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

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

📘 Get NCERT Solutions for Class 11 Chemistry Chapter 3 – Classification of Elements and Periodicity in Properties
with step-by-step explanations, solved NCERT questions, and practice MCQs.
These solutions are prepared according to the latest CBSE syllabus and are useful for
Class 11 Board Exams, NEET, JEE and other competitive exams.
Important topics covered in this chapter include
Modern Periodic Law, Periodic Table, Atomic Radius, Ionization Enthalpy, Electron Gain Enthalpy and Electronegativity.
All concepts are explained in simple language so students can easily understand periodic trends in the periodic table. ✅

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

Showing all questions

Q.1: What is the
basic theme of organisation
in the
periodic table
?

Answer
✅ The
basic theme of organisation
in the periodic table is that elements having
similar physical and chemical properties
are placed together in the
same group (or column).

Explanation (Step by Step)
Step 1: Main purpose of the periodic table
🧩 Elements with
similar properties
are arranged in the same
group of the periodic table.

Step 2: Why do elements show similar properties?
👉 The
physical and chemical properties
of elements mainly depend on the
electronic configuration of the valence shell.

Step 3: What is common among elements of the same group?
✅ Same number of
valence electrons
✅ Similar
valence shell electronic configuration

Step 4: Conclusion
🔥 Because of this similarity in valence shell configuration, elements in the same group show
similar chemical behaviour,
such as forming similar oxides, having similar valency, and undergoing similar chemical reactions.

Did You Know?
👉 By looking at the
group number,
we can often predict the number of
valence electrons in an element.

Examples:
✅
Group 1
elements (Li, Na, K) have
1 valence electron,
so they easily form
+1 ions:
• Na → Na⁺
• K → K⁺

✅
Group 17
elements (F, Cl, Br) have
7 valence electrons,
so they tend to gain one electron to form
−1 ions:
• Cl → Cl⁻

✨ That is why elements in the same group often show
similar reactivity
and form similar types of compounds.

Q.2: Which important property did
Mendeleev
use to classify the elements in his
Periodic Table?
Did he stick to that?

Answer
✅ Mendeleev used
Atomic Mass (Atomic Weight)
as the
basis for classification
of elements.

✅ Yes, he mostly followed this rule.
But in some cases he gave priority to
chemical properties
instead of strictly following atomic mass.

Explanation (Step by Step)
👉 (i) Basis of classification
📌 Mendeleev arranged elements in order of
increasing atomic mass.

👉 (ii) Main objective
🎯 Elements with
similar physical and chemical properties
should come in the same
group
of the periodic table.

👉 (iii) Blank spaces
Mendeleev left
blank spaces
for
undiscovered elements.

👉 (iv) Prediction of elements
He predicted the
atomic masses
and
properties
of those elements before they were discovered.

👉 (v) Later discoveries
When those elements were discovered later, their properties were found to be
very similar to Mendeleev’s predictions.

Did You Know?
👉 Mendeleev
was so confident about his
Periodic Law
that he sometimes placed elements slightly out of strict
atomic mass order
so that elements with
similar properties
stayed in the correct group.

✨ This idea later proved to be
scientifically correct.

Q.3: What is the
basic difference
in approach between
Mendeleev’s Periodic Law
and the
Modern Periodic Law?

Answer
✅ The main difference is the
basis of classification.

👉 Mendeleev’s Periodic Law
is based on
Atomic Mass.
👉 Modern Periodic Law
is based on
Atomic Number (Z).

Explanation (Step by Step)
👉 (i) Mendeleev’s Periodic Law
According to Mendeleev, the properties of elements are a
periodic function
of their
atomic mass.
This means that when atomic mass increases, the properties of elements repeat periodically.

👉 (ii) Modern Periodic Law
According to the modern periodic law, the properties of elements are a
periodic function
of their
atomic number (Z).
When atomic number increases, the properties repeat in a regular pattern.

👉 (iii) Why Modern Law is more accurate
The
atomic number
determines the
electronic configuration
of an element, and electronic configuration controls the
chemical properties.

Did You Know?
👉 When the
Modern Periodic Law
was adopted, several problems of Mendeleev’s table were solved.

Example:
Ar–K anomaly
and
Co–Ni anomaly
were correctly explained using
atomic number.

Q.4: On the basis of
quantum numbers,
justify that the
sixth period
of the periodic table should have
32 elements.

Answer
✅ The
sixth period
contains
32 elements
because electrons are filled in the following subshells with their maximum capacities:

6s (2) + 4f (14) + 5d (10) + 6p (6) =
32

Explanation (Step by Step)
Step 1: A new period starts with a new shell
📌 In the long form of the periodic table, each period begins with a new
principal quantum number (n).
✅ The sixth period begins with
n = 6.

Step 2: Possible subshells for n = 6
For n = 6, the
azimuthal quantum number (ℓ)
can have values:
ℓ = 0, 1, 2, 3

These correspond to the following subshells:
👉 ℓ = 0 → 6s
👉 ℓ = 1 → 6p
👉 ℓ = 2 → 6d
👉 ℓ = 3 → 6f

Step 3: Subshells actually filled in the sixth period
⚡ According to the
Aufbau principle,
electrons fill orbitals in order of increasing energy.

👉 The
6d and 6f
subshells have higher energy than
7s,
so they do not fill immediately.
👉 In the sixth period,
4f and 5d
have lower energy than
6p,
so they fill first.

✅ Therefore the filling order is:

6s → 4f → 5d → 6p

Step 4: Maximum electron capacity
👉 s subshell = 2 electrons
👉 p subshell = 6 electrons
👉 d subshell = 10 electrons
👉 f subshell = 14 electrons

Therefore for the sixth period:
👉 6s = 2
👉 4f = 14
👉 5d = 10
👉 6p = 6

Total = 2 + 14 + 10 + 6 =
32

👉 Hence the
sixth period
contains
32 elements.

Did You Know?
💡 Only the
6th and 7th periods
of the periodic table contain
32 elements
because they include the filling of the
f-block elements
(lanthanides and actinides).

Q.5: In terms of
period
and
group,
where would you locate the element with
Z = 114?

Answer
✅ The element with
Z = 114
is
Flerovium (Fl).

✅ It is located in the
7th period
and
Group 14
of the periodic table.
✅ It belongs to the
p-block.

Explanation (Step by Step)
Step 1: Electronic configuration of Z = 114
[Rn] 5f¹⁴ 6d¹⁰ 7s² 7p²

Step 2: Determining the period
👉 The
highest principal quantum number (n)
present in the electronic configuration determines the period.
Here the highest value is
n = 7.
✅ Therefore the element belongs to the
7th period.

Step 3: Determining the group
Valence shell configuration =
7s² 7p²

👉 Total
valence electrons = 4.

For
p-block elements:
Group number = 10 + valence electrons

10 + 4 =
14

✅ Hence the element belongs to
Group 14.

Did You Know?
👉
Flerovium (Fl)
is a
superheavy element
discovered in
1998.

