Class 11 · Chemistry · Chapter 2
Structure of Atom
This chapter builds the atom from the discovery of the electron, proton and neutron through Thomson's and Rutherford's models to Bohr's model and finally the quantum mechanical model that you will use for the rest of chemistry. It is the foundation for periodicity, bonding and everything in Class 12.
Before you startWhat the board asks from this chapter
Class 11 has no board exam, but this chapter is one of the most heavily tested in the school final, and every idea here — quantum numbers, orbitals, electronic configuration, Aufbau/Pauli/Hund — is used directly in Class 12 boards, JEE and NEET. Treat it as a high-priority, high-return chapter.
| Question type | Marks | Usually asked |
|---|---|---|
| MCQ | 1 | Quantum number rules, orbital shapes, nodes, subatomic particle facts, Aufbau order. |
| Assertion–Reason | 1 | Energy ordering of orbitals, degeneracy, why Bohr's model succeeds/fails, uncertainty principle scope. |
| Very short answer | 2 | State a principle/postulate; name subshells for a given n; quick numerical. |
| Short answer | 3 | Aufbau exceptions (Cr, Cu); protons/neutrons/electrons in an ion; a de Broglie or energy numerical. |
| Long answer | 5 | Full Bohr model derivation (radius and energy); de Broglie hypothesis with derivation; quantum numbers in detail. |
| Case-based | 4 | A passage on the photoelectric effect or on quantum numbers/orbitals with 4 sub-questions. |
Hot topics, in order
- Bohr's postulates and the derivation of radius and energy of the nth orbit
- Photoelectric effect and Einstein's equation
- Quantum numbers — rules and permitted values
- Shapes of s, p and d orbitals; radial and angular nodes
- Aufbau principle, Pauli exclusion principle, Hund's rule
- Electronic configurations, especially the Cr and Cu exceptions
- de Broglie relation and Heisenberg's uncertainty principle
- Line spectrum of hydrogen and the Rydberg formula
The whole chapter, in orderNotes
01 2.1 Discovery of Sub-atomic Particles
The atom is not indivisible — it is built of three fundamental particles. Each was found through a specific experiment, and examiners like to test which scientist, which apparatus and which conclusion go together.
Electron — J.J. Thomson, using a cathode ray discharge tube at low pressure and high voltage. Cathode rays travel from cathode to anode in straight lines, are deflected by electric and magnetic fields toward the positive plate, and their properties (charge/mass ratio, e/m = 1.758820 × 1011 C kg−1) do not depend on the gas or electrode material — so the electron is a universal constituent of matter. Millikan's oil drop experiment later gave the charge on the electron, 1.6 × 10−19 C, which combined with Thomson's e/m gives the mass of the electron, 9.1 × 10−31 kg.
Proton — Goldstein, using a perforated cathode in a discharge tube, observed rays travelling opposite to cathode rays: anode rays (canal rays). Unlike cathode rays, their charge-to-mass ratio depends on the gas used, and the smallest, lightest positive particle obtained (from hydrogen gas) is the proton, charge +1.6 × 10−19 C, mass 1.67 × 10−27 kg (about 1837 times the electron's mass).
Neutron — Chadwick, by bombarding beryllium with α-particles: ⁹Be + ⁴He → ¹²C + neutron. The neutron carries no charge and has a mass almost equal to the proton's (1.675 × 10−27 kg).
02 2.2 Atomic Models — Thomson and Rutherford
Thomson's model (plum pudding / watermelon model): the atom is a uniform sphere of positive charge with electrons embedded in it like seeds in a watermelon, the positive and negative charges being equal so the atom is neutral overall. It explained the overall neutrality of the atom but could not explain the results of Rutherford's scattering experiment.
Rutherford's α-particle scattering experiment: a beam of fast α-particles was directed at a thin gold foil (about 100 nm thick), surrounded by a circular fluorescent zinc sulphide screen. Observations: (i) most α-particles passed through the foil undeflected; (ii) a small fraction were deflected by small angles; (iii) a very small fraction (about 1 in 20,000) bounced back at angles close to 180°.
Rutherford concluded: (a) most of the atom is empty space, since most particles pass straight through; (b) a small region of the atom is very densely positively charged and contains almost all the mass — the nucleus, since only a heavy, concentrated positive charge could repel a fast, massive α-particle straight back; (c) the electrons revolve around this nucleus in circular paths (orbits), the size of the atom (~10−10 m) being about 10,000 times the size of the nucleus (~10−15 m).
Drawback: an electron moving in a circular orbit is accelerating, and an accelerating charged particle must radiate energy (electromagnetic theory). It should therefore spiral into the nucleus, making the atom unstable — but atoms are stable. This is the drawback that Bohr's model was built to fix.
03 2.3 Nature of Electromagnetic Radiation
Before Bohr's model, two lines of evidence — the nature of light and the atomic spectrum of hydrogen — had to come together. Electromagnetic radiation, in the wave picture (Maxwell), consists of oscillating electric and magnetic fields perpendicular to each other and to the direction of propagation, and it does not need a medium to travel (unlike sound). It is characterised by wavelength (λ), frequency (ν) and velocity, related by c = νλ, where c is the speed of light in vacuum, 3 × 108 m s−1. Wavenumber (𝜈̄), the number of wavelengths per unit length, is defined as 𝜈̄ = 1/λ, with SI unit m−1 (commonly cm−1). The electromagnetic spectrum ranges, in order of increasing frequency (decreasing wavelength), from radio waves through microwaves, infrared, visible light, ultraviolet, X-rays to γ-rays.
