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NCERT Solutions · Class 11 Chemistry · NCERT Chemistry Part I and II Part I · Chapter 3

Chapter 3: Classification of Elements and Periodicity in Properties (Periodicity)

Step-by-step solutions to all 40 exercises of Unit 3, Classification of Elements and Periodicity in Properties (NCERT Class 11 Chemistry Part I, 2026-27): Mendeleev's and the modern periodic law, locating elements by configuration, atomic and ionic radii, isoelectronic species, ionisation enthalpy, electron gain enthalpy, electronegativity, metallic character and the MCQs. All 40 questions are answered, with the key answer highlighted.

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Across a period: atomic radius ↓, ionisation enthalpy ↑, electron gain enthalpy more negative, electronegativity ↑, metallic character ↓. Down a group: the reverse. Block = last subshell filled (s, p, d, f); period = principal quantum number of the valence shell; group of p-block = 10 + number of valence electrons. Isoelectronic species: size falls as nuclear charge rises.

Exercises

3.1
What is the basic theme of organisation in the periodic table?
Solution

To arrange the elements so that elements with similar properties fall in the same group. In the modern table they are placed in order of increasing atomic number, and similar valence configurations (hence similar properties) recur at regular intervals. This makes the study of over a hundred elements systematic: we study the trends of a group or a period instead of each element separately.

Similar elements are grouped together (by increasing atomic number), so their study becomes systematic.

3.2
Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?
Solution

He used atomic weight (mass). He did not stick to it strictly: where the order by weight would put an element among unlike elements, he gave priority to similar properties. For example, he placed iodine (126.9) after tellurium (127.6) so that I fell with the halogens. He also left gaps for elements not yet discovered.

Atomic weight; no, he gave priority to similarity of properties (e.g. Te before I).

3.3
What is the basic difference in approach between the Mendeleev's Periodic Law and the Modern Periodic Law?
Solution

Mendeleev: the properties of elements are a periodic function of their atomic weights. Modern law: the properties are a periodic function of their atomic numbers. (Moseley showed that Z, not the mass, is the fundamental property; this also removes anomalies such as Te–I.)

Atomic weight (Mendeleev) versus atomic number (modern law).

3.4
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Solution

In the 6th period (n = 6) the orbitals filled, in order of energy, are 6s, 4f, 5d and 6p: 1 + 7 + 5 + 3 = 16 orbitals. Each holds 2 electrons, so 32 electrons can be accommodated, i.e. 32 elements.

6s + 4f + 5d + 6p = 16 orbitals = 32 electrons = 32 elements.

3.5
In terms of period and group where would you locate the element with Z = 114?
Solution

Configuration: [Rn]5f¹⁴ 6d¹⁰ 7s² 7p². Valence shell n = 7; p-block with 4 valence electrons: group 10 + 4 = 14.

7th period, group 14 (below lead: flerovium, Fl)

3.6
Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Solution

Third period, group 17: 1s² 2s² 2p⁶ 3s² 3p⁵, which is chlorine.

Z = 17

3.7
Which element do you think would have been named by (i) Lawrence Berkeley Laboratory (ii) Seaborg's group?
Solution

(i) Lawrencium, Lr (Z = 103) (also berkelium, Bk, 97) (ii) Seaborgium, Sg (Z = 106)

3.8
Why do elements in the same group have similar physical and chemical properties?
Solution

Properties depend mainly on the outer (valence) electrons. Elements of a group have the same valence shell configuration and number of valence electrons (e.g. group 1: ns¹), so they behave alike.

Same valence-shell electronic configuration.

3.9
What does atomic radius and ionic radius really mean to you?
Solution

Atomic radius: the distance from the centre of the nucleus to the outermost electrons, found as half the internuclear distance: the covalent radius (half the bond length between two like atoms, e.g. Cl₂: 99 pm) or the metallic radius (half the distance between neighbouring metal atoms, e.g. Cu: 128 pm).

Ionic radius: the radius of a cation or anion, estimated from the distances between cations and anions in ionic crystals.

Size of an atom (from bond lengths) and size of an ion (from ionic crystals).

3.10
How do atomic radius vary in a period and in a group? How do you explain the variation?
Solution

Across a period it decreases: electrons enter the same shell while the nuclear charge rises by one each step, so the effective nuclear charge increases and the electron cloud is pulled in.

Down a group it increases: each step adds a new shell (n increases), and the extra inner electrons shield the outer ones; this outweighs the rise in nuclear charge.

Decreases across a period (greater effective nuclear charge); increases down a group (new shells).

3.11
What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions. (i) F⁻ (ii) Ar (iii) Mg²⁺ (iv) Rb⁺
Solution

Isoelectronic species are atoms or ions with the same number of electrons.

