Periodic Classification JEE Notes 2026 | Formulas, PYQs & Trends for JEE Main & Advanced

1

Genesis of Periodic Classification

The need to classify elements arose because more than 100 elements were discovered by the 19th century, making it impossible to study each individually. Classification helps in studying the properties of elements in a systematic manner. This chapter traces the historical development of the periodic table from Dobereiner to Moseley.

1.1 Dobereiner's Triads (1817)

Johann Wolfgang Dobereiner, a German chemist, observed that when elements with similar properties were arranged in groups of three (triads) in order of increasing atomic masses, the atomic mass of the middle element was roughly the arithmetic mean of the other two.

Examples of Dobereiner's Triads

Triad	Element 1	Element 2 (Mean)	Element 3
Alkali metals	Li (6.9)	Na (23.0)	K (39.1)
Alkaline earth	Ca (40.1)	Sr (87.6)	Ba (137.3)
Halogens	Cl (35.5)	Br (79.9)	I (126.9)

Limitation: Only a few elements could be arranged in triads; the law was not universal.

1.2 Newlands' Law of Octaves (1865)

John Alexander Newlands, an English chemist, arranged the elements in order of increasing atomic masses and observed that every eighth element had properties similar to the first, like musical notes.

Important for JEE

Newlands' Law of Octaves was applicable only up to calcium (Z=20). It failed for heavier elements and also placed dissimilar elements together (e.g., Co and Ni with halogens). It was the first attempt to show periodicity.

1.3 Mendeleev's Periodic Table (1869)

Dmitri Ivanovich Mendeleev, a Russian chemist, stated the Periodic Law: "The physical and chemical properties of elements are a periodic function of their atomic masses."

Key Features of Mendeleev's Table

Elements arranged in increasing order of atomic masses
7 periods (horizontal rows) and 8 groups (vertical columns)
Groups I-VII further divided into subgroups A and B
Group VIII contained 3 triads (9 elements)
Left gaps for undiscovered elements (e.g., eka-boron → Sc, eka-aluminium → Ga, eka-silicon → Ge)

✅ Merits of Mendeleev's Table

Predicted properties of undiscovered elements
Corrected atomic masses of some elements (Be, In, Os)
Systematic study of elements became possible
Placed noble gases without disturbing the table

❌ Demerits

Hydrogen's position was not fixed
Isotopes had no place
Anomalous pairs: Ar (39.9) before K (39.1), Co (58.9) before Ni (58.7), Te (127.6) before I (126.9), Th (232) before Pa (231)
Position of lanthanoids and actinoids was unclear

PYQ — JEE Main 2022

Question: The element predicted by Mendeleev as "eka-aluminium" was later identified as:

Solution:

Mendeleev predicted the properties of eka-aluminium (meaning "beyond aluminium"). When gallium was discovered by Lecoq de Boisbaudran in 1875, its properties matched exactly with Mendeleev's predictions:

Eka-aluminium \to Gallium (Ga, Z=31)

Similarly: eka-boron → Scandium (Sc), eka-silicon → Germanium (Ge)

Answer: Gallium (Ga)

2

Modern Periodic Law & Periodic Table

The modern periodic table is based on atomic number rather than atomic mass. This fundamental change resolved all anomalies of Mendeleev's table and forms the basis of all chemical understanding today.

2.1 Moseley's Modern Periodic Law (1913)

Most Important — High PYQ Frequency

Henry Moseley studied X-ray spectra of elements and found that the frequency (ν) of X-rays is related to atomic number (Z) by:

\sqrt{\nu} = a(Z - b)

This led to the Modern Periodic Law: "The physical and chemical properties of elements are a periodic function of their atomic numbers (not atomic masses)."

2.2 Structure of Modern Periodic Table

Key Structural Features

Periods (Horizontal Rows): 7

Period 1: 2 elements (H, He) — shortest
Period 2: 8 elements (Li to Ne)
Period 3: 8 elements (Na to Ar)
Period 4: 18 elements (K to Kr)
Period 5: 18 elements (Rb to Xe)
Period 6: 32 elements (Cs to Rn) — longest
Period 7: 32 elements (Fr to Og)

Groups (Vertical Columns): 18

Group 1: Alkali metals (except H)
Group 2: Alkaline earth metals
Groups 3-12: Transition metals
Group 13: Boron family
Group 14: Carbon family
Group 15: Pnicogens (Nitrogen family)
Group 16: Chalcogens
Group 17: Halogens
Group 18: Noble gases

JEE Pro Tip — Period Types

Each period starts with filling of a new principal quantum number:

1st period: 1s (2 elements)
2nd period: 2s, 2p (8 elements)
3rd period: 3s, 3p (8 elements)
4th period: 4s, 3d, 4p (18 elements)
5th period: 5s, 4d, 5p (18 elements)
6th period: 6s, 4f, 5d, 6p (32 elements)
7th period: 7s, 5f, 6d, 7p (32 elements)

2.3 IUPAC Nomenclature of Elements with Z > 100

IUPAC Naming Rules

Elements with Z > 100 are named using roots for digits:

Digit	Root	Symbol
0	nil	n
1	un	u
2	bi	b
3	tri	t
4	quad	q
5	pent	p
6	hex	h
7	sept	s
8	oct	o
9	enn	e

Rule: Combine roots + "ium". Symbol = first letters of each root.

