KAPPA CLASSES

Class 11 Chemistry

Chapter 3 – Classification of Elements and Periodicity in Properties

NCERT Exemplar Solutions – Questions 1 to 55

Complete Chapter 3 Exemplar Question Answers

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Multiple Choice Type-I Questions

1. Consider the isoelectronic species Na+, Mg2+, F and O2−. Which of the following represents the correct order of increasing radii?
(i) F < O2− < Mg2+ < Na+
(ii) Mg2+ < Na+ < F < O2−
(iii) O2− < F < Na+ < Mg2+
(iv) O2− < F < Mg2+ < Na+
Solution:
All four species are isoelectronic and contain 10 electrons. For an isoelectronic series, ionic radius decreases as nuclear charge increases. Mg2+ has the highest nuclear charge and O2− has the lowest.
Answer: (ii) Mg2+ < Na+ < F < O2−
2. Which of the following is not an actinoid?
(i) Curium (Z = 96)
(ii) Californium (Z = 98)
(iii) Uranium (Z = 92)
(iv) Terbium (Z = 65)
Solution:
Actinoids are elements from actinium (Z = 89) to lawrencium (Z = 103). Terbium (Z = 65) belongs to the lanthanoid series.
Answer: (iv) Terbium
3. The correct order of screening effect of electrons in s, p, d and f orbitals is:
(i) s > p > d > f
(ii) f > d > p > s
(iii) p < d < s > f
(iv) f > p > s > d
Solution:
The shielding or screening ability depends on the penetration of the orbital. The penetration power follows s > p > d > f. Therefore, s-electrons shield most effectively.
Answer: (i) s > p > d > f
4. The first ionisation enthalpies of Na, Mg, Al and Si are related as:
(i) Na < Mg > Al < Si
(ii) Na > Mg > Al > Si
(iii) Na < Mg < Al < Si
(iv) Na > Mg > Al < Si
Solution:
Ionisation enthalpy generally increases across a period. However, Mg has a filled 3s subshell, so its ionisation enthalpy is higher than that of Al. Thus the order follows Na < Mg > Al < Si.
Answer: (i)
5. The electronic configuration of Gd (Z = 64) is:
(i) [Xe] 4f3 5d5 6s2
(ii) [Xe] 4f7 5d2 6s1
(iii) [Xe] 4f7 5d1 6s2
(iv) [Xe] 4f8 5d6 6s2
Solution:
Gadolinium has a half-filled 4f subshell and one electron in the 5d subshell. Its configuration is [Xe] 4f7 5d1 6s2.
Answer: (iii)
6. Which of the following statements is incorrect?
(i) The properties of elements are a periodic function of their atomic numbers.
(ii) Non-metallic elements are less in number than metallic elements.
(iii) For transition elements, 3d orbitals are filled after 3p and before 4s.
(iv) First ionisation enthalpy generally increases with atomic number across a period.
Solution:
The 4s orbital is filled before the 3d orbitals in the Aufbau order. Therefore, statement (iii) is incorrect.
Answer: (iii)
7. Among the halogens, the amount of energy released during electron gain follows:
(i) F > Cl > Br > I
(ii) F < Cl < Br < I
(iii) F < Cl > Br > I
(iv) F < Cl < Br < I
Solution:
Chlorine has a more negative electron gain enthalpy than fluorine because the very small size of fluorine causes greater electron-electron repulsion in its compact 2p subshell. Hence chlorine releases more energy on gaining an electron.
Answer: (iii)
8. The period number of an element is equal to:
(i) Magnetic quantum number
(ii) Atomic number
(iii) Maximum principal quantum number
(iv) Maximum azimuthal quantum number
Solution:
The period number corresponds to the highest principal quantum number (n) present in the ground-state electronic configuration of the element.
Answer: (iii) Maximum principal quantum number
9. The elements in which electrons are progressively filled in 4f orbitals are called:
(i) Actinoids
(ii) Transition elements
(iii) Lanthanoids
(iv) Halogens
Solution:
The lanthanoid series involves progressive filling of the 4f subshell.
Answer: (iii) Lanthanoids
10. The correct order of size of I, I and I+ is:
(i) I > I > I+
(ii) I+ > I > I
(iii) I > I+ > I
(iv) I > I > I+
Solution:
An anion is larger than its neutral atom because of increased electron-electron repulsion, whereas a cation is smaller because it has lost an electron.
Answer: (iv) I > I > I+
11. Formation of O2− from oxygen in the gas phase requires energy because:
(i) Oxygen is more electronegative.
(ii) Addition of an electron results in larger size.
(iii) Electron-electron repulsion outweighs the stability gained by noble-gas configuration.
(iv) O is smaller than O atom.
Solution:
The first electron addition to oxygen is favourable, but adding a second electron to O introduces a negative electron into an already negatively charged species. Strong electron-electron repulsion makes the second electron gain process endothermic.
Answer: (iii)
12. Read the following information and answer the questions.
Consider the modern periodic table and the position of elements according to their atomic numbers.

