NCERT Solutions for Class 11 Chemistry Chapter 3: Classification of Elements and Periodicity in Properties
These Class 11 Chemistry Chapter 3 solutions cover Classification of Elements and Periodicity in Properties with step-by-step, exam-ready answers to all 40 NCERT exercises (3.1–3.40) reproduced verbatim from the textbook, updated for session 2026–27. The chapter explains the Modern Periodic Law, the long form of the periodic table, s-, p-, d- and f-blocks, and periodic trends in atomic radius, ionization enthalpy, electron gain enthalpy, electronegativity and valence.
Class: 11Subject: ChemistryUnit: 3Chapter: Classification of Elements & PeriodicityExercises: 3.1–3.40Session: 2026–27
Unit 3 traces the development of the periodic table from Döbereiner’s triads and Newlands’ law of octaves to Mendeleev’s periodic table (based on atomic masses) and finally the Modern Periodic Law: the physical and chemical properties of elements are periodic functions of their atomic numbers. The long form of the table has 7 periods and 18 groups. The period number equals the highest principal quantum number (n) of its elements, and elements are classified into s-, p-, d- and f-blocks according to the subshell that receives the last electron. The chapter then explains how atomic and ionic radii, ionization enthalpy, electron gain enthalpy, electronegativity and valence vary periodically, and links these trends to metallic/non-metallic character and chemical reactivity, along with the anomalous behaviour of second-period elements and diagonal relationships.
Key Concepts & Definitions
Modern Periodic Law: the physical and chemical properties of the elements are periodic functions of their atomic numbers (Z).
Period: a horizontal row; its number = highest principal quantum number n of the valence shell. Periods hold 2, 8, 8, 18, 18, 32 and (incomplete) elements.
Group: a vertical column (1–18); members have similar valence-shell electronic configurations and hence similar properties.
Atomic radius: covalent radius (non-metals) or metallic radius (metals) — half the internuclear distance.
Isoelectronic species: atoms/ions having the same number of electrons.
Ionization enthalpy (ΔiH): energy needed to remove the most loosely bound electron from one mole of isolated gaseous atoms in their ground state, X(g) → X+(g) + e−.
Electron gain enthalpy (ΔegH): enthalpy change when an electron is added to a neutral gaseous atom, X(g) + e− → X−(g).
Electronegativity: the ability of an atom in a compound to attract the shared pair of electrons toward itself (Pauling scale most common).
Periodic Trends at a Glance
Across a period (left → right): atomic/ionic radius decreases; ionization enthalpy increases; electron gain enthalpy becomes more negative; electronegativity increases; metallic character decreases, non-metallic character increases.
Down a group (top → bottom): atomic/ionic radius increases; ionization enthalpy decreases; electron gain enthalpy becomes less negative; electronegativity decreases; metallic character increases.
NCERT Exercises (3.1–3.40) — Solutions
Questions are reproduced verbatim from the NCERT textbook (Unit 3). All answers are original, step-by-step and exam-ready.
3.1 What is the basic theme of organisation in the periodic table?
ANSWERThe basic theme is to classify the large number of known elements into groups and periods so that their chemistry can be studied easily and systematically. Elements with similar properties are placed together in vertical columns (groups), because they have similar valence-shell electronic configurations. This organisation lets us correlate known facts, recognise trends and even predict the properties of elements.
3.2 Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?
ANSWERMendeleev used the atomic weight (atomic mass) of the elements to arrange them. He did not stick to it strictly: wherever following the order of atomic weight placed an element among dissimilar ones, he ignored that order and placed it with chemically similar elements (e.g. iodine, with lower atomic weight than tellurium, was placed with the halogens). He also left gaps for undiscovered elements (Eka-aluminium, Eka-silicon).
