NCERT Class 11 Chemistry Chapter 3 Solutions – Classification of Elements and Periodicity in Properties. This page provides clear and step-by-step solutions to all NCERT exercise questions from 3.1 to 3.. The answers are explained in simple language to help Class 11 students understand periodic classification, electronic configuration, atomic and ionic radii, ionisation enthalpy and other important periodic properties.

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

NCERT Exercise Solutions: Questions 3.1 to 3.16

Question 3.1

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

Answer

The basic theme of organisation of elements in the modern periodic table is the periodic repetition of similar physical and chemical properties when elements are arranged in increasing order of their atomic numbers.

Explanation

Elements in the modern periodic table are arranged in increasing order of their atomic numbers.
Elements having similar outer electronic configurations are placed in the same group.
Elements belonging to the same group generally show similar chemical properties.
The properties of elements are repeated periodically with increasing atomic number.

Therefore, the modern periodic table is based on the periodic repetition of properties with increasing atomic number.

Question 3.2

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

Answer

Mendeleev mainly used atomic mass as the basis for arranging the elements in his periodic table.

Step-by-step explanation

Mendeleev arranged the elements mainly in the order of increasing atomic masses.
He also considered the chemical properties of the elements.
Elements having similar chemical properties were placed in the same group.
He did not always strictly follow the increasing order of atomic mass.
In some cases, he changed the order so that elements with similar chemical properties could be placed together.

Therefore, Mendeleev used atomic mass as an important criterion, but he did not strictly follow increasing atomic mass when it conflicted with the chemical properties of the elements.

Question 3.3

What is the basic difference in approach between Mendeleev’s Periodic Law and the Modern Periodic Law?

Answer

The basic difference is the property used as the basis for arranging the elements.

Mendeleev’s Periodic Law	Modern Periodic Law
Elements are arranged mainly according to increasing atomic mass.	Elements are arranged according to increasing atomic number.
Properties of elements are a periodic function of their atomic masses.	Properties of elements are a periodic function of their atomic numbers.

Mendeleev’s Periodic Law:
Properties of elements are a periodic function of their atomic masses.

Modern Periodic Law:
Properties of elements are a periodic function of their atomic numbers.

Thus, Mendeleev’s Periodic Law was based on atomic mass, whereas the Modern Periodic Law is based on atomic number.

Question 3.4

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

Answer

In the sixth period, electrons are filled into the 6s, 4f, 5d and 6p subshells.

Step 1: 6s subshell

An s-subshell contains one orbital and each orbital can accommodate a maximum of two electrons.

6s = 2 electrons

Step 2: 4f subshell

An f-subshell contains 7 orbitals and each orbital can accommodate 2 electrons.

4f = 7 × 2 = 14 electrons

Step 3: 5d subshell

A d-subshell contains 5 orbitals.

5d = 5 × 2 = 10 electrons

Step 4: 6p subshell

A p-subshell contains 3 orbitals.

6p = 3 × 2 = 6 electrons

Step 5: Total number of elements

Total = 2 + 14 + 10 + 6

Total = 32

Hence, the sixth period of the periodic table contains 32 elements.

Question 3.5

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

Answer

Given:

Z = 114

Step 1: Electronic configuration

[Rn] 5f¹⁴ 6d¹⁰ 7s² 7p²

Step 2: Determine the period

The highest principal quantum number is n = 7. Therefore, the element belongs to the 7th period.

Step 3: Determine the group

The valence-shell configuration is:

7s² 7p²

This configuration corresponds to Group 14.

Answer: The element with Z = 114 belongs to Period 7 and Group 14.

Question 3.6

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

Answer

The third period contains the elements from sodium to argon.

Group 17 contains the halogens.

The Group 17 element in the third period is chlorine (Cl).

Atomic number of chlorine = 17

Therefore, the required atomic number is 17.

Question 3.7

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

Answer

(i) Lawrence Berkeley Laboratory

The element is Berkelium (Bk).

Its atomic number is 97.