(a) The element with atomic number 57 belongs to which block?

(i) s-block
(ii) p-block
(iii) d-block
(iv) f-block
Answer: (iii) d-block

(b) The last p-block element of the sixth period has the configuration:

(i) 7s2 7p6
(ii) 5f14 6d10 7s2 7p0
(iii) 4f14 5d10 6s2 6p6
(iv) 4f14 5d10 6s2 6p4
Answer: (iii)

(c) Which atomic number cannot fit in the present long-form periodic table?

(i) 107
(ii) 118
(iii) 126
(iv) 102
Answer: (iii) 126

(d) The element just above Z = 43 and belonging to the same group has the configuration:

(i) 1s2 2s2 2p6 3s2 3p6 3d5 4s2
(ii) 1s2 2s2 2p6 3s2 3p6 3d5 4s3 4p6
(iii) 1s2 2s2 2p6 3s2 3p6 3d6 4s2
(iv) 1s2 2s2 2p6 3s2 3p6 3d7 4s2
Answer: (i)

(e) Elements having atomic numbers 35, 53 and 85 are:

(i) Noble gases
(ii) Halogens
(iii) Heavy metals
(iv) Light metals
Answer: (ii) Halogens
13. Four elements A, B, C and D have the following electronic configurations: A = Ne, B = O, C = Na and D = F. The increasing order of tendency to gain an electron is:
(i) A < C < B < D
(ii) A < B < C < D
(iii) D < B < C < A
(iv) D < A < B < C
Solution:
Neon has a stable noble-gas configuration and has the least tendency to gain an electron. Sodium also has a low tendency, while oxygen and fluorine have greater tendencies. Thus the order is A < C < B < D.
Answer: (i)