Visible light (400–750 nm) is only a tiny slice of this spectrum. The wave picture explained phenomena like interference and diffraction well, but it could not explain three observations: black body radiation, the photoelectric effect, and the variation of heat capacity of solids with temperature. These needed a particle picture of light.
04 2.3 Particle Nature of Radiation — Planck's Quantum Theory and the Photoelectric Effect
Black body radiation could not be explained by wave theory (it predicted radiation to increase indefinitely with frequency — the 'ultraviolet catastrophe'). Max Planck resolved this by proposing that energy is emitted or absorbed by atoms/molecules only in discrete packets called quanta, and for light, a quantum is called a photon. The energy of a quantum is proportional to its frequency: E = hν, where h is Planck's constant, 6.626 × 10−34 J s. Energy is always an integral multiple of hν (E = nhν); it cannot be emitted or absorbed continuously.
The photoelectric effect (Hertz, Lenz): when light of sufficiently high frequency strikes a metal surface, electrons are ejected instantly (photoelectrons). Wave theory predicted that intensity, not frequency, should control ejection — but experiment showed: (i) electrons are ejected only above a minimum frequency, the threshold frequency (ν0), which is characteristic of the metal, regardless of intensity; (ii) the kinetic energy of the ejected electrons increases linearly with frequency of incident light, not with intensity; (iii) intensity only increases the number of photoelectrons (the photoelectric current), not their individual kinetic energy; (iv) emission is instantaneous, with no time lag.
Einstein explained this using Planck's quantum idea: each photon of energy hν transfers all its energy to one electron. Part of this energy (h ν0, the work function, W0) is used to overcome the attraction holding the electron in the metal; the rest becomes the electron's kinetic energy.
05 2.3 Evidence for Quantised Electronic Energy Levels — Atomic Spectra
When white light is passed through a prism, it splits into a continuous rainbow — a continuous spectrum, because all wavelengths of visible light are present. But when atoms of an element are excited (by heat or an electric discharge) and the light they emit is passed through a prism, only specific, sharp coloured lines appear against a dark background, at fixed wavelengths — this is a line (atomic) emission spectrum, and it is unique to each element, like a fingerprint. A line absorption spectrum is obtained when white light is passed through a vapour of the element first — dark lines appear at exactly the wavelengths that would have been emitted, on the continuous background.
The hydrogen spectrum is the simplest and most important. Its emission lines fall into series named after their discoverers — Lyman (UV, lines end at n = 1), Balmer (visible, n = 2), Paschen, Brackett and Pfund (all infrared, n = 3, 4, 5). Each line's wavenumber fits the empirical Rydberg formula. This was the single biggest clue that electrons in an atom exist only in certain fixed, discrete energy levels — not at any arbitrary energy — because only transitions between fixed levels can produce a few sharp lines instead of a continuous smear. Bohr's model was built precisely to explain why this is so.
06 2.4 Bohr's Model for the Hydrogen Atom
Bohr combined Rutherford's nuclear model with Planck's quantum idea to fix the stability problem, using three postulates:
- The electron moves in circular orbits of fixed energy called stationary states (or orbits), and while in a stationary state it does not radiate energy, despite being in circular motion.
- The angular momentum of the electron is quantised: it can only take values that are integral multiples of h/2π — mvr = nh/2π, where n = 1, 2, 3, ... (the principal quantum number).
- Energy is absorbed or emitted only when the electron jumps from one stationary state to another, and the energy difference equals the energy of the photon absorbed or emitted: ΔE = hν = E2 − E1.
Derivation of radius: for an electron of mass m, charge −e, orbiting a nucleus of charge +Ze at radius r with speed v, the electrostatic force of attraction supplies the centripetal force: kZe2/r2 = mv2/r (k = 1/4πε0). Combining this with the quantisation condition mvr = nh/2π and eliminating v gives rn = n2h2ε0/(πmZe2), which numerically works out to rn = 0.529 (n2/Z) Å.
Derivation of energy: total energy = kinetic + potential = ½mv2 − kZe2/r. Using mv2 = kZe2/r from the force balance, this simplifies to E = −kZe2/2r. Substituting rn gives En = −2.18 × 10−18 (Z2/n2) J = −13.6 (Z2/n2) eV per electron. The negative sign shows the electron is bound to the nucleus (energy is taken as zero when the electron is completely free, at n = ∞).
Bohr's model explains the hydrogen line spectrum exactly: when an electron falls from a higher orbit n2 to a lower orbit n1, the emitted photon's energy equals En2 − En1, and working this out reproduces the Rydberg formula with RH calculated from fundamental constants — a major triumph.
07 2.4 Limitations of Bohr's Model
Bohr's model was a huge step forward but it has clear limits, and examiners like a crisp list:
- It explains the spectrum of hydrogen and hydrogen-like species (He+, Li2+, single-electron ions) very well, but it fails for multi-electron atoms — it could not account for their more complex spectra.
