Top Structure of Atom PYQs for JEE & NEET | Chemca

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Structure of Atom: The Ultimate PYQ Collection for JEE & NEET 2026

Updated June 2026 8 min read By Chemca Expert Faculty

The "Structure of Atom" (Chapter 2 of Class 11 Chemistry) is arguably one of the most foundational topics for both the JEE (Main & Advanced) and NEET examinations. Mastering concepts like Bohr's Model, Quantum Numbers, and the Dual Nature of Matter is critical not just for direct questions, but as a prerequisite for Chemical Bonding, Thermodynamics, and Organic Chemistry.

Why Practice PYQs for Atomic Structure?

Analyzing Previous Year Questions (PYQs) is the most effective way to understand the exam pattern. The NTA frequently tests numericals on Rydberg's formula, de Broglie wavelength, and conceptual questions on quantum mechanical models. Below, we have compiled the most high-yield PYQs to accelerate your preparation.

Table of Contents

• 1. Essential Revision Resources & Flashcards
• 2. Bohr's Model & Atomic Spectra
• 3. Quantum Numbers & Electronic Configuration
• 4. De Broglie, Heisenberg & Photoelectric Effect
• 5. Advanced Mixed Concepts

1. Essential Revision Resources

Before diving into the complex numericals, ensure your theoretical foundation is solid. Use these curated notes and flashcards to quickly revise the chapter:

Complete Chapter Notes

Detailed Class 11 theory and formulas.

Structure of Atom Flashcards

Quick revision for definitions & constants.

100 Short Q&A

Test your fundamental understanding.

Mistake Bank

Avoid common pitfalls in atomic structure.

2. Bohr's Model, Energy & Atomic Spectra

Questions from this section heavily feature the Rydberg equation, ionization enthalpies, and calculations relating to radius, velocity, and energy of orbits.

• Determine the highest kinetic energy in Bohr's Model
• Calculate the kinetic energy of an electron in the second Bohr orbit
• Find the radius of the third Bohr orbit of He+ ion (JEE)
• Calculate the shortest wavelength of H in the Lyman series
• Find the wavelength of the second line of the Paschen series
• Number of spectral lines from 5th excited state to first excited state
• Calculate based on the ionization enthalpy of a Hydrogen atom

3. Quantum Numbers & Electronic Configuration

Understanding the four quantum numbers (n, l, m, s) and rules like Aufbau, Pauli, and Hund's rule is essential. These questions often test exceptions and orbital structures.

• Determine the quantum numbers of the 19th electron in Chromium (Cr)
• Find the number of orbitals with n=5 and ml=2 (JEE)
• Calculate the degeneracy of the second excited state of H atom
• Count the completely filled orbitals with ml=0 in Germanium (Ge)

4. De Broglie, Heisenberg & Photoelectric Effect

This section covers the quantum mechanical approach. Focus on the mathematical relationships between momentum, uncertainty, wavelength, and threshold energy.

Matter Waves & Uncertainty

• Ratio of de Broglie wavelengths of Li3+ and proton
• Calculate wavelength associated with a proton moving at specific speed
• Wavelength of electrons accelerated from rest
• Solve uncertainty in finding the location of an electron
• Calculate minimum uncertainty in the speed of an electron
• Electron moving with a speed of 600 m/s with an accuracy of 0.005%

Photoelectric Effect & Photons

• Determine velocity of ejected electron in photoelectric effect
• Identify the number of metals showing photoelectric effect under given conditions
• Calculate number of photons emitted per second by a 50W bulb
• A gas absorbs a photon of 355 nm and emits two wavelengths
• Energy required to break one mole of Cl-Cl bonds and photon wavelength

5. Advanced & Mixed Concepts (JEE Advanced)

These questions blend multiple concepts or require a deeper understanding of theoretical postulates, common in JEE Advanced patterns.

• Wavelength of emitted photon in He+ transition (JEE Advanced PYQ)
• Calculate the change in velocity of He atom after photon absorption
• Determine the lowest possible value of a/b for hydrogen-like species (JEE Advanced)
• Conceptual analysis: Number of incorrect statements about a black body

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How to Use This Resource Effectively

1. Attempt First: Always try to solve the PYQ on your own before looking at the solution on Chemca.
2. Note Formulas: Maintain a formula sheet specifically for Atomic Structure (constants like Planck's, Rydberg's, and electron mass are crucial).
3. Analyze Mistakes: If you get a question wrong, use the Mistake Bank to understand why. Was it a conceptual error or a calculation error?

Ready for more?

Explore our complete hub for Class 11 Chemistry to access chapter-wise notes, quizzes, and more PYQs.