(i) F⁻ (10 e⁻): Ne, Na⁺, O²⁻, Mg²⁺ (ii) Ar (18 e⁻): Cl⁻, K⁺, S²⁻, Ca²⁺ (iii) Mg²⁺ (10 e⁻): Na⁺, Ne, F⁻, Al³⁺ (iv) Rb⁺ (36 e⁻): Kr, Br⁻, Sr²⁺

Same number of electrons; e.g. (i) Na⁺ (ii) K⁺ (iii) Ne (iv) Kr

3.12
Consider the following species: N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺ and Al³⁺ (a) What is common in them? (b) Arrange them in the order of increasing ionic radii.
Solution

(a) All are isoelectronic: 10 electrons each (the neon configuration 1s² 2s² 2p⁶).

(b) The nuclear charge rises from N (7) to Al (13), pulling the same 10 electrons closer, so size falls as Z rises.

(a) 10 electrons each (b) Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻

3.13
Explain why cation are smaller and anions larger in radii than their parent atoms?
Solution

Cation: it has fewer electrons but the same nuclear charge, so the effective nuclear charge per electron is larger and the cloud contracts (often the whole outer shell is lost: Na 186 pm → Na⁺ 95 pm).

Anion: the added electron(s) increase electron–electron repulsion and lower the effective nuclear charge per electron, so the cloud expands (F 64 pm → F⁻ 136 pm).

Cations: more nuclear pull per electron; anions: more repulsion, less pull per electron.

3.14
What is the significance of the terms — 'isolated gaseous atom' and 'ground state' while defining the ionization enthalpy and electron gain enthalpy? Hint: Requirements for comparison purposes.
Solution

Isolated gaseous atom: in a solid or liquid, neighbouring atoms affect the energy. In a dilute gas the atoms are far apart and do not interact, so the measured value belongs to the atom alone.

Ground state: an excited atom needs less energy to lose an electron. Taking the most stable (lowest-energy) state gives one well-defined value for every element, so the elements can be compared.

Both conditions make the values belong to the atom alone and comparable between elements.

3.15
Energy of an electron in the ground state of the hydrogen atom is –2.18 × 10⁻¹⁸ J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol⁻¹.
Solution

Ionisation takes the electron to E = 0, so J is needed per atom.

Per mole: J mol⁻¹.

1.312 × 10⁶ J mol⁻¹ (1312 kJ mol⁻¹)

3.16
Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why (i) Be has higher ΔᵢH than B (ii) O has lower ΔᵢH than N and F?
Solution

(i) Be (2s²) loses a 2s electron while B (2s² 2p¹) loses a 2p electron. The 2s electron penetrates closer to the nucleus and is held more tightly, whereas the 2p electron of B is shielded by the 2s electrons. Also 2s² is a completely filled subshell. So ΔᵢH(Be) > ΔᵢH(B).

(ii) N (2p³) has three singly occupied 2p orbitals: a stable half-filled subshell. In O (2p⁴) two electrons share one orbital and repel each other, so one of them is removed more easily: ΔᵢH(O) < ΔᵢH(N). F has a higher nuclear charge and smaller size than O, so its ΔᵢH is higher.

(i) 2s penetrates more than 2p (and 2s² is filled) (ii) half-filled 2p³ of N is stable; paired electron in O is easier to remove; F has higher Z.

3.17
How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Solution

First: Na (3s¹) and Mg (3s²) both lose a 3s electron, but Mg has a higher nuclear charge and smaller size (and a filled 3s), so ΔᵢH₁(Na) < ΔᵢH₁(Mg).

Second: Na⁺ has the stable neon configuration; its next electron comes from the inner 2p shell, close to the nucleus, which needs very high energy. Mg⁺ (3s¹) still has a valence electron to lose. So ΔᵢH₂(Na) > ΔᵢH₂(Mg).

Second electron of Na comes from a noble-gas core; of Mg from the valence shell.

3.18
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Solution

(i) Atomic size increases (a new shell each step), so the outer electron is farther from the nucleus. (ii) Shielding by inner electrons increases. These outweigh (iii) the increase in nuclear charge, so the effective attraction on the outer electron falls and ΔᵢH decreases.

Larger size and greater shielding outweigh the higher nuclear charge.

3.19
The first ionization enthalpy values (in kJ mol⁻¹) of group 13 elements are: B 801, Al 577, Ga 579, In 558, Tl 589. How would you explain this deviation from the general trend?
Solution

B → Al: falls as expected (larger size).

Ga ≥ Al: Ga has ten 3d electrons, which shield the nucleus poorly, so the effective nuclear charge on its 4p electron is higher than expected and ΔᵢH does not fall.

In: drops slightly (larger size).

