JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
Mathematics · Physics · Chemistry
This MCQ module is based on: Bohr Model Quantum Mechanics
TOPIC 7 OF 28
Bohr Model Quantum Mechanics
🎓 Class 11
Chemistry
CBSE
Theory
Ch 2 – Structure of Atom
⏱ ~14 min
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Bohr Model, Quantum Numbers and Electron Configuration
Introduction: From Quanta to Orbitals
In Part 2 we saw that light comes in packets of energy hν, and that atoms emit light at a discrete set of wavelengths. In 1913 Niels Bohr fused these two ideas into the first quantitatively successful model of an atom — hydrogen. His orbits, though later superseded, give exact answers for H-like ions and explain every line of the hydrogen spectrum on the back of an envelope.
The Bohr picture has limits of its own, and Parts 2.5 and 2.6 take the story into modern quantum mechanics: matter waves, the uncertainty principle, Schrödinger's orbitals, four quantum numbers, and the rules (Aufbau, Pauli, Hund) that predict the electron configuration of every element in the periodic table.
2.4 Bohr's Model for the Hydrogen Atom (1913)
Postulates
- An electron revolves around the nucleus only in certain permitted circular paths called stationary orbits or shells. While in such an orbit the electron does not radiate energy (breaks classical electromagnetism — this is a postulate).
- The angular momentum of the electron in a stationary orbit is quantised:
mvr = n (h/2π), n = 1, 2, 3, …
- Energy is absorbed or emitted only when the electron jumps between two stationary orbits. The frequency of the photon satisfies
ΔE = E2 − E1 = hν
Derived Expressions (H-like species, nuclear charge Z)
| Quantity | Formula | Value for H (Z = 1), n = 1 |
|---|---|---|
| Radius rn | 52.9 n2/Z pm | 52.9 pm (the Bohr radius a0) |
| Energy En | −2.18 × 10−18 (Z2/n2) J/atom = −13.6 (Z2/n2) eV | −2.18 × 10−18 J = −13.6 eV |
| Velocity vn | 2.188 × 106 (Z/n) m s−1 | 2.188 × 106 m s−1 |
Explanation of the Hydrogen Spectrum
When an electron drops from orbit n2 to orbit n1, the photon's frequency is
\( h\nu = E_{n_2}-E_{n_1}=-2.18\times10^{-18}\left[\dfrac{1}{n_2^2}-\dfrac{1}{n_1^2}\right]\text{ J}\)
Dividing by hc gives the wave number — the Rydberg formula of Part 2 drops out, and RH comes out as 1.09677 × 105 cm−1, exactly the empirical value. Each series (Lyman, Balmer, …) corresponds to a fixed landing shell n1.
2.4.1 Limitations of Bohr's Model
- Accurate only for one-electron species (H, He+, Li2+). Fails quantitatively for He, Li, …
- Cannot explain fine structure (close pairs of lines) seen at high resolution.
- Cannot explain splitting of lines in a magnetic field (Zeeman effect) or an electric field (Stark effect).
- Violates Heisenberg's uncertainty principle — a well-defined circular orbit implies precise simultaneous position and momentum.
- Cannot account for three-dimensional shapes of bonds and molecular geometry.
Worked Numericals on Bohr's Model
Problem — Energy of electron in 2nd Bohr orbit of H
Use \(E_n=-2.18\times10^{-18}(Z^2/n^2)\) J with Z = 1, n = 2:
\[ E_2=-2.18\times10^{-18}\cdot \dfrac{1}{4}=-5.45\times10^{-19}\ \text{J} \]
In eV: \(E_2=-13.6/4=-3.40\) eV.
Problem — Radius of 3rd Bohr orbit of He+
\(r_n = 52.9\,n^2/Z\) pm. For He+, Z = 2, n = 3:
\[ r_3 = 52.9 \times \dfrac{9}{2} = 238.05\ \text{pm} \]
Because nuclear charge is double that of H, the He+ orbits are half the size at each n.