- It could not explain the splitting of spectral lines in a magnetic field (Zeeman effect) or in an electric field (Stark effect).
- It could not explain the finer details (fine structure) of the hydrogen spectrum seen with better resolution.
- It treats the electron only as a particle in a fixed circular orbit and ignores its wave nature (de Broglie), and it violates Heisenberg's uncertainty principle by assuming both a definite position (on a fixed orbit) and a definite momentum can be known simultaneously.
These gaps needed two more ideas — the dual nature of matter and the uncertainty principle — before a fully correct (quantum mechanical) model of the atom could be built.
08 2.5 Towards Quantum Mechanical Model — Dual Behaviour of Matter
Louis de Broglie proposed, in 1924, that if light (normally thought of as a wave) can behave like particles (photons, as in the photoelectric effect), then matter (normally thought of as particles) should also show wave-like behaviour. He derived a relation for the wavelength associated with any moving particle by combining Einstein's E = mc2 with Planck's E = hν: the wavelength of a particle of mass m moving with velocity v is λ = h/mv.
This dual behaviour is significant only for very small, fast-moving particles like electrons — for macroscopic objects (a cricket ball, a car), the associated wavelength works out to be immeasurably small because of the large mass in the denominator, so wave behaviour is never observed for everyday objects. De Broglie's hypothesis was confirmed experimentally by the Davisson–Germer experiment, in which a beam of electrons was diffracted by a nickel crystal exactly as X-rays would be, proving electrons show wave-like diffraction.
De Broglie waves are the reason the quantum mechanical model replaces Bohr's fixed circular orbits with regions of high probability (orbitals) — an electron's exact path cannot be traced the way a planet's can.
09 2.5 Heisenberg's Uncertainty Principle
Werner Heisenberg stated, in 1927, that it is impossible to determine simultaneously and with perfect accuracy both the exact position and the exact momentum (or velocity) of a microscopic moving particle such as an electron. The uncertainties in position (Δx) and momentum (Δp) are related by Δx · Δp ≥ h/4π.
This is a fundamental limitation of nature, not a flaw in measurement technique — measuring an electron's position (say, using a photon of light) necessarily disturbs its momentum, because the electron is so light that even a single photon striking it changes its velocity appreciably. The principle is significant only for very small particles with very small mass, like electrons; for macroscopic objects the uncertainty, calculated from the same formula, is far too small to be measurable (because h is extremely small compared to the momentum of everyday objects), so classical mechanics works perfectly well for them.
The direct consequence for atomic structure is enormous: since an electron's exact position and momentum cannot both be known, the idea of a well-defined orbit (a fixed path, as Bohr assumed) has to be abandoned. It is replaced by the idea of an orbital — a three-dimensional region around the nucleus where the probability of finding the electron is high. This is exactly the shift that leads into the quantum mechanical model.
10 2.6 Quantum Mechanical Model of Atom — Orbitals and Quantum Numbers
The quantum mechanical model treats the electron as a wave and describes it using a mathematical function ψ (psi), the wave function, obtained by solving the Schrödinger wave equation. ψ itself has no direct physical meaning, but ψ2 (or |ψ|2) at a point gives the probability density of finding the electron at that point — this is the modern replacement for a 'fixed orbit'. An orbital is the three-dimensional region in space around the nucleus where the probability of finding an electron is maximum (typically taken as 90–95%). This is fundamentally different from a Bohr orbit, which was a fixed, well-defined circular path.
Each orbital, and the electron in it, is completely described by a set of four quantum numbers:
- Principal quantum number (n) = 1, 2, 3, ... — decides the size and energy of the shell; the shell with a given n is denoted K, L, M, N for n = 1, 2, 3, 4.
- Azimuthal (angular momentum) quantum number (l) = 0 to (n − 1) for a given n — decides the shape of the subshell/orbital: l = 0, 1, 2, 3 are called s, p, d, f respectively.
- Magnetic quantum number (ml) = −l to +l (including 0), giving (2l + 1) values — decides the orientation of the orbital in space.
- Spin quantum number (ms) = +½ or −½ — describes the spin (clockwise or anticlockwise) of the electron about its own axis; it was not part of Schrödinger's original equation but was added to explain fine spectral detail.
For a given shell n: the number of subshells = n; the number of orbitals = n2; the maximum number of electrons = 2n2. For a given subshell l, the number of orbitals = (2l + 1), so s has 1 orbital (2 electrons max), p has 3 (6 electrons max), d has 5 (10 electrons max), f has 7 (14 electrons max).
11 2.6 Shapes of Atomic Orbitals, Nodes and Energy Ordering
s-orbitals (l = 0) are spherically symmetric — probability of finding the electron is the same in all directions at a given distance from the nucleus; 1s, 2s, 3s all differ only in size (2s, 3s are bigger and have a spherical node inside them).
p-orbitals (l = 1) have a dumb-bell shape with two lobes on either side of the nucleus and a node (zero probability) at the nucleus itself. There are three degenerate p-orbitals per shell (px, py, pz), oriented along the three perpendicular axes, and all three have identical energy and shape but different orientation.
d-orbitals (l = 2) are more complex: four of the five (dxy, dyz, dxz, dx²−y²) have a four-lobed 'double dumb-bell' (cloverleaf) shape, while dz² has two lobes along the z-axis plus a torus (doughnut) of electron density around the middle.
