Chapter 5 of 9

Updated 6 August 2026

NCERT Solutions

Classification of Elements and Periodicity in Properties — NCERT Solutions

CBSE · Class 11 · Chemistry

NCERT Solutions for Classification of Elements and Periodicity in Properties, CBSE Class 11 Chemistry: 40 textbook questions solved step by step.

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Exercises

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

The basic theme of organisation in the periodic table is to classify the elements in such a way that elements with similar properties are grouped together. In the Modern Periodic Table, elements are arranged in order of increasing atomic numbers (Z) in horizontal rows called periods and vertical columns called groups. Elements in the same group have similar valence shell electronic configurations and hence exhibit similar physical and chemical properties. This arrangement reflects the periodicity of properties with increasing atomic number.

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

Mendeleev used atomic mass as the basis for classifying elements in his periodic table. His Periodic Law stated: 'The properties of elements are a periodic function of their atomic masses.'

No, he did not always stick strictly to this property. In certain cases, Mendeleev placed elements in positions that violated the order of increasing atomic masses in order to group elements with similar properties together. For example:

Cobalt (Co, atomic mass = 58.93) was placed before Nickel (Ni, atomic mass = 58.69), even though Co has a higher atomic mass than Ni.
Similarly, Tellurium (Te, atomic mass = 127.6) was placed before Iodine (I, atomic mass = 126.9).

Thus, Mendeleev gave priority to chemical properties over strict atomic mass ordering in such cases.

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

Mendeleev's Periodic Law: The properties of elements are a periodic function of their atomic masses. Elements were arranged in increasing order of atomic masses.

Modern Periodic Law: The properties of elements are a periodic function of their atomic numbers (Z). Elements are arranged in increasing order of atomic numbers.

Key Difference: Mendeleev used atomic mass (a physical property that can vary with isotopes) as the basis, whereas the Modern Periodic Law uses atomic number (the number of protons in the nucleus), which is a more fundamental and invariable property of an element. The use of atomic number resolved the anomalies present in Mendeleev's table (e.g., the positions of Co–Ni, Ar–K, Te–I).

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

The period number corresponds to the principal quantum number (n) of the outermost shell being filled.

For the sixth period, n = 6. The subshells that are filled during the sixth period are determined by the Aufbau principle:

Subshell	n	l	Number of orbitals	Max. electrons
6s	6	0	1	2
4f	4	3	7	14
5d	5	2	5	10
6p	6	1	3	6

Total number of electrons that can be accommodated:
$= 2 + 14 + 10 + 6 = 32$

Since each element in a period corresponds to one additional electron, the sixth period should contain 32 elements. This is confirmed by the actual periodic table (from Cs, Z = 55 to Rn, Z = 86).

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

Given: Atomic number Z = 114

We write the electronic configuration of the element with Z = 114 using the Aufbau principle:

$[\text{Rn}]\, 5f^{14}\, 6d^{10}\, 7s^2\, 7p^2$

(Rn has Z = 86; remaining electrons = 114 − 86 = 28, which fill 5f¹⁴, 6d¹⁰, 7s², 7p²)

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

Group: The last electron enters the 7p subshell. The element has the outer configuration $7s^2\,7p^2$ , which is similar to Carbon (2s²2p²) and Silicon (3s²3p²). These belong to Group 14.

Conclusion: The element with Z = 114 is located in Period 7, Group 14. (This element is Flerovium, Fl.)

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

Given: Period 3, Group 17

Group 17 elements are halogens with the general outer electronic configuration $ns^2\,np^5$ .

For Period 3, n = 3, so the configuration is $3s^2\,3p^5$ .

The full electronic configuration is:
$1s^2\,2s^2\,2p^6\,3s^2\,3p^5$

Total number of electrons = 2 + 2 + 6 + 2 + 5 = 17

Atomic number = 17 (This element is Chlorine, Cl.)

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

(i) Lawrence Berkeley Laboratory: The element named by Lawrence Berkeley Laboratory is Lawrencium (Lr, Z = 103). It was named in honour of Ernest O. Lawrence, the inventor of the cyclotron, and after the Lawrence Berkeley National Laboratory where it was discovered.

(ii) Seaborg's group: The element named by Seaborg's group is Seaborgium (Sg, Z = 106). It was named in honour of Glenn T. Seaborg, who was a key member of the team that synthesised many transuranium elements. It is one of the few elements named after a living person at the time of naming.

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

Elements in the same group have similar valence shell electronic configurations (same number and type of electrons in the outermost shell), differing only in the value of the principal quantum number (n).

For example, all Group 1 elements have the configuration $ns^1$ , and all Group 17 elements have the configuration $ns^2\,np^5$ .

Since chemical properties depend primarily on the number and arrangement of valence electrons:

They have the same valency.
They form similar types of compounds.
They exhibit similar reactivity patterns.

Physical properties also show similar trends because the nature of bonding and interatomic interactions are governed by valence electrons. Hence, elements in the same group show similar physical and chemical properties, with gradual variation due to increasing atomic size down the group.

