CBSE · NCERT · Class 11 Chemistry · Chapter 3

NCERT Solutions: Class 11 Chemistry Chapter 3 - Classification of Elements and Periodicity in Properties

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Chapter-wise NCERT intext questions and exercise answers for Classification of Elements and Periodicity in Properties, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercises 39
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1Exercises39 questions
Q.3.1What is the basic theme of organisation in the periodic table?v
Solution

Modern periodic organisation is based on increasing atomic number. Since atomic number determines electronic configuration, similar valence-shell configurations recur periodically and elements with similar physical and chemical properties fall in the same groups.

Answer:

The basic theme is to arrange elements so that elements with similar electronic configurations and similar properties occur together, making periodic trends clear.

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

Mendeleev arranged elements largely in increasing atomic weights. However, he placed some elements out of strict mass order to keep elements with similar properties in the same group, and he left gaps for undiscovered elements.

Answer:

Mendeleev used atomic weight as the main basis, but he did not stick to it strictly when chemical properties required a different placement.

Q.3.3What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?v
Solution

Mendeleev stated that properties are periodic functions of atomic weights. Moseley showed that atomic number is more fundamental, so the modern law states that physical and chemical properties are periodic functions of atomic numbers.

Answer:

Mendeleev’s Periodic Law is based on atomic weights; the Modern Periodic Law is based on atomic numbers.

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

In the sixth period, electrons enter subshells associated with n = 6 and nearby lower-energy subshells in the order 6s, 4f, 5d and 6p. Their capacities are 2, 14, 10 and 6 electrons respectively. Total capacity = 32, so the sixth period should contain 32 elements.

Answer:

The sixth period can accommodate 6s, 4f, 5d and 6p electrons: 2 + 14 + 10 + 6 = 32 elements.

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

Element Z = 114 is in the p-block after filling 7s, 5f and 6d subshells. It belongs to the carbon family, group 14, and lies in the seventh period.

Answer:

Period 7, group 14.

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

The element in the third period and group 17 is chlorine. Chlorine has atomic number 17.

Answer:

17.

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

Berkelium is named after Berkeley, the site of Lawrence Berkeley Laboratory. Seaborgium is named in honour of Glenn T. Seaborg, whose group worked on transuranium elements.

Answer:

(i) Berkelium. (ii) Seaborgium.

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

Chemical properties depend mainly on valence electrons. Elements in the same group have the same number of valence electrons and similar outer electronic configurations, so they show similar bonding patterns and related physical and chemical properties.

Answer:

They have similar valence-shell electronic configurations.

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

The boundary of an atom or ion is not sharp, so radius is defined by measurable distances. Atomic radius may be covalent, van der Waals or metallic depending on bonding. Ionic radius is inferred from interionic distances in ionic crystals.

Answer:

Atomic radius is an operational measure of atomic size, commonly half the distance between nuclei of two bonded identical atoms. Ionic radius is the effective size of an ion in a crystal lattice.

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

Across a period, electrons enter the same principal shell while nuclear charge increases, so effective nuclear charge increases and pulls electrons closer. Down a group, new shells are added and shielding increases, so atomic radius increases despite increased nuclear charge.

Answer:

Atomic radius decreases from left to right across a period and increases down a group.

Q.3.11What 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+v
Solution

F- has 10 electrons, the same as Ne. Ar has 18 electrons, the same as K+ or Ca2+. Mg2+ has 10 electrons, the same as Ne. Rb+ has 36 electrons, the same as Kr.

Answer:

Isoelectronic species have the same number of electrons. Examples: (i) Ne, (ii) K+, (iii) Ne, (iv) Kr.

Q.3.12Consider 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.v
Solution

All listed ions have 10 electrons. In an isoelectronic series, radius decreases as nuclear charge increases. The nuclear charges are N = 7, O = 8, F = 9, Na = 11, Mg = 12 and Al = 13. Therefore the smallest ion is Al3+ and the largest is N3-.

Answer:

(a) They are isoelectronic, each having 10 electrons. (b) Al3+ < Mg2+ < Na+ < F- < O2- < N3-.

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

Formation of a cation reduces the number of electrons and may remove the outer shell, so the remaining electrons are pulled closer by the nucleus. Formation of an anion adds electrons to the valence shell, increasing repulsions and lowering effective attraction per electron, so the radius increases.

Answer:

Cations are smaller because electrons are removed and effective attraction per electron increases; anions are larger because added electrons increase electron-electron repulsion.

Q.3.14What 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.v
Solution

Ionization enthalpy and electron gain enthalpy are compared for single atoms, so the atom must be isolated and gaseous to avoid interactions with neighbouring atoms or ions. Ground state means the atom is in its lowest-energy electronic state, so the measured enthalpy is not affected by prior excitation.

Answer:

They provide a standard reference state free from bonding, lattice and excitation effects.