Nodes are regions where the probability of finding the electron is zero. There are two types: radial nodes (spherical shells within the orbital), given by n − l − 1, and angular nodes (planes or cones through the nucleus), given by l. Total nodes = n − 1.
Energy ordering in a multi-electron atom is not simply by n; it follows the (n + l) rule: lower (n + l) means lower energy; if two subshells have equal (n + l), the one with lower n has lower energy. This is why 4s (n + l = 4) fills before 3d (n + l = 5), even though n = 3 < 4 — a very frequently tested point.
12 2.6 Filling of Orbitals — Aufbau Principle, Pauli Exclusion Principle, Hund's Rule and Electronic Configurations
Three rules govern how electrons fill orbitals in the ground state of an atom:
Aufbau principle: orbitals are filled in order of increasing energy, i.e., increasing (n + l); for equal (n + l), the orbital with lower n fills first. The resulting standard filling order is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d ...
Pauli exclusion principle: no two electrons in the same atom can have all four quantum numbers identical. The practical consequence is that any orbital can hold a maximum of two electrons, and those two must have opposite spins (ms = +½ and −½).
Hund's rule of maximum multiplicity: electron pairing in orbitals of the same subshell (degenerate orbitals, e.g. the three p orbitals) does not start until each orbital of that subshell has received one electron, and these singly-occupying electrons all have parallel (same-direction) spin. This gives the atom maximum stability.
Exceptions to Aufbau — the extra stability of half-filled and fully-filled subshells: chromium (Z = 24) is [Ar] 3d5 4s1, not the 'expected' [Ar] 3d4 4s2; copper (Z = 29) is [Ar] 3d10 4s1, not [Ar] 3d9 4s2. This happens because exactly half-filled (d5) and fully-filled (d10) subshells have extra stability, arising from (i) symmetrical distribution of electrons and (ii) greater exchange energy between electrons of parallel spin in degenerate orbitals — an electron moves from 4s to 3d to achieve this more stable arrangement.
Electronic configuration is written using the notation nlx (e.g. 1s2 2s2 2p6...), or in shorthand using the previous noble gas core in square brackets, e.g. sodium (Z = 11) = [Ne] 3s1.
VocabularyKey terms
Keep this on one pageFormula sheet
| Name | Expression | Notes |
|---|---|---|
| Speed of light | c = ν × λ | c = 3 × 10⁸ m s⁻¹; relates frequency and wavelength of any electromagnetic radiation |
| Wavenumber | 𝜈̄ = 1/λ | unit m⁻¹ or cm⁻¹; not the same as frequency |
| Energy of a photon | E = hν = hc/λ | h = 6.626 × 10⁻³⁴ J s; used for photon energy and photoelectric problems |
| Einstein's photoelectric equation | hν = hν₀ + ½mv²ₘₐₓ | KEₘₐₓ = h(ν − ν₀); applies only when ν ≥ ν₀ |
| Rydberg formula | 1/λ = R_H(1/n₁² − 1/n₂²) | R_H = 1.097 × 10⁷ m⁻¹; gives wavelength of a spectral line in the hydrogen spectrum, n₂ > n₁ |
| Bohr radius | rₙ = 0.529 (n²/Z) Å | for hydrogen (Z = 1), r₁ = 0.529 Å (the Bohr radius, a₀) |
| Bohr energy | Eₙ = −2.18 × 10⁻¹⁸ (Z²/n²) J = −13.6 (Z²/n²) eV | negative because the electron is bound; E₁ for H = −13.6 eV = ionisation energy magnitude |
| de Broglie wavelength | λ = h/mv | significant only for very small mass particles like electrons; for an electron through potential V, λ(Å) = 12.27/√V |
| Heisenberg uncertainty principle | Δx · Δp ≥ h/4π | minimum, not exact, uncertainty; significant only for microscopic particles |
| Orbitals and electrons per shell | orbitals = n²; electrons(max) = 2n² | e.g. for n = 3: 9 orbitals, 18 electrons max |
| Nodes | radial nodes = n − l − 1; angular nodes = l; total = n − 1 | used to identify shape complexity of an orbital |
Practise with a pencilDiagrams that earn marks
Rutherford's α-particle scattering experiment
Draw a source of α-particles aimed at a thin gold foil, with a circular fluorescent ZnS screen around it. Show three paths: most particles going straight through undeflected, a few deflected at small angles near the foil, and a very small number bouncing straight back near the source. Label the tiny dense nucleus at the centre of an atom and the mostly empty space around it.
Energy level diagram of hydrogen atom with spectral series
Draw horizontal lines for n = 1, 2, 3, 4, 5, ∞ with energy increasing (less negative) upward, spacing decreasing as n increases. Draw downward arrows ending at n = 1 and label this group 'Lyman (UV)'; arrows ending at n = 2 labelled 'Balmer (visible)'; arrows ending at n = 3 labelled 'Paschen (IR)'. Mark E = 0 at n = ∞.