👉 It belongs to the
carbon family (Group 14),
which includes
C, Si, Ge, Sn, Pb.

Q.6: Write the
atomic number
of the element present in the
third period
and
seventeenth group
of the periodic table.

Answer
✅ The element present in the
third period
and
group 17
is
Chlorine (Cl).

✅ The
atomic number
of chlorine is
17.

✅ Chlorine belongs to the
p-block.

Explanation (Step by Step)
Step 1: Elements in the third period
The
third period
contains the following elements:

Na (11), Mg (12), Al (13), Si (14), P (15), S (16),
Cl (17),
Ar (18)

Step 2: Identify Group 17
👉
Group 17
elements are known as
Halogens.
Examples include:

F,
Cl,
Br, I, At

Step 3: Element common to both
👉 In the
third period,
the element belonging to
Group 17
is
Chlorine (Cl).

✅ Therefore its
atomic number
is
17.

Did You Know?
💡
Group 17 elements
are called
Halogens,
which means
“salt-forming elements.”

👉 For example,
Chlorine (Cl)
reacts with
Sodium (Na)
to form common salt:

NaCl (Sodium Chloride)

Q.7: Which element do you think would have been named by:
(i) Lawrence Berkeley Laboratory?
(ii) Seaborg’s group?

Answer
✅ (i) Elements named after the
Lawrence Berkeley Laboratory:
• Lawrencium (Lr), Z = 103
• Berkelium (Bk), Z = 97

✅ (ii) Element named after
Seaborg’s group:
• Seaborgium (Sg), Z = 106

Explanation (Step by Step)
Step 1: Basis of naming elements
🧪 Many
synthetic (man-made) elements
are named after
scientists,
places,
or
laboratories
where they were discovered.

Step 2: Elements related to Lawrence Berkeley Laboratory
✅ Lawrencium
— named in honour of
Ernest O. Lawrence.
✅ Berkelium
— named after the city
Berkeley in California.

Step 3: Element related to Seaborg’s group
✅ Seaborgium
was named in honour of
Glenn T. Seaborg,
a famous nuclear chemist who contributed greatly to the discovery of many transuranium elements.

Did You Know?
💡 Many
superheavy elements
are named after
scientists,
cities,
or
research laboratories.

This helps preserve the
history of scientific discoveries.

Q.8: Why do elements in the same
group
have similar
physical
and
chemical properties?

Answer
✅ Because elements in the same
group
have the same number of
electrons in their valence shell.
Therefore their
physical
and
chemical properties
are very similar.

Explanation (Step by Step)
Step 1: What do the properties of an element depend on?
🔍 The properties of an element mainly depend on its
outermost (valence) electrons.

Step 2: What is common in the same group?
✅ All elements of the same group have similar
valence shell electronic configuration.
👉 This means the
number of valence electrons
is the same.

Step 3: Effect of same valence electrons
👉 They usually have the same
valency.
👉 They show similar tendency to form
ions
(for example +1, +2, −1).
👉 They form similar types of
compounds.

Step 4: Conclusion
✅ Therefore elements of the same group show similar
chemical properties
and many
physical properties.

Did You Know?
💡 Elements of
Group 1
(Li, Na, K) all have
1 valence electron.

👉 Because of this they usually form
+1 ions
and react vigorously with water.

✅ The rule of
same valence electrons
is the main reason for the similarity of elements in a group.

Q.9: What do you understand by
Atomic Radius
and
Ionic Radius?

Answer
✅ Atomic Radius:
It is the measure of the size of an atom and is generally defined as the distance from the
centre of the nucleus
to the
outermost electron shell.

✅ Ionic Radius:
It is the measure of the size of an
ion,
defined as the distance from the nucleus of the ion to the effective boundary of its
electron cloud.

Explanation (Step by Step)
Step 1: What does Atomic Radius indicate?
👉 It represents the size of a
neutral atom.
✅ In simple terms: distance from the
nucleus
to the
outermost electron
is called
Atomic Radius.

Step 2: What is Ionic Radius?
👉 When an atom
loses
or
gains electrons,
it forms an ion and its size changes.
✅ The effective distance from the
nucleus
to the boundary of the
electron cloud
in an ion is called
Ionic Radius.

Step 3: Important rule to remember
👉 When a
cation (positive ion)
is formed, the size
decreases
because electrons are lost.
👉 When an
anion (negative ion)
is formed, the size
increases
because electrons are gained.

Did You Know?
💡 When a
Na atom
forms
Na⁺,
its radius becomes much smaller because:

✅
electron–electron repulsion
decreases.
✅ The
nuclear attraction
becomes stronger.

Q.10: How do atomic radius vary in a period and in a group? How do you explain the variation?

Answer
✅ In a
period,
atomic radius
decreases
from left to right.

✅ In a
group,
atomic radius
increases
from top to bottom.

Explanation (Step by Step)
(A) Why does atomic radius decrease across a period? (Left → Right)

Step 1: As we move across a period, the
atomic number (Z)
increases, which means the
nuclear charge
also increases.

Step 2: Electrons are added to the
same outer shell.

Step 3:
Shielding effect
does not increase much because electrons are added in the same shell.

Step 4: Therefore the
effective nuclear charge (Z_eff)
increases and the outer electrons are pulled
closer to the nucleus.
✅ Result: Atomic radius
decreases.
Example: In the 3rd period, Na has a larger radius while Cl has a smaller radius.

(B) Why does atomic radius increase down a group? (Top → Bottom)

Step 1: As we move down a group, a new
electron shell
is added for each element.

Step 2: The outer electrons move
farther from the nucleus.

Step 3: Inner electrons increase the
shielding effect,
which reduces the nuclear attraction on outer electrons.
✅ Result: Atomic radius
increases.
Example: In Group 1 → Li < Na < K < Rb < Cs (radius increases).

Did You Know?
👉
Noble gases
(He, Ne, Ar…) are often compared differently because they generally do not form
chemical bonds.

✅ Therefore, instead of covalent radius, their size is often measured using
Van der Waals radius.

Q.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
✅
Isoelectronic species
are atoms or ions that have the
same number of electrons
but different
nuclear charge (Z).

Examples:
(i) F⁻ →
Ne
(ii) Ar →
K⁺
(iii) Mg²⁺ →
O²⁻
(iv) Rb⁺ →
Kr

Explanation (Step by Step)
Idea: First find the
number of electrons,
then choose a species having the same number of electrons. ✅

(i) F⁻
Atomic number of F = 9
Gains 1 electron → 9 + 1 = 10 e⁻
✅ Therefore
Ne
(10 electrons) is isoelectronic.

(ii) Ar
Ar has 18 electrons.
✅ Therefore
K⁺
(19 − 1 = 18 electrons) is isoelectronic.

(iii) Mg²⁺
Atomic number of Mg = 12
Loses 2 electrons → 12 − 2 = 10 e⁻
✅ Therefore
O²⁻
(8 + 2 = 10 electrons) is isoelectronic.