Tl > In: Tl has fourteen 4f and ten 5d electrons, which shield even more poorly, so the effective nuclear charge rises.

Poor shielding by d (Ga) and f + d (Tl) electrons raises the effective nuclear charge.

3.20
Which of the following pairs of elements would have a more negative electron gain enthalpy? (i) O or F (ii) F or Cl
Solution

(i) F: it is one electron short of a noble-gas configuration and has a higher nuclear charge and smaller size than O, so adding an electron releases more energy.

(ii) Cl: in F the incoming electron enters the small, compact 2p subshell and is strongly repelled by the electrons already there; in the larger 3p subshell of Cl the repulsion is less.

(i) F (ii) Cl

3.21
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
Solution

Positive. The first step (O + e⁻ → O⁻) releases energy. The second electron must be added to the negative ion O⁻ against strong electrostatic repulsion, so energy has to be supplied (about +780 kJ mol⁻¹).

Positive: the electron is added to a negative ion.

3.22
What is the basic difference between the terms electron gain enthalpy and electronegativity?
Solution

Electron gain enthalpy: the enthalpy change when an isolated gaseous atom gains an electron; an experimental, measurable energy (kJ mol⁻¹).

Electronegativity: the tendency of an atom in a molecule to attract the shared electron pair of a bond; a relative number (Pauling scale), not directly measurable, and it varies with the bonding situation.

Electron gain enthalpy: energy for an isolated atom; electronegativity: relative pull on a shared pair in a molecule.

3.23
How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
Solution

The statement is not correct. Electronegativity is not a fixed property of an atom: it depends on the atoms it is bonded to, its oxidation state and its hybridisation (an sp nitrogen is more electronegative than an sp³ one; N in NO₂ differs from N in NH₃). 3.0 is only an average value.

Not correct; it varies with bonding, oxidation state and hybridisation.

3.24
Describe the theory associated with the radius of an atom as it (a) gains an electron (b) loses an electron
Solution

(a) Gains: the anion has more electrons than protons; electron–electron repulsion increases and the effective nuclear charge per electron falls, so the radius increases.

(b) Loses: the cation has fewer electrons but the same nuclear charge, so each electron is pulled harder (and a whole shell may be lost): the radius decreases.

(a) Radius increases (b) radius decreases

3.25
Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Solution

The same. Isotopes have the same nuclear charge and the same electronic configuration, and ionisation enthalpy depends on these, not on the number of neutrons (the tiny effect of different nuclear mass is negligible).

The same

3.26
What are the major differences between metals and non-metals?
Solution

Metals: low ionisation enthalpy; lose electrons to form cations (electropositive, reducing); lustrous, malleable, ductile, good conductors of heat and electricity; mostly solids with high melting points; form basic oxides.

Non-metals: high ionisation enthalpy and very negative electron gain enthalpy; gain electrons to form anions (electronegative, oxidising); dull, brittle, poor conductors (except graphite); solids or gases with low melting points (bromine is a liquid); form acidic oxides.

Metals lose electrons, are lustrous, malleable conductors with basic oxides; non-metals gain electrons, are brittle non-conductors with acidic oxides.

3.27
Use the periodic table to answer the following questions. (a) Identify an element with five electrons in the outer subshell. (b) Identify an element that would tend to lose two electrons. (c) Identify an element that would tend to gain two electrons. (d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.
Solution

(a) A p⁵ element (ns² np⁵): F or Cl. (b) Group 2: Mg or Ca. (c) Group 16: O or S. (d) Group 17: F₂, Cl₂ are gases, Br₂ a liquid, I₂ a solid non-metal, and astatine is metallic in character.

(a) Cl (b) Mg (c) O (d) group 17

3.28
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > Cl > Br > I. Explain.
Solution

Alkali metals react by losing their ns¹ electron. Down the group the size increases and ΔᵢH falls, so the electron is lost more easily: reactivity increases from Li to Cs.

Halogens react by gaining an electron. Down the group the size increases and the attraction for an extra electron (electronegativity, oxidising power) falls, so reactivity decreases from F to I.

Alkali metals: ease of losing an electron increases down; halogens: ease of gaining one decreases down.

3.29
Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Solution

s: ns¹⁻²; p: ns² np¹⁻⁶ (He 1s² is an exception); d: (n−1)d¹⁻¹⁰ ns⁰⁻²; f: (n−2)f¹⁻¹⁴ (n−1)d⁰⁻¹ ns²

3.30
Assign the position of the element having outer electronic configuration (i) ns² np⁴ for n = 3 (ii) (n–1)d² ns² for n = 4, and (iii) (n–2)f⁷ (n–1)d¹ ns² for n = 6, in the periodic table.
Solution

(i) 3s² 3p⁴: period 3, group 10 + 6 = 16, p-block (sulphur).