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

Atomic Radius: The atomic radius is defined as the distance from the centre of the nucleus to the outermost shell of electrons in an isolated atom. Since an isolated atom does not have a sharp boundary, atomic radius is practically measured as:

Covalent radius: Half the distance between the nuclei of two identical atoms bonded by a single covalent bond (for non-metals).
Metallic radius (crystal radius): Half the internuclear distance between two adjacent atoms in a metallic crystal.
van der Waals radius: Half the distance between the nuclei of two adjacent non-bonded atoms of the same element in the solid state.

Ionic Radius: The ionic radius is the effective distance from the nucleus of an ion to the point up to which it has influence on its electron cloud. It is the radius of an atom in its ionic form (cation or anion). It is determined from X-ray crystallography by measuring the interionic distances in ionic crystals.

In essence, atomic radius gives us an idea of the size of a neutral atom, while ionic radius gives us the size of a charged species (ion).

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

Variation in a Period (left to right):
Atomic radius decreases from left to right across a period.

Explanation: As we move across a period, the atomic number (nuclear charge, Z) increases while the number of shells remains the same. The increased nuclear charge pulls the electron cloud closer to the nucleus more strongly, resulting in a decrease in atomic radius. The shielding effect remains nearly constant across a period (since electrons are added to the same shell), so the effective nuclear charge ( $Z_{eff}$ ) increases, causing contraction.

Variation in a Group (top to bottom):
Atomic radius increases from top to bottom in a group.

Explanation: As we move down a group, a new shell of electrons is added with each successive element. The increase in the number of shells increases the distance of the outermost electrons from the nucleus. Although nuclear charge also increases, the shielding effect of the inner core electrons is significant, so the effective nuclear charge experienced by the outermost electrons does not increase proportionally. The net result is an increase in atomic radius down the group.

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⁺Show solution

Isoelectronic Species: Species (atoms, molecules, or ions) that have the same number of electrons and the same electronic configuration are called isoelectronic species.

(i) F⁻:
F has Z = 9; F⁻ has 9 + 1 = 10 electrons. Configuration: $1s^2\,2s^2\,2p^6$
Isoelectronic species: Ne (Z = 10, 10 electrons), Na⁺, O²⁻, Mg²⁺, Al³⁺

(ii) Ar:
Ar has Z = 18; 18 electrons. Configuration: $1s^2\,2s^2\,2p^6\,3s^2\,3p^6$
Isoelectronic species: K⁺ (Z = 19, loses 1e⁻ → 18 electrons), Ca²⁺, Cl⁻, S²⁻

(iii) Mg²⁺:
Mg has Z = 12; Mg²⁺ has 12 − 2 = 10 electrons. Configuration: $1s^2\,2s^2\,2p^6$
Isoelectronic species: Ne (Z = 10, 10 electrons), Na⁺, F⁻, O²⁻, Al³⁺

(iv) Rb⁺:
Rb has Z = 37; Rb⁺ has 37 − 1 = 36 electrons. Configuration: $[\text{Kr}]$ i.e., $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^2\,4p^6$
Isoelectronic species: Kr (Z = 36, 36 electrons), Sr²⁺, Br⁻

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.Show solution

Given species: N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺, Al³⁺

(a) What is common in them?

All these species are isoelectronic — they all have 10 electrons and the same electronic configuration: $1s^2\,2s^2\,2p^6$ .

Species	Z	Electrons
N³⁻	7	7+3 = 10
O²⁻	8	8+2 = 10
F⁻	9	9+1 = 10
Na⁺	11	11−1 = 10
Mg²⁺	12	12−2 = 10
Al³⁺	13	13−3 = 10

(b) Order of increasing ionic radii:

For isoelectronic species, as the nuclear charge (Z) increases, the electrons are pulled closer to the nucleus, so the ionic radius decreases with increasing Z.

$\text{Al}^{3+} < \text{Mg}^{2+} < \text{Na}^{+} < \text{F}^{-} < \text{O}^{2-} < \text{N}^{3-}$

(Z: 13 > 12 > 11 > 9 > 8 > 7, so radius increases in the reverse order.)

3.13Explain why cations are smaller and anions larger in radii than their parent atoms?Show solution

Cations are smaller than parent atoms:

When an atom loses one or more electrons to form a cation:

The number of electrons decreases while the nuclear charge (Z) remains the same.
The effective nuclear charge per electron increases.
The remaining electrons are pulled closer to the nucleus.
Also, if the outermost shell is completely emptied, the cation has one fewer shell than the parent atom.

All these factors result in a decrease in size, so cations are smaller than their parent atoms.

Example: Na (186 pm) → Na⁺ (102 pm)

Anions are larger than parent atoms:

When an atom gains one or more electrons to form an anion:

The number of electrons increases while the nuclear charge (Z) remains the same.
The effective nuclear charge per electron decreases.
Electron–electron repulsions increase in the outermost shell.
The electron cloud expands outward.

All these factors result in an increase in size, so anions are larger than their parent atoms.