(iv) Rb⁺
Atomic number of Rb = 37
Loses 1 electron → 37 − 1 = 36 e⁻
✅ Therefore
Kr
(36 electrons) is isoelectronic.

Did You Know?
👉 In an
isoelectronic series,
as the
nuclear charge (Z)
increases, the
atomic/ionic radius decreases.

Example (size decreases):
O²⁻ > F⁻ > Ne > Na⁺ > Mg²⁺

Q.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) Common feature: All these species have
the same number of electrons (10),
therefore they are
Isoelectronic species.

✅ (b) Increasing order of ionic radii:

Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻

Explanation (Step by Step)
(a) Why are they isoelectronic?
Let us calculate the number of electrons in each species:

👉 N³⁻ : 7 + 3 = 10
👉 O²⁻ : 8 + 2 = 10
👉 F⁻ : 9 + 1 = 10
👉 Na⁺ : 11 − 1 = 10
👉 Mg²⁺ : 12 − 2 = 10
👉 Al³⁺ : 13 − 3 = 10

✅ All species have
10 electrons,
therefore they are
isoelectronic.

(b) How to arrange ionic radii?
For
isoelectronic species,
the rule is:
👉 As the
atomic number (Z)
increases, the
nuclear attraction
increases and the
ionic radius decreases.

Atomic numbers:
N = 7, O = 8, F = 9, Na = 11, Mg = 12, Al = 13

✅ Largest Z →
smallest radius → Al³⁺
✅ Smallest Z →
largest radius → N³⁻

Therefore the increasing order of ionic radii is:

Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻

Did You Know?
👉 Because of the
Isoelectronic rule,
ions with the
same number of electrons
can still have very different sizes.

✅ Just remember the simple rule:
Higher nuclear charge (Z) →
stronger attraction →
smaller ionic radius.

Q.13: Explain why
cations
are smaller and
anions
are larger in radii than their
parent atoms.

Answer
✅ When a
cation
is formed, the atomic size
decreases,
therefore cations are
smaller
than their parent atoms.

✅ When an
anion
is formed, the atomic size
increases,
therefore anions have
larger radii
than their parent atoms.

Explanation (Step by Step)
(A) Why are cations smaller?

Step 1: A cation is formed when an atom
loses one or more electrons.

Step 2: Removal of electrons decreases
electron–electron repulsion.

Step 3: The same nuclear charge now acts on
fewer electrons,
so the effective nuclear charge
Z_eff
increases.

Step 4: The remaining electrons are pulled
closer to the nucleus.
✅ Conclusion: Radius
decreases,
so the cation becomes smaller.
Example: Na → Na⁺ (Na⁺ is smaller than Na)

(B) Why are anions larger?

Step 1: An anion is formed when an atom
gains one or more electrons.

Step 2: Addition of electrons increases
electron–electron repulsion.

Step 3: The nuclear attraction remains the same but has to act on
more electrons,
so the effective attraction per electron decreases.

Step 4: The electron cloud
expands,
increasing the atomic radius.
✅ Conclusion: Anions are
larger
than their parent atoms.
Example: Cl → Cl⁻ (Cl⁻ is larger than Cl)

Did You Know?
👉 For the same element the size relationship is always:
Anion >
Atom >
Cation

Example:
Cl⁻ > Cl > Cl⁺

Q.14: What is the significance of the terms — ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy?

Answer
✅ The term
“isolated gaseous atom”
is used so that the atom is not influenced by
other atoms or neighbouring particles,
ensuring the measured values are
true and independent.

✅ The term
“ground state”
is important because it represents the
most stable
and
lowest energy
state of an atom. If the atom were in an excited state, the values of ionization enthalpy or electron gain enthalpy would change.

Explanation (Step by Step)
1) Why “isolated gaseous atom”?

Step 1: The attraction between the nucleus and electrons in an atom can be affected by
other atoms or chemical bonds nearby.

Step 2: The value of
ionization enthalpy (IE)
and
electron gain enthalpy (EA)
depends on this nuclear attraction.

Step 3: Therefore, the atom is considered
isolated
in the definition.

Step 4: In the
gaseous state,
atoms are far apart from each other and can be treated as nearly isolated.
✅ Hence the term
isolated gaseous atom is used.

2) Why “ground state”?

Step 1: The
ground state
is the
lowest energy and most stable
state of an atom.

Step 2: If the atom is in an
excited state,
its energy is already higher.

Step 3: In that case, the energy required to remove an electron (IE) or the energy change during electron gain (EA) would be
different.
✅ Therefore, to maintain a
standard and comparable definition,
the atom is always considered in the
ground state.

Did You Know?
👉 That is why the definitions of
Ionization Enthalpy
and
Electron Gain Enthalpy
always include the phrase:

“isolated gaseous atom in the ground state.”

✅ This ensures that values of different elements can be compared under the
same standard conditions.

Q.15: The 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
J mol⁻¹.
(Hint: Use the mole concept.)

Answer
✅ Ionization enthalpy of atomic hydrogen =
1.313 × 10⁶ J mol⁻¹

≈
1313 kJ mol⁻¹

Explanation (Step by Step)
Step 1: Energy required for ionization (per atom)
Energy of electron in ground state:
E₁ = −2.18 × 10⁻¹⁸ J

After ionization the electron moves to
infinity,
where energy is:
E_∞ = 0

Ionization energy per atom:
ΔE = E_∞ − E₁
ΔE = 0 − (−2.18 × 10⁻¹⁸)
✅
ΔE = 2.18 × 10⁻¹⁸ J atom⁻¹

Step 2: Convert to J mol⁻¹ (Mole concept)
Number of particles in 1 mol =
Avogadro number
N_A = 6.022 × 10²³ mol⁻¹

Ionization enthalpy:
ΔH = (2.18 × 10⁻¹⁸) × (6.022 × 10²³)

= (2.18 × 6.022) × 10⁵
= 13.12796 × 10⁵
✅
ΔH = 1.312796 × 10⁶ J mol⁻¹

≈
1.313 × 10⁶ J mol⁻¹

Did You Know?
💡 The same value per atom is approximately
13.6 eV
,
which is known as the
first ionization energy of hydrogen
.

Q.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) Be has
higher ionization enthalpy
than B because Be has the electronic configuration
2s² (completely filled).
The 2s electron has greater
penetration,
so it is held more strongly by the nucleus.

✅ (ii) O has
lower ionization enthalpy
than N and F because in oxygen the 2p orbitals contain
paired electrons.
This increases
electron–electron repulsion,
making it easier to remove an electron.

Explanation (Step by Step)
(i) Why does Be have higher Δ_iH than B?

Step 1: Electronic configuration
👉 Be (Z = 4): 1s² 2s²
👉 B (Z = 5): 1s² 2s² 2p¹

Step 2: From which subshell is the electron removed?
👉 In Be, the electron is removed from the
2s orbital.
👉 In B, the electron is removed from the
2p orbital.

