Atomic Structure

This unit follows one long argument. Rutherford showed the atom has a tiny positive nucleus but could not say where the electrons are.

Bohr placed them in fixed orbits and successfully predicted the hydrogen spectrum — then failed for every other atom.

De Broglie and Heisenberg showed why: an electron is not a particle on a track, and the best we can state is a probability.

What survives is the quantum mechanical model, four quantum numbers and the three filling rules you will use in every later unit.

3.1 Rutherford's atomic model and its limitations

In 1909 Geiger and Marsden, working under Rutherford, fired a narrow beam of α-particles (He²⁺, mass 4 u, charge +2) at a gold foil about 0.00004 cm thick and recorded the flashes on a movable zinc sulphide screen. Rutherford explained their results in 1911.

Observation	Conclusion
Most α-particles passed straight through	The atom is mostly empty space
A small fraction were deflected by small angles	There is a concentrated positive charge that repels them
Roughly 1 in 8,000 was turned through more than 90°	All the positive charge and almost all the mass is packed into a very small nucleus

The model. The atom has a small, dense, positively charged nucleus containing practically the whole mass; electrons revolve around it at relatively large distances, and the centripetal force is supplied by electrostatic attraction.

The nuclear radius is of the order of 10⁻¹⁵ m against an atomic radius of 10⁻¹⁰ m — the nucleus is about 10⁵ times smaller than the atom.

Limitations.

1. It predicts an unstable atom. By Maxwell's electromagnetic theory a charged body moving in a circle is accelerating, so it must radiate energy continuously. The electron would spiral into the nucleus in about 10⁻⁸ s. Matter would not exist.
2. It predicts the wrong spectrum. A continuously spiralling electron would emit a continuous spectrum, whereas atoms give sharp line spectra.
3. It says nothing about the arrangement of electrons — no shells, no energies, so it cannot explain chemical behaviour or valency.

3.2 Postulates of Bohr's atomic model and its application

Niels Bohr (1913) kept the nuclear atom and added quantisation.

1. Electrons revolve round the nucleus only in certain permitted circular paths called stationary states or energy levels, in which they do not radiate energy.
2. Only those orbits are allowed for which the angular momentum is an integral multiple of h/2\pi:

3. Energy is emitted or absorbed only when an electron jumps from one stationary state to another, in a packet of size

4. The electrostatic attraction between nucleus and electron provides the centripetal force needed for the circular motion.

The energy is negative because the electron is bound; zero energy corresponds to a free electron at infinity. The orbital speed works out as v_n = 2.188\times10^{6}\,(Z/n)\ \text{m s}^{-1}.

3.3 Spectrum of hydrogen atom

When hydrogen gas at low pressure is excited in a discharge tube and the light is passed through a prism, the result is not a continuous rainbow but a set of sharp coloured lines on a dark background — an emission line spectrum.

The wavenumber of every line fits one empirical equation:

where R_H = 1.097\times10^{7}\ \text{m}^{-1} is the Rydberg constant. Bohr's model derives this equation: putting \Delta E = h c/\lambda with E_n = -13.6/n^2 eV reproduces R_H to three figures. That was its triumph.

Series	n_1	n_2	Region
Lyman	1	2, 3, 4 …	Ultraviolet
Balmer	2	3, 4, 5 …	Visible
Paschen	3	4, 5, 6 …	Infrared
Brackett	4	5, 6, 7 …	Infrared
Pfund	5	6, 7, 8 …	Far infrared

3.5 Elementary idea of the quantum mechanical model: de Broglie's wave equation

Light behaves as a wave (diffraction) and as a particle (photoelectric effect). In 1924 Louis de Broglie proposed that all matter has this dual character.

The wavelength is inversely proportional to mass, which is why the effect is invisible for everyday objects: a 100 g ball thrown at 10 m s⁻¹ has \lambda \approx 6.6\times10^{-34} m, far too small to detect.

For an electron it is about 10⁻¹⁰ m — the size of an atom — and Davisson and Germer confirmed it in 1927 by diffracting electrons from a nickel crystal.

3.6 Heisenberg's Uncertainty Principle

The same limit applies to energy and time, \Delta E\cdot\Delta t \geq h/4\pi. The principle is not a limitation of our instruments: any measurement of an electron requires a photon, and the photon inevitably disturbs it.

3.7 Concept of probability

Because a path cannot be defined, Erwin Schrödinger (1926) replaced it with a wave function \psi, obtained by solving his wave equation for the electron.