3.3 What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?
ANSWERMendeleev’s Periodic Law stated that the properties of elements are a periodic function of their atomic weights. The Modern Periodic Law (after Moseley) states that the properties of elements are a periodic function of their atomic numbers. Atomic number (nuclear charge / number of electrons) is a more fundamental property than atomic mass, and using it removed anomalies such as the Ar–K and Te–I orderings.
3.4 On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
ANSWERFor the sixth period n = 6. As we move across this period, electrons fill the 6s, 4f, 5d and 6p subshells in order of increasing energy.Orbitals available: 6s (1) + 4f (7) + 5d (5) + 6p (3) = 16 orbitals.Each orbital holds 2 electrons, so maximum electrons = 16 × 2 = 32. Therefore the sixth period accommodates 32 elements.
3.5 In terms of period and group where would you locate the element with Z = 114?
ANSWERElectronic configuration of Z = 114 is [Rn] 5f146d107s27p2. The highest n = 7, so it lies in Period 7. The valence configuration ends in 7s27p2 (4 valence electrons of a p-block element), so it belongs to Group 14 (the carbon family).
3.6 Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
ANSWERThird period, Group 17 means the configuration is 1s22s22p63s23p5. Counting the electrons: 2 + 2 + 6 + 2 + 5 = 17. The element is chlorine (Z = 17).
3.7 Which element do you think would have been named by (i) Lawrence Berkeley Laboratory (ii) Seaborg’s group?
ANSWER(i) Lawrencium (Lr, Z = 103) — named after Ernest O. Lawrence, founder of the Lawrence Berkeley Laboratory.(ii) Seaborgium (Sg, Z = 106) — named in honour of Glenn T. Seaborg, head of the group that discovered it.
3.8 Why do elements in the same group have similar physical and chemical properties?
ANSWERElements in the same group have the same number of valence electrons and the same valence-shell electronic configuration. Since chemical and most physical properties are governed by the valence electrons, members of a group show similar bonding behaviour, valence and reactivity (e.g. all alkali metals have ns1 and form +1 ions).
3.9 What does atomic radius and ionic radius really mean to you?
ANSWERAtomic radius is a measure of the size of an atom. As the electron cloud has no sharp boundary, it is estimated indirectly: as the covalent radius (half the internuclear distance between two like atoms joined by a single bond) for non-metals, or the metallic radius (half the internuclear distance in the metallic crystal) for metals.Ionic radius is the effective size of an ion in a crystal, found from the internuclear distance between cations and anions in ionic crystals. A cation is smaller than its parent atom; an anion is larger.
3.10 How do atomic radius vary in a period and in a group? How do you explain the variation?
ANSWERIn a period (L → R): atomic radius decreases. Electrons are added to the same valence shell while nuclear charge increases, so the effective nuclear charge rises and pulls the electron cloud closer to the nucleus.In a group (top → bottom): atomic radius increases. The principal quantum number n increases (a new shell is added) and the inner electrons shield the valence electrons from the nucleus, so the outer electrons lie farther out.
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+
ANSWERIsoelectronic species are atoms and ions that contain the same number of electrons.(i) F– has 10 electrons → isoelectronic with Na+ (also Ne, O2−, Mg2+).(ii) Ar has 18 electrons → isoelectronic with K+ (also Cl−, Ca2+).(iii) Mg2+ has 10 electrons → isoelectronic with Ne (also Na+, F−, O2−).(iv) Rb+ has 36 electrons → isoelectronic with Kr (also Sr2+, Br−).
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.
ANSWER(a) All six are isoelectronic — each has 10 electrons (the same as neon).(b) Among isoelectronic species, radius decreases as nuclear charge (Z) increases. Nuclear charges: N (7) < O (8) < F (9) < Na (11) < Mg (12) < Al (13). Higher Z → smaller radius, so increasing ionic radii:Al3+ < Mg2+ < Na+ < F– < O2– < N3–.