(ii) Seaborg’s group

The element is Seaborgium (Sg).

Its atomic number is 106.

Lawrencium (Lr) and Berkelium (Bk). Lawrencium (Lr) → Atomic number 103 Berkelium (Bk) → Atomic number 97 Both names are associated with the Lawrence Berkeley Laboratory.

Question 3.8

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

Answer

Elements belonging to the same group generally have similar valence-shell electronic configurations.

Step-by-step explanation

The chemical properties of an element are largely determined by its valence electrons.
Elements in the same group generally have the same number of valence electrons.
Therefore, they have similar valence-shell electronic configurations.
As a result, they show similar chemical properties.

Therefore, elements in the same group have similar properties mainly because of their similar valence-shell electronic configurations.

Question 3.9

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

Answer

Atomic Radius

Atomic radius is a measure of the size of an atom. Since an atom does not have a sharply defined boundary, its radius is generally defined in terms of the distance between the nuclei of two bonded atoms.

Atomic radius = ½ × internuclear distance

Ionic Radius

Ionic radius is the effective distance from the nucleus of an ion to the outermost region of its electron cloud.

Thus, atomic radius represents the size of an atom, whereas ionic radius represents the effective size of a cation or anion.

Question 3.10

How do atomic radii vary in a period and in a group? How do you explain the variation?

Answer

Across a period

Atomic radius generally decreases from left to right across a period.

Reason

Atomic number increases from left to right.
The number of protons increases.
Electrons are added to the same principal shell.
Effective nuclear charge increases.
The nucleus attracts the electrons more strongly.
Therefore, atomic radius decreases.

Down a group

Atomic radius generally increases from top to bottom in a group.

Reason

A new electron shell is added at each step down the group.
The distance of the outermost electrons from the nucleus increases.
Shielding effect also increases.
Therefore, atomic radius increases.

Across a period: Atomic radius decreases.
Down a group: Atomic radius increases.

Question 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⁺

Answer

Species having the same number of electrons are called isoelectronic species.

(i) F⁻

Fluorine has atomic number 9.

F⁻ = 9 + 1 = 10 electrons

Neon (Ne) has 10 electrons.

Answer: Ne

(ii) Ar

Argon has 18 electrons. Chlorine has atomic number 17.

Cl⁻ = 17 + 1 = 18 electrons

Answer: Cl⁻

(iii) Mg²⁺

Mg²⁺ = 12 − 2 = 10 electrons

Answer: Ne

(iv) Rb⁺

Rb⁺ = 37 − 1 = 36 electrons

Krypton (Kr) has 36 electrons.

Answer: Kr

(i) F⁻ → Ne
(ii) Ar → Cl⁻
(iii) Mg²⁺ → Ne
(iv) Rb⁺ → Kr

Question 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) What is common in them?

All the given ions contain 10 electrons. Therefore, they are isoelectronic species.

Ion	Number of electrons
N³⁻	7 + 3 = 10
O²⁻	8 + 2 = 10
F⁻	9 + 1 = 10
Na⁺	11 − 1 = 10
Mg²⁺	12 − 2 = 10
Al³⁺	13 − 3 = 10

(b) Increasing order of ionic radii

Since all the ions have the same number of electrons, their ionic radii depend mainly on nuclear charge.

Greater nuclear charge attracts the electrons more strongly and therefore produces a smaller ionic radius.

Nuclear charge: N < O < F < Na < Mg < Al

Increasing order of ionic radii:

Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻

Question 3.13

Explain why cations are smaller and anions larger in radii than their parent atoms.

Answer

Why are cations smaller?

A cation is formed when an atom loses one or more electrons.
The nuclear charge remains unchanged.
Electron-electron repulsion decreases.
Sometimes the entire outermost shell is lost.
The remaining electrons are attracted more strongly by the nucleus.
Therefore, the cation has a smaller radius.

Example:

Na → Na⁺ + e⁻

Why are anions larger?