Example: Cl (99 pm) → Cl⁻ (181 pm)

3.14What is the significance of the terms — 'isolated gaseous atom' and 'ground state' while defining the ionization enthalpy and electron gain enthalpy?Show solution

The terms 'isolated gaseous atom' and 'ground state' are used to define standard reference conditions for comparison purposes:

'Isolated gaseous atom':

In the gaseous state, atoms are far apart and there are no interatomic or intermolecular interactions (no bonding, no lattice energy effects).
'Isolated' ensures that the measurement is for a single, free atom unaffected by neighbouring atoms.
If atoms were in solid or liquid state, energy values would be influenced by intermolecular forces, making comparison unreliable.
This ensures that the measured ionization enthalpy or electron gain enthalpy is an intrinsic property of the atom alone.

'Ground state':

The ground state is the lowest energy state of an atom where electrons occupy the lowest available orbitals according to Aufbau principle.
If the atom were in an excited state, it would already have extra energy, and the measured ionization enthalpy would be lower than the true value.
Specifying ground state ensures that we always start from the same energy reference point, making values comparable across different elements.

In summary, both conditions are necessary to obtain reproducible, comparable, and intrinsic values of ionization enthalpy and electron gain enthalpy.

3.15Energy 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.Show solution

Given: Energy of electron in ground state of H atom = $-2.18 \times 10^{-18}$ J

Concept: Ionization enthalpy is the energy required to remove the electron completely from the atom (i.e., to take the electron from the ground state to $n = \infty$ , where energy = 0).

For one atom:
$\Delta_i H = E_{\infty} - E_1 = 0 - (-2.18 \times 10^{-18}) = 2.18 \times 10^{-18} \text{ J}$

For one mole of hydrogen atoms (using Avogadro's number, $N_A = 6.022 \times 10^{23}$ mol⁻¹):
$\Delta_i H = 2.18 \times 10^{-18} \times 6.022 \times 10^{23} \text{ J mol}^{-1}$

$\Delta_i H = 2.18 \times 6.022 \times 10^{(-18+23)} \text{ J mol}^{-1}$

$\Delta_i H = 13.128 \times 10^{5} \text{ J mol}^{-1}$

$\boxed{\Delta_i H = 1.312 \times 10^{6} \text{ J mol}^{-1} = 1312 \text{ kJ mol}^{-1}}$

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?Show solution

The general trend across a period is that ionization enthalpy increases from left to right due to increasing nuclear charge. However, there are exceptions:

(i) Be has higher Δᵢ H than B:

Be has the electronic configuration: $1s^2\,2s^2$ . The outermost electron is in the 2s orbital.
B has the electronic configuration: $1s^2\,2s^2\,2p^1$ . The outermost electron is in the 2p orbital.
The 2p orbital has higher energy and is farther from the nucleus than the 2s orbital. Also, the 2p electron is shielded by the 2s electrons.
Therefore, it is easier to remove the 2p electron of B than the 2s electron of Be.
Hence, $\Delta_i H(\text{Be}) > \Delta_i H(\text{B})$ .

(ii) O has lower Δᵢ H than N and F:

N has the configuration: $1s^2\,2s^2\,2p^3$ . The three 2p electrons are in half-filled orbitals ( $2p_x^1, 2p_y^1, 2p_z^1$ ), which is an extra stable configuration due to exchange energy.
O has the configuration: $1s^2\,2s^2\,2p^4$ . One of the 2p orbitals has a paired electron ( $2p_x^2, 2p_y^1, 2p_z^1$ ). The electron–electron repulsion in the paired orbital makes it easier to remove one electron from O.
F has higher nuclear charge (Z = 9) than O (Z = 8), which more than compensates, so $\Delta_i H(\text{F}) > \Delta_i H(\text{O})$ .
Hence, $\Delta_i H(\text{O}) < \Delta_i H(\text{N})$ and $\Delta_i H(\text{O}) < \Delta_i H(\text{F})$ .

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?Show solution

Electronic configurations:

Na: $1s^2\,2s^2\,2p^6\,3s^1$ (1 valence electron)
Mg: $1s^2\,2s^2\,2p^6\,3s^2$ (2 valence electrons)

First Ionization Enthalpy ( $\Delta_i H_1$ ):

Na has only one electron in the 3s orbital. The effective nuclear charge experienced by this electron is lower (more shielding from inner electrons).
Mg has two electrons in the 3s orbital and a higher nuclear charge (Z = 12 vs Z = 11 for Na).
Therefore, Mg holds its outermost electron more tightly.
Hence, $\Delta_i H_1(\text{Na}) < \Delta_i H_1(\text{Mg})$ . ✓

Second Ionization Enthalpy ( $\Delta_i H_2$ ):

After losing one electron:
Na⁺ has configuration $1s^2\,2s^2\,2p^6$ — a stable noble gas configuration. Removing the second electron requires breaking into the core, which needs very high energy.
Mg⁺ has configuration $1s^2\,2s^2\,2p^6\,3s^1$ — still has one valence electron in the 3s orbital, which is relatively easy to remove.
Therefore, the second ionization enthalpy of Na is much higher than that of Mg.
Hence, $\Delta_i H_2(\text{Na}) > \Delta_i H_2(\text{Mg})$ . ✓

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

The ionization enthalpy of main group elements decreases down a group due to the following factors:

Increase in atomic size (atomic radius): As we move down a group, new shells are added, increasing the distance between the nucleus and the outermost electrons. Greater distance means weaker nuclear attraction, making it easier to remove the outermost electron.