Multiple Choice Type-II Questions

14. Which of the following elements can show covalency greater than 4?
(i) Be
(ii) P
(iii) S
(iv) B
Solution:
Phosphorus and sulphur can show expanded valence shells and therefore can exhibit covalencies greater than 4.
Answers: (ii), (iii)
15. The elements belonging to which groups commonly give characteristic flame colours?
(i) Group 2
(ii) Group 13
(iii) Group 1
(iv) Group 17
Solution:
Alkali and alkaline earth metal compounds commonly produce characteristic colours in flame tests.
Answers: (i), (iii)
16. Which of the following sets contains only representative elements?
(i) 3, 33, 53, 87
(ii) 2, 10, 22, 36
(iii) 7, 17, 25, 37, 48
(iv) 9, 35, 51, 88
Solution:
Representative elements belong to the s- and p-blocks. Sets (i) and (iv) contain only s- and p-block elements.
Answers: (i), (iv)
17. Which of the following atoms can gain one electron more readily than the other members of their respective groups?
(i) S(g)
(ii) Na(g)
(iii) O(g)
(iv) Cl(g)
Solution:
Sulphur gains an electron more readily than oxygen in its group because the larger size of sulphur reduces electron-electron repulsion. Chlorine has a highly favourable electron gain enthalpy among halogens.
Answers: (i), (iv)
18. Which of the following statements are correct?
(i) He has the highest ionisation enthalpy.
(ii) Cl has less negative electron gain enthalpy than F.
(iii) Hg and Br2 are liquids at room temperature.
(iv) An alkali metal has the highest atomic radius in any period.
Solution:
Helium has the highest ionisation enthalpy. Mercury and bromine are liquids at room temperature. Alkali metals have the largest atomic radii in their respective periods.
Answers: (i), (iii), (iv)
19. Which of the following sets contains only isoelectronic species?
(i) Zn2+, Ca2+, Ga3+, Al3+
(ii) K+, Ca2+, Sc3+, Cl
(iii) P3−, S2−, Cl, K+
(iv) Ti4+, Ar, Cr3+, V5+
Solution:
K+, Ca2+, Sc3+ and Cl each contain 18 electrons. Similarly, P3−, S2−, Cl and K+ each contain 18 electrons.
Answers: (ii), (iii)
20. Which of the following sequences do not agree with the general periodic trends?
(i) Al3+ < Mg2+ < Na+ < F in increasing ionic size
(ii) B < C < N < O in increasing first ionisation enthalpy
(iii) I < Br < Cl < F in increasing electron gain enthalpy
(iv) Li < Na < K < Rb in increasing metallic radius
Solution:
The sequence in (ii) ignores the known exception between nitrogen and oxygen because nitrogen has a half-filled 2p3 configuration. The sequence in (iii) also does not correctly represent the electron gain enthalpy trend because chlorine has a more negative value than fluorine.
Answers: (ii), (iii)
21. Which of the following properties have no unit?
(i) Electronegativity
(ii) Electron gain enthalpy
(iii) Ionisation enthalpy
(iv) Metallic character
Solution:
Electronegativity and metallic character are relative properties and are expressed without units.
Answers: (i), (iv)
22. Ionic radii vary:
(i) In inverse proportion to effective nuclear charge
(ii) In inverse proportion to the square of effective nuclear charge
(iii) In direct proportion to screening effect
(iv) In direct proportion to the square of screening effect
Solution:
An increase in effective nuclear charge pulls electrons closer to the nucleus, reducing atomic or ionic radius. Greater screening reduces the effective attraction and increases the radius.
Answers: (i), (iii)
23. The element belonging to period 3 and group 13 is characterised by which properties?
(i) Good conductor of electricity
(ii) Liquid metallic
(iii) Solid metallic
(iv) Solid non-metallic
Solution:
The period 3, group 13 element is aluminium. Aluminium is a solid metal and is a good conductor of electricity.
Answers: (i), (iii)

Short Answer Questions – 24 to 42

24. Explain why the electron gain enthalpy of fluorine is less negative than that of chlorine.
Solution:
Fluorine has a very small atomic size. The incoming electron has to enter the compact 2p subshell where electron-electron repulsion is relatively high. In chlorine, the electron enters the larger 3p orbital, where repulsion is lower. Therefore, more energy is released when chlorine gains an electron, making its electron gain enthalpy more negative than that of fluorine.
25. All transition elements are d-block elements, but all d-block elements are not transition elements. Explain.
Solution:
Transition elements are those elements whose atoms or at least one of their ions have partially filled d-orbitals. They belong to the d-block. However, elements such as Zn, Cd and Hg have completely filled d10 configurations in their atoms and common ions, so they are d-block elements but are not transition elements.
26. An element has atomic number 119. Identify its group, valency, outer electronic configuration and formula of its oxide.
Solution:
The element with Z = 119 would begin the eighth period and have the outer configuration 8s1.

