Modern Periodic Law and Electronic Configuration

Master the modern periodic law, periodic table structure, and electronic configuration patterns for JEE Chemistry.

The Hook: Why Did Mendeleev Leave Gaps?

Connect: The Detective Work of Chemistry

In Sherlock Holmes movies, the detective predicts criminals before meeting them. In 1869, Dmitri Mendeleev did something similar - he left blank spaces in his periodic table for elements that hadn’t been discovered yet! He predicted their properties so accurately that when Gallium was discovered 6 years later, it matched his predictions perfectly.

But here’s the twist: Mendeleev arranged elements by atomic mass. Sometimes this failed. Modern periodic law fixed this by using atomic number instead. Why does atomic number work better? That’s what makes chemistry predictable!


The Core Concept: Modern Periodic Law

The Revolutionary Change

Mendeleev’s Periodic Law (1869):

Properties of elements are periodic functions of their atomic masses.

Modern Periodic Law (1913 - Moseley):

Properties of elements are periodic functions of their atomic numbers.

$$\boxed{\text{Properties} = f(\text{Atomic Number})}$$

In simple terms: When you arrange elements by atomic number (number of protons), their properties repeat in a pattern. It’s like a musical scale - the notes repeat after every octave!

Why Atomic Number, Not Mass?

The Problem with Atomic Mass

Mendeleev’s Challenges:

  1. Isotopes: Same element, different masses (Cl-35 and Cl-37)
  2. Inversions: Ar (39.9) came before K (39.1) by mass, but properties didn’t match
  3. Cobalt-Nickel: Co (58.9) should come after Ni (58.7) by mass, but properties reversed

Moseley’s Solution: Atomic number = number of protons = defines the element

  • Same atomic number → same element (regardless of mass)
  • Explains all anomalies perfectly!

Structure of Modern Periodic Table

The Layout

graph TD
    A[Periodic Table: 7 Periods, 18 Groups] --> B[s-block: Groups 1-2]
    A --> C[p-block: Groups 13-18]
    A --> D[d-block: Groups 3-12]
    A --> E[f-block: Lanthanoids + Actinoids]

    B --> B1[Alkali Metals]
    B --> B2[Alkaline Earth Metals]
    C --> C1[Halogens]
    C --> C2[Noble Gases]
    D --> D1[Transition Metals]

Key Numbers to Remember

FeatureCount
Periods7 horizontal rows
Groups18 vertical columns
s-block elements14 (Groups 1, 2)
p-block elements36 (Groups 13-18)
d-block elements40 (Groups 3-12)
f-block elements28 (Lanthanoids + Actinoids)
Total elements118 (as of 2023)

Interactive Demo: Explore the Periodic Table

See the organization of elements, blocks, and groups in an interactive periodic table.


Electronic Configuration and Periodic Position

The SPDF Rule

Electronic configuration determines where an element sits in the periodic table!

General Pattern:

$$\text{1s}^2 \, \text{2s}^2 \, \text{2p}^6 \, \text{3s}^2 \, \text{3p}^6 \, \text{4s}^2 \, \text{3d}^{10} \, \text{4p}^6 \, ...$$

Block Classification

The last electron determines the block:

Last Electron inBlockGroupsExample
ns¹⁻²s-block1, 2Na: [Ne] 3s¹
np¹⁻⁶p-block13-18Cl: [Ne] 3s² 3p⁵
(n-1)d¹⁻¹⁰d-block3-12Fe: [Ar] 3d⁶ 4s²
(n-2)f¹⁻¹⁴f-blockLanthanoids/ActinoidsCe: [Xe] 4f¹ 5d¹ 6s²
Memory Trick: S-P-D-F Stands For...

“Smart People Don’t Forget”

  • S = s-block (Groups 1-2)
  • P = p-block (Groups 13-18)
  • D = d-block (Groups 3-12)
  • F = f-block (separate rows below)

Finding Period and Group from Electronic Configuration

Step-by-Step Method

Step 1: Find Period Period number = Highest principal quantum number (n)

Step 2: Find Group

  • s-block: Group = number of valence electrons (1 or 2)
  • p-block: Group = 10 + number of p electrons
  • d-block: Group = number of d electrons + number of s electrons

Examples

Example 1: Sodium (Na)

Electronic configuration: 1s² 2s² 2p⁶ 3s¹

Period: Highest n = 3 → Period 3 Group: Last electron in 3s¹ → s-block → 1 valence electron → Group 1

Example 2: Chlorine (Cl)

Electronic configuration: 1s² 2s² 2p⁶ 3s² 3p⁵

Period: Highest n = 3 → Period 3 Group: Last electron in 3p⁵ → p-block → Group = 10 + 5 = Group 15… WAIT!