Increase in shielding (screening) effect: As the number of inner shells increases down the group, the inner electrons shield the outermost electrons from the full nuclear charge more effectively. This reduces the effective nuclear charge ( $Z_{eff}$ ) experienced by the valence electrons.

Decrease in effective nuclear charge ( $Z_{eff}$ ): Due to increased shielding, the effective nuclear charge felt by the outermost electrons decreases down the group, reducing the force of attraction.

The combined effect of increased atomic size and increased shielding results in the outermost electrons being held less tightly, and hence ionization enthalpy decreases down a group.

3.19The 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?Show solution

General trend expected: Ionization enthalpy should decrease steadily from B to Tl down Group 13.

Observed values: B (801) > Al (577) < Ga (579) > In (558) < Tl (589)

Deviations observed:

Ga has slightly higher $\Delta_i H$ than Al.
Tl has higher $\Delta_i H$ than In.

Explanation:

(i) Ga vs Al (Ga has slightly higher $\Delta_i H$ than Al):

Between Al and Ga, the d-block elements (Sc to Zn, 10 elements) are introduced for the first time.
The 3d electrons in Ga are poor shielders of nuclear charge compared to s and p electrons.
As a result, the effective nuclear charge experienced by the 4p electron in Ga is higher than expected.
This makes it slightly harder to remove the outermost electron from Ga than from Al.

(ii) Tl vs In (Tl has higher $\Delta_i H$ than In):

Between In and Tl, the f-block elements (lanthanides, 14 elements) are introduced.
The 4f electrons in Tl are very poor shielders of nuclear charge.
This leads to the lanthanide contraction, causing Tl to have a smaller atomic radius and higher effective nuclear charge than expected.
Hence, Tl has a higher ionization enthalpy than In.

Thus, the poor shielding by d and f electrons explains the deviation from the expected decreasing trend.

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

(i) O or F:

F has a more negative electron gain enthalpy than O.

F (Z = 9) has a higher nuclear charge than O (Z = 8) and a smaller atomic size.
When an electron is added to F, it enters the 2p subshell and experiences a stronger nuclear attraction.
O already has a half-filled 2p subshell tendency is less pronounced, but the main reason is that F has higher $Z_{eff}$ .
$\Delta_{eg}H$ : O = −141 kJ mol⁻¹; F = −328 kJ mol⁻¹
F has more negative electron gain enthalpy.

(ii) F or Cl:

Cl has a more negative electron gain enthalpy than F.

Although F has a higher nuclear charge and smaller size, the electron being added to F enters the very compact 2p orbital, where electron–electron repulsion is very high due to the small size of F.
In Cl, the incoming electron enters the larger 3p orbital, where repulsion is less.
$\Delta_{eg}H$ : F = −328 kJ mol⁻¹; Cl = −349 kJ mol⁻¹
Cl has more negative electron gain enthalpy than F.

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

The second electron gain enthalpy of O would be positive.

First electron gain enthalpy of O:
$\text{O}(g) + e^- \rightarrow \text{O}^-(g) \quad \Delta_{eg}H_1 = -141 \text{ kJ mol}^{-1} \text{ (negative, energy released)}$

Second electron gain enthalpy of O:
$\text{O}^-(g) + e^- \rightarrow \text{O}^{2-}(g) \quad \Delta_{eg}H_2 = +780 \text{ kJ mol}^{-1} \text{ (positive, energy absorbed)}$

Justification:

O⁻ is already a negatively charged ion. Adding a second electron to O⁻ means forcing an electron into a species that is already negative.
The incoming electron experiences strong electrostatic repulsion from the negative charge already present on O⁻.
Energy must be supplied to overcome this repulsion and force the second electron onto O⁻.
Therefore, the second electron gain enthalpy is positive (endothermic process).

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

Property	Electron Gain Enthalpy	Electronegativity
Definition	The enthalpy change when an electron is added to an isolated gaseous atom in its ground state.	The tendency of a bonded atom in a molecule to attract the shared pair of electrons towards itself.
Nature	It is a measurable, quantitative thermodynamic property with units (kJ mol⁻¹).	It is a relative, dimensionless concept (no units); it is not directly measurable.
Condition	Refers to an isolated gaseous atom (no bonding).	Refers to an atom in a chemical bond within a molecule.
Constancy	It is a fixed value for a given element.	It can vary slightly depending on the oxidation state and hybridization of the atom in different compounds.
Scale	Measured in energy units.	Expressed on arbitrary scales (Pauling scale, Mulliken scale, etc.).

In essence, electron gain enthalpy measures the energy change for electron addition to a free atom, while electronegativity measures the relative ability of a bonded atom to attract electrons in a covalent bond.