Q.3.15Energy of an electron in the ground state of the hydrogen atom is -2.18x10^-18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol^-1. Hint: Apply the idea of mole concept to derive the answer.v
Solution

Ionizing one hydrogen atom from ground state requires 2.18 x 10^-18 J. For one mole of atoms, ionization enthalpy = (2.18 x 10^-18)(6.022 x 10^23) = 1.31 x 10^6 J mol^-1.

Answer:

1.31 x 10^6 J mol^-1.

Q.3.16Among 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 Δi H than B (ii) O has lower Δi H than N and F?v
Solution

Be: 1s2 2s2 has a completely filled 2s subshell. B: 1s2 2s2 2p1 loses a higher-energy 2p electron, so its ionization enthalpy is lower. N has 2p3 half-filled stability; O has 2p4, where one p orbital contains paired electrons and electron-electron repulsion makes removal easier. F has greater nuclear charge than O, so its ionization enthalpy is higher.

Answer:

(i) Be has a stable filled 2s2 subshell, while B loses a 2p electron more easily. (ii) O has paired-electron repulsion in 2p4, while N has stable half-filled 2p3; F has higher effective nuclear charge.

Q.3.17How 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?v
Solution

Sodium is [Ne]3s1, while magnesium is [Ne]3s2 with higher nuclear charge. Thus first ionization enthalpy of Na is lower. After first ionization, Na+ is [Ne], a stable noble-gas configuration; the second ionization removes a core electron. Mg+ is [Ne]3s1, so its second ionization still removes a valence electron.

Answer:

Na loses its single 3s electron more easily than Mg, but Na+ has a noble-gas core, so removing a second electron from Na+ is much harder than removing the second 3s electron from Mg+.

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

Down a group, the valence electron occupies a shell with higher principal quantum number and is farther from the nucleus. Inner-shell electrons provide greater shielding. These effects outweigh the increase in nuclear charge, so the electron is removed more easily and ionization enthalpy decreases.

Answer:

Increase in atomic size and shielding down a group reduce attraction for the valence electron.

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

Ionization enthalpy generally decreases down a group, but Ga has filled 3d electrons that shield poorly, so its valence electron feels higher effective nuclear charge and its value is slightly higher than Al. Tl is affected by poor shielding by both d and f electrons, giving a higher value than expected after In.

Answer:

The deviations arise from poor shielding by inner d and f electrons and the resulting higher effective nuclear charge.

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

Across the second period, electron gain enthalpy generally becomes more negative, so F is more negative than O. However, Cl has a more negative electron gain enthalpy than F because the added electron enters the larger 3p orbital of Cl with less electron-electron repulsion than in compact 2p orbital of F.

Answer:

(i) F. (ii) Cl.

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

The first electron gain forms O-. Adding a second electron to an already negative ion is opposed by strong electron-electron repulsion, so energy must be supplied. Therefore the second electron gain enthalpy of oxygen is positive.

Answer:

It is positive.

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

Electron gain enthalpy is measurable in kJ mol^-1 for isolated gaseous atoms. Electronegativity is a dimensionless relative property used for bonded atoms and depends on bonding environment.

Answer:

Electron gain enthalpy is an enthalpy change for adding an electron to an isolated gaseous atom; electronegativity is the relative tendency of an atom in a molecule to attract shared electrons.

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

The Pauling value 3.0 is a useful tabulated value for nitrogen, but electronegativity depends on the atom's chemical environment, oxidation state and hybridisation. Therefore it should not be treated as exactly 3.0 in all nitrogen compounds.

Answer:

The statement is not strictly correct; electronegativity is not a fixed constant in every compound.

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

On gaining an electron, the number of electrons increases while nuclear charge is unchanged, increasing repulsion and producing an anion larger than the parent atom. On losing an electron, electron-electron repulsion decreases and sometimes the outer shell is removed, producing a smaller cation.

Answer:

(a) Radius increases when an atom gains an electron. (b) Radius decreases when an atom loses an electron.

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

Isotopes of an element have the same atomic number and electronic configuration. Ionization enthalpy depends mainly on nuclear charge and electronic arrangement, not on neutron number. Isotope mass effects are negligible for this comparison.

Answer:

They are expected to be essentially the same.

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

Metals usually have low ionization enthalpies and form cations and basic oxides. Non-metals have higher ionization enthalpies and electronegativities, form anions or covalent compounds, and often form acidic oxides.

Answer:

Metals are generally lustrous, malleable, ductile, good conductors and tend to lose electrons; non-metals are generally poor conductors, often brittle in solid form and tend to gain or share electrons.

Q.3.27Use 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.v
Solution

Elements with p5 outer subshells are halogens, such as chlorine. Group 2 metals such as magnesium tend to lose two electrons. Group 16 non-metals such as oxygen tend to gain two electrons. Group 17 includes gaseous F2 and Cl2, liquid Br2, solid iodine and metalloid/metallic heavier members such as astatine.

Answer:

(a) Chlorine, for example, has 3p5. (b) Magnesium. (c) Oxygen. (d) Group 17, the halogens.