Step 3: Main reason (2s vs 2p)
✅ The 2s electron has greater
penetration
than a 2p electron.
✅ Therefore it is closer to the nucleus and
less shielded.
✅ Also Be has a
fully filled stable configuration
(2s²).

✅ Hence removing an electron from Be is more difficult →
Δ_iH(Be) > Δ_iH(B)

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

Step 1: Electronic configuration
👉 N (Z = 7): 1s² 2s² 2p³
(half-filled p orbitals, extra stability)

👉 O (Z = 8): 1s² 2s² 2p⁴
(one p orbital contains a pair of electrons)

Step 2: Effect of pairing in oxygen
👉 In O, two electrons occupy the same 2p orbital.
✅ This increases
electron–electron repulsion.
✅ Therefore removing one of the paired electrons becomes
easier.

✅ Hence
Δ_iH(O) < Δ_iH(N)

Comparison with F
👉 In F the
nuclear charge
and
effective nuclear charge (Z_eff)
are higher.
✅ Therefore removing an electron from F is harder than from O.
✅ Hence
Δ_iH(O) < Δ_iH(F)

Did You Know?
👉 In the second period there are two famous
exceptions
in the ionization enthalpy trend:

✅ Be > B
(2s vs 2p electron)
✅ N > O
(half-filled stability vs electron pairing repulsion)

✅ Remembering these two exceptions helps in understanding the entire trend easily.

Q.17: How would you explain the fact that the
first ionization enthalpy
of
sodium (Na)
is lower than that of
magnesium (Mg),
but its
second ionization enthalpy
is higher than that of magnesium?

Answer
✅ The
first ionization enthalpy
of Na is lower than that of Mg because Na has a lower nuclear charge and its
3s¹ electron
is more loosely held.

✅ However, after losing one electron,
Na⁺
attains a very stable
noble gas configuration.
Removing another electron from this stable configuration requires very high energy, so the
second ionization enthalpy
of Na becomes higher than that of Mg.

Explanation (Step by Step)
Step 1: Electronic configurations
👉 Na (Z = 11):
1s² 2s² 2p⁶ 3s¹
👉 Mg (Z = 12):
1s² 2s² 2p⁶ 3s²

(A) Why is the first ionization enthalpy of Na lower?
1️⃣ Sodium has only one electron in the
3s orbital,
which is relatively easy to remove.

2️⃣ Magnesium has
higher nuclear charge (+12)
compared to sodium (+11), so it holds its electrons more strongly.

✅ Therefore:
Δ_iH₁ (Na) < Δ_iH₁ (Mg)

(B) Why is the second ionization enthalpy of Na higher?
After removal of the first electron:
👉 Na → Na⁺ + e⁻
Na⁺ configuration:
1s² 2s² 2p⁶
(noble gas configuration like Ne)

👉 Mg → Mg⁺ + e⁻
Mg⁺ configuration:
1s² 2s² 2p⁶ 3s¹

Now the second electron removal:
• Removing an electron from
Na⁺
means breaking a stable noble gas configuration → requires very high energy.

• Removing an electron from
Mg⁺
still involves removing the outer 3s electron → comparatively easier.

✅ Therefore:
Δ_iH₂ (Na) > Δ_iH₂ (Mg)

Did You Know?
💡 A large
jump in ionization enthalpy
occurs when removing an electron would disturb a
noble gas configuration.

This is why the
second ionization energy of Na
is extremely high.

Q.18: What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?

Answer
✅ As we move
down a group,
the
ionization enthalpy decreases
because removing the outermost electron becomes
progressively easier.

Explanation (Step by Step)
1. Increase in nuclear charge (Z)
👉 As we move down a group, the number of
protons increases, so
Z increases.
👉 However, this factor alone does not dominate because other effects become more significant.

2. Increase in atomic size (addition of new shells)
👉 With each successive period, a new
electron shell is added.
👉 The outermost electron moves
farther from the nucleus.
👉 Greater distance reduces the
attractive force between the nucleus and the electron.

3. Increase in shielding effect
👉 Down the group, the number of
inner electrons increases.
👉 These inner electrons
shield the outer electron from the nuclear attraction.
👉 Thus the nucleus holds the outermost electron
less strongly.

4. Final understanding
✅ Although
nuclear charge increases, the combined effect of
larger atomic size and greater shielding
reduces the attraction between the nucleus and the outermost electron.
👉 Therefore the
ionization enthalpy decreases
down a group.

Did You Know?
👉 Because of this trend, the metals in
Group 1 (Li, Na, K, Rb, Cs)
become more
reactive
as we move down the group.

✅ This happens because losing the outer electron becomes
easier.

Q.19: The first
ionization enthalpy
values (in kJ mol⁻¹) of
Group 13 elements
are given below:

B	Al	Ga	In	Tl
801	577	579	558	589

Unit: kJ mol⁻¹ (First Ionization Enthalpy)

How would you explain this
deviation from the general trend?

Answer
✅ Normally in
Group 13,
the first ionization enthalpy
decreases
on moving down the group.

✅ However, the values for
Ga
and
Tl
are slightly
higher than expected.

✅ This occurs because
d and f electrons provide poor shielding,
which increases the
effective nuclear charge (Z_eff)
on the outer electron.

Explanation (Step by Step)
1) Expected general trend
Down the group:
👉
Atomic size increases
👉
Shielding effect increases
👉 The outer electron is farther from the nucleus.
✅ Therefore the
ionization enthalpy should decrease.
This trend is roughly observed from B → Al → In.

2) Why is IE of Ga slightly higher than Al?
👉 Al:
[Ne] 3s² 3p¹

👉 Ga:
[Ar] 3d¹⁰ 4s² 4p¹

The presence of
3d¹⁰ electrons
in Ga causes
poor shielding.
➜ This increases the
effective nuclear charge
on the outer electron.
✅ Hence Ga holds its outer electron more strongly and its IE becomes slightly higher than Al.

3) Why is IE of Tl higher than In?
👉 In:
[Kr] 4d¹⁰ 5s² 5p¹

👉 Tl:
[Xe] 4f¹⁴ 5d¹⁰ 6s² 6p¹

The presence of
4f¹⁴ and 5d¹⁰ electrons
causes very
poor shielding.
➜ This increases the
effective nuclear charge
on the outer electron.
✅ Therefore Tl has a slightly higher ionization enthalpy than In.

Did You Know?
👉
d and f electrons
are called
poor shielders.

Because of this, small
exceptions
appear in periodic trends, such as
Ga > Al and Tl > In.

Q.20: Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F
(ii) F or Cl

Answer
✅ (i)
F
has
more negative electron gain enthalpy.

✅ (ii)
Cl
has
more negative electron gain enthalpy.

Explanation (Step by Step)
(i) Why does F have more negative EA than O?
1. From O to F, the
atomic size decreases.
2. The
nuclear charge (Z)
also increases.
3. Therefore the attraction between the nucleus and the incoming electron becomes stronger.
✅ Result:
Fluorine (F)
accepts an electron more easily → electron gain enthalpy is more negative.