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

The statement is not entirely correct.

Electronegativity is not an absolute, fixed property of an element. It depends on:

Hybridization state: The electronegativity of N varies with its hybridization. sp-hybridized N is more electronegative than sp²-hybridized N, which is more electronegative than sp³-hybridized N. This is because s-orbitals are closer to the nucleus, and higher s-character means greater electronegativity.
Oxidation state: The effective nuclear charge experienced by the bonding electrons changes with oxidation state, affecting electronegativity.
Nature of bonded atoms: The electronegativity of an atom can be influenced by the electronegativities of the atoms it is bonded to.

The value of 3.0 on the Pauling scale is an average or representative value for nitrogen. In reality, the electronegativity of N can vary in different compounds (e.g., in N₂, NH₃, HCN, NO₂, etc.).

Conclusion: The electronegativity of N is approximately 3.0 on the Pauling scale as a general reference value, but it is not exactly 3.0 in all nitrogen compounds. The statement is an oversimplification.

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

(a) When an atom gains an electron (forms an anion):

When an atom gains one or more electrons, the nuclear charge (Z) remains the same but the number of electrons increases.
The increased electron–electron repulsion in the outermost shell causes the electron cloud to expand.
The effective nuclear charge per electron decreases, so each electron is held less tightly.
As a result, the ionic radius (anion) is larger than the atomic radius of the parent atom.
The more electrons gained, the larger the anion.
Example: O (66 pm) → O²⁻ (140 pm); Cl (99 pm) → Cl⁻ (181 pm)

(b) When an atom loses an electron (forms a cation):

When an atom loses one or more electrons, the nuclear charge (Z) remains the same but the number of electrons decreases.
The electron–electron repulsion decreases, and the remaining electrons experience a greater effective nuclear charge.
The electron cloud contracts towards the nucleus.
If the outermost shell is completely emptied, the cation has one fewer shell.
As a result, the ionic radius (cation) is smaller than the atomic radius of the parent atom.
The more electrons lost, the smaller the cation.
Example: Na (186 pm) → Na⁺ (102 pm); Mg (160 pm) → Mg²⁺ (72 pm)

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

The first ionization enthalpies for two isotopes of the same element would be the same (or virtually the same).

Justification:

Isotopes of an element have the same atomic number (Z) (same number of protons) and the same number of electrons, but differ in the number of neutrons (and hence atomic mass).
The electronic configuration of isotopes is identical.
Ionization enthalpy depends on:

The nuclear charge (Z) — same for isotopes.
The electronic configuration — same for isotopes.
The atomic radius — essentially the same for isotopes (neutrons do not significantly affect atomic size).

Neutrons do not participate in chemical bonding and do not affect the electron–nucleus interactions significantly.
Therefore, the first ionization enthalpy is essentially the same for isotopes of the same element.

Note: There is a very slight difference due to the mass effect on nuclear motion (isotope effect), but this is negligibly small and not chemically significant.

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

Property	Metals	Non-metals
Physical state	Mostly solids at room temperature (except Hg which is liquid)	Solids, liquids, or gases at room temperature
Lustre	Have metallic lustre (shiny surface)	Generally dull (except iodine and graphite)
Conductivity	Good conductors of heat and electricity	Poor conductors (except graphite)
Malleability/Ductility	Malleable and ductile	Brittle (when solid)
Density	Generally high density	Generally low density
Melting/Boiling point	Generally high	Generally low (except diamond, Si)
Ionization enthalpy	Low ionization enthalpy	High ionization enthalpy
Electronegativity	Low electronegativity	High electronegativity
Electron gain enthalpy	Low (less negative or positive)	High (more negative)
Valence electrons	1–3 valence electrons (generally)	4–8 valence electrons
Oxide nature	Form basic oxides	Form acidic oxides
Chemical behaviour	Act as reducing agents; lose electrons	Act as oxidizing agents; gain electrons
Position in periodic table	Left and centre of the periodic table	Upper right of the periodic table
Number	More than 78% of known elements	Less than 20 in number

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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.

(a) Element with five electrons in the outer subshell:

Five electrons in the outer subshell means the outer subshell is a p-subshell with 5 electrons (p⁵) or a d-subshell with 5 electrons (d⁵).

For p⁵: Configuration is $ns^2\,np^5$ → Group 17 (Halogens). Example: Cl ( $3s^2\,3p^5$ ) or F ( $2s^2\,2p^5$ ).
For d⁵: Example: Mn ( $3d^5\,4s^2$ ).

Most commonly, the answer refers to Group 17 elements (e.g., Cl, F, Br).

(b) Element that would tend to lose two electrons:

Elements that tend to lose two electrons have two valence electrons → Group 2 (Alkaline Earth Metals).

Example: Mg ( $1s^2\,2s^2\,2p^6\,3s^2$ ) loses 2 electrons to form Mg²⁺.

(c) Element that would tend to gain two electrons:

Elements that tend to gain two electrons have six valence electrons (need 2 more to complete the octet) → Group 16 (Chalcogens).