Example: Z=101 → Unnilnilium (Unu), Z=109 → Unnilennium (Une)

PYQ — JEE Main 2023

Question: According to IUPAC nomenclature, the element with atomic number 107 is named as:

Solution:

Z = 107: digits are 1, 0, 7

1 → un (u)

0 → nil (n)

7 → sept (s)

Name = Un + nil + sept + ium = Unnilseptium (Uns)

Note: Now officially named Bohrium (Bh) after Niels Bohr.

Answer: Unnilseptium (Uns) / Bohrium (Bh)

Common Mistake Alert

Modern periodic law is based on atomic number, NOT atomic mass. Many students confuse this. Moseley proved that atomic number is the fundamental property, not atomic mass. This resolved the anomalous pairs of Mendeleev's table (Ar-K, Co-Ni, Te-I, Th-Pa).

3

Electronic Configuration & Periodic Table Blocks

The periodic table is divided into four blocks (s, p, d, f) based on the orbital into which the last electron enters. Understanding the block classification is essential for predicting properties of elements.

3.1 The Four Blocks of Periodic Table

Block Classification

Block	Last e⁻ enters	Groups	Outer config	Elements
s-block	ns orbital	1, 2	ns¹⁻²	Alkali & alkaline earth metals
p-block	np orbital	13-18	ns²np¹⁻⁶	Metalloids, non-metals, noble gases
d-block	(n-1)d orbital	3-12	(n-1)d¹⁻¹⁰ns⁰⁻²	Transition metals
f-block	(n-2)f orbital	3 (separate)	(n-2)f¹⁻¹⁴(n-1)d⁰⁻¹ns²	Lanthanoids & Actinoids

s-block Elements

Groups 1 (ns¹) and 2 (ns²)
All metals, highly reactive
Low IE, form basic oxides
H and He also placed here
Fixed oxidation states (+1, +2)

p-block Elements

Groups 13-18 (ns²np¹⁻⁶)
Contains metals, non-metals, metalloids
Variable oxidation states
Most electronegative elements
Noble gases (ns²np⁶) are inert

3.2 Types of Elements

Special Element Categories

Representative Elements: s and p block (main group)
Transition Elements: d-block (Groups 3-12)
Inner Transition Elements: f-block (4f and 5f series)
Noble Gases: Group 18 (He, Ne, Ar, Kr, Xe, Rn, Og)
Metals: ~78% of all elements
Non-metals: Upper right of periodic table
Metalloids: B, Si, Ge, As, Sb, Te, At — along zig-zag line

PYQ — JEE Main 2024

Question: An element X has atomic number 35. To which block, group, and period does it belong?

Solution:

Step 1: Write electronic configuration

1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^5

Step 2: Last electron enters 4p → p-block

Step 3: Group number = 10 + (ns + np electrons)

ns = 2, \quad np = 5 \Rightarrow 10 + 2 + 5 = 17

Step 4: Period = highest principal quantum number = 4

\text{Block: p-block}, \quad \text{Group: 17 (Halogens)}, \quad \text{Period: 4}

Element is Bromine (Br).

Quick Formula for Group & Period

s-block: Group = ns electrons (1 or 2)
p-block: Group = 10 + (ns + np electrons)
d-block: Group = (n-1)d + ns electrons
f-block: Group = 3
Period = Highest n value in configuration

4

Atomic & Ionic Radius

Atomic radius is one of the most fundamental periodic properties. The size of an atom determines many of its physical and chemical properties. Understanding the trends in atomic and ionic radii is essential for JEE.