Shapes of s, p and d orbitals
Draw a single sphere for 1s. Draw three dumb-bell shapes along x, y, z axes for 2pₓ, 2p_y, 2p_z, each with a node at the nucleus. Draw the four-lobed cloverleaf shape for d_xy (lobes between the axes) and the two-lobes-plus-collar shape for d_z². Label the nucleus at the centre of each.
Aufbau (n + l) filling order diagram
Draw a grid with rows n = 1 to 7 and columns s, p, d, f, placing each subshell box (1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f, ...) in its cell. Draw diagonal arrows from top-right to bottom-left through the boxes to show the filling order (1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p ...). Label this the '(n+l) rule' diagram.
Orbital diagram showing Hund's rule (example: nitrogen, 2p³)
Draw three boxes side by side for the three 2p orbitals. Place one upward arrow (↑) in each box (all parallel spin, no pairing), and write '2p³, all unpaired, parallel spin — Hund's rule' below.
TextbookNCERT questions the board keeps asking
2.1 (i) Calculate the number of electrons which will together weigh one gram. (ii) Calculate the mass and charge of one mole of electrons.3 m
(i) Mass of one electron = 9.10939 × 10−31 kg = 9.10939 × 10−28 g.
Number of electrons = 1 g ÷ 9.10939 × 10−28 g = 1.098 × 1027 electrons.
(ii) Mass of one mole of electrons = 6.022 × 1023 × 9.10939 × 10−31 kg = 5.48 × 10−7 kg = 5.48 × 10−4 g.
Charge on one mole of electrons = 6.022 × 1023 × 1.602 × 10−19 C = 9.65 × 104 C (this is 1 Faraday).
2.2 (i) Calculate the total number of electrons present in one mole of methane (CH₄). (ii) Find the total number and total mass of neutrons in 7 mg of ¹⁴C. (Assume mass of a neutron = 1.675 × 10⁻²⁷ kg.)3 m
(i) Electrons in one CH₄ molecule = 6 (from C) + 4 × 1 (from 4 H) = 10.
Electrons in 1 mole CH₄ = 6.022 × 1023 × 10 = 6.022 × 1024 electrons.
(ii) ¹⁴C has mass number 14, atomic number 6, so neutrons per atom = 14 − 6 = 8.
Moles of ¹⁴C in 7 mg = 7 × 10−3 g ÷ 14 g mol−1 = 5 × 10−4 mol.
Atoms = 5 × 10−4 × 6.022 × 1023 = 3.011 × 1020.
Total neutrons = 8 × 3.011 × 1020 = 2.409 × 1021.
Total mass of neutrons = 2.409 × 1021 × 1.675 × 10−27 kg = 4.04 × 10−6 kg ≈ 4.04 mg.
2.3 Give the number of protons, electrons and neutrons in ³⁵₁₇Cl.2 m
For ³⁵₁₇Cl: atomic number Z = 17, mass number A = 35.
Protons = Z = 17. Electrons (neutral atom) = 17. Neutrons = A − Z = 35 − 17 = 18.
2.4 Write the complete symbol for the atom with atomic number 17 and mass number 35.1 m
³⁵₁₇Cl — mass number written as a left superscript, atomic number as a left subscript.
2.5 Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency and wavenumber of this radiation.3 m
λ = 580 nm = 580 × 10−9 m.
Frequency, ν = c/λ = (3 × 108) ÷ (580 × 10−9) = 5.17 × 1014 s−1.
Wavenumber, 𝜈̄ = 1/λ = 1 ÷ (580 × 10−9 m) = 1.72 × 106 m−1 = 1.72 × 104 cm−1.
2.6 Calculate the energy of one mole of photons of radiation whose frequency is 5 × 10¹⁴ Hz.2 m
Energy of one photon, E = hν = 6.626 × 10−34 × 5 × 1014 = 3.313 × 10−19 J.
Energy of one mole of photons = 3.313 × 10−19 × 6.022 × 1023 = 1.995 × 105 J mol−1 ≈ 199.5 kJ mol−1.
2.7 What is the maximum number of emission lines when the excited electron of a hydrogen atom in n = 6 drops to the ground state?2 m
Number of spectral lines = n(n − 1)/2 = 6 × 5/2 = 15 lines.
2.8 The threshold frequency (ν₀) for a metal is 7.0 × 10¹⁴ s⁻¹. Calculate the kinetic energy of an electron emitted when radiation of frequency ν = 1.0 × 10¹⁵ s⁻¹ hits the metal.3 m
KE = h(ν − ν₀) = 6.626 × 10−34 × (1.0 × 1015 − 7.0 × 1014) = 6.626 × 10−34 × 3.0 × 1014 = 1.99 × 10−19 J.
2.9 Calculate the de Broglie wavelength of an electron moving with a velocity of 2.05 × 10⁷ m s⁻¹.2 m
λ = h/mv = (6.626 × 10−34) ÷ (9.109 × 10−31 × 2.05 × 107) = (6.626 × 10−34) ÷ (1.868 × 10−23) = 3.55 × 10−11 m = 35.5 pm.