3.13 Explain why cation are smaller and anions larger in radii than their parent atoms?
ANSWERA cation forms by loss of electron(s). It has fewer electrons but the same nuclear charge, so the effective nuclear charge per electron rises, electron–electron repulsion falls, and the remaining electrons are pulled in — often an entire outer shell is lost. Hence a cation is smaller than its parent atom.An anion forms by gain of electron(s). The added electrons increase electron–electron repulsion while nuclear charge stays the same, so the effective nuclear charge per electron falls and the electron cloud expands. Hence an anion is larger than its parent atom.
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.
ANSWER‘Isolated gaseous atom’ ensures the atom is free from the influence of neighbouring atoms (no interatomic forces, as exist in liquids/solids). Only then does the measured energy refer to a single atom and not to lattice or intermolecular interactions.‘Ground state’ specifies the lowest, most stable energy arrangement of the electrons. If the atom were excited, less energy would be needed and the value would not be comparable. Both terms standardise the conditions so that values for different elements can be compared meaningfully.
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. Hint: Apply the idea of mole concept to derive the answer.
ANSWERIonization removes the electron from n = 1 to n = ∞ (E = 0). Energy needed per atom:ΔE = E∞ − E1 = 0 − (−2.18×10–18 J) = +2.18×10–18 J per atom.For one mole, multiply by Avogadro’s number (6.022×1023):ΔiH = 2.18×10–18 × 6.022×1023 = 1.313×106 J mol–1= 1.313×106 J mol–1 ≈ 1313 kJ mol–1 (matches the standard value for hydrogen).
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?
ANSWER(i) Be (1s22s2) > B (1s22s22p1): In Be the electron is removed from a fully filled, stable 2s orbital, which is highly penetrating and tightly held. In B the electron is removed from a higher-energy 2p orbital that is more shielded by the 2s electrons, so less energy is needed. Hence ΔiH(Be) > ΔiH(B).(ii) O (1s22s22p4) < N (2p3) and F (2p5): Nitrogen has a half-filled, extra-stable 2p3 configuration, so removing an electron is hard (high ΔiH). In oxygen, two of the four 2p electrons share one orbital, causing extra electron–electron repulsion; removing one relieves this repulsion, so it is easier (lower ΔiH). Across the period F has a still higher nuclear charge than O, so ΔiH(F) > ΔiH(O).
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?
ANSWERFirst IE: Na (3s1) loses one electron to attain the stable [Ne] noble-gas configuration, which is easy. Mg (3s2) has a higher nuclear charge and a filled 3s orbital, so its first electron is held more tightly. Hence ΔiH1(Na) < ΔiH1(Mg).Second IE: After losing one electron, Na+ already has the stable [Ne] configuration; removing a second electron means breaking this very stable noble-gas core (a 2p electron), needing huge energy. Mg+ (3s1) still has one easily removable 3s electron giving the stable [Ne] core. Hence ΔiH2(Na) > ΔiH2(Mg).
3.18 What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
ANSWERTwo main factors lower the ionization enthalpy down a group:(i) Increase in atomic size — the valence electron lies in a higher shell, farther from the nucleus, so it is held less tightly.(ii) Increase in shielding (screening) effect — the growing number of inner-shell electrons screens the valence electron from the nuclear charge.Although nuclear charge also increases down a group, the combined effect of larger size and greater shielding outweighs it, so ΔiH decreases.
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?
ANSWERFrom B to Al, ΔiH falls (size increase) as expected. But from Al onwards the values do not fall regularly — Ga (579) > Al (577), and Tl (589) > In (558).Ga > Al: Gallium follows the first d-block series; the poor shielding by the intervening 3d10 electrons causes a larger effective nuclear charge on the valence electron, so its ΔiH does not decrease (slightly increases) relative to Al.Tl > In: Thallium follows both 4f14 and 5d10 electrons, which shield very poorly (lanthanoid contraction + poor d/f screening). The valence electron experiences a higher effective nuclear charge, raising its ΔiH above that of In.