Problem — Wavelength of H line for 3 → 2 transition
\[ \bar{\nu}=109677\Big[\dfrac{1}{4}-\dfrac{1}{9}\Big]=109677\cdot\dfrac{5}{36}=15\,233\ \text{cm}^{-1} \]
λ = 1/\(\bar{\nu}\) = 656.5 nm (red Hα line). Alternatively using ΔE:
\[ \Delta E = -2.18\times10^{-18}\big(1/9-1/4\big)=3.03\times10^{-19}\ \text{J} \]
\[ \lambda = hc/\Delta E = (6.626\times10^{-34}\cdot 3\times10^{8})/3.03\times10^{-19}=656\ \text{nm}\ \checkmark \]
Problem — Ionisation energy of hydrogen from Bohr
IE = energy to remove the electron from n = 1 to n = ∞:
\[ IE = 0 - E_1 = +2.18\times10^{-18}\ \text{J/atom} = +13.6\ \text{eV} \]
Per mole: IE = 2.18 × 10−18 × 6.022 × 1023 = 1312 kJ mol−1, matching the experimental value.
2.5 Towards the Quantum-Mechanical Model
Dual Behaviour of Matter (de Broglie, 1924)
If light — classically a wave — carries momentum p = h/λ, Louis de Broglie proposed that every moving particle should have an associated wavelength:
λ = h/p = h/(mv)
Macroscopic bodies have impossibly tiny wavelengths; for electrons moving at 106 m s−1, λ ≈ 700 pm — measurable. Davisson and Germer (1927) confirmed electron diffraction from a nickel crystal. The electron microscope is a practical consequence.
Example — Wavelength of a ball vs an electron
(a) A 0.1 kg ball at 10 m s−1:
\( \lambda=h/mv=6.626\times10^{-34}/(0.1\cdot10)=6.6\times10^{-34}\) m — undetectable.
(b) An electron at 106 m s−1:
\( \lambda=6.626\times10^{-34}/(9.11\times10^{-31}\cdot10^6)=7.27\times10^{-10}\) m = 7.27 Å.
Wave properties matter only for light, fast-moving microscopic particles.
Heisenberg's Uncertainty Principle (1927)
For any particle, position (x) and momentum (p) cannot be known simultaneously with arbitrary precision:
Δx · Δp ≥ h/4π
This is not a limitation of our instruments but a fundamental property of waves. A "sharp" particle (small Δx) must be built from a wide range of momenta (large Δp), and vice-versa.
Consequence: Bohr's neat circular orbit — with a definite radius and a definite speed — cannot describe a real electron. We abandon "orbits" and adopt probability clouds.
2.6 The Quantum-Mechanical Model of the Atom
Erwin Schrödinger (1926) wrote down a wave equation whose solutions — the wavefunctions ψ — give every allowed energy state of an electron in an atom. |ψ|2 is the probability density of finding the electron at that point. A region of 3-D space where the probability of finding the electron is appreciable (say 90 %) is called an atomic orbital.
Quantum Numbers
Four quantum numbers uniquely identify every electron in an atom.
| Symbol | Name | Values | What it tells us |
|---|---|---|---|
| n | Principal | 1, 2, 3, … | Shell; size and energy (n = 1 is K, 2 = L, 3 = M, 4 = N) |
| ℓ | Azimuthal (angular momentum) | 0 to n−1 | Subshell; shape. ℓ = 0 (s), 1 (p), 2 (d), 3 (f) |
| mℓ | Magnetic | −ℓ … 0 … +ℓ | Spatial orientation of the orbital (2ℓ + 1 values) |
| ms | Spin | +½ or −½ | Intrinsic angular momentum ("up/down" spin) |
For a given n: there are n subshells, n2 orbitals, and up to 2n2 electrons. E.g. n = 3 → subshells 3s (1 orbital), 3p (3 orbitals), 3d (5 orbitals) = 9 orbitals = 18 electrons maximum.
Shapes of Atomic Orbitals
Boundary surface diagrams show the surface within which there is 90% probability of locating the electron.
Energy of Orbitals and Electron Configuration
For hydrogen (a one-electron atom) orbital energies depend only on n — so 2s and 2p are degenerate. In multi-electron atoms, electron–electron repulsion and nuclear shielding split the degeneracy; the energy depends on the combination (n + ℓ).
(n + ℓ) Rule (Madelung rule). (i) The orbital with the lower (n + ℓ) value has the lower energy. (ii) If two orbitals have the same (n + ℓ), the one with the lower n is lower in energy.
The Three Filling Rules
- Aufbau principle. Electrons occupy orbitals in order of increasing energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.
