Why this chapter is worth double credit
The Bohr energy levels, the hydrogen spectral series, de Broglie wavelengths and the uncertainty principle all appear in the Modern Physics chapter of the physics paper. They are the same results with the same formulas, and a student who prepares them properly for one paper has prepared them for the other.
What Atomic Structure adds is the quantum-mechanical model - four quantum numbers, orbital shapes, and the filling rules that generate the periodic table. That part is chemistry-specific, and it is where the chapter's distinctive questions live.
How JEE actually asks Atomic Structure
NTA publishes no chapter-wise weightage, so any figure online is a coaching estimate from past papers; confirm your session's syllabus and pattern in the official NTA information bulletin.
Questions divide between numerical work on the hydrogen atom - energies, radii, spectral transitions - and recall-based questions on quantum numbers, orbital capacity and electronic configurations. The chromium and copper exceptions appear regularly, as do questions distinguishing an orbit from an orbital.
Key concepts, compressed
- ✓Bohr quantised angular momentum, which produced discrete energy levels and explained the hydrogen spectrum - but only for one-electron systems.
- ✓The quantum model replaces definite orbits with orbitals: regions where an electron is likely to be found.
- ✓Four quantum numbers specify an electron completely - shell, subshell, orientation and spin.
- ✓The Pauli exclusion principle forbids any two electrons in an atom from sharing all four quantum numbers.
- ✓Aufbau filling follows increasing n + l, with lower n breaking ties.
- ✓Hund's rule requires degenerate orbitals to be filled singly with parallel spins before any pairing.
Relations you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| Bohr energy | E(n) = −13.6 Z²/n² eV | hydrogen-like only |
| Bohr radius | r(n) = 0.529 n²/Z Å | |
| Rydberg equation | 1/λ = R Z²(1/n₁² − 1/n₂²) | |
| Values of l | 0 to n − 1 | n values in total |
| Values of m | −l to +l | 2l + 1 values |
| Orbitals in shell n | n² | |
| Maximum electrons in shell n | 2n² | |
| Subshell capacity | s 2, p 6, d 10, f 14 | |
| Filling order | increasing n + l, then increasing n | |
| de Broglie wavelength | λ = h/mv | |
| Accelerated electron | λ = 12.27/√V Å | V in volts |
| Uncertainty principle | Δx · Δp ≥ h/4π | |
| Radial nodes | n − l − 1 | total nodes = n − 1 |
| Magnetic moment | μ = √(n(n + 2)) BM | n unpaired electrons |
The five mistakes that cost the most marks
- ✓Applying the Bohr formula to multi-electron atoms. It holds only for hydrogen and one-electron ions such as He⁺ and Li²⁺.
- ✓Filling orbitals by n alone rather than by n + l. That ordering is why 4s precedes 3d and why the transition series appears where it does.
- ✓Forgetting the chromium and copper exceptions, which are asked far more often than their rarity suggests.
- ✓Confusing an orbit with an orbital. The first is a definite path, the second a probability region, and only the second is physically meaningful.
- ✓Assuming transition metals lose 3d electrons first when forming ions. Once occupied, 4s is higher in energy and is emptied first.
Practice set 1: atomic models and the Bohr atom
1. What did the Thomson, Rutherford and Bohr models each contribute?
Thomson proposed electrons embedded in a positive sphere. Rutherford's scattering experiment showed the positive charge is concentrated in a tiny nucleus, but his model could not explain why orbiting electrons do not spiral in. Bohr added quantised orbits, which fixed that problem and explained the hydrogen spectrum - at the cost of an assumption with no classical justification.
2. What is the energy of an electron in the n = 2 level of hydrogen?
−3.4 eV. Using E(n) = −13.6/n² = −13.6/4 = −3.4 eV. The negative sign means the electron is bound; zero corresponds to a free electron at rest infinitely far from the nucleus. Levels crowd together as n increases, converging on zero.
3. What is the ionisation energy of hydrogen from its ground state?
13.6 eV. Ionisation means moving the electron from n = 1, where E = −13.6 eV, to n = ∞ where E = 0, so the energy required is the difference. This single number anchors the whole chapter and should be known without calculation.
4. What is the radius of the n = 2 orbit in hydrogen?
About 2.116 Å. Using r(n) = 0.529 n²/Z Å with n = 2 and Z = 1 gives 0.529 × 4 = 2.116 Å. Radius grows as n² while energy scales as 1/n², so outer orbits are much larger and much less tightly bound - which is why outer electrons do the chemistry.