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
ANSWER(i) F has the more negative electron gain enthalpy than O. Fluorine needs only one electron to reach the stable noble-gas configuration and has higher effective nuclear charge across the period.(ii) Cl has the more negative electron gain enthalpy than F. Although F is smaller, its compact 2p subshell causes strong electron–electron repulsion on the incoming electron. Chlorine’s larger 3p subshell accommodates the extra electron with less repulsion, so ΔegH of Cl (−349) is more negative than that of F (−328).
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.
ANSWERThe second electron gain enthalpy of oxygen is positive. The first electron is added to a neutral O atom (energy released, negative value): O(g) + e− → O−(g).The second electron must be forced onto the already negatively charged O− ion against strong electrostatic repulsion: O−(g) + e− → O2−(g). Energy has to be supplied to overcome this repulsion, so ΔegH2 is positive (endothermic).
3.22 What is the basic difference between the terms electron gain enthalpy and electronegativity?
ANSWERElectron gain enthalpy is the enthalpy change when an isolated gaseous atom gains an electron to form an anion. It is a measurable quantity with definite units (kJ mol–1) and refers to a free atom.Electronegativity is the tendency of an atom in a chemically bonded molecule to attract the shared electron pair toward itself. It is a relative, qualitative property (no units), and its value can change depending on the atom to which the element is bonded.
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?
ANSWERThe statement is incorrect. Electronegativity is not a fixed, constant property of an element; it varies with the atom’s oxidation state and the element to which it is bonded, and also with hybridisation. The Pauling value 3.0 is only an approximate/average figure. So nitrogen does not have the same electronegativity of exactly 3.0 in all its compounds.
3.24 Describe the theory associated with the radius of an atom as it (a) gains an electron (b) loses an electron.
ANSWER(a) Gaining an electron forms an anion. The extra electron increases electron–electron repulsion in the valence shell while nuclear charge is unchanged, so the effective nuclear charge per electron decreases and the electron cloud expands. The radius increases (anion > parent atom).(b) Losing an electron forms a cation. With fewer electrons but the same nuclear charge, the effective nuclear charge per electron increases, repulsion decreases, and the remaining electrons are pulled in (often a whole shell is removed). The radius decreases (cation < parent atom).
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.
ANSWERThe first ionization enthalpies would be the same. Isotopes of an element have the same atomic number, same nuclear charge and identical electronic configuration; they differ only in the number of neutrons (i.e. in mass). Since ionization enthalpy depends on the attraction between the nucleus and the valence electrons (an electronic property), and not on mass, isotopes have equal first ionization enthalpies.
3.26 What are the major differences between metals and non-metals?
ANSWER
Metals
Non-metals
Found on the left side of the periodic table (> 78% of elements)
Found on the top right side (less than 20 elements)
Usually solids with high melting/boiling points
Usually solids or gases with low melting/boiling points
Good conductors of heat and electricity
Poor conductors (insulators)
Malleable and ductile
Brittle; neither malleable nor ductile
Low ionization enthalpy; lose electrons to form cations; form basic oxides
High electron gain enthalpy/electronegativity; gain electrons to form anions; form 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.
ANSWER(a) Five electrons in the outer (p) subshell → ns2np5, i.e. a Group 17 halogen such as fluorine (or chlorine).(b) An element that loses two electrons easily → ns2, a Group 2 alkaline earth metal such as magnesium (or calcium).(c) An element that gains two electrons → ns2np4, a Group 16 element such as oxygen (or sulphur).(d) The group containing a metal, a non-metal, a liquid and a gas at room temperature is Group 17 (halogens): astatine (metal/metalloid, solid), iodine (non-metal solid), bromine (liquid), chlorine and fluorine (gases).
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.
ANSWERGroup 1: Reactivity depends on the ease of losing the single valence electron, i.e. on ionization enthalpy. Down the group atomic size increases and ionization enthalpy decreases, so the electron is lost more easily. Hence reactivity increases: Li < Na < K < Rb < Cs.Group 17: Reactivity depends on the tendency to gain an electron, which is greatest for the smallest atom with highest effective nuclear charge. Down the group atomic size increases and the attraction for an incoming electron decreases, so reactivity decreases: F > Cl > Br > I.