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

In group 1, atomic size increases and ionization enthalpy decreases down the group, so alkali metals lose their valence electron more readily. In group 17, atomic size increases and the attraction for an added electron decreases down the group, so oxidizing power and reactivity decrease from F to I.

Answer:

Group 1 reactivity increases down the group because electron loss becomes easier; group 17 reactivity decreases down the group because electron gain becomes less favourable.

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

The block name is determined by the subshell receiving the differentiating electron. Thus s-block ends in ns, p-block in np, d-block in the penultimate d subshell and f-block in the anti-penultimate f subshell.

Answer:

s-block: ns1-2; p-block: ns2 np1-6; d-block: (n-1)d1-10 ns0-2; f-block: (n-2)f1-14 (n-1)d0-1 ns2.

Q.3.30Assign 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) f 7 (n-1)d1ns2 for n=6, in the periodic table.v
Solution

(i) n = 3 gives 3s2 3p4, the configuration of sulphur, group 16. (ii) n = 4 gives 3d2 4s2, titanium, period 4 group 4. (iii) n = 6 gives 4f7 5d1 6s2, corresponding to a lanthanoid such as gadolinium in period 6 f-block.

Answer:

(i) Period 3, group 16, p-block. (ii) Period 4, group 4, d-block. (iii) Period 6, f-block, lanthanoid series.

Q.3.31The 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: Elements ΔH1 ΔH2 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 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)?v
Solution

V has very high ionization enthalpy and positive electron gain enthalpy, typical of a noble gas, so it is least reactive. II has the lowest first ionization enthalpy, so it is the most reactive metal. III has the most negative electron gain enthalpy and high ionization enthalpy, so it is the most reactive non-metal. IV is a non-metal but less reactive than III. VI has comparatively low first and second ionization enthalpies, fitting a divalent metal that forms MX2. I has a very high second ionization enthalpy after losing one electron, fitting a monovalent metal whose MX may have predominantly covalent character.

Answer:

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

Q.3.32Predict 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 fluorinev
Solution

Use common valencies/oxidation states: Li is +1 and O is -2, giving Li2O. Mg is +2 and N is -3, giving Mg3N2. Al is +3 and I is -1, giving AlI3. Si is +4 and O is -2, giving SiO2. Phosphorus forms fluorides PF3 and PF5. Element 71 is lutetium, commonly +3, so with fluoride it forms LuF3.

Answer:

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

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

The period number corresponds to the highest principal quantum number n of the valence shell in that period.

Answer:

(c) principal quantum number.

Q.3.34Which 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.v
  1. a. The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
  2. b. The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
  3. c. Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
  4. 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 five orbitals and can hold 10 electrons, so the d-block has 10 columns, not 8. Therefore statement (b) is incorrect.

Answer:

(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.

Q.3.35Anything 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.v
  1. a. Valence principal quantum number (n)
  2. b. Nuclear charge (Z )
  3. c. Nuclear mass
  4. d. Number of core electrons.
Solution

Valence-shell behaviour depends on shell number, nuclear charge and shielding by core electrons. Nuclear mass changes mainly with neutron number and has negligible direct effect on valence electrons.

Answer:

(c) Nuclear mass.

Q.3.36The 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.v
  1. a. nuclear charge (Z )
  2. b. valence principal quantum number (n)
  3. c. electron-electron interaction in the outer orbitals
  4. d. none of the factors because their size is the same.
Solution

F-, Ne and Na+ are isoelectronic, so they have the same number of electrons and the same principal shell. Their size difference is mainly due to increasing nuclear charge: Na+ is smallest and F- is largest.

Answer:

(a) nuclear charge (Z ).

Q.3.37Which 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.v
  1. a. Ionization enthalpy increases for each successive electron.
  2. b. The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
  3. c. End of valence electrons is marked by a big jump in ionization enthalpy.
  4. d. Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Solution

Electrons with lower n are closer to the nucleus and more strongly held, so they are harder, not easier, to remove. Therefore option (d) is incorrect.

Answer:

(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.

Q.3.38Considering 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 > Bv
  1. a. B > Al > Mg > K
  2. b. Al > Mg > B > K
  3. c. Mg > Al > K > B
  4. d. K > Mg > Al > B
Solution

Metallic character increases down a group and decreases from left to right across a period. K is an alkali metal and most metallic. In period 3, Mg is more metallic than Al. B is a metalloid and least metallic among these.

Answer:

(d) K > Mg > Al > B.

Q.3.39Considering 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 > Bv
  1. a. B > C > Si > N > F
  2. b. Si > C > B > N > F
  3. c. F > N > C > B > Si
  4. d. F > N > C > Si > B
Solution

Non-metallic character increases across a period and decreases down a group. Among the second-period elements listed, F is highest, followed by N, C and B. Silicon lies below carbon and is less non-metallic than boron.

Answer:

(c) F > N > C > B > Si.