Actually: 3s² 3p⁵ = 7 valence electrons → Group 17

Trap: p-block Group Number

WRONG: Group = 10 + number of p electrons RIGHT: Group = 10 + total valence electrons (s + p)

For Cl: 3s² 3p⁵ → 2 + 5 = 7 valence electrons → Group = 10 + 7 = 17

Example 3: Iron (Fe)

Electronic configuration: [Ar] 3d⁶ 4s²

Period: Highest n = 4 → Period 4 Group: d-block → Group = d electrons + s electrons = 6 + 2 = Group 8


Memory Tricks & Patterns

Pattern 1: Period Length Pattern

PeriodNumber of ElementsWhy?
12Only 1s
282s + 2p (2 + 6)
383s + 3p (2 + 6)
4184s + 3d + 4p (2 + 10 + 6)
5185s + 4d + 5p (2 + 10 + 6)
6326s + 4f + 5d + 6p (2 + 14 + 10 + 6)
732*7s + 5f + 6d + 7p (incomplete)

Pattern: 2, 8, 8, 18, 18, 32, 32

Memory trick: “2 Ate 8 At 18 In 32” (completely made up, but it works!)

Pattern 2: Aufbau Order

The Diagonal Rule

Remember the filling order using diagonals:

1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p

Fill diagonally from top-right to bottom-left!

Or use: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p

Pattern 3: Exceptions to Remember

Half-filled and Fully-filled d orbitals are stable!

ElementExpectedActualReason
Cr (24)[Ar] 3d⁴ 4s²[Ar] 3d⁵ 4s¹Half-filled d⁵ stable
Cu (29)[Ar] 3d⁹ 4s²[Ar] 3d¹⁰ 4s¹Fully-filled d¹⁰ stable
Mo (42)[Kr] 4d⁴ 5s²[Kr] 4d⁵ 5s¹Half-filled d⁵ stable
Ag (47)[Kr] 4d⁹ 5s²[Kr] 4d¹⁰ 5s¹Fully-filled d¹⁰ stable

Memory trick: “Crazy Copper Makes Silver” (Cr, Cu, Mo, Ag)


Common Mistakes to Avoid

Trap #1: d-block Group Number

Mistake: Thinking Group = only d electrons Correct: Group = d electrons + s electrons

Example: Fe [Ar] 3d⁶ 4s² Wrong: Group 6 Right: Group 8 (6 + 2)

Trap #2: Period Number for d-block

Mistake: Using (n-1) from d orbital for period Correct: Use highest n value (from s orbital)

Example: Fe [Ar] 3d⁶ 4s² Wrong: Period 3 (from 3d) Right: Period 4 (from 4s)

Trap #3: Noble Gas Core

Mistake: Forgetting to account for noble gas core Correct: Expand it when counting electrons

Example: Fe [Ar] 3d⁶ 4s² Ar = 18 electrons, then add 6 + 2 = 8 more Total = 26 electrons (atomic number of Fe)


When to Use This

Decision Tree

Use electronic configuration to:

  1. Determine element’s position in periodic table
  2. Predict chemical properties (valency, oxidation states)
  3. Understand periodic trends
  4. Identify elements from their properties

Quick identification:

  • Last electron in → Group 1 (alkali metal)
  • Last electron in → Group 2 (alkaline earth)
  • Last electron in p⁵ → Group 17 (halogen)
  • Last electron in p⁶ → Group 18 (noble gas)

Practice Problems

Level 1: Foundation (NCERT)

Problem 1

Question: An element has electronic configuration [Ne] 3s² 3p³. Identify its period and group.