3.29 Write the general outer electronic configuration of s-, p-, d- and f- block elements.
3.30 Assign the position of the element having outer electronic configuration (i) ns2np4 for n = 3 (ii) (n−1)d2ns2 for n = 4, and (iii) (n−2)f7(n−1)d1ns2 for n = 6, in the periodic table.
ANSWER(i) n = 3, 3s23p4: total e− = 2+2+6+2+4 = 16 → sulphur (Z = 16); Period 3, Group 16, p-block.(ii) n = 4, 3d24s2: configuration [Ar]3d24s2 → titanium (Z = 22); Period 4, Group 4, d-block.(iii) n = 6, 4f75d16s2: configuration [Xe]4f75d16s2 → gadolinium (Z = 64); Period 6, Group 3 (lanthanoid series), f-block.
3.31 The first (ΔiH1) and the second (ΔiH2) ionization enthalpies (in kJ mol–1) and the (ΔegH) electron gain enthalpy (in kJ mol–1) of a few elements are given below:
DATA & ANSWER
Element
ΔiH1
ΔiH2
ΔegH
I
520
7300
−60
II
419
3051
−48
III
1681
3374
−328
IV
1008
1846
−295
V
2372
5251
+48
VI
738
1451
−40
(a) Least reactive element → V. Highest ΔiH1 (2372) and positive ΔegH (+48); it neither loses nor gains electrons — a noble gas.(b) Most reactive metal → II. Lowest ΔiH1 (419), so it loses an electron most easily — an alkali metal.(c) Most reactive non-metal → III. Very high ΔiH1 and the most negative ΔegH (−328) — a halogen.(d) Least reactive non-metal → IV. Non-metallic (negative ΔegH = −295) but less so than III; a halogen below III.(e) Metal forming a stable binary halide MX2 → VI. Low ΔiH1 (738) and a relatively low ΔiH2 (1451) — it readily loses two electrons (alkaline earth metal, e.g. Mg/Ca).(f) Metal forming a predominantly covalent halide MX → I. A small alkali metal (low ΔiH1 = 520 but a huge jump to ΔiH2 = 7300, so only one electron is lost) with high charge/size ratio — lithium, whose halide LiX is largely covalent.
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
ANSWER(a) Li (valence 1) & O (valence 2) → Li2O.(b) Mg (valence 2) & N (valence 3) → Mg3N2.(c) Al (valence 3) & I (valence 1) → AlI3.(d) Si (valence 4) & O (valence 2) → SiO2.(e) P (valence 3 or 5) & F (valence 1) → PF3 or PF5.(f) Element 71 is lutetium (Lu, valence 3) & F (valence 1) → 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.
ANSWER(c) Principal quantum number (n). The period number equals the highest principal quantum number of the elements in that period.
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.
ANSWER(b) is incorrect. The d-subshell has 5 orbitals and can hold a maximum of 10 electrons, so the d-block has 10 columns, not 8. (The statement that it holds only 8 electrons / 8 columns is wrong.)
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.
ANSWER(c) Nuclear mass. Chemistry is governed by electronic factors (n, nuclear charge, and shielding by core electrons). The nuclear mass (number of neutrons) has no effect on the valence shell or chemical behaviour.
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.
ANSWER(a) Nuclear charge (Z). All three have 10 electrons and the same n, so the differing nuclear charges (F = 9, Ne = 10, Na = 11) decide the size: higher Z pulls electrons in more → smaller size, giving the order F– > Ne > Na+.
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.
ANSWER(d) is incorrect. Electrons in orbitals with a lower n are closer to the nucleus and held more tightly, so they are harder (not easier) to remove. Removal from a higher-n orbital is easier. Statements (a), (b) and (c) are correct.