An anion is formed by gaining one or more electrons.
The nuclear charge remains unchanged.
Electron-electron repulsion increases.
The electron cloud expands.
Therefore, the anion has a larger radius.

Example:

Cl + e⁻ → Cl⁻

Therefore, cations are smaller and anions are larger than their respective parent atoms.

Question 3.14

What is the significance of the terms ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy?

Answer

Isolated gaseous atom

The atom is considered in the gaseous state and isolated from other atoms so that interactions with neighbouring atoms or molecules do not affect the energy change.

Ground state

The atom must be in its ground state, in which its electrons occupy the lowest possible energy levels.

Ionization enthalpy

X(g) → X⁺(g) + e⁻

Ionization enthalpy is the enthalpy required to remove an electron from an isolated gaseous atom in its ground state.

Electron gain enthalpy

X(g) + e⁻ → X⁻(g)

Electron gain enthalpy is the enthalpy change when an electron is added to an isolated gaseous atom in its ground state.

Thus, “isolated gaseous atom” and “ground state” specify definite conditions under which ionization enthalpy and electron gain enthalpy are measured.

Question 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.

Answer

Given

E = −2.18 × 10⁻¹⁸ J atom⁻¹

Step 1: Energy required to remove the electron

The negative sign indicates that the electron is bound to the hydrogen nucleus. Therefore, the energy required to remove it is:

Ionization energy = +2.18 × 10⁻¹⁸ J atom⁻¹

Step 2: Convert into J mol⁻¹

Avogadro constant:

N_A = 6.022 × 10²³ mol⁻¹

Therefore,

Ionization enthalpy = 2.18 × 10⁻¹⁸ × 6.022 × 10²³

= 1.313 × 10⁶ J mol⁻¹

Therefore, ionization enthalpy of atomic hydrogen ≈ 1.31 × 10⁶ J mol⁻¹

= 1312 kJ mol⁻¹ approximately.

Question 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 Δ_iH than B
(ii) O has lower Δ_iH than N and F

Answer

(i) Why does Be have higher ionization enthalpy than B?

Electronic configurations:

Be: 1s² 2s²

B: 1s² 2s² 2p¹

Explanation

In Be, the electron removed during ionization is from the 2s orbital.
In B, the electron removed is from the 2p orbital.
The 2p orbital has higher energy than the 2s orbital.
Therefore, the electron in B is easier to remove.
Hence, B has lower ionization enthalpy than Be.

Therefore, Be has higher ionization enthalpy than B because B loses a higher-energy 2p electron, whereas Be loses a more strongly held 2s electron.

(ii) Why does O have lower ionization enthalpy than N and F?

Electronic configurations:

N: 1s² 2s² 2p³

O: 1s² 2s² 2p⁴

F: 1s² 2s² 2p⁵

Reason for N > O

Nitrogen has a stable half-filled 2p³ configuration.

In oxygen, the 2p⁴ configuration contains one pair of electrons in one of the 2p orbitals. The repulsion between the paired electrons makes it easier to remove one of them.

Therefore, oxygen has lower ionization enthalpy than nitrogen.

Reason for O < F

Fluorine has a greater nuclear charge than oxygen. Therefore, its electrons are attracted more strongly towards the nucleus.

Hence, fluorine has higher ionization enthalpy than oxygen.

Therefore, oxygen has lower ionization enthalpy than nitrogen because of electron-pair repulsion, while fluorine has higher ionization enthalpy because of its greater effective nuclear attraction.

Quick Revision – Chapter 3

Modern Periodic Law: Properties of elements are a periodic function of their atomic numbers.
Across a period: Atomic radius generally decreases.
Down a group: Atomic radius generally increases.
Isoelectronic species: Species having the same number of electrons.
Cations: Smaller than their parent atoms.
Anions: Larger than their parent atoms.
Sixth period: Contains 32 elements.
Z = 114: Period 7, Group 14.

Question 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?