- Pauli's exclusion principle. No two electrons in an atom can have all four quantum numbers identical. Consequence: an orbital holds at most 2 electrons, with opposite spins.
- Hund's rule of maximum multiplicity. In a given subshell, electrons occupy orbitals singly with parallel spins before pairing starts. (It minimises electron–electron repulsion and maximises exchange energy.)
Worked Electron Configurations
| Element (Z) | Expected | Actual | Reason |
|---|---|---|---|
| Cl (17) | 1s² 2s² 2p⁶ 3s² 3p⁵ = [Ne] 3s² 3p⁵ | Standard | |
| Fe (26) | [Ar] 3d⁶ 4s² | 4s fills before 3d | |
| Cr (24) | [Ar] 3d⁴ 4s² | [Ar] 3d⁵ 4s¹ | Half-filled 3d⁵ + half-filled 4s¹ = extra exchange stability |
| Cu (29) | [Ar] 3d⁹ 4s² | [Ar] 3d¹⁰ 4s¹ | Completely filled 3d¹⁰ is extra stable |
Example — Number of electrons with n = 3, ℓ = 2, mℓ = +1
n = 3, ℓ = 2 identifies the 3d(mℓ = +1) orbital — a single orbital.
Such a single orbital can hold at most 2 electrons (with opposite spins), by Pauli.
Example — Total electrons with n = 4 in a fully-filled atom
Maximum electrons = 2n² = 2(16) = 32.
Breakdown: 4s (2) + 4p (6) + 4d (10) + 4f (14) = 32.
Example — Write the electron configuration of Mn (Z = 25) and count unpaired electrons
[Ar] 3d⁵ 4s² — 3d⁵ = one electron in each of five d-orbitals (Hund).
Unpaired electrons = 5. Mn(II) with 3d⁵ is famously half-filled and stable.
Activity 2.3 — Standing-Wave Interpretation of Bohr's Quantisation L4 Analyse
Objective: Show that Bohr's quantisation of angular momentum is equivalent to fitting an integer number of de Broglie wavelengths around the orbit.
- Write de Broglie's expression λ = h/(mv).
- A standing wave requires the circumference of the orbit to be a whole-number multiple of λ: 2πr = nλ.
- Substitute λ from (1) into (2) and rearrange for mvr.
Predict: What expression do you get for angular momentum mvr?
From 2πr = nλ = n(h/mv), we get mvr = nh/(2π) — exactly Bohr's postulate. So the second postulate is not an arbitrary guess but the requirement that the electron's matter-wave close on itself after one revolution.
Electron Configuration Builder
Enter an atomic number (1 – 54). The tool returns the ground-state electron configuration using the Aufbau order and flags Cr / Cu exceptions.
Competency-Based Questions (with Bohr numericals)
A student is studying hydrogen-like ions He+ and Li2+. She also prepares orbital diagrams for Cr and Cu.
Q1. The energy of an electron in the ground state of He+ is:
C. E = −13.6 Z²/n² = −13.6 (4)/1 = −54.4 eV.
Q2. Calculate the radius of the 1st Bohr orbit of Li2+.
r₁ = 52.9 × 1²/3 ≈ 17.6 pm. Higher Z pulls the orbit closer to the nucleus.
Q3. State why the ground-state electron configuration of Cr is [Ar] 3d⁵ 4s¹ rather than [Ar] 3d⁴ 4s².
A half-filled 3d⁵ set together with a half-filled 4s¹ gives the atom additional exchange-energy stabilisation. Promotion of one 4s electron to 3d costs very little energy because 3d and 4s are almost degenerate, and the gain in exchange energy more than compensates.
Q4. Fill in the blank: The set of quantum numbers (n = 3, ℓ = 3, mℓ = 0, ms = +½) is ______ (valid / not valid).
Not valid. ℓ can only go up to n−1 = 2, so ℓ = 3 is impossible when n = 3.
Q5. Short answer: An electron jumps from n = 5 to n = 2 in H. What is the wavelength of the emitted photon?
\(\bar{\nu}=109677(1/4 - 1/25) = 109677(21/100) = 23\,032\) cm−1. λ ≈ 434 nm (violet Hγ line, Balmer series).