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
ANSWER(d) K > Mg > Al > B. Metallic character increases down a group and decreases across a period. K (Period 4, Group 1) is the most metallic; among Period-3/2 elements Mg (Gr 2) > Al (Gr 13) > B (Period 2, Gr 13) is the least metallic.
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
ANSWER(c) F > N > C > B > Si. Non-metallic character increases across a period and decreases down a group. In Period 2: F > N > C > B; Si (just below C in Period 3) is less non-metallic than B, giving F > N > C > B > Si.
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
ANSWER(b) F > O > Cl > N. Oxidising power follows electronegativity/non-metallic character. Order: F (4.0) > O (3.5) > Cl (3.0) > N (3.0 but a Period-2 element less oxidising than Cl in practice), giving F > O > Cl > N.
Extra Practice Questions
Short Answer Type Questions
Q1. State the Modern Periodic Law.
ANSWERThe physical and chemical properties of the elements are periodic functions of their atomic numbers.
Q2. Why is the atomic radius of a noble gas not compared with the covalent radii of other elements?
ANSWERNoble gases are monoatomic and do not form normal covalent bonds, so only their non-bonded (van der Waals) radii can be measured. These are much larger than covalent radii, so noble-gas radii are compared only with van der Waals radii, not covalent radii.
Q3. Define diagonal relationship with one example.
ANSWERThe similarity in properties between an element and the element diagonally placed to its right in the next period is called a diagonal relationship, e.g. Li resembles Mg, and Be resembles Al. It arises from similar charge/size ratios and electronegativities.
Q4. Why does the second period element of a group differ from the rest of its group?
ANSWERBecause of its small size, large charge/radius ratio, high electronegativity and absence of d-orbitals (only four valence orbitals, so maximum covalency 4). This makes it behave anomalously, e.g. forming more covalent compounds and pπ–pπ multiple bonds.
Q5. Classify the oxides Na2O, Cl2O7, Al2O3 and CO as acidic, basic, amphoteric or neutral.
Q1. Explain the variation of ionization enthalpy across a period and give two reasons for irregularities observed.
ANSWERAcross a period, the nuclear charge increases while electrons are added to the same shell, so the effective nuclear charge rises and the atomic radius falls; the outer electrons are held more tightly and ionization enthalpy generally increases. However, irregularities arise from sub-shell stability: (i) fully filled s-subshells (e.g. Be 2s2) and half-filled p-subshells (e.g. N 2p3) are extra stable, so their ΔiH is higher than the next element (B, O respectively); (ii) removing an electron from a singly-occupied higher-energy p-orbital (B) or from a paired 2p orbital (O, with relief of repulsion) needs less energy. Hence the order Li < B < Be < C < O < N < F < Ne.
Q2. Describe how metallic and non-metallic character vary in the periodic table and relate them to oxide nature.
ANSWERMetallic character (tendency to lose electrons, low ionization enthalpy) decreases across a period and increases down a group; non-metallic character (tendency to gain electrons, high electronegativity) shows the opposite trend. Metals on the left form basic oxides (e.g. Na2O), non-metals on the right form acidic oxides (e.g. Cl2O7), and elements in the centre form amphoteric (Al2O3, As2O3) or neutral (CO, NO, N2O) oxides. Thus the acid–base nature of oxides mirrors the metallic–non-metallic trend.
Q3. Explain why, among isoelectronic species, the cation with the greater positive charge has the smallest radius, using O2–, F–, Na+ and Mg2+.
ANSWERAll four have 10 electrons but different nuclear charges: O (8), F (9), Na (11), Mg (12). For the same number of electrons, a greater nuclear charge attracts the electron cloud more strongly, giving a smaller radius. Therefore Mg2+ (highest Z) is the smallest and O2– (lowest Z) is the largest, the order being O2– > F– > Na+ > Mg2+. The cation with the greater positive charge thus has the smallest radius.