Answer

Step 1: Electronic configurations

Na (Z = 11): 1s² 2s² 2p⁶ 3s¹

Mg (Z = 12): 1s² 2s² 2p⁶ 3s²

First ionization enthalpy

The first electron in both Na and Mg is removed from the 3s orbital. However, Mg has a greater nuclear charge than Na. Therefore, the electron in Mg is held more strongly by the nucleus.

First ionization enthalpy: Na < Mg

Second ionization enthalpy

After losing one electron, sodium forms Na⁺, which has the stable noble gas configuration of neon.

Na⁺: 1s² 2s² 2p⁶

Therefore, the second electron of Na has to be removed from a completely filled inner shell. This requires a very large amount of energy.

On the other hand, Mg⁺ still has one electron in the 3s orbital, which can be removed more easily.

Second ionization enthalpy: Na > Mg

Hence, Na has lower first ionization enthalpy than Mg, but its second ionization enthalpy is much higher because Na⁺ has a stable noble gas configuration.

Question 3.18

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

Answer

The ionization enthalpy generally decreases from top to bottom in a group because of the following factors:

1. Increase in atomic size

As we move down a group, a new electron shell is added at each step. Therefore, the outermost electron is farther away from the nucleus.

2. Increase in shielding effect

The number of inner-shell electrons increases down the group. These electrons shield the valence electrons from the attractive force of the nucleus.

3. Decrease in effective nuclear attraction

Due to the increased distance and shielding effect, the effective attraction between the nucleus and the outermost electron decreases.

Therefore, increased atomic size, increased shielding effect and reduced effective nuclear attraction cause ionization enthalpy to generally decrease down a group.

Question 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?

Answer

Generally, ionization enthalpy is expected to decrease down a group because atomic size and shielding effect increase.

However, the given values show some deviations from this general trend.

Gallium compared with Aluminium

Al = 577 kJ mol⁻¹
Ga = 579 kJ mol⁻¹

Gallium has completely filled 3d orbitals. The 3d electrons do not shield the nuclear charge effectively. Therefore, the outer electrons in Ga experience a relatively greater effective nuclear charge.

Hence, Ga has slightly higher ionization enthalpy than Al.

Thallium compared with Indium

In = 558 kJ mol⁻¹
Tl = 589 kJ mol⁻¹

In Tl, the filled 4f electrons shield the nuclear charge poorly. This results in increased effective nuclear charge, known as lanthanide contraction.

Consequently, the outer electrons in Tl are held more strongly, resulting in higher ionization enthalpy.

The deviations are mainly due to the poor shielding of d and f electrons and the resulting increase in effective nuclear charge.

Question 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) O or F

Fluorine has a higher nuclear charge and greater tendency to accept an electron than oxygen.

Therefore, F has more negative electron gain enthalpy than O.

(ii) F or Cl

Although fluorine is more electronegative than chlorine, the electron gain enthalpy of chlorine is more negative.

Fluorine has a very small atomic size. When an additional electron enters its compact 2p orbital, strong electron-electron repulsion occurs.

In chlorine, the added electron enters the larger 3p orbital where electron-electron repulsion is comparatively less.

Therefore, Cl has more negative electron gain enthalpy than F.

Question 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.

Answer

The second electron gain enthalpy of oxygen is expected to be positive.

First electron gain

O(g) + e⁻ → O⁻(g)

The first electron is added to a neutral oxygen atom and energy is released.

Second electron gain

O⁻(g) + e⁻ → O²⁻(g)

The second electron has to be added to an already negatively charged O⁻ ion. The incoming electron experiences strong electron-electron repulsion.

Therefore, energy must be supplied to add the second electron.

Hence, the second electron gain enthalpy of oxygen is positive, whereas the first electron gain enthalpy is negative.

Question 3.22

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

Answer

Electron Gain Enthalpy	Electronegativity
It is the enthalpy change when an electron is added to an isolated gaseous atom.	It is the tendency of an atom in a chemical bond to attract the shared electron pair towards itself.
It is an experimentally measurable quantity.	It is a relative quantity.
It has units of energy, usually kJ mol⁻¹.	It has no unit.