4.1 Types of Atomic Radius

Three Types of Radius

Covalent Radius

r_{cov} = \frac{d_{A-A}}{2}

Half the distance between nuclei of two covalently bonded identical atoms

Metallic Radius

r_{met} = \frac{d_{M-M}}{2}

Half the distance between nuclei of two adjacent metal atoms in metallic crystal

van der Waals Radius

Half the distance between nuclei of two non-bonded atoms of same element in adjacent molecules

Order: van der Waals radius > Metallic radius > Covalent radius

4.2 Trends in Atomic Radius

Most Repeated PYQ Topic

Across a Period (→)

\text{Decreases } \rightarrow

Nuclear charge increases, electrons added to same shell → stronger attraction → smaller size

Down a Group (↓)

\text{Increases } \downarrow

New shells added → screening effect increases → outer electrons farther from nucleus → larger size

Screening Effect & Effective Nuclear Charge

Z_{eff} = Z - \sigma

Where Z = atomic number, σ = screening constant

Order of screening effect: s > p > d > f

Reason: s electrons are closest to nucleus, f electrons are farthest and poorly screen

4.3 Ionic Radius

Cation (M⁺)

Formed by loss of electrons → smaller than parent atom

r_{cation} < r_{atom}

Reason: Same nuclear charge holds fewer electrons; Z_eff increases

Anion (M⁻)

Formed by gain of electrons → larger than parent atom

r_{anion} > r_{atom}

Reason: Same nuclear charge holds more electrons; repulsion increases

Isoelectronic Species Trick

Species with same number of electrons but different nuclear charge:

\text{For isoelectronic species: higher } Z \Rightarrow \text{smaller size}

Example: O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺ (all have 10 electrons)

PYQ — JEE Main 2024

Question: Which of the following is the correct order of ionic radii?

(a) N³⁻ < O²⁻ < F⁻ < Na⁺

(b) N³⁻ > O²⁻ > F⁻ > Na⁺

(c) Na⁺ > F⁻ > O²⁻ > N³⁻

(d) F⁻ > O²⁻ > N³⁻ > Na⁺

Solution:

All four species are isoelectronic (10 electrons each):

N³⁻ (Z=7), O²⁻ (Z=8), F⁻ (Z=9), Na⁺ (Z=11)

For isoelectronic species, as Z increases, Z_eff increases → electrons pulled closer → size decreases

Z_{eff}: \text{Na}^+ > \text{F}^- > \text{O}^{2-} > \text{N}^{3-}

\text{Size}: \text{N}^{3-} > \text{O}^{2-} > \text{F}^- > \text{Na}^+

\text{Correct order: N}^{3-} > \text{O}^{2-} > \text{F}^- > \text{Na}^+

Answer: (b) N³⁻ > O²⁻ > F⁻ > Na⁺

Exception Alert (JEE Favorite!)

In transition metals, atomic size first decreases then becomes nearly constant (due to d-electron screening)
Noble gases have the largest atomic radii in their periods (van der Waals radius is measured, not covalent)
Li (152 pm) is smaller than Mg (160 pm) due to diagonal relationship (not always intuitive)
Lanthanoid contraction: 4f electrons poorly screen → Hf (144 pm) ≈ Zr (160 pm) despite being below

5

Ionization Enthalpy — The Rank Decider

Ionization enthalpy is the most frequently tested topic from this chapter in JEE. Every year 1-2 questions directly test ionization enthalpy trends, exceptions, and successive ionization energies. Master this section for guaranteed marks.

5.1 Definition & Key Concepts

Highest PYQ Frequency Topic

Ionization Enthalpy (IE): The minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state.

M(g) \rightarrow M^+(g) + e^- \quad \Delta H = IE_1

Units: kJ/mol or eV/atom (1 eV = 96.485 kJ/mol)

Always positive (endothermic process)

5.2 Successive Ionization Enthalpies

Successive Ionization Enthalpy

\[M(g) \rightarrow M^+(g) + e^- \quad IE_1\]

\[M^+(g) \rightarrow M^{2+}(g) + e^- \quad IE_2\]

\[M^{2+}(g) \rightarrow M^{3+}(g) + e^- \quad IE_3\]

IE_1 < IE_2 < IE_3 < IE_4 \ldots

Reason: After each removal, remaining electrons are held more tightly by the same nucleus → more energy needed each time

JEE Shortcut — Jump in IE

A large jump in successive IE indicates removal of a core electron (new shell). This helps determine group number:

1st large jump after IE₁: Group 1 (1 valence e⁻)
Large jump after IE₂: Group 2 (2 valence e⁻)
Large jump after IE₃: Group 13 (3 valence e⁻)
Large jump after IE₄: Group 14 (4 valence e⁻)

5.3 Periodic Trends in Ionization Enthalpy

Across a Period (→)

\text{Increases } \rightarrow

Nuclear charge ↑, atomic size ↓ → electrons held more tightly → IE ↑

Down a Group (↓)

\text{Decreases } \downarrow

Atomic size ↑, screening ↑ → outer electrons held loosely → IE ↓

5.4 Exceptions (JEE Favorites — Must Memorize!)

Top 4 Exceptions — 80% of JEE Questions

1. Be > B (Group 2 > Group 13)

Be: 1s² 2s² (fully filled 2s) → stable → higher IE

B: 1s² 2s² 2p¹ (2p electron easier to remove, less penetrating)