2.10 The uncertainty in the velocity of an electron is 5.7 × 10⁵ m s⁻¹. Calculate the uncertainty in its position (Δx), given h = 6.626 × 10⁻³⁴ J s and mass of electron = 9.109 × 10⁻³¹ kg.3 m
Δx ≥ h/(4πmΔv) = (6.626 × 10−34) ÷ (4 × 3.1416 × 9.109 × 10−31 × 5.7 × 105)
= (6.626 × 10−34) ÷ (6.52 × 10−24) = 1.02 × 10−10 m ≈ 102 pm.
2.11 Write the electronic configurations of chromium (Z = 24) and copper (Z = 29), and explain why they differ from the configuration predicted by the simple Aufbau order.3 m
Chromium: [Ar] 3d5 4s1 (not [Ar] 3d4 4s2).
Copper: [Ar] 3d10 4s1 (not [Ar] 3d9 4s2).
An exactly half-filled (d5) or fully-filled (d10) d-subshell has extra stability, due to the symmetrical distribution of electrons and greater exchange energy among electrons of parallel spin in degenerate orbitals. One electron shifts from 4s to 3d to reach this more stable configuration.
2.12 Which of the following sets of quantum numbers is not possible for an electron, and why: (a) n = 3, l = 3, mₗ = 0 (b) n = 2, l = 1, mₗ = −1?2 m
(a) Not possible. For a given n, l can only take values from 0 to (n − 1); for n = 3, the maximum allowed value of l is 2, so l = 3 is not permitted.
(b) Possible. For n = 2, l can be 0 or 1; for l = 1, mₗ can be −1, 0 or +1, so mₗ = −1 is allowed. This describes an electron in a 2p orbital.
Previous years and frequently askedBoard questions with model answers
MCQ The number of radial nodes present in a 3p orbital is:- 0
- 1
- 2
- 3
Frequently asked1 m
MCQ According to the Aufbau principle, which orbital is filled before the 3d orbital in a multi-electron atom?- 3p
- 4s
- 4p
- 4d
Frequently asked1 m
MCQ The maximum number of electrons that a subshell with azimuthal quantum number l = 3 can hold is:- 6
- 10
- 14
- 18
Frequently asked1 m
MCQ If a proton, a neutron, an electron and an α-particle all move with the same velocity, which one has the shortest de Broglie wavelength?- Electron
- Proton
- Neutron
- α-particle
Frequently asked1 m
MCQ Which set of quantum numbers is NOT possible for an electron in an atom?- n = 3, l = 2, mₗ = −2
- n = 4, l = 0, mₗ = 0
- n = 2, l = 2, mₗ = 0
- n = 3, l = 0, mₗ = 0
Frequently asked1 m
MCQ In Rutherford's α-particle scattering experiment, the observation that a very small fraction of α-particles bounced back at large angles suggested that:- The atom is a uniform sphere of positive charge
- The nucleus is very small, dense and positively charged
- Electrons are embedded uniformly in the atom
- Most of the atom's mass is due to electrons
Frequently asked1 m
MCQ The kinetic energy of photoelectrons ejected from a metal surface primarily depends on:- The intensity of incident light only
- The frequency of incident light only
- Both intensity and frequency equally
- The distance of the source from the metal
Frequently asked1 m
Assertion–Reason Assertion (A): Bohr's model successfully explains the line spectrum of the hydrogen atom. Reason (R): Bohr's model could not explain the spectra of multi-electron atoms.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): An electron in a 3p orbital has higher energy than an electron in a 3s orbital of the same atom. Reason (R): For a given value of n, the energy of orbitals in a multi-electron atom increases in the order s < p < d < f.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): Heisenberg's uncertainty principle is significant for the motion of a cricket ball as well as an electron. Reason (R): Planck's constant h is so small that the uncertainty predicted by Δx·Δp ≥ h/4π is negligible for objects of large mass, but significant for particles as light as an electron.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Very short answer State Heisenberg's uncertainty principle. Write the mathematical expression.Frequently asked2 m
Very short answer What is the physical significance of ψ² in the quantum mechanical model of the atom?Frequently asked2 m
Very short answer How many subshells are associated with n = 4? Name them.Frequently asked2 m
Very short answer Arrange s, p, d and f orbitals of the same shell in increasing order of their shielding effect.Frequently asked2 m
Very short answer Why is the electronic configuration of copper (Z = 29) written as [Ar] 3d¹⁰ 4s¹ and not [Ar] 3d⁹ 4s²?Frequently asked2 m
Short answer State (i) the Aufbau principle, (ii) the Pauli exclusion principle and (iii) Hund's rule of maximum multiplicity, in one line each.Frequently asked3 m
Short answer Calculate the number of protons, neutrons and electrons in the ion ⁵⁶₂₆Fe³⁺.Frequently asked3 m
Short answer Explain why the electronic configuration of chromium (Z = 24) is [Ar] 3d⁵ 4s¹ and not [Ar] 3d⁴ 4s².Frequently asked3 m
Short answer The threshold wavelength for a metal is 230 nm. Calculate (i) the work function of the metal in joules, and (ii) the kinetic energy of the photoelectron ejected when light of wavelength 180 nm strikes the metal.Frequently asked3 m
(ii) Energy of incident photon, E = hc/λ = (1.988 × 10⁻²⁵) ÷ (180 × 10⁻⁹) = 1.104 × 10⁻¹⁸ J.