(ii) Why does Cl have more negative EA than F?
This is a
famous exception.

1.
Fluorine
has a very small size and its outer
2p orbital is very compact.
2. When an extra electron enters F, strong
electron–electron repulsion
occurs in this small orbital.
3.
Chlorine
is larger and its outer
3p orbital
has more space, so repulsion is less.

✅ Therefore adding an electron to
Cl
is easier and more energy is released.
➜ Hence the electron gain enthalpy of
Cl
is more negative.

Did You Know?
👉 In
Halogens,
the general trend is that electron gain enthalpy becomes
less negative
down the group.

✅ However, between
F and Cl,
chlorine shows a
more negative value
because the very small size of F causes stronger
electron–electron repulsion.

Q.21: Would you expect the
second electron gain enthalpy
of
oxygen (O)
to be positive, more negative, or less negative than the
first electron gain enthalpy?
Justify your answer.

Answer
✅ The
second electron gain enthalpy
of oxygen is
positive.

This means it is not just “less negative” than the first electron gain enthalpy, but actually
energy must be supplied
to add the second electron.

Explanation (Step by Step)
Step 1: First electron gain enthalpy
When a neutral oxygen atom gains one electron:
O(g) + e⁻ → O⁻(g)

Energy is released, so the value is
negative.
Example:
ΔegH₁ ≈ −141 kJ mol⁻¹

Step 2: Second electron gain enthalpy
Now oxygen is already a negatively charged ion:
O⁻
Adding another electron:
O⁻(g) + e⁻ → O²⁻(g)

The incoming electron experiences strong
electrostatic repulsion
from the already negatively charged ion.

Therefore
energy must be absorbed
to add the second electron.

✅ Hence the value becomes
positive
.
Example value:
ΔegH₂ ≈ +780 kJ mol⁻¹

Did You Know?
👉 This is why the
oxide ion (O²⁻)
does not form easily in the
gaseous state
.

In reality it becomes stable in
ionic solids
such as
MgO and CaO, where the
lattice energy
stabilizes the ion.

Q.22: What is the basic difference between the terms
electron gain enthalpy
and
electronegativity
?

Answer
✅
Electron Gain Enthalpy
:
It refers to the
energy change
when an
isolated gaseous atom
gains an additional electron.

✅
Electronegativity
:
It is the tendency of an atom in a
covalent bond
to attract the
shared electrons
towards itself.

Explanation (Step by Step)
1) What does Electron Gain Enthalpy represent?

👉 It applies to an
isolated gaseous atom
.
👉 The process can be written as:
X(g) + e⁻ → X⁻(g)

👉 Energy may be released
(negative value)
or absorbed
(positive value).
✅ Thus it is a
measurable energy change
expressed in kJ mol⁻¹.

2) What does Electronegativity represent?

👉 It is a property observed
within a chemical bond
.
👉 When two atoms share electrons, electronegativity indicates which atom pulls the electrons more strongly.
✅ It is a
relative scale
(for example the
Pauling scale
)
rather than a direct enthalpy value.

3) Simple way to remember

👉 Electron Gain Enthalpy =
“accepting an electron”
(isolated atom in gaseous state).
👉 Electronegativity =
“pulling shared electrons”
inside a chemical bond.

Did You Know?
👉
Noble gases
(Ne, Ar etc.) usually do not have electronegativity values because they rarely form
covalent bonds
.

However, the concept of
electron gain enthalpy
can still be discussed for them, and it is often
positive or less favorable
.

Q.23: “The
electronegativity
of nitrogen (N) on the
Pauling scale
is
3.0
in all nitrogen compounds.”
How would you react to this statement?

Answer
✅ This statement is
not completely correct
.

The electronegativity value
3.0
for nitrogen is only an
average value
on the Pauling scale.
In different compounds, the actual electronegativity may vary depending on
hybridisation
and
oxidation state
.

Explanation (Step by Step)
1️⃣ On the
Pauling scale
,
the electronegativity of nitrogen is approximately
3.0
.
However, this value represents a
general or average value
.

2️⃣ In different compounds, the environment around nitrogen changes:
👉 The
oxidation state
of nitrogen may change.
👉 The
hybridisation
of nitrogen may change.
👉 Therefore the ability of nitrogen to
attract electrons
also changes slightly.

3️⃣ Hence it is not correct to say that nitrogen always has exactly
3.0
electronegativity in every compound.

Example:
In
NO and
NO₂,
nitrogen has different
oxidation states
,
so its electron-attracting behavior changes slightly.

Did You Know?
👉 Electronegativity is often taught as a
fixed number
,
but in reality it depends on the
bonding environment
of the atom.

That is why slightly different values may appear in different chemical situations.

Q.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, it forms an
anion
and its radius
increases.

✅ (b) When an atom
loses an electron, it forms a
cation
and its radius
decreases.

Explanation (Step by Step)
(a) When an atom gains an electron (Why does radius increase in an anion?)

1. The atom gains one or more electrons and forms an
anion.
2. The
nuclear charge remains the same, but the number of electrons in the
valence shell increases.
3. Increased electrons lead to stronger
electron–electron repulsion and shielding.
4. As a result, the effective nuclear attraction
(Z_eff)
experienced by each electron becomes less effective.
✅ Result: The electron cloud
expands → the
atomic radius increases.
Example:
Cl → Cl⁻
(Cl⁻ is larger)

(b) When an atom loses an electron (Why does radius decrease in a cation?)

1. The atom loses one or more electrons and forms a
cation.
2. The
electron–electron repulsion decreases because there are fewer electrons.
3. The nuclear charge remains the same but now attracts fewer electrons, so the
effective nuclear charge (Z_eff) increases.
4. Sometimes the entire outer shell may be removed, causing an even greater reduction in size.
✅ Result: Electrons move
closer to the nucleus → the
atomic radius decreases.
Example:
Na → Na⁺
(Na⁺ is smaller)

Did You Know?
👉 For the same element, the size rule is always:

✅
Anion
>
Atom
>
Cation

Example:
Cl⁻
> Cl >
Cl⁺

Q.25: Would you expect the
first ionization enthalpy
for two
isotopes
of the same element to be the
same or different
?
Justify your answer.

Answer
✅ The
first ionization enthalpy
of two isotopes of the same element is generally
the same
.

Explanation (Step by Step)
1️⃣ In isotopes, the
atomic number (Z)
is the same.
👉 This means the number of
protons
in the nucleus is identical.

2️⃣ Therefore isotopes have the same:

👉
number of electrons
(in neutral atoms)
👉
nuclear charge

👉
electronic configuration

3️⃣ Ionization enthalpy mainly depends on the
attraction between the nucleus and the outer electron
and the
electronic configuration
.

✅ Since these factors are the same for isotopes, their
first ionization enthalpy is also nearly the same
.