2. N > O (Group 15 > Group 16)

N: 1s² 2s² 2p³ (half-filled 2p, extra stable due to exchange energy)

O: 1s² 2s² 2p⁴ (removal gives half-filled 2p³, easier)

3. Mg > Al (Group 2 > Group 13)

Mg: [Ne] 3s² (fully filled, more penetrating)

Al: [Ne] 3s² 3p¹ (3p electron less penetrating, easier to remove)

4. P > S (Group 15 > Group 16)

P: [Ne] 3s² 3p³ (half-filled 3p, stable)

S: [Ne] 3s² 3p⁴ (removal gives half-filled 3p³)

PYQ — JEE Main 2023

Question: The first ionization enthalpy of Na, Mg, Al, and Si are in the order:

(a) Na < Mg > Al < Si

(b) Na > Mg > Al > Si

(c) Na < Mg < Al < Si

(d) Na < Al < Mg < Si

Solution:

General trend: IE increases across a period → Na < Mg < Al < Si

But exception: Mg > Al (Mg has fully-filled 3s², Al has 3p¹)

Mg: [Ne] 3s² → fully filled, more stable

Al: [Ne] 3s² 3p¹ → 3p electron easier to remove

IE_1: \text{Na} < \text{Al} < \text{Mg} < \text{Si}

Values: Na (496) < Al (577) < Mg (738) < Si (786) kJ/mol

Answer: (d) Na < Al < Mg < Si

5.5 Factors Affecting Ionization Enthalpy

Factor	Effect on IE	Reason
Atomic size ↑	IE ↓	Electrons farther from nucleus
Nuclear charge ↑	IE ↑	Stronger attraction
Screening effect ↑	IE ↓	Inner electrons shield outer electrons
Half-filled/Full-filled	IE ↑	Extra stability (exchange energy)
Penetration: s > p > d > f	IE: s > p > d > f	s electrons held most tightly

PYQ — JEE Main 2021

Question: The successive ionization enthalpies of an element are 740, 1500, 7700, 10500, 13600 kJ/mol. The element belongs to which group?

Solution:

IE₁ = 740, IE₂ = 1500, IE₃ = 7700

Identify the large jump:

IE_2 \rightarrow IE_3: 1500 \rightarrow 7700 \quad (\text{jump of } 6200)

This large jump after IE₂ means the 3rd electron comes from a new (inner) shell

Therefore, the element has 2 valence electrons

\text{Group 2 (Alkaline Earth Metals)}

The element is likely Mg (IE₁ = 738, IE₂ = 1451, IE₃ = 7733)

Answer: Group 2

6

Electron Gain Enthalpy

Electron gain enthalpy (EGE) measures the tendency of an atom to accept electrons. This property is crucial for understanding chemical bonding and reactivity of non-metals. JEE questions on EGE often test exceptions and the difference between EGE and electronegativity.

6.1 Definition

Important Definition

Electron Gain Enthalpy: The enthalpy change when an electron is added to an isolated gaseous atom in its ground state to form a gaseous anion.

X(g) + e^- \rightarrow X^-(g) \quad \Delta H = \Delta_{eg}H

Can be positive or negative (unlike IE which is always positive)

Sign Convention (JEE Trap!)

Negative EGE (Exothermic)

Energy is released when electron is added → atom wants the electron

Example: Cl (ΔegH = −349 kJ/mol)

Positive EGE (Endothermic)

Energy is absorbed → atom resists adding electron

Example: Noble gases (positive EGE)

6.2 Periodic Trends

Across a Period (→)

\text{More negative } \rightarrow

Z_eff ↑, size ↓ → easier to add electron → EGE more negative

Down a Group (↓)

\text{Less negative } \downarrow

Size ↑, screening ↑ → less attraction → EGE less negative

6.3 Exceptions (JEE Critical!)

Must-Memorize Exceptions

1. Cl > F (Chlorine > Fluorine)

F is small → added electron experiences strong e⁻-e⁻ repulsion in small 2p orbital → less energy released

Cl is larger → 3p has more space → less repulsion → more energy released

ΔegH: Cl (−349) > F (−328) kJ/mol

2. S > O (Sulfur > Oxygen)

Same reason: O is small, strong repulsion in 2p

ΔegH: S (−200) > O (−141) kJ/mol

3. Noble gases have positive EGE

Fully-filled configuration → adding electron requires energy

4. Alkaline earth metals (Group 2) have near-zero/positive EGE

Fully-filled ns² → don't want extra electron

5. N has less negative EGE than expected

Half-filled 2p³ is stable → adding electron disturbs stability

PYQ — JEE Main 2024

Question: Which of the following has the most negative electron gain enthalpy?