KE of photoelectron = E − W₀ = 1.104 × 10⁻¹⁸ − 8.64 × 10⁻¹⁹ = 2.40 × 10⁻¹⁹ J.
Short answer Calculate the wavelength of the spectral line obtained when an electron in a hydrogen atom jumps from n = 4 to n = 2. Identify the series this line belongs to.Frequently asked3 m
λ = 1 ÷ 2.057 × 10⁶ = 4.86 × 10⁻⁷ m = 486 nm.
Since the transition ends at n₁ = 2, this line belongs to the Balmer series (visible region).
Long answer State the postulates of Bohr's model of the hydrogen atom. Derive an expression for the radius of the nth orbit and the energy of an electron in the nth orbit.Frequently asked5 m
Postulates: (1) The electron revolves around the nucleus in circular orbits of fixed energy called stationary states, and does not radiate energy while in a stationary state, despite being in circular (accelerated) motion. (2) The angular momentum of the electron is quantised: mvr = nh/2π, n = 1, 2, 3, ... (3) Energy is emitted or absorbed only when an electron jumps between stationary states, with ΔE = hν = E₂ − E₁.
Radius: the electrostatic force of attraction provides the centripetal force: kZe²/r² = mv²/r, so mv² = kZe²/r ... (i), where k = 1/4πε₀.
From the quantisation condition, v = nh/2πmr ... (ii). Substituting (ii) into (i) and solving for r gives rₙ = n²h²ε₀/(πmZe²), which evaluates to rₙ = 0.529 (n²/Z) Å.
Energy: total energy E = KE + PE = ½mv² − kZe²/r. Using mv² = kZe²/r from (i), ½mv² = kZe²/2r, so E = kZe²/2r − kZe²/r = −kZe²/2r. Substituting rₙ gives Eₙ = −2.18 × 10⁻¹⁸ (Z²/n²) J = −13.6 (Z²/n²) eV. The negative sign shows the electron is bound to the nucleus, with zero energy taken at n = ∞ (a free electron).
Long answer State de Broglie's hypothesis. Derive the de Broglie relation and briefly describe the experimental evidence that confirmed it. Why is the wave nature of matter not observed for everyday (macroscopic) objects?Frequently asked5 m
Hypothesis: de Broglie proposed that just as radiation shows dual (wave and particle) behaviour, matter — normally considered to be made of particles — should also exhibit wave-like properties.
Derivation: for a photon, energy E = hν = hc/λ (Planck), and by Einstein's mass–energy relation E = mc². Equating, hc/λ = mc², so λ = h/mc. Generalising c to the velocity v of any particle of mass m, de Broglie proposed λ = h/mv, where mv is the particle's momentum p, so λ = h/p.
Experimental confirmation: the Davisson–Germer experiment directed a beam of electrons at a nickel crystal and observed a diffraction pattern exactly like that produced by X-rays of comparable wavelength, confirming that electrons behave as waves.
Why not observed macroscopically: λ = h/mv has Planck's constant h (an extremely small number, 6.626 × 10⁻³⁴ J s) in the numerator and mass in the denominator. For a macroscopic object (large mass, e.g. a cricket ball), the resulting wavelength is so vanishingly small that it cannot be detected or has any observable effect; wave behaviour becomes significant only for particles of very small mass, such as electrons.
Long answer Define the four quantum numbers used to describe an electron in an atom. State the permitted values and physical significance of each.Frequently asked5 m
Principal quantum number (n): values 1, 2, 3, ...; decides the size and energy of the shell (larger n → larger, higher-energy shell); shells are also labelled K, L, M, N for n = 1, 2, 3, 4.
Azimuthal (angular momentum) quantum number (l): values 0 to (n − 1) for a given n; decides the shape of the orbital/subshell — l = 0, 1, 2, 3 correspond to s, p, d, f subshells respectively.
Magnetic quantum number (mₗ): values from −l to +l, including 0, giving (2l + 1) values for a given l; decides the orientation of the orbital in space, e.g. for l = 1 (p), mₗ = −1, 0, +1 correspond to pₓ, p_y, p_z.
Spin quantum number (mₛ): values +½ or −½; describes the two possible spin orientations (clockwise or anticlockwise) of an electron about its own axis; not derived from the Schrödinger equation but added to account for finer spectral details and the Pauli exclusion principle.
Together, the four quantum numbers uniquely specify every electron in an atom, since by the Pauli exclusion principle no two electrons can share the same set of all four.
Case-based Read the passage and answer the questions that follow.
When light of a certain minimum frequency (called the threshold frequency, ν₀) falls on a clean metal surface, electrons are instantly ejected from the surface — this is called the photoelectric effect. Below this threshold frequency, no electrons are ejected however intense the light is. Einstein explained this using Planck's quantum theory: each photon of energy hν interacts with a single electron, part of the energy (hν₀, the work function) frees the electron from the metal, and the rest appears as the kinetic energy of the ejected electron.
(i) What happens if the frequency of incident light is less than ν₀, no matter how intense the light is?