5. What energy is released in a transition from n = 3 to n = 2 in hydrogen?
About 1.89 eV. E₃ = −13.6/9 ≈ −1.51 eV and E₂ = −3.4 eV, so the photon carries the difference of 1.89 eV. This is the first line of the Balmer series and lies in the visible red region, which is why hydrogen discharge tubes glow pink.
6. What does the Rydberg equation give?
The wavelength of a spectral line from a transition between two levels: 1/λ = R Z²(1/n₁² − 1/n₂²), with n₁ the lower level. Transitions ending at n = 1 give the Lyman series in the ultraviolet, at n = 2 the Balmer series in the visible, and at n = 3 the Paschen series in the infrared.
7. What were the limitations of the Bohr model?
It works only for one-electron systems, cannot explain fine spectral structure or the splitting of lines in magnetic and electric fields, and it assumes a definite orbit, which the uncertainty principle forbids. It was a necessary step rather than a final answer, and the quantum-mechanical model replaced it.
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Practice set 2: quantum numbers
8. What do the four quantum numbers describe?
The principal quantum number n gives the shell and main energy level; the azimuthal number l gives the subshell and orbital shape; the magnetic number m gives the orbital's orientation in space; and the spin number s gives the electron's spin. Together they specify an electron completely.
9. What values can l take when n = 3?
0, 1 and 2, corresponding to the 3s, 3p and 3d subshells. In general l runs from 0 to n − 1, giving n possible values. This is why the first shell has only an s subshell and why d orbitals cannot appear before the third shell.
10. How many values of m are possible when l = 2?
Five: −2, −1, 0, +1 and +2. In general there are 2l + 1 values, which is why there are five d orbitals, three p orbitals and one s orbital. Each value corresponds to a distinct spatial orientation of the same subshell.
11. How many orbitals are there in the n = 3 shell?
Nine. The total is n² = 9, made up of one 3s, three 3p and five 3d orbitals. Checking the breakdown against the formula is a quick way to confirm you have the subshells right.
12. What is the maximum number of electrons in the n = 3 shell?
Eighteen. Each orbital holds two electrons, so the capacity is 2n² = 18. Note that the third period of the periodic table contains only eight elements, because 3d fills after 4s - the shell capacity and the period length are different things.
13. State the Pauli exclusion principle.
No two electrons in an atom can have the same set of all four quantum numbers. Since an orbital fixes the first three, the two electrons in it must differ in spin - which is why an orbital holds a maximum of two electrons, and why the 2n² capacity arises.
14. What are the electron capacities of the s, p, d and f subshells?
2, 6, 10 and 14 respectively. Each follows from 2(2l + 1): the s subshell has one orbital, p has three, d has five and f has seven, and each orbital holds two electrons. These four numbers are worth knowing instantly, since they determine the shape of the periodic table.
Practice set 3: electronic configuration
15. What are the shapes of the s and p orbitals?
The s orbital is spherical, and each p orbital is dumbbell-shaped with two lobes along one axis and a node at the nucleus. The three p orbitals are identical in shape but oriented along the x, y and z axes. Orbital shape is determined by l, which is why all s orbitals are spherical regardless of n.
16. State the Aufbau principle and the n + l rule.
Electrons occupy the lowest-energy orbitals first. The order is determined by increasing n + l, with the lower n taking precedence when two subshells have the same n + l value. This produces the familiar sequence 1s, 2s, 2p, 3s, 3p, 4s, 3d and so on.
17. Why does 4s fill before 3d?
Because 4s has n + l = 4 + 0 = 4 while 3d has 3 + 2 = 5, so 4s is lower at the point of filling. Once both are occupied, the ordering reverses and 3d drops below 4s - which is why transition metals lose their 4s electrons first when forming ions.
18. State Hund's rule of maximum multiplicity.
Electrons occupy degenerate orbitals singly with parallel spins before any pairing begins. This minimises electron-electron repulsion and maximises exchange energy. It is why nitrogen has three unpaired electrons in its 2p subshell rather than one pair and one single electron, and it determines magnetic behaviour throughout the chapter.
19. What is the electronic configuration of chromium, and why is it anomalous?
[Ar]3d⁵4s¹ rather than the expected [Ar]3d⁴4s². Promoting one 4s electron into 3d produces a half-filled d subshell, which is unusually stable because of symmetric charge distribution and maximised exchange energy. The 4s and 3d energies are close enough that this small gain outweighs the promotion cost.