MCQs & Assertion–Reason
1. The Modern Periodic Law is based on:
(a) atomic mass (b) atomic number (c) atomic volume (d) density
2. The number of elements in the sixth period is:
(a) 18 (b) 8 (c) 32 (d) 50
3. The element with Z = 114 belongs to:
(a) Period 6, Group 14 (b) Period 7, Group 14 (c) Period 7, Group 18 (d) Period 6, Group 18
4. Which of the following has the largest size?
(a) Na+ (b) Mg2+ (c) Al3+ (d) F–
5. Across a period from left to right, atomic radius:
(a) increases (b) decreases (c) stays constant (d) first increases then decreases
6. The element with the highest electronegativity is:
(a) oxygen (b) chlorine (c) fluorine (d) nitrogen
7. The most negative electron gain enthalpy is shown by:
(a) F (b) Cl (c) O (d) S
8. The general outer electronic configuration of d-block elements is:
For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.
A-R 1. Assertion: A cation is smaller than its parent atom.
Reason: On losing an electron, the effective nuclear charge per electron increases and the electron cloud contracts.
A-R 2. Assertion: Ionization enthalpy decreases down a group.
Reason: Atomic size and shielding effect increase down a group, outweighing the increase in nuclear charge.
A-R 3. Assertion: The electron gain enthalpy of chlorine is more negative than that of fluorine.
Reason: The small 2p subshell of fluorine causes strong electron–electron repulsion on the incoming electron.
A-R 4. Assertion: Isotopes of an element have different first ionization enthalpies.
Reason: Isotopes have the same nuclear charge and identical electronic configuration.
A-R 5. Assertion: The d-block of the periodic table has 10 columns.
Reason: A d-subshell has five orbitals and can hold a maximum of ten electrons.
Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).
Common Mistakes to Avoid
Watch out for these
Saying the Modern Periodic Law is based on atomic mass — it is based on atomic number.
Forgetting that, among isoelectronic species, size depends only on nuclear charge (higher Z → smaller).
Writing ΔiH(F) > ΔegH(Cl) — actually Cl has the more negative electron gain enthalpy than F.
Saying the d-block has 8 columns — it has 10 columns (10 electrons).
Ignoring half-filled/fully-filled stability when explaining IE anomalies (Be > B, N > O).
Treating electronegativity as a fixed, measurable quantity — it is relative and varies with the bonded atom.
How to score full marks in this chapter
Always justify trend questions with the two key factors: effective nuclear charge and shielding/atomic size. For numericals like 3.15, show units at every step and use Avogadro’s number for per-mole answers. For configuration-based questions, write the full configuration first, then read off the period (highest n), group and block. Memorise the standard data tables (radii, IE, electron gain enthalpy, electronegativity) — they make ordering questions instant. State exceptions (Be/B, N/O, Ga/Al, Tl/In) precisely.
Frequently Asked Questions
What is Class 11 Chemistry Chapter 3 about?
Chapter 3, Classification of Elements and Periodicity in Properties, covers the development of the periodic table, the Modern Periodic Law, the long form of the table with s-, p-, d- and f-blocks, IUPAC naming of elements with Z > 100, and periodic trends in atomic and ionic radii, ionization enthalpy, electron gain enthalpy, electronegativity and valence.
How many exercises are solved on this page?
All 40 NCERT end-of-chapter exercises (3.1 to 3.40) are reproduced verbatim and solved step by step, including the ionization-enthalpy numerical (3.15) and the data-based questions (3.31).
Why is the second ionization enthalpy of sodium so high?
After losing one electron, Na+ attains the very stable [Ne] noble-gas configuration. Removing a second electron means breaking this stable core, which requires a very large amount of energy — hence the big jump from 520 to 7300 kJ mol–1.
Are these Class 11 Chemistry Chapter 3 solutions free?
Yes. All solutions are free and follow the official NCERT Chemistry textbook for session 2026–27.