(a) F (b) Cl (c) Br (d) I

Solution:

General trend: EGE becomes less negative down a group → F should be most negative

But exception: Cl > F

Fluorine has small size → 2p electrons create strong repulsion for incoming electron

Chlorine has larger 3p orbital → less repulsion → more energy released

\Delta_{eg}H: \text{Cl}(-349) > \text{F}(-328) > \text{Br}(-325) > \text{I}(-295) \text{ kJ/mol}

\text{Most negative: Cl}

Answer: (b) Cl

EGE vs Electronegativity

Don't confuse these two:

EGE: Property of isolated atom (gaseous state) — can be measured directly
Electronegativity: Property of atom in a molecule (bonded state) — relative, cannot be measured directly
EGE: Can be +ve or −ve
Electronegativity: Always positive

7

Electronegativity

Electronegativity is the measure of an atom's ability to attract shared electrons in a chemical bond. It is a key concept that explains bond polarity, bond character, and molecular properties. Different scales exist for measuring electronegativity.

7.1 Definition & Scales

Important for JEE

Electronegativity (χ): The ability of an atom in a chemical bond to attract the shared pair of electrons towards itself.

It is a relative property — no units, always positive, measured on different scales.

Electronegativity Scales

Scale	Scientist	Basis	Range
Pauling	Linus Pauling	Bond dissociation energies	0.7 (Cs) to 4.0 (F)
Mulliken	Robert Mulliken	(IE + EGE)/2	Varies
Allred-Rochow	Allred & Rochow	Z_eff/r²	Varies
Allen	L.C. Allen	Spectroscopic energies	Varies

Mulliken Scale Formula

\chi_M = \frac{IE_1 + \Delta_{eg}H}{2}

Where IE₁ and ΔegH are in eV. To convert to Pauling scale:

\chi_P = \frac{\chi_M}{2.8} \quad \text{(approximate)}

7.2 Periodic Trends

Across a Period (→)

\text{Increases } \rightarrow

Z_eff ↑, size ↓ → greater attraction for bonding electrons

Down a Group (↓)

\text{Decreases } \downarrow

Size ↑, bonding electrons farther from nucleus

Most & Least Electronegative Elements

Most EN

\text{F} = 4.0

Followed by O (3.5), N (3.0), Cl (3.0)

Least EN

\text{Cs} = 0.7

Followed by Fr (0.7), Rb (0.8)

7.3 Applications of Electronegativity

Bond Character Prediction

Pauling's formula for percentage ionic character:

\% \text{Ionic} = 100\left(1 - e^{-\frac{1}{4}(\chi_A - \chi_B)^2}\right)

Δχ = 0: Pure covalent bond (e.g., H₂, Cl₂)
Δχ < 1.7: Predominantly covalent
Δχ = 1.7: 50% ionic, 50% covalent
Δχ > 1.7: Predominantly ionic
Δχ > 2.0: Ionic bond (e.g., NaCl, Δχ = 2.23)

PYQ — JEE Main 2022

Question: Which of the following bonds is the most polar?

(a) C−F (b) C−Cl (c) C−O (d) C−N

Solution:

Polarity depends on electronegativity difference (Δχ):

χ values: C = 2.5, N = 3.0, O = 3.5, F = 4.0, Cl = 3.0

\text{C-F}: \Delta\chi = 4.0 - 2.5 = 1.5

\text{C-Cl}: \Delta\chi = 3.0 - 2.5 = 0.5

\text{C-O}: \Delta\chi = 3.5 - 2.5 = 1.0

\text{C-N}: \Delta\chi = 3.0 - 2.5 = 0.5

\text{Most polar: C-F} (\Delta\chi = 1.5)

Answer: (a) C−F

8

Anomalous Properties & Diagonal Relationship

The second period elements (Li, Be, B) show anomalous behavior compared to other members of their groups. They also show resemblance with elements of the third period placed diagonally — this is called the diagonal relationship. JEE Advanced frequently tests these concepts.

8.1 Anomalous Properties of Second Period Elements

JEE Advanced Favorite

Reasons for anomaly:

Very small atomic size
High electronegativity
High charge density
Non-availability of d-orbitals (cannot expand octet)
Maximum covalency = 4

Specific Anomalies

Lithium (Li) vs Other Alkali Metals:

Li forms covalent compounds (others form ionic)
Li reacts with N₂ to form Li₃N (others don't)
Li₂CO₃ decomposes on heating (others stable)
LiNO₃ decomposes to Li₂O, NO₂, O₂ (others give nitrite)
LiF, Li₂O are less soluble (high lattice energy)

Beryllium (Be) vs Other Alkaline Earth Metals:

Be forms covalent compounds
BeO is amphoteric (others basic)
Be doesn't react with water (others do)
BeCl₂ is covalent and polymeric
Be can form [BeF₄]²⁻ (covalent complexes)

Boron (B) vs Other Group 13:

B is a non-metal (others are metals)
B forms only covalent compounds
B₂O₃ is acidic (others amphoteric/basic)
B shows max covalency of 4 (others can exceed)

8.2 Diagonal Relationship

High PYQ Frequency

Elements of 2nd period show resemblance with elements of 3rd period placed diagonally:

\text{Li} \sim \text{Mg}, \quad \text{Be} \sim \text{Al}, \quad \text{B} \sim \text{Si}

Reason: Similar charge/radius ratio (ionic potential) and electronegativity

Li ∼ Mg Similarities

Both react with N₂ to form nitrides (Li₃N, Mg₃N₂)
Both form oxides on heating (Li₂O, MgO) — normal oxides
Both carbonates decompose on heating
Both fluorides are sparingly soluble
Both hydroxides are weak bases
Both form organometallic compounds

Be ∼ Al Similarities

Both form amphoteric oxides (BeO, Al₂O₃)
Both don't react with water
Both form covalent halides (BeCl₂, AlCl₃)
Both form complex ions: [BeF₄]²⁻, [AlF₆]³⁻
Both nitrides hydrolyze to give NH₃
Both carbonates are unstable

PYQ — JEE Advanced 2023

Question: The correct statement(s) about the diagonal relationship between Be and Al is/are:

(a) BeO and Al₂O₃ are both amphoteric

(b) BeCl₂ and AlCl₃ are both covalent

(c) Both Be and Al react with water to produce H₂

(d) Both form carbides that give methane on hydrolysis

Solution:

(a) ✅ True — Both BeO and Al₂O₃ are amphoteric

(b) ✅ True — Both BeCl₂ and AlCl₃ are covalent (BeCl₂ is polymeric, AlCl₃ dimeric)

(c) ❌ False — Neither Be nor Al react with water (unlike other group members)

(d) ✅ True — Be₂C + 4H₂O → 2Be(OH)₂ + CH₄; Al₄C₃ + 12H₂O → 4Al(OH)₃ + 3CH₄

\text{Correct: (a), (b), (d)}

9

Periodic Trends & Chemical Reactivity

The periodic trends in atomic properties directly influence chemical reactivity. Understanding how reactivity varies across periods and groups helps predict chemical behavior of elements.

9.1 Valency

Valency Trends

Across a Period (→)

Valency w.r.t. oxygen: 1 → 2 → 3 → 4 → 5 → 6 → 7 → 0

Valency w.r.t. hydrogen: 1 → 2 → 3 → 4 → 3 → 2 → 1 → 0

Increases from 1 to 4, then decreases to 0 (for hydrogen valency)

Down a Group (↓)

Valency remains the same (equals group number for s and p block)

Example: Group 1 → valency 1; Group 17 → valency 1

9.2 Chemical Reactivity Trends

Metallic Character

Increases down a group
Decreases across a period
Best metals: bottom left (Cs, Fr)
Related to IE: lower IE → more metallic

Non-metallic Character

Decreases down a group
Increases across a period
Best non-metals: top right (F, O)
Related to EN: higher EN → more non-metallic

9.3 Reactivity of Metals vs Non-metals

Key Reactivity Patterns

Metals (Low IE):

Reactivity ∝ 1/IE → More reactive down a group, less reactive across a period

Example: K > Na > Li (Group 1); Na > Mg > Al (Period 3)

Non-metals (High EN):

Reactivity ∝ EN → More reactive across a period, less reactive down a group

Example: F > Cl > Br > I (Group 17); F > O > N > C (Period 2)

9.4 Inert Pair Effect

JEE Advanced Concept

The tendency of ns² electrons to remain unshared in compounds of p-block elements as we go down the group.

Group 13: B (+3) → Al (+3) → Ga (+3, +1) → In (+3, +1) → Tl (+1 dominant)
Group 14: C (+4) → Si (+4) → Ge (+4, +2) → Sn (+4, +2) → Pb (+2 dominant)
Group 15: N (+5) → P (+5) → As (+5, +3) → Sb (+5, +3) → Bi (+3 dominant)

Reason: Poor shielding by d and f electrons → ns² electrons held more tightly → harder to remove

PYQ — JEE Main 2023

Question: The most stable oxidation state of thallium (Tl) is:

(a) +3 (b) +1 (c) +2 (d) +5

Solution:

Thallium (Tl) is in Group 13, Period 6

Due to inert pair effect, the 6s² electrons are reluctant to participate in bonding