(ii) Write Einstein's photoelectric equation relating KEₘₐₓ, ν and ν₀.
(iii) Does increasing the intensity of light (keeping frequency fixed and above ν₀) increase the kinetic energy of the ejected electrons, or something else? Explain.
(iv) A photon of frequency 6.0 × 10¹⁴ Hz strikes a metal whose threshold frequency is 5.0 × 10¹⁴ Hz. Calculate the kinetic energy of the ejected electron.Frequently asked4 m
(ii) KEₘₐₓ = hν − hν₀ = h(ν − ν₀).
(iii) Increasing intensity increases the number of photons striking the metal per second, and hence increases the number of photoelectrons ejected (the photoelectric current) — it does not increase the kinetic energy of each individual electron, which depends only on the frequency of the light.
(iv) KE = h(ν − ν₀) = 6.626 × 10⁻³⁴ × (6.0 × 10¹⁴ − 5.0 × 10¹⁴) = 6.626 × 10⁻³⁴ × 1.0 × 10¹⁴ = 6.626 × 10⁻²⁰ J.
Case-based Read the passage and answer the questions that follow.
The quantum mechanical model describes an electron using four quantum numbers. The principal quantum number (n) decides the size and energy of the shell. The azimuthal quantum number (l), which ranges from 0 to (n − 1), decides the shape of the subshell. The magnetic quantum number (mₗ) decides the orientation of the orbital, ranging from −l to +l. Electrons are filled into these orbitals following the Aufbau principle, the Pauli exclusion principle and Hund's rule, though a few atoms such as chromium and copper show exceptions to the simple Aufbau order because of the extra stability of half-filled and fully-filled subshells.
(i) For n = 3, list all the possible values of l and name the corresponding subshells.
(ii) How many orbitals and how many electrons (maximum) can the n = 3 shell hold in total?
(iii) State Hund's rule of maximum multiplicity in one line.
(iv) Write the expected electronic configuration of chromium (Z = 24) by the simple Aufbau order, and its actual configuration, with a one-line reason for the difference.Frequently asked4 m
(ii) Orbitals = n² = 9; maximum electrons = 2n² = 18.
(iii) Degenerate orbitals of a subshell are singly occupied by electrons with parallel spins before any orbital receives a second, paired electron.
(iv) Expected (simple Aufbau): [Ar] 3d⁴ 4s². Actual: [Ar] 3d⁵ 4s¹, because a half-filled 3d⁵ subshell is extra stable due to symmetrical distribution and higher exchange energy among parallel-spin electrons.
Test yourselfPractice quiz
Marks lost every yearCommon mistakes
- Confusing atomic number (Z, number of protons) with mass number (A, protons + neutrons) — always check which one a question is asking for before substituting values.
- Writing the electronic configuration of chromium or copper by strict Aufbau order ([Ar] 3d⁴ 4s² or [Ar] 3d⁹ 4s²) instead of the actual, more stable configurations ([Ar] 3d⁵ 4s¹ and [Ar] 3d¹⁰ 4s¹).
- Forgetting that the azimuthal quantum number l can only range from 0 to (n − 1) — writing an invalid combination like n = 2, l = 2 costs the mark instantly.
- Treating a Bohr 'orbit' (a fixed, well-defined circular path) and a quantum mechanical 'orbital' (a probability region) as the same thing — they are conceptually different, and examiners test this distinction directly.
- Writing the uncertainty relation as Δx · Δp ≥ h instead of Δx · Δp ≥ h/4π, or dropping the ≥ sign and treating it as an exact equality.
- Dropping the negative sign in the Bohr energy expression Eₙ = −13.6(Z²/n²) eV, which then gives a wrong sign when computing ionisation energy or transition energy (always take the higher-n orbit as less negative, i.e., higher energy).
- Mixing up the formulas for radial nodes (n − l − 1) and angular nodes (l) — learn them as a pair, not separately.
- Believing that increasing the intensity of light increases the kinetic energy of photoelectrons in the photoelectric effect — intensity changes only the number of photoelectrons (current); frequency alone decides kinetic energy.
How to prepareStrategy for this chapter
This chapter rewards precision more than memory. Start with the historical run — Thomson, Rutherford, and why Rutherford's model failed on stability — because it is asked almost every year as a short, factual question and takes little time to lock down. Then give your heaviest practice time to Bohr's model: be able to write the three postulates and derive both the radius and energy formulas from scratch, without looking, in under eight minutes. This single derivation covers the long-answer slot in most papers.
Next, drill the photoelectric effect and Einstein's equation with numericals — these numbers repeat with only the given values changed, so once you can do one, you can do all of them. The de Broglie relation and Heisenberg's uncertainty principle are shorter but just as testable; know both formulas exactly and be ready to explain, in words, why each matters for the shift from 'orbit' to 'orbital'.
Save real time for quantum numbers, orbital shapes and the filling rules — Aufbau, Pauli, Hund — because these come back in every later chapter of chemistry, not just this one. Memorise the Cr and Cu exceptions with the reason (half-filled/fully-filled stability), not just the answer. Finish with 15–20 minutes writing full electronic configurations of atoms and ions from Z = 1 to 30 without help; this single drill catches almost every careless mistake before the exam does.