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Tags: #JEE2026 #NEET2026 #StructureOfAtom #ChemistryPYQs #Class11

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n-Factor in Intramolecular Redox Reactions | Chemca

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Physical Chemistry Redox Reactions Study Material Chemca Hindi

Advanced Redox Concepts

n-Factor in Intramolecular Redox Reactions

What do you do when one element in a compound gets oxidized and another element in the same compound gets reduced? Master this advanced equivalent weight concept for JEE and NEET.

By Abhishek Sengar 10 Min Read

Table of Contents

• What is Intramolecular Redox?
• The Golden Rule for n-Factor
• Case Study 1: Decomposition of KClO₃
• Case Study 2: Ammonium Dichromate
• The Deadly Addition Trap
• Frequently Asked Questions

Standard redox reactions usually involve two distinct reactants: an oxidizing agent and a reducing agent. But sometimes, a single chemical compound acts as both. When different elements within the same molecule undergo simultaneous oxidation and reduction, the process is called an Intramolecular Redox Reaction.

Calculating the equivalent weight of such compounds in JEE Main, JEE Advanced, and NEET frequently causes panic because students don't know which element to use for the n-factor calculation.

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The Golden Rule for n-Factor

In any balanced redox reaction, the total number of electrons lost must equal the total number of electrons gained. Therefore, for an intramolecular redox compound, the n-factor is:

n-factor = | Total e^- lost by oxidized element |
OR
n-factor = | Total e^- gained by reduced element |

CRUCIAL: You calculate one OR the other. You NEVER add them together!

Case Study 1: Decomposition of Potassium Chlorate (KClO₃)

This is a classic NTA favorite. When KClO₃ is heated, it decomposes to form potassium chloride and oxygen gas.

Step-by-Step Calculation

KClO₃ → KCl + 3/2 O₂

The Reduction Part

• Element: Chlorine (Cl)
• Initial State: +5 (in KClO₃)
• Final State: -1 (in KCl)
• Change per atom: Gains 6 e^-
• Atoms per molecule: 1
• Total e^- gained = 6

The Oxidation Part

• Element: Oxygen (O)
• Initial State: -2 (in KClO₃)
• Final State: 0 (in O₂)
• Change per atom: Loses 2 e^-
• Atoms per molecule: 3
• Total e^- lost = 3 × 2 = 6

n-factor of KClO₃ = 6

Case Study 2: Decomposition of Ammonium Dichromate ((NH₄)₂Cr₂O₇)

This reaction (the "chemical volcano") is the ultimate test of your understanding. Here, Nitrogen is oxidized, and Chromium is reduced.

Step-by-Step Calculation

(NH₄)₂Cr₂O₇ → N₂ + Cr₂O₃ + 4 H₂O

The Oxidation Part

• Element: Nitrogen (N)
• Initial State: -3 (in NH₄⁺)
• Final State: 0 (in N₂)
• Change per atom: Loses 3 e^-
• Atoms per molecule: 2
• Total e^- lost = 2 × 3 = 6

The Reduction Part

• Element: Chromium (Cr)
• Initial State: +6 (in Cr₂O₇^2-)
• Final State: +3 (in Cr₂O₃)
• Change per atom: Gains 3 e^-
• Atoms per molecule: 2
• Total e^- gained = 2 × 3 = 6

n-factor of (NH₄)₂Cr₂O₇ = 6

The Deadly Addition Trap

In both examples above, the electrons lost (6) perfectly match the electrons gained (6), confirming the reaction is balanced internally.

Equivalent Weight Calculation

Equivalent Weight = Molecular Weight (M)

                    n-factor

WRONG Answer

Eq. Wt = M / 12

(Student added 6 lost + 6 gained)

CORRECT Answer

Eq. Wt = M / 6

(Student took the magnitude of transfer)

Frequently Asked Questions

The n-factor for a compound undergoing intramolecular redox is equal to either the total number of moles of electrons lost by the oxidized element OR the total number of moles of electrons gained by the reduced element. You must account for all atoms of that element in the molecule, but do NOT add the oxidation and reduction electrons together.

The n-factor of Potassium Chlorate (KClO₃) is 6. The single chlorine atom gains 6 electrons (reducing from +5 to -1), while the three oxygen atoms collectively lose 6 electrons (each oxidizing from -2 to 0).

No. In an intramolecular redox reaction, different elements in the same molecule are oxidized and reduced (e.g., N and Cr in Ammonium Dichromate). In a disproportionation reaction, the exact same element in a molecule is simultaneously oxidized and reduced (e.g., H₂O₂ decomposing into H₂O and O₂).

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n-Factor When Multiple Elements are Oxidized: Ferrous Oxalate & Cu2S | Chemca

Chemca

Physical Chemistry Redox Titrations Study Material Chemca Hindi

Advanced Equivalent Concept

n-Factor: When Multiple Elements are Oxidized in the Same Compound

What happens when an oxidizing agent attacks two different elements in the same molecule? Master the "Addition Rule" to conquer the hardest titration questions in JEE and NEET.