The poor shielding by 4f and 5d electrons increases Z_eff on 6s² electrons

This makes +1 oxidation state more stable than +3

\text{Most stable oxidation state of Tl: } +1

Answer: (b) +1

Common Mistakes in Periodic Trends

IE exceptions: Always remember Be > B, N > O, Mg > Al, P > S
EGE exception: Cl > F (not F > Cl)
Noble gases: Don't have zero EN — they don't form bonds, so EN is undefined
Atomic radius: Noble gases have largest radii (van der Waals radius measured)
Inert pair effect: Increases down the group, not across period

Previous Year Questions Analysis

JEE Main (Last 10 Years)

✓ Ionization Enthalpy Trends: 30%
✓ Atomic/Ionic Radius: 20%
✓ Electron Gain Enthalpy: 18%
✓ Electronegativity & Bond Character: 15%
✓ Periodic Table Position: 10%
✓ Anomalous/Diagonal: 7%

JEE Advanced (Last 10 Years)

✓ Successive IE & Group ID: 25%
✓ Anomalous Properties: 20%
✓ Diagonal Relationship: 20%
✓ Inert Pair Effect: 15%
✓ Isoelectronic Species: 10%
✓ IUPAC Nomenclature: 10%

Weightage Analysis

JEE Main: 8-12 marks (2-3 questions)

JEE Advanced: 6-12 marks (1-3 questions)

Difficulty Level: Easy to Moderate

Time Required: 3-4 hours practice

Download Complete PYQ PDF (2015-2024)

Practice Problem Set

Level 1: Basic (JEE Main Standard)

Arrange in order of increasing atomic radius: Na, Mg, Al, Si, P, S, Cl, Ar
Which has higher IE₁: Na or Mg? Why?
Arrange isoelectronic species in decreasing ionic radii: O²⁻, F⁻, Na⁺, Mg²⁺
Identify block, group, period for element with Z = 20
Why does nitrogen have higher IE₁ than oxygen?

Level 2: Intermediate (JEE Main/Advanced)

The successive IEs of an element are 520, 7300, 10800, 14000 kJ/mol. Identify the group.
Why does Cl have more negative EGE than F despite being below F?
Arrange in order of IE₃: Na, Mg, Al, Si
Predict the most stable oxidation state of Pb and explain inert pair effect.
Which is more covalent: BeCl₂ or MgCl₂? Justify using Fajan's rules.

Level 3: Advanced (JEE Advanced/Olympiad)

Explain why BeO is amphoteric while MgO is basic using periodic trends.
Calculate % ionic character of HCl bond using Pauling's formula (χH=2.1, χCl=3.0).
An element has IE₁=577, IE₂=1817, IE₃=2745, IE₄=11580. Identify the element.
Prove that the diagonal relationship between Li and Mg is due to similar charge/radius ratio.
Why does the atomic radius not decrease much from Zr to Hf despite increasing Z?

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> FAQ

Frequently Asked Questions

What is the modern periodic law?

The modern periodic law states that the physical and chemical properties of elements are a periodic function of their atomic numbers. It was proposed by Henry Moseley in 1913, replacing Mendeleev's law which was based on atomic masses.

What is the trend of ionization enthalpy in the periodic table?

Ionization enthalpy increases across a period (left to right) due to increasing nuclear charge and decreases down a group due to increasing atomic size and screening effect. Exceptions occur at half-filled and fully-filled configurations (e.g., Be > B, N > O).

Why does electron gain enthalpy of noble gases become positive?

Noble gases have stable fully-filled electronic configurations. Adding an electron requires energy to overcome the repulsion, making the electron gain enthalpy positive (endothermic).

What is the diagonal relationship in the periodic table?

Elements of second period (Li, Be, B) show resemblance with elements of third period placed diagonally (Mg, Al, Si) due to similar charge/radius ratio and electronegativity. This is called diagonal relationship.

Why is the first ionization enthalpy of boron less than beryllium?

Boron has electronic configuration 1s²2s²2p¹. The 2p electron is less penetrating and easier to remove. Beryllium has fully-filled 2s² which is more stable, requiring more energy to remove an electron.

What is screening effect or shielding effect?

The screening effect is the decrease in attractive force between nucleus and valence electrons due to presence of inner shell electrons. The order of screening effect is s > p > d > f. Effective nuclear charge Z_eff = Z - σ, where σ is the screening constant.

How many questions come from periodic classification in JEE Mains?

On average 2-3 questions (8-12 marks) appear from periodic classification in every JEE Main session. Most questions are from ionization enthalpy trends, electron gain enthalpy, and periodic table positions.

What is the weightage of periodic classification in JEE Advanced?

Periodic classification carries roughly 4-6% weightage in JEE Advanced. Advanced questions often combine periodic trends with chemical bonding and require understanding of exceptions and anomalous behavior.

Why does nitrogen have higher ionization enthalpy than oxygen?