Group: 1
Valency: 1
Outer electronic configuration: 8s1
Oxide: M2O

It would be an alkali-metal type element.
27. The first ionisation enthalpies of the elements of the second period are given below. Match the values with the elements and complete the graph. Also write the symbols and atomic numbers of the elements.
ElementFirst ionisation enthalpyAtomic number
Li520 kJ mol−13
Be899 kJ mol−14
B801 kJ mol−15
C1086 kJ mol−16
N1402 kJ mol−17
O1314 kJ mol−18
F1681 kJ mol−19
Ne2080 kJ mol−110
Solution:
The first ionisation enthalpy generally increases from Li to Ne because effective nuclear charge increases and atomic size decreases. Two important deviations occur:

Be > B: B loses an electron from the higher-energy 2p orbital, whereas Be loses an electron from the stable 2s orbital.
N > O: N has a stable half-filled 2p3 configuration, while O has paired electrons in one 2p orbital.
28. Among B, Al, C and Si, which has the highest first ionisation enthalpy? Which is the most metallic? Give reasons.
Solution:
Highest first ionisation enthalpy: Carbon (C).

Carbon is smaller than silicon and therefore its valence electron is held more strongly by the nucleus. Between B and C, carbon has greater effective nuclear charge.

Most metallic: Aluminium (Al).

Metallic character increases down a group and decreases from left to right across a period. Aluminium therefore has the greatest tendency to lose electrons among the four.
29. Write four characteristic properties of p-block elements.
Solution:
The important characteristics of p-block elements are:

1. Their valence-shell electronic configuration is generally ns2np1–6.
2. The p-block contains metals, metalloids and non-metals.
3. They show variable oxidation states.
4. They form a large number of covalent compounds and molecular compounds.
30. Which of the following represents the correct order of atomic radii of F and Ne in pm?
(i) F = 72 pm, Ne = 160 pm
(ii) F = 160 pm, Ne = 160 pm
(iii) F = 72 pm, Ne = 72 pm
(iv) F = 160 pm, Ne = 72 pm
Solution:
The van der Waals radius of neon is much larger than the covalent radius of fluorine. Therefore, the values are approximately 72 pm for F and 160 pm for Ne.
Answer: (i)
31. Illustrate the different oxidation states shown by elements on the basis of their electronic configurations. Give suitable examples from transition and non-transition elements.
Solution:
Transition elements show variable oxidation states because the energies of ns and (n−1)d electrons are comparable. For example, iron shows +2 and +3 oxidation states, while manganese shows +2, +3, +4, +6 and +7.

Non-transition elements can also show variable oxidation states because both s and p electrons may participate in bonding. For example, phosphorus shows +3 and +5 oxidation states, while chlorine shows +1, +3, +5 and +7 oxidation states.
32. Nitrogen has positive electron gain enthalpy whereas oxygen has negative electron gain enthalpy. Also, the first ionisation enthalpy of oxygen is lower than that of nitrogen. Explain.
Solution:
Nitrogen has the stable half-filled configuration 2p3. Addition of an electron disturbs this stability, so the process is less favourable.

Oxygen has configuration 2p4, so gaining an electron produces a more stable arrangement relative to nitrogen and its electron gain enthalpy is negative.

For ionisation enthalpy, nitrogen has a stable half-filled 2p3 configuration. Oxygen has 2p4, with one pair of electrons. Repulsion between paired electrons makes removal of one electron easier. Therefore, IE1(O) < IE1(N).
33. The first member of a representative group shows anomalous behaviour. Illustrate with two examples.
Solution:
The first member of a group often behaves differently because of its very small size, high ionisation enthalpy and high electronegativity.

Example 1 – Lithium: Lithium differs from other alkali metals and shows diagonal relationship with magnesium.

Example 2 – Boron: Boron is a non-metallic/metalloid element, whereas the other members of group 13 are more metallic. Boron also forms mainly covalent compounds.
34. Explain the nature of oxides of p-block elements. Give two examples each of acidic, basic and amphoteric oxides and write their reactions with water.
Solution:
Across a period, the acidic character of oxides generally increases, while the basic character decreases.