The wave function itself has no physical meaning, but Max Born showed that \psi^{2} at any point gives the probability density — the chance of finding the electron in unit volume at that point.

An orbital is therefore a charge cloud, not a track. Solving the Schrödinger equation automatically produces three quantum numbers n, l and m; the fourth, s, comes from experiment.

l	Subshell	Values of m	Orbitals	Maximum electrons
0	s	0	1	2
1	p	−1, 0, +1	3	6
2	d	−2 … +2	5	10
3	f	−3 … +3	7	14

For a shell of principal quantum number n there are n^{2} orbitals and at most 2n^{2} electrons. Angular momentum of the electron in an orbital is \sqrt{l(l+1)}\,(h/2\pi).

	s orbital	p orbital
Value of l	0	1
Shape	Spherical	Dumb-bell
Number per shell	1	3 (p_x, p_y, p_z)
Orientation	Non-directional	Directional, along the three axes, mutually perpendicular
Nodal planes through the nucleus	0	1
Present from shell	n = 1	n = 2

The three p orbitals of a given shell are degenerate — equal in energy — in an isolated atom. A 1s orbital is smaller than a 2s orbital, and a 2s orbital has one spherical node inside it, but both are spheres.

3.10 Aufbau Principle, Pauli's exclusion principle and Hund's rule

The order of energies follows the (n + l) rule: the subshell with the lower value of (n+l) fills first, and if two subshells have the same (n+l), the one with the lower n fills first. For 4s, n+l = 4+0 = 4; for 3d, n+l = 3+2 = 5; so 4s fills before 3d.

The two anomalies. Chromium is [Ar]3d⁵4s¹ and copper is [Ar]3d¹⁰4s¹, not 3d⁴4s² and 3d⁹4s².

A half-filled (d⁵) or completely filled (d¹⁰) subshell is extra stable, because such an arrangement is symmetrical and gives the maximum exchange energy between electrons of parallel spin.

Ions. For an anion, add electrons in the usual Aufbau order; for a cation, remove electrons from the shell of highest n first.

Ion	Derived from	Configuration
Na⁺	Na [Ne]3s¹	1s²2s²2p⁶ = [Ne]
Cl⁻	Cl [Ne]3s²3p⁵	[Ne]3s²3p⁶ = [Ar]
O²⁻	O [He]2s²2p⁴	1s²2s²2p⁶ = [Ne]
Fe²⁺	Fe [Ar]3d⁶4s²	[Ar]3d⁶
Fe³⁺	Fe [Ar]3d⁶4s²	[Ar]3d⁵ (half-filled, hence the stability of Fe³⁺)
Cu²⁺	Cu [Ar]3d¹⁰4s¹	[Ar]3d⁹
Zn²⁺	Zn [Ar]3d¹⁰4s²	[Ar]3d¹⁰

(a) A hydrogen discharge tube glowing pink-red

(b) Balmer lines of hydrogen from the Rydberg formula

Chapter summary

• Rutherford's α-scattering proved the nuclear atom but predicted an unstable atom and a continuous spectrum, and said nothing about electron arrangement.
• Bohr quantised angular momentum, mvr = nh/2\pi, giving r_n = 0.529\,n^2/Z Å and E_n = -13.6\,Z^2/n^2 eV, and radiation only on transitions, \Delta E = h\nu.
• Hydrogen line spectrum: 1/\lambda = R_H(1/n_1^{2} - 1/n_2^{2}) with R_H = 1.097\times10^{7}\ \text{m}^{-1}; Lyman (UV, n_1=1), Balmer (visible, n_1=2), Paschen, Brackett, Pfund (IR).
• Bohr fails for many-electron atoms, fine structure, Zeeman and Stark effects, and contradicts both wave–particle duality and the uncertainty principle.
• De Broglie: \lambda = h/mv. A Bohr orbit is a standing wave, 2\pi r = n\lambda.
• Heisenberg: \Delta x\cdot\Delta p \geq h/4\pi, so orbits give way to probability; \psi^{2} is the probability density and an orbital is the 90 % probability region.
• Quantum numbers: n (shell, size), l = 0 to n-1 (shape), m = -l to +l (orientation), s = \pm\tfrac{1}{2} (spin); a shell holds n^{2} orbitals and 2n^{2} electrons.
• s orbitals are spherical, p orbitals are dumb-bell shaped with one nodal plane and three orientations.
• Fill by the Aufbau (n+l) rule, two per orbital with opposite spins (Pauli), singly first with parallel spins (Hund). Cr and Cu are anomalous; cations lose their 4s electrons before their 3d electrons.