20. What is the electronic configuration of copper?
[Ar]3d¹⁰4s¹ rather than [Ar]3d⁹4s². The same reasoning as chromium applies, with a fully-filled d subshell providing the stability instead of a half-filled one. Chromium and copper are the two exceptions asked about most often, and they are worth memorising together with the reason.
21. Which electrons does a transition metal lose first when forming a cation?
The 4s electrons, despite 4s having been filled first. Once 3d is occupied it drops below 4s in energy, so 4s becomes the outermost and highest-energy subshell. This is why iron forms Fe²⁺ as [Ar]3d⁶ rather than [Ar]3d⁴4s², and it catches students who assume removal reverses the filling order.
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Practice set 4: wave nature of matter
22. State the de Broglie relation.
λ = h/mv - every particle has an associated wavelength inversely proportional to its momentum. For an electron the wavelength is comparable to atomic dimensions and therefore observable; for any everyday object it is far too small to measure, which is why wave behaviour is invisible at ordinary scales.
23. What is the de Broglie wavelength of an electron accelerated through 100 V?
About 1.227 Å. Using λ = 12.27/√V Å = 12.27/10 = 1.227 Å. This is comparable to interatomic spacing in crystals, which is exactly why electron diffraction works and why the Davisson-Germer experiment could confirm matter waves experimentally.
24. State the Heisenberg uncertainty principle.
Δx · Δp ≥ h/4π - position and momentum cannot both be known to arbitrary precision at the same time. It is a statement about nature rather than about measurement technique, and it is why the Bohr model's definite orbits cannot be correct.
25. What is the difference between an orbit and an orbital?
An orbit is a definite circular path of fixed radius, as in the Bohr model. An orbital is a three-dimensional region where an electron is likely to be found, described by a wave function. The uncertainty principle rules out definite paths, so only the orbital concept survives - and this is a standard one-line question.
26. How many radial and angular nodes does a 3p orbital have?
One radial node and one angular node, two in total. Radial nodes are n − l − 1 = 3 − 1 − 1 = 1, angular nodes equal l = 1, and the total is always n − 1 = 2. Checking that the two parts sum to n − 1 is a reliable way to confirm you have not miscounted.
Practice set 5: mixed
27. What are isotopes, isobars and isotones?
Isotopes share an atomic number but differ in mass number - same element, different neutron count. Isobars share a mass number but differ in atomic number. Isotones share a neutron count. Only isotopes are chemically alike, since chemistry depends on electron count, which follows the atomic number.
28. How is magnetic moment calculated from unpaired electrons?
μ = √(n(n + 2)) Bohr magnetons, where n is the number of unpaired electrons. So one unpaired electron gives 1.73 BM, two give 2.83, three give 3.87 and five give 5.92. This spin-only formula ignores orbital contribution, which is a reasonable approximation for first-row transition metals.
29. What are shielding and effective nuclear charge?
Inner electrons screen outer ones from the full nuclear charge, so an outer electron experiences an effective nuclear charge Zeff that is less than Z. Shielding is strongest from s electrons and weakest from d and f electrons, which is why poor d-orbital shielding causes the lanthanide contraction.
30. How does the photoelectric effect connect to this chapter?
It established that light delivers energy in discrete quanta, which is the same quantisation that Bohr applied to electron energy levels. The photon energy E = hf appears in both, and the spectral transitions of question 5 are photons being emitted. It is the clearest point where the chemistry and physics treatments meet.
How to study this chapter efficiently
- ✓Prepare it alongside Modern Physics. The Bohr, de Broglie and spectral material is shared, so the effort counts for both papers.
- ✓Memorise 13.6 eV and 0.529 Å as anchors, and derive the rest from n² and 1/n² scaling.
- ✓Write the n + l values when deciding filling order rather than trying to recall the sequence.
- ✓Learn chromium and copper together with the half-filled and fully-filled stability reason, so you can reconstruct them.
- ✓Check node counts by confirming radial plus angular equals n − 1.
- ✓Note where every formula stops applying - the Bohr expressions fail beyond one-electron systems, and that boundary is examined directly.
Turn this into active practice
The recall half of this chapter - quantum number ranges, subshell capacities, configuration exceptions - consolidates through short frequent testing far better than through one long reading session. There is nothing to derive, only to make retrievable.
The JEE Atomic Structure quiz on QUFF generates fresh questions across the Bohr model, quantum numbers, orbital filling and wave-particle behaviour, marks them instantly and explains each answer. Do short sets spaced across several days rather than one long block - spacing is what makes recall stick.