Basic oxides: Na2O and MgO.
Na2O + H2O → 2NaOH
MgO + H2O → Mg(OH)2

Acidic oxides: SO2 and SO3.
SO2 + H2O → H2SO3
SO3 + H2O → H2SO4

Amphoteric oxides: Al2O3 and ZnO.
Al2O3 and ZnO react with both acids and bases.
35. The first ionisation enthalpy of sodium is lower than that of magnesium, but its second ionisation enthalpy is higher than that of magnesium. Explain.
Solution:
Na has configuration [Ne]3s1, while Mg has [Ne]3s2.

The first electron is easier to remove from Na, so IE1(Na) < IE1(Mg).

After losing one electron, Na+ has the stable noble-gas configuration [Ne]. Removing another electron from Na+ requires breaking this stable configuration. Therefore, IE2(Na) is very high.

Mg+ still has a 3s electron, so its second electron is easier to remove than that of Na+.
36. Define exothermic and endothermic processes with one example of each.
Solution:
Exothermic process: A process in which energy is released to the surroundings. Example: electron gain by chlorine:
Cl(g) + e → Cl(g)

Endothermic process: A process in which energy is absorbed from the surroundings. Example:
Cl(g) → Cl(g) + e
37. Arrange N, P, O and S in:
(i) Increasing order of first ionisation enthalpy.
(ii) Increasing order of non-metallic character.
Solution:
(i) Increasing first ionisation enthalpy:
S < P < N < O

P has higher ionisation enthalpy than S due to its smaller size, while N has higher ionisation enthalpy than O because of its stable half-filled 2p3 configuration.

(ii) Increasing non-metallic character:
P < S < N < O

Non-metallic character increases from left to right across a period and from bottom to top in a group.
38. Explain the deviations in the first ionisation enthalpy values represented in the periodic trend.
Solution:
The general trend is an increase in first ionisation enthalpy from left to right across a period. However, there are exceptions.

Be > B: B loses a 2p electron, which is higher in energy and easier to remove than the 2s electron of Be.

N > O: Nitrogen has a stable half-filled 2p3 configuration. Oxygen has one paired set of electrons in 2p orbitals, causing additional repulsion and making electron removal easier.
39. Explain:
(a) Electronegativity increases from left to right across a period.
(b) Ionisation enthalpy decreases from top to bottom in a group.
Solution:
(a) Across a period, effective nuclear charge increases and atomic size decreases. Therefore, the attraction between the nucleus and the shared pair of electrons increases, so electronegativity increases.

(b) Down a group, the number of electron shells increases and atomic size increases. Shielding also increases. Therefore, the attraction between the nucleus and the valence electron decreases, making electron removal easier. Hence ionisation enthalpy decreases.
40. Explain the variation of metallic and non-metallic character across a period.
Solution:
Across a period from left to right, effective nuclear charge increases and atomic size decreases. The tendency to lose electrons decreases, so metallic character decreases.

At the same time, the tendency to gain electrons increases, so non-metallic character increases.
41. Explain why the radius of Na+ is less than the radius of Na atom.
Solution:
Na has electronic configuration [Ne]3s1. When it loses its 3s electron to form Na+, the third shell is completely removed. Na+ therefore has fewer occupied shells and a smaller radius than the Na atom.
42. Which is the least electronegative alkali metal? Explain why.
Solution:
Cesium (Cs) is the least electronegative alkali metal among the commonly considered stable alkali metals. Down group 1, atomic size increases and shielding increases. The valence electron is therefore held less strongly by the nucleus, so the tendency to attract shared electrons decreases.