(ii) 3d² 4s²: period 4, group 2 + 2 = 4, d-block (titanium).

(iii) 4f⁷ 5d¹ 6s²: period 6, group 3, f-block lanthanoid (gadolinium, Z = 64).

(i) Period 3, group 16 (S) (ii) period 4, group 4 (Ti) (iii) period 6, group 3, f-block (Gd)

3.31
The first (ΔᵢH₁) and the second (ΔᵢH₂) ionization enthalpies (in kJ mol⁻¹) and the electron gain enthalpy (ΔegH, in kJ mol⁻¹) of a few elements are: I 520, 7300, –60; II 419, 3051, –48; III 1681, 3374, –328; IV 1008, 1846, –295; V 2372, 5251, +48; VI 738, 1451, –40. Which of the above elements is likely to be: (a) the least reactive element. (b) the most reactive metal. (c) the most reactive non-metal. (d) the least reactive non-metal. (e) the metal which can form a stable binary halide of the formula MX₂ (X = halogen). (f) the metal which can form a predominantly stable covalent halide of the formula MX (X = halogen)?
Solution

(a) V: highest ΔᵢH and positive ΔegH, a noble gas (He).

(b) II: lowest ΔᵢH₁ (like K).

(c) III: high ΔᵢH and the most negative ΔegH (like F).

(d) IV: high ΔᵢH and negative ΔegH, but less so than III (like I).

(e) VI: low ΔᵢH₁ and a ΔᵢH₂ that is not too high: an alkaline earth metal (like Mg), forming MgX₂.

(f) I: ΔᵢH₂ very high (alkali metal) but ΔᵢH₁ the highest among the metals (like Li); the small Li⁺ polarises the anion, so LiX is largely covalent.

(a) V (b) II (c) III (d) IV (e) VI (f) I

3.32
Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements. (a) Lithium and oxygen (b) Magnesium and nitrogen (c) Aluminium and iodine (d) Silicon and oxygen (e) Phosphorus and fluorine (f) Element 71 and fluorine
Solution

Valencies: Li 1, O 2, Mg 2, N 3, Al 3, I 1, Si 4, P 3 or 5, F 1, Lu (Z = 71, a lanthanoid) 3.

(a) Li₂O (b) Mg₃N₂ (c) AlI₃ (d) SiO₂ (e) PF₃ or PF₅ (f) LuF₃

3.33
In the modern periodic table, the period indicates the value of: (a) atomic number (b) atomic mass (c) principal quantum number (d) azimuthal quantum number.
Solution

The period number equals the principal quantum number n of the valence shell.

(c)

3.34
Which of the following statements related to the modern periodic table is incorrect? (a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell. (b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell. (c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell. (d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration.
Solution

A d-subshell has 5 orbitals and holds 10 electrons, so the d-block has 10 columns (groups 3–12).

(b)

3.35
Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell? (a) Valence principal quantum number (n) (b) Nuclear charge (Z) (c) Nuclear mass (d) Number of core electrons.
Solution

Neutrons add mass but no charge, so nuclear mass does not change the attraction on the valence electrons.

(c)

3.36
The size of isoelectronic species — F⁻, Ne and Na⁺ is affected by (a) nuclear charge (Z) (b) valence principal quantum number (n) (c) electron-electron interaction in the outer orbitals (d) none of the factors because their size is the same.
Solution

All have 10 electrons in the same shells; the size falls as Z rises (F 9, Ne 10, Na 11): F⁻ > Ne > Na⁺.

(a)

3.37
Which one of the following statements is incorrect in relation to ionization enthalpy? (a) Ionization enthalpy increases for each successive electron. (b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration. (c) End of valence electrons is marked by a big jump in ionization enthalpy. (d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Solution

An electron with lower n is closer to the nucleus and held more tightly, so it is harder to remove.

(d)

3.38
Considering the elements B, Al, Mg, and K, the correct order of their metallic character is: (a) B > Al > Mg > K (b) Al > Mg > B > K (c) Mg > Al > K > B (d) K > Mg > Al > B
Solution

Metallic character increases down a group (B → Al) and decreases across a period (K > Mg > Al).

(d)

3.39
Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is: (a) B > C > Si > N > F (b) Si > C > B > N > F (c) F > N > C > B > Si (d) F > N > C > Si > B
Solution

Non-metallic character increases across period 2 (B < C < N < F); Si (period 3) is the least non-metallic.

(c)

3.40
Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is: (a) F > Cl > O > N (b) F > O > Cl > N (c) Cl > F > O > N (d) O > F > N > Cl
Solution

Oxidising power follows electronegativity: F (4.0) > O (3.5) > Cl (3.0) > N (3.0); chlorine is a stronger oxidant than nitrogen.

(b)

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