Assertion–Reason Questions
Options: A. Both true, R explains A. B. Both true, R does not explain A. C. A true, R false. D. A false, R true.
Assertion: Bohr's model fails to describe multi-electron atoms.
Reason: Bohr's model neglects electron–electron repulsion.
A. Both true; the omitted repulsion becomes significant the moment a second electron is added, so Bohr's single-electron derivation is no longer exact.
Assertion: It is impossible to know simultaneously the exact position and momentum of an electron.
Reason: Electrons behave like particles only.
C. Assertion is Heisenberg's principle — true. Reason is false: electrons also show wave behaviour.
Assertion: Cu has electron configuration [Ar] 3d¹⁰ 4s¹ rather than [Ar] 3d⁹ 4s².
Reason: Completely filled d-subshells give additional stability.
A. Both true; the extra exchange energy of the fully-filled 3d¹⁰ set drives the 4s → 3d promotion.
Frequently Asked Questions — Bohr Model, Quantum Numbers and Electron Configuration
What are the postulates of Bohr's atomic model?
Bohr's atomic model (1913) is built on four postulates studied in NCERT Class 11 Chemistry: (1) electrons revolve around the nucleus only in certain fixed circular orbits called stationary states where they do not radiate energy; (2) the angular momentum of an electron in these orbits is quantised: mvr = nh/2π; (3) electrons can jump between orbits by absorbing or emitting energy equal to the difference in orbit energies: ΔE = hν; (4) the lowest energy state is the ground state. The model successfully explained the hydrogen spectrum but failed for multi-electron atoms.
What is the de Broglie hypothesis?
The de Broglie hypothesis (1924) states that every moving particle has an associated wave with wavelength λ = h/(mv), where h is Planck's constant, m is mass and v is velocity. This dual wave-particle nature is significant only for subatomic particles like electrons. For example, an electron at 10⁶ m/s has λ ≈ 7.27 × 10⁻¹⁰ m — comparable to atomic dimensions, so wave behaviour matters. In NCERT Class 11 Chemistry, de Broglie's idea led to electron diffraction experiments and motivated Schrödinger's wave mechanics, which replaced Bohr's fixed-orbit picture.
What is the Heisenberg uncertainty principle?
The Heisenberg uncertainty principle (1927) states that it is impossible to simultaneously measure both the exact position and exact momentum of a microscopic particle. Mathematically: Δx · Δp ≥ h/4π. This is not a measurement limitation but a fundamental property of matter at quantum scales. In NCERT Class 11 Chemistry, the principle invalidates Bohr's idea of well-defined electron orbits with fixed radii. Instead, we describe regions where electrons are likely to be found, called orbitals. The principle is foundational to all of quantum mechanics.
What are atomic orbitals?
An atomic orbital is a three-dimensional region around the nucleus where the probability of finding an electron is maximum (usually drawn to enclose 90–95% probability). Orbitals are solutions to the Schrödinger wave equation and are characterised by three quantum numbers (n, l, m_l). Different orbital types — s (spherical), p (dumbbell), d (clover-leaf), f (complex) — have different shapes and orientations. In NCERT Class 11 Chemistry, students learn to draw and label these orbitals and to use them to write electron configurations. Each orbital can hold a maximum of two electrons with opposite spins.
What is the Aufbau principle and how is it used?
The Aufbau principle (German for 'building up') states that electrons fill atomic orbitals starting from the lowest energy orbital and moving to higher energy ones. The order is determined by the (n + l) rule: orbitals with lower n + l fill first; if two orbitals have the same n + l, the one with lower n fills first. The standard sequence is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d. NCERT Class 11 Chemistry uses this principle along with Pauli's exclusion principle and Hund's rule to write ground-state electron configurations of elements.
What is Hund's rule of maximum multiplicity?
Hund's rule states that in a set of degenerate (equal-energy) orbitals such as the three 2p orbitals or five 3d orbitals, electrons occupy each orbital singly with parallel spins before any orbital receives a second electron. This minimises electron-electron repulsion and maximises total spin. For example, nitrogen (Z = 7) has the configuration 1s² 2s² 2p³ where each of the three 2p orbitals holds one electron with parallel spins, not two electrons in one orbital. NCERT Class 11 Chemistry uses Hund's rule along with the Aufbau principle and Pauli's exclusion principle to correctly write electron configurations.
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