Solution:

  • Period: Highest n = 3 → Period 3
  • Group: p-block, valence electrons = 2 + 3 = 5 → Group = 10 + 5 = Group 15
  • Element: Phosphorus (P)
Problem 2

Question: How many elements are present in the 5th period?

Solution: Period 5 contains: 5s + 4d + 5p = 2 + 10 + 6 = 18 elements

Level 2: JEE Main

Problem 3

Question: An element with atomic number 26 belongs to which block, period, and group?

Solution: Z = 26 → Electronic configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶ Or: [Ar] 3d⁶ 4s²

  • Block: Last electron in d → d-block
  • Period: Highest n = 4 → Period 4
  • Group: d + s = 6 + 2 = Group 8
  • Element: Iron (Fe)
Problem 4

Question: The electronic configuration of an element is [Ar] 3d¹⁰ 4s¹. Identify the element and explain any exception.

Solution: Expected: [Ar] 3d⁹ 4s² Actual: [Ar] 3d¹⁰ 4s¹ ← This is given

Reason: Fully-filled d¹⁰ is more stable than d⁹ Element has total electrons: 18 + 10 + 1 = 29 → Copper (Cu)

Group: 10 + 1 = 11, Period: 4

Level 3: JEE Advanced

Problem 5

Question: How many elements in the periodic table have their valence electrons in d-orbitals?

Solution: d-block elements = Groups 3-12 × 4 periods (4, 5, 6, 7) However, Period 7 is incomplete.

  • Period 4: 10 elements (Sc to Zn)
  • Period 5: 10 elements (Y to Cd)
  • Period 6: 10 elements (La, Hf to Hg) - Note: La has 5d¹
  • Period 7: 10 elements (Ac, Rf to Cn) - Note: Ac has 6d¹

Total = 40 elements

But the question asks for “valence electrons in d-orbitals” - this means only elements where d is the outermost orbital. Elements like Pd [Kr] 4d¹⁰ have no s electrons, making d the valence orbital.

Answer: Most d-block elements, specifically 40 (standard answer for JEE)

Problem 6

Question: An element X forms X³⁺ ion which has electronic configuration [Ar] 3d⁵. Identify element X and its position.

Solution: X³⁺ has configuration: [Ar] 3d⁵ Electrons in X³⁺ = 18 + 5 = 23

Since X lost 3 electrons to form X³⁺: Neutral X has 23 + 3 = 26 electrons

Z = 26 → Iron (Fe) Configuration of Fe: [Ar] 3d⁶ 4s²

Position:

  • Period 4 (highest n = 4)
  • Group 8 (6 + 2 = 8)
  • d-block

Quick Revision Box

QuestionQuick Answer
What determines period?Highest principal quantum number (n)
What determines group in s-block?Number of valence electrons (1 or 2)
What determines group in p-block?10 + total valence electrons
What determines group in d-block?d electrons + s electrons
Which elements have exceptions?Cr, Cu, Mo, Ag (half/full d orbitals)
Period length pattern?2, 8, 8, 18, 18, 32, 32
Modern periodic law basis?Atomic number (not mass)

JEE Strategy: High-Yield Points

Exam Time-Savers

What JEE Loves to Ask:

  1. Electronic configuration → Find period/group (direct 1-marker)
  2. Exceptions in d-block (Cr, Cu especially)
  3. Number of elements in a period
  4. Identifying element from ion’s configuration
  5. Comparing Mendeleev vs Modern periodic law

Time-saving tip: Memorize configurations of elements 1-30. Questions become 10-second solves!

Common trap: d-block group number - always add s electrons!


Teacher’s Summary

Key Takeaways
  1. Modern Periodic Law: Properties depend on atomic number, not mass - this is why the periodic table works!

  2. Electronic configuration determines position: Last electron tells you the block, highest n gives period, counting rules give group.

  3. Remember the exceptions: Cr, Cu, Mo, Ag prefer half-filled or fully-filled d orbitals.

  4. Period length pattern: 2, 8, 8, 18, 18, 32, 32 - comes from orbital filling capacity.

  5. For JEE: Master configurations 1-30, know the traps (d-block group calculation), practice ion-based questions.

“The periodic table is like a city map - once you know how the streets are organized, you can find any address!”


Prerequisites

Next Steps in Periodic Classification

Cross-Subject Connections