Matching Type Questions – 43 to 45

43. Match the elements in Column I with the appropriate atomic radii in Column II.
Column I Column II – Atomic radius (pm)
Be74
C88
O111
B77
N66
Solution:
The correct matching is:
ElementMatched value
Be111
O66
C77
B88
N74
44. Match the elements in Column I with their appropriate properties given in Column II.
Element IE1 (kJ mol−1) IE2 (kJ mol−1) Electron gain enthalpy
A4193051−48
B16813374−328
C7381451−40
D23725251+48
Solution:
The values correspond to elements having different tendencies to lose or gain electrons.

(i) Most reactive non-metal: B
(ii) Most reactive metal: A
(iii) Least reactive element: D
(iv) Metal forming binary halide: C
Answer: (i) B   (ii) A   (iii) D   (iv) C
45. Match the following electronic configurations with the appropriate electron gain enthalpy values.
Electronic configuration Matching value
(i) 1s2 2s2 2p6 D: +48
(ii) 1s2 2s2 2p6 3s1 A: −53
(iii) 1s2 2s2 2p5 B: −328
(iv) 1s2 2s2 2p4 C: −141
Solution:
A noble-gas configuration is highly stable and therefore has an unfavourable electron gain enthalpy. A halogen with 2p5 configuration has a strongly negative electron gain enthalpy. Oxygen has 2p4, while sodium has 3s1.
Answer: (i) D, (ii) A, (iii) B, (iv) C

Assertion and Reason Questions – 46 to 48

46. Assertion (A): The first ionisation enthalpy of nitrogen is greater than that of oxygen.

Reason (R): Nitrogen has a stable half-filled 2p3 electronic configuration.
(i) Assertion is correct but Reason is wrong.
(ii) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
(iii) Both Assertion and Reason are wrong.
(iv) Assertion is wrong but Reason is correct.
Solution:
Nitrogen has the stable half-filled 2p3 configuration, whereas oxygen has paired electrons in one of its 2p orbitals. Electron-electron repulsion makes electron removal from oxygen easier. Hence IE1(N) > IE1(O).
Answer: (ii)
47. Assertion (A): The atomic radius decreases from left to right across a period.

Reason (R): Effective nuclear charge increases from left to right across a period.
(i) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
(ii) Assertion is correct but Reason is wrong.
(iii) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
(iv) Both Assertion and Reason are wrong.
Solution:
Across a period, nuclear charge increases while electrons are added to the same principal shell. Therefore, effective nuclear charge increases and attracts the electrons more strongly towards the nucleus. Consequently, atomic radius decreases.
Answer: (iii)
48. Assertion (A): Electron gain enthalpy becomes less negative down a group.

Reason (R): Atomic size increases down a group.
(i) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
(iii) Both Assertion and Reason are wrong.
(iv) Assertion is wrong but Reason is correct.
Solution:
Down a group, atomic size increases and the incoming electron experiences weaker attraction from the nucleus. Consequently, less energy is released on electron addition and electron gain enthalpy generally becomes less negative.
Answer: (ii)

Long Answer Questions – 49 to 55

49. Discuss the factors affecting electron gain enthalpy and explain its periodic trends.
Solution:
Electron gain enthalpy is the enthalpy change accompanying the addition of an electron to an isolated gaseous atom.

Factors affecting electron gain enthalpy:

1. Atomic size: As atomic size increases, the incoming electron is farther from the nucleus and the attraction decreases. Therefore, electron gain enthalpy generally becomes less negative.

2. Effective nuclear charge: Higher effective nuclear charge attracts the incoming electron more strongly and makes electron gain enthalpy more negative.

3. Electronic configuration: Stable configurations such as completely filled and half-filled subshells oppose the addition of an electron.

Periodic trend: Across a period, electron gain enthalpy generally becomes more negative because effective nuclear charge increases and atomic size decreases.

Down a group, electron gain enthalpy generally becomes less negative because atomic size and shielding increase.

Exception: The electron gain enthalpy of chlorine is more negative than that of fluorine because fluorine's very small 2p orbital produces considerable electron-electron repulsion.
50. Define ionisation enthalpy. Discuss the factors affecting it and explain its periodic trends.
Solution:
Ionisation enthalpy is the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state.