Example: O ( $1s^2\,2s^2\,2p^4$ ) gains 2 electrons to form O²⁻.

(d) Group having metal, non-metal, liquid as well as gas at room temperature:

Group 17 (Halogens):

F₂ — gas at room temperature
Cl₂ — gas at room temperature
Br₂ — liquid at room temperature
I₂ — solid non-metal at room temperature
At (Astatine) — considered to have metallic character

Thus, Group 17 contains elements that are gas (F₂, Cl₂), liquid (Br₂), solid non-metal (I₂), and the heaviest member (At) shows some metallic character.

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3.28The 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.

Group 1 Elements (Alkali Metals) — Reactivity increases down the group:

Group 1 elements react by losing electrons (acting as reducing agents). Their reactivity depends on how easily they can lose the valence electron.

As we go down Group 1 (Li → Cs), atomic size increases and ionization enthalpy decreases.
The outermost electron is farther from the nucleus and more shielded by inner electrons.
Therefore, it becomes progressively easier to lose the valence electron.
Hence, reactivity increases from Li to Cs: $\text{Li} < \text{Na} < \text{K} < \text{Rb} < \text{Cs}$ .

Group 17 Elements (Halogens) — Reactivity decreases down the group:

Group 17 elements react by gaining electrons (acting as oxidizing agents). Their reactivity depends on how easily they can gain an electron.

As we go down Group 17 (F → I), atomic size increases and electron gain enthalpy becomes less negative.
The incoming electron is added to a shell farther from the nucleus with more shielding.
Therefore, the tendency to gain an electron decreases down the group.
Hence, reactivity decreases from F to I: $\text{F} > \text{Cl} > \text{Br} > \text{I}$ .

In summary, the opposite trends arise because Group 1 reactivity is based on electron loss (favoured by low ionization enthalpy) while Group 17 reactivity is based on electron gain (favoured by high electron gain enthalpy/electronegativity).

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

Block	General Outer Electronic Configuration	Remarks
s-block	$ns^{1-2}$	n = 1 to 7; includes Groups 1 and 2
p-block	$ns^2\,np^{1-6}$	n = 2 to 6; includes Groups 13 to 18
d-block	$(n-1)d^{1-10}\,ns^{0-2}$	n = 4 to 7; includes Groups 3 to 12 (transition elements)
f-block	$(n-2)f^{1-14}\,(n-1)d^{0-1}\,ns^2$	n = 6 to 7; includes lanthanides and actinides (inner transition elements)

Note: In the d-block, the ns electrons may be 0, 1, or 2 due to special stability of half-filled and fully-filled d-orbitals (e.g., Cr: $3d^5\,4s^1$ ; Cu: $3d^{10}\,4s^1$ ).

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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)f⁷(n-1)d¹ns² for n=6, in the periodic table.

(i) $ns^2\,np^4$ for n = 3:

Configuration: $3s^2\,3p^4$

Full configuration: $1s^2\,2s^2\,2p^6\,3s^2\,3p^4$ → Z = 16 (Sulfur, S)

Period: n = 3 → Period 3
Group: The element has 6 valence electrons ( $3s^2\,3p^4$ ) → Group 16
Block: Last electron enters p-subshell → p-block

Position: Period 3, Group 16 (Sulfur, S)

(ii) $(n-1)d^2\,ns^2$ for n = 4:

Configuration: $3d^2\,4s^2$

Full configuration: $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^2\,4s^2$ → Z = 22 (Titanium, Ti)

Period: n = 4 → Period 4
Group: The element has 2 electrons in d-subshell + 2 in s = 4 valence electrons → Group 4
Block: Last electron enters d-subshell → d-block

Position: Period 4, Group 4 (Titanium, Ti)

(iii) $(n-2)f^7\,(n-1)d^1\,ns^2$ for n = 6:

Configuration: $4f^7\,5d^1\,6s^2$

Full configuration: $[\text{Xe}]\,4f^7\,5d^1\,6s^2$ → Z = 64 (Gadolinium, Gd)

Period: n = 6 → Period 6
Group: f-block element → f-block (lanthanides)
The element belongs to the f-block

Position: Period 6, f-block (Lanthanide series — Gadolinium, Gd, Z = 64)

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3.31The first (ΔᵢH₁) and the second (ΔᵢH₂) ionization enthalpies (in kJ mol⁻¹) and the (ΔeqH) electron gain enthalpy (in kJ mol⁻¹) of a few elements are given below: [Table with Elements I–VI and their values]. 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 MX₃ (X=halogen)?