Did You Know?
👉 The main difference between isotopes is their
mass number
.

Because of this, some
physical properties
(such as diffusion rate or density) may differ,
but most
chemical properties
and
ionization enthalpy
remain almost the same.

Q.26: What are the major differences between
metals
and
non-metals
?

Answer
✅
Metals generally
lose electrons
and form
cations
.

✅
Non-metals generally
gain electrons
and form
anions
.

Explanation (Step by Step)
1) Tendency to lose or gain electrons

• Metals are
electropositive in nature.
👉 They lose one or more valence electrons:
M → M⁺ + e⁻

• Non-metals are
electronegative in nature.
👉 They gain electrons:
X + e⁻ → X⁻

2) Redox nature

• Metals lose electrons and therefore usually act as
reducing agents
.
• Non-metals gain electrons and therefore usually act as
oxidising agents
.

3) Energy and property trends

• Metals usually have:
✅
low ionization enthalpy

✅
low electron gain enthalpy
(less negative or sometimes positive)
✅
low electronegativity

• Non-metals usually have:
✅
high ionization enthalpy

✅
more negative electron gain enthalpy

✅
high electronegativity

4) Nature of oxides

• Metals generally form
basic oxides
.
Example:
Na₂O, CaO

• Non-metals generally form
acidic oxides
.
Example:
CO₂, SO₃

Did You Know?
👉 Some elements such as
Al and Zn
form
amphoteric oxides
,
which can behave both as
acids
and
bases
under different conditions.

Q.27: Use the
periodic table
to answer the following questions:
(a) Identify an element with
five electrons in the outer subshell
.
(b) Identify an element that would tend to
lose two electrons
.
(c) Identify an element that would tend to
gain two electrons
.
(d) Identify the group that contains
solid, liquid and gaseous elements
at room temperature.

Answer
✅ (a)
Fluorine (F)

✅ (b)
Magnesium (Mg)

✅ (c)
Oxygen (O)

✅ (d)
Group 17 (Halogens)

Explanation (Step by Step)
(a) Element with five electrons in outer subshell
• This corresponds to the configuration
np⁵
which belongs to the
halogen group
.
Example:
Fluorine configuration:
1s² 2s² 2p⁵

✅ Therefore the element is
Fluorine (F)

(b) Element that tends to lose two electrons
Elements of
Group 2
easily lose two electrons to form
M²⁺ ions
.
Magnesium configuration:
1s² 2s² 2p⁶ 3s²

Reaction:
Mg → Mg²⁺ + 2e⁻
✅ Therefore the element is
Magnesium (Mg)

(c) Element that tends to gain two electrons
Elements of
Group 16 (chalcogens)
tend to gain two electrons to complete the
octet
.
Oxygen configuration:
1s² 2s² 2p⁴

Reaction:
O + 2e⁻ → O²⁻
✅ Therefore the element is
Oxygen (O)

(d) Group containing solid, liquid and gas at room temperature
The
halogen group (Group 17)
shows different physical states:

• Gas:
F₂, Cl₂
• Liquid:
Br₂
• Solid:
I₂
✅ Therefore the required group is
Group 17 (Halogens).

Did You Know?
👉
Group 17
is one of the rare groups where elements exist in all three states at room temperature:

Gas: F₂, Cl₂

Liquid: Br₂

Solid: I₂

Q.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
✅ In
Group 1 (Alkali metals),
reactivity
increases
down the group, therefore:
Li < Na < K < Rb < Cs

✅ In
Group 17 (Halogens),
reactivity
decreases
down the group, therefore:
F > Cl > Br > I

Explanation (Step by Step)
(A) Why does reactivity increase down Group 1?

1. Group 1 elements have
one valence electron and easily lose it to form
M⁺.
2. Their reactivity depends on how easily this electron can be removed.
3. Down the group:
👉
atomic size increases

👉
shielding effect increases

👉 the outer electron becomes farther from the nucleus and loosely held.
✅ Therefore
ionization enthalpy decreases
→ electron removal becomes easier →
reactivity increases.

Hence:
Li < Na < K < Rb < Cs

(B) Why does reactivity decrease down Group 17?

1. Group 17 elements have
seven valence electrons and tend to gain one electron to form
X⁻.
2. Their reactivity depends on their ability to
attract an extra electron
(oxidising power).
3. Down the group:
👉
atomic size increases

👉 the incoming electron is added farther from the nucleus.
👉 the
nuclear attraction decreases
.
✅ Therefore the tendency to gain electrons decreases →
reactivity decreases.

Hence:
F > Cl > Br > I

Did You Know?
👉 Simple trick to remember:

👉 Metals react by
“losing electrons”
→ down the group it becomes easier →
reactivity increases ✅

👉 Halogens react by
“gaining electrons”
→ down the group it becomes harder →
reactivity decreases ✅

Q.29: Write the
general outer electronic configuration
of
s-, p-, d- and f-block elements
.

Answer
✅
s-block
:
ns^1–2

✅
p-block
:
ns² np^1–6

✅
d-block (transition elements)
:
(n−1)d^1–10 ns^0–2

✅
f-block (inner transition elements)
:
(n−2)f^1–14 (n−1)d^0–1 ns²

Explanation (Step by Step)
1️⃣ The term
block
in the periodic table refers to the type of
subshell being filled
by the last electron.

2️⃣ Therefore:

• In the
s-block, the
s-subshell is filled →
ns^1–2

• In the
p-block, the
p-subshell is filled →
ns² np^1–6

• In the
d-block, the
(n−1)d subshell is filled →
(n−1)d^1–10 ns^0–2

• In the
f-block, the
(n−2)f subshell is filled →
(n−2)f^1–14 (n−1)d^0–1 ns²

Did You Know?
👉 In
d-block elements
,
the
ns electrons
are often removed first during oxidation, even though they appear outside in the configuration.

This behavior makes the chemistry of
transition elements
particularly interesting.

Q.30: Assign the position of the element having outer electronic configuration
(i)
ns² np⁴
for
n = 3
(ii)
(n−1)d² ns²
for
n = 4
(iii)
(n−2)f⁷ (n−1)d¹ ns²
for
n = 6, in the
periodic table
.

Answer
✅ (i)
Period 3,
Group 16,
p-block
(Element:
Sulfur – S)

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

✅ (iii)
Period 6,
f-block (Lanthanide),
under
Group 3
(Element:
Gadolinium – Gd)

Explanation (Step by Step)
(i) ns² np⁴, n = 3

1. n = 3 →
3rd period
2. Configuration:
3s² 3p⁴
→ filling of p-subshell →
p-block
3. For p-block elements:
Group =
10 + valence electrons
Valence electrons = 2 + 4 =
6
Group = 10 + 6 =
16
✅ Position:
Period 3,
Group 16,
p-block
Element:
Sulfur (S)

(ii) (n−1)d² ns², n = 4

1. n = 4 →
4th period
2. (n−1)d = 3d → configuration:
3d² 4s²
→ filling of d-orbital →
d-block
3. For d-block elements:
Group =
(d electrons + s electrons)
= 2 + 2 =
4
✅ Position:
Period 4,
Group 4,
d-block
Element:
Titanium (Ti)

(iii) (n−2)f⁷ (n−1)d¹ ns², n = 6

1. n = 6 →
6th period
2. (n−2)f = 4f →
4f⁷ → filling of f-orbital →
f-block (Lanthanides)
3. Remaining configuration:
4f⁷ 5d¹ 6s²

✅ Position:
Period 6,
f-block (Lanthanides),
under
Group 3
Element:
Gadolinium (Gd)

Did You Know?
👉
f-block elements
are shown separately at the bottom of the periodic table, but they actually belong to the
6th and
7th periods
(lanthanoids and actinoids).