For example:
M(g) → M+(g) + e

Factors affecting ionisation enthalpy:

1. Atomic size: Larger atomic size means the valence electron is farther from the nucleus and is easier to remove. Thus ionisation enthalpy decreases.

2. Nuclear charge: Greater nuclear charge increases attraction for the electron and increases ionisation enthalpy.

3. Shielding effect: Greater shielding reduces the effective nuclear attraction and lowers ionisation enthalpy.

4. Electronic configuration: Half-filled and completely filled subshells are relatively stable and require more energy for electron removal.

Across a period: Ionisation enthalpy generally increases because effective nuclear charge increases and atomic radius decreases.

Down a group: Ionisation enthalpy generally decreases because atomic size and shielding increase.

Important exceptions include Be > B and N > O.
51. Explain why the modern periodic law is based on atomic number rather than atomic mass.
Solution:
The modern periodic law states that the physical and chemical properties of elements are periodic functions of their atomic numbers.

Atomic number represents the number of protons in the nucleus and determines the electronic configuration of an element. Since the electronic configuration determines chemical properties, elements with similar outer electronic configurations show similar properties.

The arrangement according to atomic number also removes several anomalies present in Mendeleev's atomic-mass-based table.

For example, isotopes have different masses but the same atomic number and therefore occupy the same position in the modern periodic table.
52. Explain why alkali metals have the general outer electronic configuration ns1 and why they are placed in group 1.
Solution:
Alkali metals have one electron in their outermost shell. Therefore, their general valence-shell configuration is ns1.

This single valence electron can be easily lost to form a stable noble-gas configuration:

M → M+ + e

As they have one valence electron and commonly show +1 oxidation state, they are placed in group 1 of the modern periodic table.

Examples include Li, Na, K, Rb and Cs.
53. Discuss the main drawbacks of Mendeleev's periodic table.
Solution:
The important limitations of Mendeleev's periodic table were:

1. Position of hydrogen: Hydrogen resembles both alkali metals and halogens, so its position was uncertain.

2. Anomalous pairs: Some elements had to be placed in positions that did not strictly follow increasing atomic mass.

3. Isotopes: Isotopes have different atomic masses but similar chemical properties. Mendeleev's table could not provide separate positions for isotopes.

4. Cause of periodicity: Mendeleev's periodic law did not explain the fundamental reason for periodic repetition of properties.

5. Rare earth elements: The proper position of lanthanoids was not clearly established.
54. Why is the long-form periodic table considered better than Mendeleev's periodic table?
Solution:
The long-form periodic table is based on atomic number and electronic configuration, making it more logical and systematic.

Its advantages include:

1. Elements are arranged according to increasing atomic number.
2. Elements with similar valence-shell configurations are placed in the same group.
3. It explains periodicity in terms of electronic configuration.
4. Isotopes occupy the same position because they have the same atomic number.
5. The table clearly separates s-, p-, d- and f-block elements.
6. It accommodates transition and inner-transition elements systematically.
7. The anomalous pairs in Mendeleev's table are eliminated.
55. Compare the first ionisation enthalpies of group 1 and group 17 elements. Explain their variation in the periodic table.
Solution:
Group 1 elements have one valence electron and have relatively low first ionisation enthalpies. Group 17 elements have seven valence electrons and are much more strongly attracted to the nucleus, so they have high first ionisation enthalpies.

Across a period: Ionisation enthalpy generally increases from group 1 to group 17 because effective nuclear charge increases and atomic size decreases.

Down group 1: Ionisation enthalpy decreases because atomic size and shielding increase. The outer electron becomes easier to remove.

Down group 17: Ionisation enthalpy also decreases because the valence shell moves farther from the nucleus and shielding increases.

Thus, group 17 elements have much higher first ionisation enthalpies than group 1 elements in the same period, while both groups show a decreasing trend down their respective groups.