Given data:

Element	ΔᵢH₁ (kJ mol⁻¹)	ΔᵢH₂ (kJ mol⁻¹)	ΔegH (kJ mol⁻¹)
I	520	7300	−60
II	419	3051	−48
III	1681	3374	−328
IV	1008	1846	−295
V	2372	5251	+48
VI	738	1451	−40

Analysis:

Element I: Very low ΔᵢH₁ (520), very high ΔᵢH₂ (7300) → loses 1 electron easily → alkali metal (likely Na, Group 1)
Element II: Lowest ΔᵢH₁ (419), moderately high ΔᵢH₂ → loses 1 electron very easily → most reactive alkali metal (likely K, Group 1)
Element III: Very high ΔᵢH₁ (1681), very negative ΔegH (−328) → gains electron readily → most reactive halogen (likely F or Cl, Group 17)
Element IV: High ΔᵢH₁ (1008), moderately high ΔᵢH₂, very negative ΔegH (−295) → could be a halogen or non-metal (likely Cl or similar, Group 17 but less reactive than III)
Element V: Highest ΔᵢH₁ (2372), positive ΔegH (+48) → neither loses nor gains electrons easily → noble gas (likely He or Ne)
Element VI: Moderate ΔᵢH₁ (738), moderate ΔᵢH₂ (1451), small negative ΔegH → loses 2 electrons → alkaline earth metal (likely Mg or Ca, Group 2)

Answers:

(a) Least reactive element:
Element V — It has the highest ionization enthalpy (2372 kJ mol⁻¹) and positive electron gain enthalpy (+48 kJ mol⁻¹), meaning it neither loses nor gains electrons. This is characteristic of a noble gas (e.g., He or Ne).

(b) Most reactive metal:
Element II — It has the lowest first ionization enthalpy (419 kJ mol⁻¹), meaning it loses an electron most easily. This is characteristic of the most reactive alkali metal (e.g., K or Cs).

(c) Most reactive non-metal:
Element III — It has a very high ionization enthalpy (1681 kJ mol⁻¹) and the most negative electron gain enthalpy (−328 kJ mol⁻¹), meaning it gains electrons most readily. This is characteristic of the most reactive halogen (e.g., F or Cl).

(d) Least reactive non-metal:
Element IV — It has a high ionization enthalpy (1008 kJ mol⁻¹) and a less negative electron gain enthalpy (−295 kJ mol⁻¹) compared to Element III. This suggests it is a less reactive non-metal/halogen (e.g., I or a heavier halogen).

(e) Metal forming stable binary halide MX₂:
Element VI — It has moderate ΔᵢH₁ (738) and ΔᵢH₂ (1451), both relatively low, indicating it can lose 2 electrons to form M²⁺. This is characteristic of an alkaline earth metal (e.g., Mg or Ca), which forms MX₂ halides.

(f) Metal forming predominantly stable covalent halide MX₃:
Element I — Wait, re-examining: Element I has ΔᵢH₁ = 520 (loses 1e easily) and ΔᵢH₂ = 7300 (very hard to lose 2nd electron). This is a Group 1 metal forming MX.

For MX₃, we need an element that loses 3 electrons. However, looking at the data again — none explicitly shows 3 low ionization enthalpies. Element VI with ΔᵢH₁ = 738 and ΔᵢH₂ = 1451 could be a Group 2 metal (MX₂).

Actually, for a covalent halide MX₃, the element should have moderate ionization enthalpies (not too low, not too high) and be able to form 3 bonds. Element IV with ΔᵢH₁ = 1008 and ΔᵢH₂ = 1846 could represent a Group 13 element (like Al or Ga) that forms AlX₃ — predominantly covalent halides.

Element IV is likely to be the metal forming predominantly stable covalent halide MX₃ (e.g., Al, forming AlCl₃).

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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 fluorine

(a) Lithium and oxygen:

Li is in Group 1 → valency = 1 (forms Li⁺)
O is in Group 16 → valency = 2 (forms O²⁻)
Formula: Li₂O (Lithium oxide)

(b) Magnesium and nitrogen:

Mg is in Group 2 → valency = 2 (forms Mg²⁺)
N is in Group 15 → valency = 3 (forms N³⁻)
Using cross-multiplication: Mg₃N₂
Formula: Mg₃N₂ (Magnesium nitride)

(c) Aluminium and iodine:

Al is in Group 13 → valency = 3 (forms Al³⁺)
I is in Group 17 → valency = 1 (forms I⁻)
Formula: AlI₃ (Aluminium iodide)

(d) Silicon and oxygen:

Si is in Group 14 → valency = 4
O is in Group 16 → valency = 2
Using cross-multiplication: Si₁O₂ → SiO₂
Formula: SiO₂ (Silicon dioxide)

(e) Phosphorus and fluorine:

P is in Group 15 → can show valency of 3 or 5
F is in Group 17 → valency = 1
P can form PF₃ (valency 3) or PF₅ (valency 5, using expanded octet)
Both are stable; the most stable binary compound is PF₅ (phosphorus pentafluoride) as F stabilises the higher oxidation state.
Formulas: PF₃ and PF₅

(f) Element 71 and fluorine:

Element 71 is Lutetium (Lu), a lanthanide in Group 3, Period 6.
Lu has the configuration $[\text{Xe}]\,4f^{14}\,5d^1\,6s^2$ → valency = 3 (forms Lu³⁺)
F is in Group 17 → valency = 1
Formula: LuF₃ (Lutetium(III) fluoride)

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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.

