Atomic Theory
Real-Life Hook: Why Gold Doesn’t Rust but Iron Does
Ever wondered why your grandmother’s gold jewelry looks as good as new after decades, while an iron gate rusts in just a few years? The answer lies in atomic theory - gold atoms are incredibly stable and unreactive, while iron atoms readily combine with oxygen. Understanding atomic theory isn’t just academic - it explains everything from why diamonds are forever to how medicine works in your body at the molecular level.
Introduction
The concept of atoms forms the foundation of modern chemistry. From ancient Greek philosophers to modern quantum mechanics, our understanding of matter’s fundamental building blocks has evolved dramatically. This journey from Dalton’s billiard ball model to today’s quantum mechanical model is essential for JEE aspirants.
Dalton’s Atomic Theory (1808)
The Five Postulates
John Dalton proposed his atomic theory based on experimental evidence from chemical reactions:
- Matter is made of atoms: All matter consists of extremely small, indivisible particles called atoms
- Atoms of same element are identical: All atoms of a given element have identical mass and properties
- Atoms of different elements differ: Atoms of different elements have different masses and properties
- Atoms combine in simple ratios: Atoms combine in simple whole-number ratios to form compounds
- Atoms are neither created nor destroyed: In chemical reactions, atoms are rearranged, not created or destroyed
Experimental Basis
Dalton’s theory successfully explained:
- Law of Conservation of Mass (Lavoisier, 1789): Mass is neither created nor destroyed in chemical reactions
- Law of Definite Proportions (Proust, 1799): A pure compound always contains the same elements in the same proportion by mass
- Law of Multiple Proportions (Dalton, 1803): When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio
Law of Multiple Proportions - Example
Consider carbon and oxygen forming CO and CO₂:
| Compound | Mass of C (g) | Mass of O (g) | Ratio of O |
|---|---|---|---|
| CO | 12 | 16 | 16 |
| CO₂ | 12 | 32 | 32 |
Ratio of oxygen masses = 32:16 = 2:1 (simple whole number)
$$\boxed{\text{Mass ratio of oxygen in CO₂ : CO = 2:1}}$$Limitations of Dalton’s Atomic Theory
What Dalton Got Wrong
- Atoms are divisible: Discovery of electrons (1897), protons (1919), and neutrons (1932) proved atoms have internal structure
- Isotopes exist: Atoms of the same element can have different masses (e.g., ¹²C and ¹⁴C)
- Isobars exist: Atoms of different elements can have the same mass (e.g., ⁴⁰Ar and ⁴⁰Ca)
- Allotropes differ: Same element can have different properties (e.g., diamond vs graphite)
- Atoms can be created/destroyed: Nuclear reactions can convert mass to energy (E = mc²)
Memory Trick: “DILAN’s Limitations”
- Divisible (atoms have parts)
- Isotopes (same element, different mass)
- Lacking explanation for bonding
- Allotropes (same element, different properties)
- Nuclear reactions (atoms can transform)
Modern Atomic Theory
Key Concepts
- Atom has structure: Consists of nucleus (protons + neutrons) and electrons
- Quantized energy levels: Electrons exist in specific energy states
- Quantum mechanical model: Electrons behave as both particles and waves
- Probability distribution: Electron location described by probability (orbitals)
- Nuclear reactions possible: Atoms can be transmuted via nuclear processes
Fundamental Particles
| Particle | Symbol | Charge | Mass (amu) | Location |
|---|---|---|---|---|
| Proton | p⁺ | +1 | 1.00728 | Nucleus |
| Neutron | n⁰ | 0 | 1.00866 | Nucleus |
| Electron | e⁻ | -1 | 0.000549 | Around nucleus |
Interactive Demo: Visualize Bohr Model
Explore the atomic structure and understand how protons, neutrons, and electrons are arranged.