Q.31: The
first (Δ_iH₁) and second (Δ_iH₂) ionization enthalpies
and the
electron gain enthalpy (Δ_egH)
of some elements are given below (kJ mol⁻¹):

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

Identify 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
MX₂
(f) the metal which can form a predominantly stable covalent halide
MX.

Answer
✅ (a) Least reactive element =
V

✅ (b) Most reactive metal =
II

✅ (c) Most reactive non-metal =
III

✅ (d) Least reactive non-metal =
IV

✅ (e) Metal forming stable
MX₂ =
VI

✅ (f) Metal forming covalent halide
MX =
I

Explanation (Step by Step)
(a) Least reactive element
Element V has a
very high ionization enthalpy (2372)
and a
positive electron gain enthalpy.
This means it neither loses nor gains electrons easily.
Hence it behaves like a
noble gas and is the least reactive element.

(b) Most reactive metal
Metals are more reactive when their
first ionization enthalpy is lowest.
Element II has the lowest value
419 kJ mol⁻¹.
Hence it easily loses an electron and behaves like an
alkali metal.

(c) Most reactive non-metal
Non-metals are more reactive when their
electron gain enthalpy is highly negative.
Element III has the most negative value
−328 kJ mol⁻¹.
Therefore it behaves like a
halogen.

(d) Least reactive non-metal
Element IV has a negative electron gain enthalpy but less negative than III.
Hence its tendency to gain electrons is weaker.
Therefore IV is the
least reactive non-metal.

(e) Metal forming MX₂
Metals forming
M²⁺ ions generally have moderate first and second ionization enthalpies.
Element VI has values
738 and 1451,
so it can lose two electrons and form a stable
MX₂ compound like a
Group 2 metal.

(f) Metal forming covalent MX
Element I shows a
very large jump between ΔH₁ and ΔH₂.
This means only one electron is easily removed.
Therefore it behaves like a
Group 1 metal and forms MX type halides.

Did You Know?
👉 A very large
jump in ionization enthalpy
usually indicates that the atom has achieved a
stable noble gas configuration
after losing one electron.

This “jump rule” is a quick trick to identify
alkali metals and alkaline earth metals in such questions.

Q.32: Predict the formulas of the
stable binary compounds
that would be formed by the combination of the following pairs of elements:
(a)
Lithium
and
oxygen
(b)
Magnesium
and
nitrogen
(c)
Aluminium
and
iodine
(d)
Silicon
and
oxygen
(e)
Phosphorus
and
fluorine
(f)
Element 71
and
fluorine

Answer
✅ (a)
Li₂O
✅ (b)
Mg₃N₂
✅ (c)
AlI₃
✅ (d)
SiO₂
✅ (e)
PF₃
or
PF₅
✅ (f)
LuF₃

Explanation (Step by Step)
(a) Li and O → Li₂O
• Li (Group 1) valency =
1
→
Li⁺
• O (Group 16) valency =
2
→
O²⁻
✅ By cross method →
Li₂O

(b) Mg and N → Mg₃N₂
• Mg (Group 2) valency =
2
→
Mg²⁺
• N (Group 15) valency =
3
→
N³⁻
✅ Cross method →
Mg₃N₂

(d) Si and O → SiO₂
• Si valency =
4
• O valency =
2
✅ Formula →
SiO₂

(e) P and F → PF₃ or PF₅
• P (Group 15) valency =
3 or 5
• F valency =
1
✅ Therefore stable compounds:
PF₃
and
PF₅

(f) Element 71 and F → LuF₃
• Element with atomic number
71 =
Lutetium (Lu)
• Lu valency =
3
• F valency =
1
✅ Formula →
LuF₃

Did You Know?
👉 The fastest way to solve such questions is:
Find the group → determine
valency/charge → apply the
cross method
to obtain the compound formula.

Q.33: In the
modern periodic table
, the
period
indicates the value of:
(a) Atomic number
(b) Atomic mass
(c) Principal quantum number
(d) Azimuthal quantum number

Answer
✅ (c)
Principal Quantum Number
(
n
)

Explanation (Step by Step)
1️⃣ In the
modern periodic table
, each
period
begins with the filling of a new
electron shell
.

2️⃣ The identity of an electron shell is determined by the
principal quantum number (n)
.

3️⃣ Therefore, the
period number
tells us the value of
n
for the outermost occupied shell.

Example:
• Elements in the
2nd period have outer shell
n = 2.
• Elements in the
3rd period have outer shell
n = 3.

✅ Hence, the
period
represents the
principal quantum number (n)
.

Did You Know?
👉 A quick shortcut to determine the
period of an element
:

Look at the
highest value of n
in its
electronic configuration
.

That value directly gives the
period number
.

Q.34: Which of the following statements related to the
modern periodic table
is
incorrect
?
(a) The
p-block
has
6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
(b) The
d-block
has
8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates the value of the
azimuthal quantum number (ℓ)
for the last subshell that received electrons while building the electronic configuration.

Answer
✅ (b)
This statement is incorrect.

Explanation (Step by Step)
1️⃣
p-subshell
Maximum electrons =
6
(3 orbitals × 2 electrons each).
✅ Therefore statement
(a) is correct, and the p-block has
6 columns.

2️⃣
d-subshell
Maximum electrons =
10
(5 orbitals × 2 electrons each).
👉 Therefore the d-block has
10 columns, not 8.
✅ Hence statement
(b) is incorrect.

3️⃣ Statement (c) is also correct because the number of columns in a block equals the electron capacity of that subshell:
•
s-block →
2
•
p-block →
6
•
d-block →
10
•
f-block →
14

4️⃣ Statement (d) is also correct because the block (s, p, d, f) corresponds to the value of the
azimuthal quantum number (ℓ)
of the subshell that receives the last electron.

Did You Know?
👉 The
d-block
has
10 columns because the d-subshell contains
5 orbitals,
and each orbital can hold
2 electrons.

5 × 2 = 10

Q.35: Anything that influences the
valence electrons
will affect the
chemical properties
of an element.
Which one of the following factors does
NOT affect the valence shell
?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons

Answer
✅ (c)
Nuclear mass
does not affect the
valence shell
.