In short, electron gain enthalpy refers to the actual energy change when an isolated gaseous atom gains an electron, whereas electronegativity refers to the tendency of a bonded atom to attract shared electrons.

Question 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?

Answer

The statement is not correct.

The value 3.0 is the approximate electronegativity assigned to nitrogen on the Pauling scale under standard conditions.

However, electronegativity is not an absolute constant for an atom. It can vary depending upon its chemical environment, oxidation state and the nature of atoms to which it is bonded.

Therefore, the electronegativity of nitrogen cannot be considered exactly 3.0 in all nitrogen compounds. The Pauling value of 3.0 is an approximate standard value.

Question 3.24

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

Answer

(a) When an atom gains an electron

When an atom gains one or more electrons, it forms an anion.

The nuclear charge remains unchanged, but the number of electrons increases. Therefore, electron-electron repulsion increases.

As a result, the electron cloud expands and the ionic radius becomes larger than the atomic radius.

Atomic radius < Anionic radius

(b) When an atom loses an electron

When an atom loses one or more electrons, it forms a cation.

The number of electrons decreases while the nuclear charge remains the same. Sometimes the entire outermost shell is lost.

Therefore, the attraction of the nucleus on the remaining electrons becomes stronger and the size decreases.

Cationic radius < Atomic radius

Thus, gaining electrons increases the radius because of increased electron-electron repulsion, whereas losing electrons decreases the radius because of increased effective nuclear attraction.

Question 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.

Answer

The first ionization enthalpies of two isotopes of the same element are expected to be nearly the same.

Reason

Isotopes of an element have the same atomic number and therefore have the same number of electrons and the same electronic configuration.

Ionization enthalpy mainly depends upon electronic configuration, effective nuclear charge and atomic size.

Since these factors are almost identical for isotopes, their first ionization enthalpies are nearly identical.

Therefore, isotopes of the same element have nearly the same first ionization enthalpy.

Question 3.26

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

Answer

Metals	Non-metals
Generally have low ionization enthalpies.	Generally have high ionization enthalpies.
Generally have low electronegativities.	Generally have high electronegativities.
Tend to lose electrons and form cations.	Tend to gain or share electrons.
Generally form basic oxides.	Generally form acidic or neutral oxides.
Most metals are good conductors of heat and electricity.	Most non-metals are poor conductors of heat and electricity.
Generally have a lustrous appearance.	Generally lack metallic lustre.

Thus, metals generally lose electrons and form cations, whereas non-metals generally gain or share electrons and form anions or covalent compounds.

Question 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.

(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.

(a) Element with five electrons in the outer subshell

An element with five electrons in the p-subshell has the general configuration ns²np⁵.

Example: Chlorine (Cl) = 3s²3p⁵

Answer: Chlorine (Cl)

(b) Element that tends to lose two electrons

Group 2 elements have two valence electrons and generally tend to lose both electrons to achieve a stable noble gas configuration.

Answer: Magnesium (Mg) — an example of a Group 2 element.

(c) Element that tends to gain two electrons

Group 16 elements have six valence electrons and generally tend to gain two electrons to complete their octet.

Answer: Oxygen (O) — an example of a Group 16 element.

(d) Group having metal, non-metal, liquid and gas

Group 17 contains halogens. It includes gaseous elements such as fluorine and chlorine and liquid bromine at room temperature. The group also shows a transition towards metallic character in its heavier members.

Answer: Group 17 (Halogens).

Question 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.

Answer

Group 1 elements

Group 1 elements are metals and their reactivity depends on their tendency to lose an electron.

Down the group, atomic size increases and ionization enthalpy decreases. Therefore, the outermost electron is lost more easily.

Li < Na < K < Rb < Cs

Thus, metallic reactivity increases down Group 1.

Group 17 elements

Group 17 elements are non-metals and their reactivity depends mainly on their tendency to gain an electron.

Down the group, atomic size increases and the attraction for an incoming electron decreases.