Correct answer: (c) principal quantum number

Justification: In the Modern Periodic Table, the period number corresponds to the highest principal quantum number (n) of the outermost shell of the elements in that period. For example, elements in Period 2 have their outermost electrons in the shell with n = 2, and elements in Period 3 have their outermost electrons in the shell with n = 3. Thus, the period number directly indicates the value of the principal quantum number of the valence shell.

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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.

Correct answer: (b)

Justification: Statement (b) is incorrect. The d-subshell has 5 orbitals and can accommodate a maximum of 10 electrons (not 8). Therefore, the d-block should have 10 columns (Groups 3 to 12), not 8. This is consistent with statement (c), which correctly states that each block contains columns equal to the number of electrons that can occupy that subshell:

s-block: 2 columns (2 electrons in s-subshell)
p-block: 6 columns (6 electrons in p-subshell)
d-block: 10 columns (10 electrons in d-subshell)
f-block: 14 columns (14 electrons in f-subshell)

Statements (a), (c), and (d) are all correct.

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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.

Correct answer: (c) Nuclear mass

Justification: The chemistry of an element is determined by its valence electrons. The factors that affect valence electrons are:

(a) Valence principal quantum number (n): Determines the energy level and size of the valence shell — directly affects valence electrons.
(b) Nuclear charge (Z): Determines the attractive force on electrons — directly affects valence electrons.
(d) Number of core electrons: Determines the shielding/screening effect, which affects the effective nuclear charge experienced by valence electrons.

Nuclear mass (option c) depends on the number of protons and neutrons. Neutrons do not carry charge and do not interact electromagnetically with electrons. Therefore, nuclear mass does not affect the valence electrons or the chemistry of the element. (Isotopes have the same chemistry despite different nuclear masses.)

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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.

Correct answer: (a) nuclear charge (Z)

Justification: F⁻, Ne, and Na⁺ are isoelectronic species — all have 10 electrons with the configuration $1s^2\,2s^2\,2p^6$ .

Since they have the same number of electrons in the same orbitals, the principal quantum number (n) is the same for all → option (b) does not differentiate them.
Electron–electron interactions are essentially the same since all have 10 electrons in the same configuration → option (c) does not differentiate them.
However, their nuclear charges differ: F⁻ (Z=9), Ne (Z=10), Na⁺ (Z=11).
Higher nuclear charge pulls the electron cloud closer, resulting in smaller size.
Therefore: Size order is F⁻ > Ne > Na⁺.

The size of these isoelectronic species is affected by nuclear charge (Z).

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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.

Correct answer: (d)

Justification: Statement (d) is incorrect. Removal of an electron from an orbital with a lower n value (inner shell, closer to nucleus) is harder, not easier, than from an orbital with a higher n value (outer shell, farther from nucleus). Electrons in inner shells (lower n) are closer to the nucleus, experience higher effective nuclear charge, and are more tightly held. Therefore, more energy is required to remove them.

The correct statement should be: "Removal of electron from orbitals bearing higher n value is easier than from orbitals having lower n value."

Statements (a), (b), and (c) are all correct:

(a) Each successive electron is removed from an increasingly positive ion, requiring more energy.
(b) After all valence electrons are removed, the next electron comes from the noble gas core, requiring a very large jump in energy.
(c) The big jump in successive ionization enthalpies marks the end of valence electrons.

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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 > B

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

Justification: Metallic character depends on the ease of losing electrons (low ionization enthalpy, large atomic size).

K (Group 1, Period 4): Largest atom, lowest ionization enthalpy → most metallic.
Mg (Group 2, Period 3): Smaller than K, higher ionization enthalpy → less metallic than K.
Al (Group 13, Period 3): Higher ionization enthalpy than Mg, smaller size → less metallic than Mg.
B (Group 13, Period 2): Smallest, highest ionization enthalpy among these four, has some non-metallic/metalloid character → least metallic.

Correct order: K > Mg > Al > B

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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 > B

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

Justification: Non-metallic character increases across a period (left to right) and decreases down a group. It is related to electronegativity and electron gain enthalpy.

F (Period 2, Group 17): Most electronegative element, highest non-metallic character.
N (Period 2, Group 15): High electronegativity, strong non-metallic character.
C (Period 2, Group 14): Moderate non-metallic character.
B (Period 2, Group 13): Borderline metalloid, less non-metallic than C.
Si (Period 3, Group 14): Metalloid, less non-metallic than B (being in a lower period).

Correct order: F > N > C > B > Si

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3.40Considering 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

Correct answer: (b) F > O > Cl > N

Justification: Oxidizing power is the ability to gain electrons. It increases with electronegativity and electron gain enthalpy.

F is the most electronegative element and the strongest oxidizing agent among all elements.
O has very high electronegativity (3.5 on Pauling scale) and is a stronger oxidizing agent than Cl in most reactions. Although Cl has a more negative electron gain enthalpy than O, the overall oxidizing power of O is higher due to its smaller size and higher electronegativity.
Cl is a strong oxidizing agent but less so than O.
N has lower electronegativity and electron gain enthalpy compared to the others, making it the weakest oxidizing agent among these four.

Correct order of oxidizing property: F > O > Cl > N

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