Atomic Notation
$$\boxed{^{A}_{Z}X^{\text{charge}}}$$Example: $^{23}_{11}\text{Na}^{+}$ means:
- Mass number A = 23
- Atomic number Z = 11
- Charge = +1 (lost 1 electron)
- Neutrons = 23 - 11 = 12
- Electrons = 11 - 1 = 10
Isotopes, Isobars, and Isotones
Isotopes
Same atomic number (Z), different mass number (A)
Examples:
- Hydrogen isotopes: ¹H (protium), ²H (deuterium), ³H (tritium)
- Carbon isotopes: ¹²C, ¹³C, ¹⁴C
- Oxygen isotopes: ¹⁶O, ¹⁷O, ¹⁸O
Isobars
Different atomic number (Z), same mass number (A)
Examples:
- ⁴⁰Ar (18 protons) and ⁴⁰Ca (20 protons)
- ¹⁴C (6 protons) and ¹⁴N (7 protons)
Isotones
Same number of neutrons (n), different Z and A
Examples:
- ¹⁴C (8 neutrons) and ¹⁵N (8 neutrons)
- ³⁰Si (16 neutrons) and ³¹P (16 neutrons)
Memory Trick: “ZIP Code”
- Z same = Isotopes
- Isobars = A (mass) same
- Protones… wait, neutrons! = Isotones
Average Atomic Mass Calculations
Formula
For an element with isotopes:
$$\boxed{M_{\text{avg}} = \frac{M_1 \times a_1 + M_2 \times a_2 + \ldots}{100}}$$Where:
- M = mass of isotope
- a = % abundance
Solved Example 1: Chlorine
Chlorine has two isotopes: ³⁵Cl (75%) and ³⁷Cl (25%). Calculate average atomic mass.
Solution:
$$M_{\text{avg}} = \frac{35 \times 75 + 37 \times 25}{100}$$ $$M_{\text{avg}} = \frac{2625 + 925}{100} = \frac{3550}{100} = 35.5 \text{ amu}$$ $$\boxed{M_{\text{avg}} = 35.5 \text{ amu}}$$Solved Example 2: Reverse Calculation
Boron has average atomic mass 10.81 amu. It has two isotopes: ¹⁰B and ¹¹B. Find their % abundances.
Solution: Let abundance of ¹⁰B = x%, then ¹¹B = (100-x)%
$$10.81 = \frac{10x + 11(100-x)}{100}$$ $$1081 = 10x + 1100 - 11x$$ $$1081 = 1100 - x$$ $$x = 19\%$$ $$\boxed{\text{¹⁰B = 19\%, ¹¹B = 81\%}}$$Common Calculation Mistakes
Mistake 1: Confusing Mass Number with Atomic Mass
❌ Wrong: Using mass number (12) directly in calculations ✓ Right: Use atomic mass (12.011 amu) from periodic table
Mistake 2: Forgetting to Convert Percentages
❌ Wrong: $M_{\text{avg}} = 35 \times 0.75 + 37 \times 0.25$ (gives 35.5, correct by luck!) ✓ Right: $M_{\text{avg}} = \frac{35 \times 75 + 37 \times 25}{100} = 35.5$
Mistake 3: Electrons in Ions
❌ Wrong: Na⁺ has 11 electrons ✓ Right: Na⁺ has 10 electrons (lost 1)
Mistake 4: Isotope vs Isobar
❌ Wrong: ¹²C and ¹⁴C are isobars ✓ Right: ¹²C and ¹⁴C are isotopes (same Z, different A)
Practice Problems
Level 1: Basic (JEE Main)
Q1. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass? a) Atoms are indivisible b) Atoms of the same element are identical c) Atoms can neither be created nor destroyed d) Atoms combine in simple ratios
Answer
c) Atoms can neither be created nor destroyed - This directly explains why mass is conserved in reactions.Q2. An element has two isotopes with mass numbers 10 and 11 in the ratio 1:4. Calculate average atomic mass.
Solution
Ratio 1:4 means 20% and 80% $$M_{\text{avg}} = \frac{10 \times 20 + 11 \times 80}{100} = \frac{200 + 880}{100} = 10.8 \text{ amu}$$Q3. How many electrons are present in Na⁺ ion? (Z = 11)
Answer
10 electrons (11 - 1 = 10, lost one electron to form cation)Level 2: Intermediate (JEE Main/Advanced)
Q4. Carbon and oxygen form two compounds. In the first, 1.5 g of carbon combines with 2 g of oxygen. In the second, 1.5 g of carbon combines with 4 g of oxygen. Show that this illustrates the law of multiple proportions.
Solution
For fixed mass of carbon (1.5 g): - Compound 1: 2 g oxygen - Compound 2: 4 g oxygenRatio = 4:2 = 2:1 (simple whole number)
This proves law of multiple proportions. The compounds are CO and CO₂.
Q5. Naturally occurring boron consists of two isotopes whose atomic masses are 10.01 and 11.01. The atomic mass of natural boron is 10.81. Calculate the percentage of each isotope.
Solution
Let x = % of ¹⁰B, then (100-x) = % of ¹¹B $$10.81 = \frac{10.01x + 11.01(100-x)}{100}$$ $$1081 = 10.01x + 1101 - 11.01x$$ $$1081 = 1101 - x$$ $$x = 20\%$$Answer: ¹⁰B = 20%, ¹¹B = 80%
Q6. An atom of an element has mass number 23 and 12 neutrons. What is the electronic configuration of the cation formed by losing 1 electron?
