its electron gain enthalpy is less negative than Cl because: (1) F has a very small atomic size (2p subshell), causing strong inter-electronic repulsions when an extra electron is added. (2) Cl has a larger size (3p subshell) with less repulsion for the incoming electron. So Cl accommodates the added electron more easily: EGE: Cl (−349 kJ/mol) > F (−328 kJ/mol).
Q7. Predict the formula of the stable compound formed by: (i) Mg and O, (ii) Al and F, (iii) K and S.
Answer
- (i) Mg (group 2, +2) + O (group 16, −2) → MgO
- (ii) Al (group 13, +3) + F (group 17, −1) → AlF₃
- (iii) K (group 1, +1) + S (group 16, −2) → K₂S
Q8. Write the general electronic configuration of s-block and p-block elements.
Answer
- s-block: ns¹–² (outermost s orbital being filled)
- p-block: ns²np¹–⁶ (outermost p orbital being filled; noble gases: ns²np⁶ except He = 1s²)
Q9. Which element in each pair has a higher electronegativity: (i) N or P, (ii) F or Cl, (iii) Na or Cl?
Answer
- (i) N > P (up a group, electronegativity increases)
- (ii) F > Cl (F is the most electronegative element)
- (iii) Cl > Na (across a period, electronegativity increases)
Q10. What is the difference between electronegativity and electron gain enthalpy?
AnswerElectron gain enthalpy is the enthalpy change when an electron is added to an isolated gaseous atom — it is a measurable thermodynamic quantity. Electronegativity is the tendency of an atom to attract shared electrons in a chemical bond — it is a relative, dimensionless property measured on various scales (Pauling, Mulliken). EGE refers to gaseous atoms; electronegativity refers to bonded atoms.
Q11. Arrange F, Cl, Br, I in order of increasing electron gain enthalpy.
AnswerI < Br < F < Cl. Cl has the most negative EGE due to its optimal size. F has less negative EGE than Cl due to inter-electronic repulsion. Order: I < Br < F < Cl (from least negative to most negative).
Q12. Why does metallic character decrease across a period?
AnswerAcross a period, effective nuclear charge increases, making it harder for atoms to lose electrons. Since metallic character is the tendency to lose electrons, it decreases from left to right across a period. Elements transition from strongly metallic (Group 1) to non-metallic (Group 17).
Q13. The atomic radii of the second period elements are: Li (152), Be (111), B (88), C (77), N (70), O (66), F (64). Explain the trend.
AnswerThe atomic radius decreases from Li to F. This is because as we move from left to right across period 2, the nuclear charge increases (from Z = 3 to Z = 9) while electrons are added to the same second shell. The increasing nuclear charge pulls electrons closer, reducing the atomic radius progressively.
Q14. Which of the following is the correct order of increasing IE₁? (a) Ba < Sr < Ca < Mg, (b) Ba < Ca < Sr < Mg.
AnswerDown group 2, IE₁ decreases: Mg > Ca > Sr > Ba. So the correct increasing order (smallest to largest) is: Ba < Sr < Ca < Mg. Answer: (a).
Q15. What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Answer(1) Increase in atomic size — the distance between the nucleus and the outermost electron increases, weakening the nuclear attraction. (2) Increase in shielding effect — more inner electron shells screen the nuclear charge from the outermost electrons. (3) Reduced effective nuclear charge on the outermost electrons due to both factors above.
Q16. The first ionization enthalpy values (in kJ mol⁻¹) of group 13 elements are: B (801), Al (577), Ga (579), In (558), Tl (589). Explain why Ga has a higher IE than Al.
AnswerAlthough Ga is below Al, Ga has a higher IE because: Ga follows the first transition series (3d¹⁰). The 3d¹⁰ electrons provide poor shielding (d-electrons shield less effectively than s/p electrons), so the effective nuclear charge on Ga's 4p electron is greater than expected. This makes removing the outer electron from Ga slightly more difficult than from Al, resulting in IE(Ga) > IE(Al).
Q17. Explain the diagonal relationship between Li and Mg.
AnswerLi (group 1, period 2) and Mg (group 2, period 3) show similar properties because they have comparable charge/size ratios. Similarities: (1) Both form normal oxides (Li₂O, MgO). (2) Both form nitrides with N₂ (Li₃N, Mg₃N₂). (3) Both carbonates decompose on heating. (4) Both hydroxides are weak bases. (5) Both form more covalent compounds compared to their group members.
Q18. Define electronegativity. On what factors does it depend?
AnswerElectronegativity is the tendency of an atom to attract the shared pair of electrons towards itself in a chemical bond. It depends on: (1) Nuclear charge — higher nuclear charge means higher electronegativity. (2) Atomic size — smaller atoms have higher electronegativity. (3) Number of inner shells — more shells mean lower electronegativity. (4) Type of hybridisation — sp > sp² > sp³ (more s-character = more electronegative).
Q19. Arrange the following in order of increasing reducing power: Na, K, Mg, Ca.
AnswerReducing power is the tendency to lose electrons, which increases down a group and decreases across a period. Order of increasing reducing power: Mg < Ca < Na < K. (Mg is least reactive, K is most reactive.)
Q20. Which of the following will have the most negative electron gain enthalpy and which the least negative? P, S, Cl, F.
AnswerMost negative EGE: Cl (optimal size for adding electron, minimal repulsion). Least negative EGE: P (half-filled 3p³, adding an electron disrupts stability). Order: P > S > F > Cl (from least negative to most negative, i.e., Cl has the most negative EGE).
Internal Links
- [Chapter 3 Guide — Classification of Elements and Periodicity](/school/cbse/class/11/chemistry/cbse-classification-periodicity)
- [All Class 11 Chemistry Solutions](/school/cbse/class/11/chemistry/solutions)
- [Class 11 Chemistry Subject Guide](/school/cbse/class/11/chemistry)