By Abhishek Sengar 10 Min Read

Table of Contents

• The Scenario
• The Addition Rule
• Case 1: Ferrous Oxalate (Most Asked)
• Case 2: Cuprous Sulfide
• The Subscript Trap
• Frequently Asked Questions

In standard redox reactions, usually, only one element in a compound undergoes a change in oxidation state. However, in advanced titration problems, you will encounter compounds where a strong oxidizing agent (like KMnO₄ or K₂Cr₂O₇) simultaneously oxidizes two or more elements within the same reactant molecule.

If you don't calculate the total electron transfer correctly, your equivalent weight calculation will be completely wrong.

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The Addition Rule for n-Factor

Unlike intramolecular redox (where one element oxidizes and another reduces, and we DO NOT add them), here both elements are undergoing the same process (oxidation). Therefore, the molecule as a whole is acting as a massive electron donor.

Total n-factor = Σ (Moles of e^- lost by Element 1) + (Moles of e^- lost by Element 2) ...

Rule of Thumb: When elements suffer the same fate (both oxidized or both reduced), ADD their individual n-factors.

Case Study 1: Ferrous Oxalate (FeC₂O₄)

This is arguably the most frequently asked question in JEE physical chemistry. When Ferrous Oxalate reacts with an acidic solution of KMnO₄, the permanganate oxidizes both the Ferrous ion and the Oxalate ion.

Step-by-Step Breakdown

FeC₂O₄ → Fe³⁺ + 2 CO₂

Element 1: Iron (Fe)

• Process: Oxidation
• Initial State: +2 (Ferrous)
• Final State: +3 (Ferric)
• Change: Loses 1 e^- per atom
• Atoms per molecule: 1
• e^- lost by Fe = 1 × 1 = 1

Element 2: Carbon (C)

• Process: Oxidation
• Initial State: +3 (in Oxalate)
• Final State: +4 (in Carbon Dioxide)
• Change: Loses 1 e^- per atom
• Atoms per molecule: 2
• e^- lost by C = 2 × 1 = 2

Total n-factor = 1 (from Fe) + 2 (from C) = 3

Therefore, Equivalent Weight of FeC₂O₄ = M / 3

Case Study 2: Cuprous Sulfide (Cu₂S)

This is a slightly more advanced example often seen in JEE Advanced. When Cuprous Sulfide is treated with a strong oxidizing agent, both Copper and Sulfur are oxidized to their higher stable states.

Step-by-Step Breakdown

Cu₂S → 2 Cu²⁺ + SO₂

Element 1: Copper (Cu)

• Initial State: +1 (Cuprous)
• Final State: +2 (Cupric)
• Change: Loses 1 e^- per atom
• Atoms per molecule: 2
• e^- lost by Cu = 2 × 1 = 2

Element 2: Sulfur (S)

• Initial State: -2 (Sulfide)
• Final State: +4 (in Sulfur Dioxide)
• Change: Loses 6 e^- per atom
• Atoms per molecule: 1
• e^- lost by S = 1 × 6 = 6

Total n-factor = 2 (from Cu) + 6 (from S) = 8

Therefore, Equivalent Weight of Cu₂S = M / 8

The Subscript Trap

The number one mistake students make is calculating the change in oxidation state correctly but forgetting to multiply by the subscript of the element in the parent molecule.

Take Ferrous Oxalate (FeC₂O₄) as an example:

The Mistake

Fe loses 1.

Carbon loses 1.

n-factor = 1 + 1 = 2

(Forgot there are TWO carbons!)

The Correct Way

Fe loses 1.

Carbon loses 1 × 2 = 2.

n-factor = 1 + 2 = 3

(Multiplied by the subscript C₂)

Frequently Asked Questions

The n-factor of Ferrous Oxalate (FeC₂O₄) is 3. When reacted with an oxidizing agent, the iron oxidizes from +2 to +3 (losing 1 electron). Simultaneously, the two carbon atoms in the oxalate ion oxidize from +3 to +4 (losing a total of 2 electrons). The total electrons lost per molecule is 1 + 2 = 3.

When multiple elements in the same compound are oxidized, you calculate the individual number of electrons lost by each element (making sure to multiply by their stoichiometric subscript in the molecule) and ADD them together. The sum is the total n-factor of the compound.

The n-factor of Cuprous Sulfide (Cu₂S) is usually 8. The two copper atoms oxidize from +1 to +2 (losing 2 electrons total), and the single sulfur atom oxidizes from -2 to +4 to form SO₂ (losing 6 electrons). The total electron loss is 2 + 6 = 8.

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