Nitrogen has half-filled 2p³ configuration which is extra stable due to exchange energy. Oxygen has 2p⁴ configuration where removal of one electron gives half-filled stability. Hence nitrogen's first IE is higher than oxygen.

What are the different blocks in the periodic table?

The periodic table is divided into four blocks based on the orbital receiving the last electron: s-block (groups 1-2), p-block (groups 13-18), d-block (groups 3-12, transition elements), and f-block (lanthanoids and actinoids).

Periodic Classification notes, trends & PYQs for JEE 2026

📑 Table of Contents

Genesis of Periodic Classification

1.1 Dobereiner's Triads (1817)

Examples of Dobereiner's Triads

1.2 Newlands' Law of Octaves (1865)

⭐ Important for JEE

1.3 Mendeleev's Periodic Table (1869)

Key Features of Mendeleev's Table

✅ Merits of Mendeleev's Table

❌ Demerits

📝 PYQ — JEE Main 2022

Modern Periodic Law & Periodic Table

2.1 Moseley's Modern Periodic Law (1913)

⭐ Most Important — High PYQ Frequency

2.2 Structure of Modern Periodic Table

Key Structural Features

💡 JEE Pro Tip — Period Types

2.3 IUPAC Nomenclature of Elements with Z > 100

IUPAC Naming Rules

📝 PYQ — JEE Main 2023

⚠️ Common Mistake Alert

Electronic Configuration & Periodic Table Blocks

3.1 The Four Blocks of Periodic Table

Block Classification

s-block Elements

p-block Elements

3.2 Types of Elements

Special Element Categories

📝 PYQ — JEE Main 2024

🎯 Quick Formula for Group & Period

Atomic & Ionic Radius

4.1 Types of Atomic Radius

Three Types of Radius

4.2 Trends in Atomic Radius

⭐ Most Repeated PYQ Topic

Screening Effect & Effective Nuclear Charge

4.3 Ionic Radius

Cation (M⁺)

Anion (M⁻)

💡 Isoelectronic Species Trick

📝 PYQ — JEE Main 2024

⚠️ Exception Alert (JEE Favorite!)

Ionization Enthalpy — The Rank Decider

5.1 Definition & Key Concepts

⭐ Highest PYQ Frequency Topic

5.2 Successive Ionization Enthalpies

Successive Ionization Enthalpy

🎯 JEE Shortcut — Jump in IE

5.3 Periodic Trends in Ionization Enthalpy

Across a Period (→)

Down a Group (↓)

5.4 Exceptions (JEE Favorites — Must Memorize!)

🔥 Top 4 Exceptions — 80% of JEE Questions

📝 PYQ — JEE Main 2023

5.5 Factors Affecting Ionization Enthalpy

📝 PYQ — JEE Main 2021

Electron Gain Enthalpy

6.1 Definition

⭐ Important Definition

Sign Convention (JEE Trap!)

6.2 Periodic Trends

Across a Period (→)

Down a Group (↓)

6.3 Exceptions (JEE Critical!)

🔥 Must-Memorize Exceptions

📝 PYQ — JEE Main 2024

💡 EGE vs Electronegativity

Electronegativity

7.1 Definition & Scales

⭐ Important for JEE

Electronegativity Scales

💡 Mulliken Scale Formula

7.2 Periodic Trends

Across a Period (→)

Down a Group (↓)

Most & Least Electronegative Elements

7.3 Applications of Electronegativity

Bond Character Prediction

📝 PYQ — JEE Main 2022

Table of Contents

Important for JEE

PYQ — JEE Main 2022

Most Important — High PYQ Frequency

JEE Pro Tip — Period Types

PYQ — JEE Main 2023

Common Mistake Alert

PYQ — JEE Main 2024

Quick Formula for Group & Period

Most Repeated PYQ Topic

Isoelectronic Species Trick

PYQ — JEE Main 2024

Exception Alert (JEE Favorite!)

Highest PYQ Frequency Topic

JEE Shortcut — Jump in IE

Top 4 Exceptions — 80% of JEE Questions

PYQ — JEE Main 2023

PYQ — JEE Main 2021

Important Definition

Must-Memorize Exceptions

PYQ — JEE Main 2024

EGE vs Electronegativity

Important for JEE

Mulliken Scale Formula

PYQ — JEE Main 2022

JEE Advanced Favorite

High PYQ Frequency

PYQ — JEE Advanced 2023

JEE Advanced Concept

PYQ — JEE Main 2023

Common Mistakes in Periodic Trends

Previous Year Questions Analysis

Practice Problem Set

Why Periodic Classification is Important

Must-Remember Trends & Formulas

5-Step Exam Strategy

7-Day Periodic Classification Mastery Roadmap