Solution
Z = A - n = 23 - 12 = 11 (Sodium)Na has 11 electrons: 1s² 2s² 2p⁶ 3s¹
Na⁺ has 10 electrons: 1s² 2s² 2p⁶
Answer: 1s² 2s² 2p⁶ or [Ne]
Level 3: Advanced (JEE Advanced - Tricky)
Q7. Three isotopes of an element have mass numbers A, (A+1), and (A+2). If their abundances are x%, y%, and z% respectively, and the average atomic mass is (A+0.5), find the relation between x, y, and z.
Solution
$$A + 0.5 = \frac{Ax + (A+1)y + (A+2)z}{100}$$ $$100A + 50 = Ax + Ay + y + Az + 2z$$ $$100A + 50 = A(x+y+z) + y + 2z$$Since x + y + z = 100:
$$100A + 50 = 100A + y + 2z$$ $$50 = y + 2z$$Answer: y + 2z = 50
Q8. Which of the following pairs are isotones? a) ³⁰Si and ³¹P b) ⁴⁰Ar and ⁴⁰Ca c) ¹²C and ¹⁴N d) ¹⁴C and ¹⁴N
Solution
Calculate neutrons (n = A - Z):a) ³⁰Si: n = 30-14 = 16; ³¹P: n = 31-15 = 16 ✓ Isotones b) ⁴⁰Ar: n = 40-18 = 22; ⁴⁰Ca: n = 40-20 = 20 (Isobars, not isotones) c) ¹²C: n = 12-6 = 6; ¹⁴N: n = 14-7 = 7 d) ¹⁴C: n = 14-6 = 8; ¹⁴N: n = 14-7 = 7 (Isobars, not isotones)
Answer: a) only
Q9. (JEE Advanced 2018 Pattern) Rubidium has two isotopes ⁸⁵Rb (72.15%) and ⁸⁷Rb. The average atomic mass is 85.47 amu. A sample contains 10²⁰ atoms of Rb. How many atoms of ⁸⁷Rb are present?
Solution
Since ⁸⁵Rb = 72.15%, then ⁸⁷Rb = 27.85%Number of ⁸⁷Rb atoms = $10^{20} \times \frac{27.85}{100}$
$$= 2.785 \times 10^{19} \text{ atoms}$$Answer: 2.785 × 10¹⁹ atoms
Q10. (Conceptual Tricky) If Dalton’s atomic theory were completely correct, which of the following would NOT exist? a) CO and CO₂ b) Diamond and graphite c) ¹²C and ¹⁴C d) CH₄
Answer
Multiple answers: b) and c)b) Diamond and graphite are allotropes - same element with different properties, which contradicts Dalton’s postulate that all atoms of same element are identical in properties.
c) ¹²C and ¹⁴C are isotopes - same element with different masses, which contradicts Dalton’s postulate that all atoms of same element have identical mass.
Dalton’s theory CAN explain: a) Multiple proportions, d) Simple compounds
Cross-Links and Further Study
Related Topics
- Atomic Structure - Detailed quantum mechanical model
- Periodic Classification - How atomic theory led to periodic table
- Stoichiometry - Applications of atomic mass in calculations
- Mole Concept - Counting atoms and molecules
For JEE Advanced Students
- Quantum mechanical atomic model (detailed in Atomic Structure)
- Mass spectrometry for isotope determination
- Nuclear chemistry and radioactive isotopes
- Average atomic mass in complex multi-isotope systems
Quick Revision Notes
Dalton’s Theory - Key Points
- Matter made of indivisible atoms ❌ (disproved)
- Same element = identical atoms ❌ (isotopes exist)
- Different elements = different atoms ✓
- Simple whole number ratios ✓
- Atoms conserved in reactions ✓ (chemical, not nuclear)
Modern Atomic Theory - Key Points
- Atoms have internal structure (nucleus + electrons)
- Isotopes: same Z, different A
- Isobars: different Z, same A
- Isotones: same neutrons
- Average atomic mass from weighted abundance
Important Formulas
- Mass number: A = Z + n
- Average atomic mass: $M_{\text{avg}} = \sum (M_i \times a_i)/100$
- Electrons in ion: e⁻ = Z - charge
Tips for JEE
- Always check units - amu for atomic mass, not mass number
- Isotope problems - Set up equations carefully with percentages
- Ion electron count - Subtract for cations, add for anions
- Multiple proportions - Fix one element’s mass, compare other
- Read carefully - Distinguish between isotopes, isobars, isotones
Last updated: May 2025 Part of JEE Chemistry - Basic Concepts series