Therefore, the tendency to gain an electron decreases down the group.

F > Cl > Br > I

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

Question 3.29

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

Answer

Block	General outer electronic configuration
s-block	ns^1–2
p-block	ns²np^1–6
d-block	(n−1)d^1–10 ns^0–2
f-block	(n−2)f^1–14(n−1)d^0–1ns²

Therefore, the general outer electronic configurations are:

s-block → ns^1–2
p-block → ns²np^1–6
d-block → (n−1)d^1–10ns^0–2
f-block → (n−2)f^1–14(n−1)d^0–1ns²

Question 3.30

Assign the position of the element having outer electronic configuration:

(i) ns²np⁴ for n = 3

(ii) (n−1)d²ns² for n = 4

(iii) (n−2)f⁷(n−1)d¹ns² for n = 6

in the periodic table.

Answer

(i) ns²np⁴ for n = 3

Put n = 3:

3s²3p⁴

This is the outer electronic configuration of sulphur (S).

Atomic number = 16

The highest principal quantum number is 3, so it belongs to Period 3.

The configuration ns²np⁴ corresponds to Group 16.

(i) Sulphur (S): Period 3, Group 16, p-block

(ii) (n−1)d²ns² for n = 4

Put n = 4:

3d²4s²

This is the outer electronic configuration of titanium (Ti).

Atomic number = 22

Since the highest principal quantum number is 4, it belongs to Period 4.

For d-block elements:

Group number = (n−1)d electrons + ns electrons

= 2 + 2

= 4

(ii) Titanium (Ti): Period 4, Group 4, d-block

(iii) (n−2)f⁷(n−1)d¹ns² for n = 6

Put n = 6:

4f⁷5d¹6s²

This configuration corresponds to gadolinium (Gd).

Atomic number = 64

The highest principal quantum number is 6. Therefore, it belongs to Period 6.

Since the differentiating electron enters the 4f subshell, the element belongs to the f-block and is a lanthanide.

(iii) Gadolinium (Gd): Period 6, f-block (lanthanide series)

Final Answers at a Glance

Part	Element	Position
(i)	Sulphur (S)	Period 3, Group 16, p-block
(ii)	Titanium (Ti)	Period 4, Group 4, d-block
(iii)	Gadolinium (Gd)	Period 6, f-block, lanthanide

These configurations can be used to identify the element, period, group/block and position of an element in the periodic table.

Question 3.31

The first (Δ_iH) and the second (Δ_iH) ionization enthalpies (in kJ mol⁻¹) and the (Δ_egH) electron gain enthalpy (in kJ mol⁻¹) of a few elements are given below:

Element	Δ_iH₁	Δ_iH₂	Δ_egH
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.

(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).

Answer

(a) The least reactive element

Element V has the highest first ionization enthalpy (2372 kJ mol⁻¹) and positive electron gain enthalpy. These are characteristic of a noble gas.

Answer: Element V

(b) The most reactive metal

Reactive metals lose electrons easily. Therefore, a metal with very low first ionization enthalpy is highly reactive.

Element II has the lowest first ionization enthalpy (419 kJ mol⁻¹).

Answer: Element II

(c) The most reactive non-metal

Reactive non-metals have a strong tendency to gain electrons. Hence, they generally have highly negative electron gain enthalpy.

Element III has the most negative electron gain enthalpy (−328 kJ mol⁻¹).

Answer: Element III

(d) The least reactive non-metal

Element V has a very high ionization enthalpy and positive electron gain enthalpy. This indicates a noble gas configuration.

Answer: Element V

(e) Metal forming MX₂

A compound of the formula MX₂ indicates that the metal forms a stable M²⁺ ion.

Element VI has ionization enthalpies corresponding to a Group 2 metal. It can lose two electrons to form M²⁺.

Answer: Element VI

(f) Metal forming predominantly covalent MX

Element IV has relatively high ionization enthalpy and can form a small, highly polarising cation. Such a metal tends to form predominantly covalent halides of the type MX.