Explanation (Step by Step)
1️⃣ The
valence shell
is related to the
arrangement of electrons
and their attraction toward the nucleus.

2️⃣ The following factors influence the valence shell:

•
Principal quantum number (n)
determines which shell is the outermost.

•
Nuclear charge (Z)
controls the attraction between nucleus and electrons
(effective nuclear charge
Z_eff).

•
Core electrons
cause the
shielding effect
, which modifies the attraction felt by valence electrons.

3️⃣ However, the
nuclear mass
does not change the
electronic configuration
or the arrangement of valence electrons.

✅ Therefore,
nuclear mass
does not influence the valence shell.

Did You Know?
👉 This is the reason why
isotopes
of the same element show almost identical
chemical properties
.

Their
atomic number
and
electronic configuration
remain the same, while only the
mass
changes.

Q.36: The size of the
isoelectronic species
F⁻, Ne and Na⁺
is affected by which of the following factors?
(a)
Nuclear charge (Z)
(b)
Valence principal quantum number (n)
(c)
Electron–electron interaction in the outer orbitals
(d) None of the factors because their size is the same.

Answer
✅ (a)
Nuclear charge (Z)

Explanation (Step by Step)
1️⃣ The species
F⁻, Ne and Na⁺
all contain
10 electrons,
therefore they are
isoelectronic species.

2️⃣ In an
isoelectronic series,
the number of electrons and the principal quantum number
(n)
remain almost the same, and electron–electron repulsion is similar.

3️⃣ The main factor that changes is the
nuclear charge (Z).
As
Z increases,
the attraction between the nucleus and electrons becomes stronger and the electron cloud is pulled closer.
✅ Therefore the
atomic radius decreases.

Nuclear charges:
• F →
Z = 9 → F⁻
• Ne →
Z = 10
• Na →
Z = 11 → Na⁺

✅ Order of size (radius):
F⁻ > Ne > Na⁺

Did You Know?
👉 Quick rule for an
isoelectronic series:

✅
Higher nuclear charge (Z)
→
smaller radius

This simple rule helps solve such questions in
just a few seconds.

Q.37: Which one of the following statements is
incorrect
in relation to
ionization enthalpy
?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from
core noble gas configuration
.
(c) End of
valence electrons
is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals having lower
n value
is easier than from orbitals having higher n value.

Answer
❌ Incorrect statement =
(d)

Explanation (Step by Step)
(a) Correct
Each successive electron is removed from an increasingly
positively charged ion
,
so the attraction between nucleus and electrons increases.
Therefore the
ionization enthalpy increases.

(b) Correct
When all
valence electrons
are removed, the atom attains a
noble gas configuration
.
Removing an electron from this stable core requires a very large amount of energy.

(c) Correct
A
big jump in ionization enthalpy
indicates that all valence electrons have been removed and the next electron must be removed from the
inner shell
.

(d) Incorrect
Electrons in orbitals with
higher n value
are farther from the nucleus and experience more
shielding
.
Therefore they are
easier to remove
.
❌ Hence statement (d) is incorrect.

Did You Know?
👉 The
largest jump
in successive ionization energies helps identify the number of
valence electrons
of an element.

Example:
If the big jump occurs after removing two electrons, the element belongs to
Group 2
.

Q.38: Considering the elements
B,
Al,
Mg, and
K,
the correct order of their
metallic character
is:
(a) B > Al > Mg > K
(b) Al > Mg > B > K
(c) Mg > Al > K > B
(d) K > Mg > Al > B

Answer
✅ (d)
K > Mg > Al > B

Explanation (Step by Step)
1️⃣ What does metallic character mean?
👉 Metallic character refers to the tendency of an element to
lose electrons easily.
👉 In other words, it shows the
electropositive nature of an element.

2️⃣ Periodic trend of metallic character
• In a
group, metallic character
increases
from top to bottom.
• In a
period, metallic character
decreases
from left to right.

3️⃣ Comparing the given elements
•
K (Potassium)
→ Group 1, Period 4
→ very low ionization enthalpy
→
most metallic.

•
Mg (Magnesium)
→ Group 2, Period 3
→ metallic but less than K.

•
Al (Aluminium)
→ Group 13, Period 3
→ more towards the right side of the period
→ metallic character decreases.

•
B (Boron)
→ Period 2 element
→ behaves like a
metalloid
→
least metallic.

✅ Therefore the correct order is:
K > Mg > Al > B

Did You Know?
👉 A quick rule to identify metallic character:

✅
Lower ionization enthalpy
→
higher metallic character.

That is why
Potassium (K)
is the
most metallic
among the given elements.

Q.39: Considering the elements
B, C, N, F and Si
, the correct order of their
non-metallic character
is:
(a) B > C > Si > N > F
(b) Si > C > B > N > F
(c) F > N > C > B > Si
(d) F > N > C > Si > B

Answer
✅ (c)
F > N > C > B > Si

Explanation (Step by Step)
1️⃣
Non-metallic character
refers to the tendency of an element to
attract or gain electrons
.
It is closely related to
electronegativity
.

2️⃣ General periodic trends:

• Across a period (left → right),
non-metallic character increases
.

• Down a group (top → bottom),
non-metallic character decreases
.

3️⃣ Applying these trends:

•
Fluorine (F) is the most electronegative element → highest non-metallic character.

•
Nitrogen (N) lies to the right of carbon → N > C.

•
Boron (B) lies to the left of carbon → C > B.

•
Silicon (Si) lies below carbon in Group 14 → non-metallic character decreases → C > Si.

✅ Therefore the correct order is
F > N > C > B > Si
.

Did You Know?
👉
Silicon (Si)
is a
metalloid
.

This means it shows properties of both
metals
and
non-metals
, which is why its non-metallic character is lower than carbon but higher than many metals.

Q.40: Considering the elements
F,
Cl,
O and
N,
the correct order of their chemical reactivity in terms of
oxidizing property
is:
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) Cl > F > O > N
(d) O > F > N > Cl

Answer
✅ (b)
F > O > Cl > N

Explanation (Step by Step)
1️⃣ Meaning of oxidising property
👉 An
oxidising agent
is a substance that
gains electrons
or removes electrons from another substance.

2️⃣ Periodic trend of oxidising power
• Across a
period, oxidising power generally
increases from left to right.
• Down a
group, oxidising power generally
decreases.

3️⃣ Comparing the given elements
•
Fluorine (F)
→ highest electronegativity
→
strongest oxidising agent.

•
Oxygen (O)
→ very high electron affinity and electronegativity
→ strong oxidising nature.

•
Chlorine (Cl)
→ strong oxidising halogen but slightly weaker than oxygen in this comparison.

•
Nitrogen (N)
→ relatively lower tendency to gain electrons
→
least oxidising among these.

✅ Therefore the correct order is:
F > O > Cl > N

Did You Know?
👉
Fluorine (F)
is the
most powerful oxidising agent
among all elements because of its
very high electronegativity
and strong tendency to gain electrons.

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

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