Answer: Element IV

Final Answers

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

Question 3.32

Predict the formulae 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

(d) Silicon and oxygen

(e) Phosphorus and fluorine

(f) Element 71 and fluorine

Answer

(a) Lithium and oxygen

Lithium forms Li⁺ and oxygen forms O²⁻.

2Li⁺ + O²⁻ → Li₂O

(a) Li₂O

(b) Magnesium and nitrogen

Magnesium forms Mg²⁺ and nitrogen forms N³⁻. Balancing the charges gives:

3Mg²⁺ + 2N³⁻ → Mg₃N₂

(b) Mg₃N₂

(c) Aluminium and iodine

Aluminium forms Al³⁺ and iodine forms I⁻.

Al³⁺ + 3I⁻ → AlI₃

(c) AlI₃

(d) Silicon and oxygen

Silicon generally exhibits valency 4 and oxygen exhibits valency 2.

Si : O = 1 : 2

(d) SiO₂

(e) Phosphorus and fluorine

Phosphorus can form a stable covalent compound with fluorine in which phosphorus shows valency 5.

P + 5F → PF₅

(e) PF₅

(f) Element 71 and fluorine

Element 71 is lutetium (Lu). It commonly forms Lu³⁺.

Fluorine forms F⁻.

Lu³⁺ + 3F⁻ → LuF₃

(f) LuF₃

Question 3.33

In the modern periodic table, the period indicates the value of:

(a) atomic number

(b) atomic mass

(d) azimuthal quantum number

Answer

The period number of an element indicates the highest principal quantum number n occupied by its electrons in the ground state.

For example, elements of Period 3 have their valence electrons in the third shell, where n = 3.

Correct Answer: (c) Principal quantum number

Question 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-subshell.

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

A d-subshell contains 5 orbitals, and each orbital can accommodate 2 electrons.

Maximum electrons in d-subshell = 5 × 2 = 10

Therefore, the d-block has 10 columns, not 8.

Correct Answer: (b) The d-block has 8 columns.

This statement is incorrect because the d-block has 10 columns.

Question 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)

(d) Number of core electrons

Answer

The chemistry of an element is mainly influenced by factors such as the principal quantum number, nuclear charge and shielding due to core electrons.

Nuclear mass does not directly affect the valence shell or the chemical behaviour of an element.

Correct Answer: (c) Nuclear mass

Question 3.36

The size of isoelectronic species — F⁻, Ne and Na⁺ — is affected by:

(a) nuclear charge (Z)

(b) valence principal quantum number (n)

(d) none of the factors because their size is the same

Answer

F⁻, Ne and Na⁺ are isoelectronic species. Each contains 10 electrons.

F⁻ = 9 + 1 = 10 electrons
Ne = 10 electrons
Na⁺ = 11 − 1 = 10 electrons

Although they have the same number of electrons, their nuclear charges are different:

F⁻ : Z = 9
Ne : Z = 10
Na⁺ : Z = 11

Greater nuclear charge attracts the same electron cloud more strongly and produces a smaller radius.

Correct Answer: (a) Nuclear charge (Z)

Size order: F⁻ > Ne > Na⁺

Question 3.37

Which 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.

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

Answer

Electrons in orbitals with a higher principal quantum number (higher n) are generally farther from the nucleus and are easier to remove.

Therefore, the statement saying that removal from an orbital with lower n is easier than from one with higher n is incorrect.

Correct Answer: (d)

Removal of an electron from an orbital having higher n is generally easier than from an orbital having lower n.

Question 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

(d) K > Mg > Al > B

Answer

Metallic character increases down a group and decreases from left to right across a period.

Among the given elements, potassium is farthest towards the metallic side of the periodic table, while boron is least metallic.

K > Mg > Al > B

Correct Answer: (d) K > Mg > Al > B

Question 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

(d) F > N > C > Si > B

Answer

Non-metallic character increases from left to right across a period and decreases down a group.

Fluorine is the most non-metallic among the given elements. Nitrogen is next, followed by carbon.

Between boron and silicon, boron is more non-metallic because it lies above silicon in the periodic table.

F > N > C > B > Si

Correct Answer: (c) F > N > C > B > Si

Question 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

(d) O > F > N > Cl

Answer

An oxidizing agent is a species that has a strong tendency to gain electrons.

Fluorine is the strongest oxidizing agent because of its very high electronegativity and strong tendency to accept an electron.

Among the halogens, oxidizing power decreases down the group:

F > Cl

Oxygen is also a strong oxidizing element, but fluorine and chlorine have stronger oxidizing tendencies in this comparison. Nitrogen has the lowest oxidizing property among the given elements.

F > Cl > O > N

Correct Answer: (a) F > Cl > O > N

More Class 11 Chemistry Resources

📘 NCERT Class 11 Chemistry Exemplar Solutions 📝 Class 11 Chemistry Practice Papers

NCERT Class 11 Chemistry Chapter 3 – Questions 3.31 to 3.40

These solutions cover the important concepts from NCERT Class 11 Chemistry Chapter 3 – Classification of Elements and Periodicity in Properties, including ionization enthalpy, electron gain enthalpy, periodic trends, metallic character, non-metallic character, oxidizing property, electronic configuration and formation of binary compounds.

Questions 3.1 to 3.40 are now covered with clear, step-by-step NCERT solutions for Class 11 Chemistry.

NCERT Class 11 Chemistry Chapter 3 – Exercise Completed

Questions 3.1 to 3.30 of NCERT Class 11 Chemistry Chapter 3, Classification of Elements and Periodicity in Properties, are covered with clear explanations and step-by-step solutions.

Periodic classification of elements
Modern periodic law
Atomic and ionic radii
Ionization enthalpy
Electron gain enthalpy
Electronegativity
Metallic and non-metallic character
Periodic trends
Reactivity of elements
Electronic configuration and block identification

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

NCERT Exercise Solutions: Questions 3.1 to 3.16

Question 3.1

Answer

Explanation

Question 3.2

Answer

Step-by-step explanation

Question 3.3

Answer

Question 3.4

Answer

Step 1: 6s subshell

Step 2: 4f subshell

Step 3: 5d subshell

Step 4: 6p subshell

Step 5: Total number of elements

Question 3.5

Answer

Step 1: Electronic configuration

Step 2: Determine the period

Step 3: Determine the group

Question 3.6

Answer

Question 3.7

Answer

(i) Lawrence Berkeley Laboratory

(ii) Seaborg’s group

Question 3.8

Answer

Step-by-step explanation

Question 3.9

Answer

Atomic Radius

Ionic Radius

Question 3.10

Answer

Across a period

Reason

Down a group

Reason

Question 3.11

Answer

(i) F−

(ii) Ar

(iii) Mg2+

(iv) Rb+

Question 3.12

Answer

(a) What is common in them?

(b) Increasing order of ionic radii

Question 3.13

Answer

Why are cations smaller?

Why are anions larger?

Question 3.14

Answer

Isolated gaseous atom

Ground state

Ionization enthalpy

Electron gain enthalpy

Question 3.15

Answer

Given

Step 1: Energy required to remove the electron

Step 2: Convert into J mol−1

Question 3.16

Answer

(i) Why does Be have higher ionization enthalpy than B?

Explanation

(ii) Why does O have lower ionization enthalpy than N and F?

Reason for N > O

Reason for O < F

Quick Revision – Chapter 3

Question 3.17

Answer

Step 1: Electronic configurations

First ionization enthalpy

Second ionization enthalpy

Question 3.18

(i) F⁻

(iii) Mg²⁺

(iv) Rb⁺

Step 2: Convert into J mol⁻¹

(i) ns²np⁴ for n = 3

(ii) (n−1)d²ns² for n = 4

(iii) (n−2)f⁷(n−1)d¹ns² for n = 6

(e) Metal forming MX₂