Practice questions

Group A — Multiple choice (1 mark each)

1. The Balmer series of the hydrogen spectrum lies in the [1]aultraviolet regionbvisible regioncinfrared regiondX-ray region
2. The maximum number of electrons in a shell with principal quantum number n is [1]an^{2}b2n^{2}c2ndn+1
3. The electronic configuration of Cr (Z = 24) is [1]a[Ar]3d⁴4s²b[Ar]3d⁵4s¹c[Ar]3d⁶d[Ar]3d³4s²
4. The number of unpaired electrons in Fe³⁺ is [1]a3b4c5d6
5. Which pair of quantum numbers is not allowed? [1]an = 3,\ l = 2bn = 2,\ l = 1cn = 2,\ l = 2dn = 4,\ l = 0

Answers to Group A

1. (b) — Balmer transitions end at n = 2 and fall at 656–410 nm.

2. (b) — n^{2} orbitals, each holding 2 electrons.

3. (b) — a half-filled 3d⁵ with 4s¹ is more stable than 3d⁴4s².

4. (c) — Fe³⁺ is [Ar]3d⁵, and by Hund's rule all five are unpaired.

5. (c) — l can only run from 0 to n-1, so l = 2 is impossible for n = 2.

Group B — Short answer (5 marks each)

1. Describe Rutherford's α-scattering experiment and state the conclusions drawn from it. Give two limitations of his model. [5]
2. State the postulates of Bohr's atomic theory and mention any three defects of the theory. [5]
3. Calculate the radius and energy of the second Bohr orbit of the hydrogen atom, and the wavelength of the first line of the Lyman series. (R_H = 1.097\times10^{7}\ \text{m}^{-1}) [5]
4. State Heisenberg's uncertainty principle. Calculate the uncertainty in the position of an electron whose velocity is known to within 1.0\times10^{5}\ \text{m s}^{-1}. [5]
5. Define the four quantum numbers and write the electronic configurations of Sc (Z = 21), Cu (Z = 29) and Zn²⁺. [5]

Answers to Group B

3. r_2 = 0.529\times2^{2} = 2.116\ \text{Å} = 2.116\times10^{-10} m and E_2 = -13.6/4 = -3.40 eV.

First Lyman line is n = 2 \rightarrow n = 1: 1/\lambda = 1.097\times10^{7}(1 - 1/4) = 8.23\times10^{6}\ \text{m}^{-1}, so \lambda = 1.215\times10^{-7}\ \text{m} = 121.5 nm (ultraviolet).

4. \Delta x = h/(4\pi m \Delta v) = 6.626\times10^{-34}/(4\times3.1416\times9.11\times10^{-31}\times1.0\times10^{5}) = 6.626\times10^{-34}/(1.145\times10^{-24}) = 5.79\times10^{-10} m, i.e. about 5.8 Å — several atomic diameters.

5. Sc: [Ar]3d¹4s². Cu: [Ar]3d¹⁰4s¹ (filled 3d is extra stable). Zn²⁺: Zn is [Ar]3d¹⁰4s², so removing the two 4s electrons gives [Ar]3d¹⁰.

Group C — Long answer (8 marks each)

1. (a) State Bohr's postulates and derive expressions for the radius and the energy of the n^{\text{th}} orbit of a hydrogen-like atom. [5](b) Using these, show how the Rydberg equation for the hydrogen spectrum follows, and name the series obtained for n_1 = 1, 2 and 3. [3]
2. (a) State the Aufbau principle, Pauli's exclusion principle and Hund's rule. [3](b) Write the ground-state electronic configurations of ₂₄Cr, ₂₆Fe, ₂₉Cu and ₃₀Zn, explaining the two anomalies. [3](c) Calculate the de Broglie wavelength of an electron moving with a velocity of 1.0\times10^{7}\ \text{m s}^{-1}. [2]

Answer outline for Group C

1 (b). For a jump from n_2 to n_1, \Delta E = 13.6(1/n_1^{2} - 1/n_2^{2}) eV.

Since \Delta E = hc/\lambda, 1/\lambda = (13.6\times1.602\times10^{-19})/(hc)\times(1/n_1^{2} - 1/n_2^{2}), and the constant evaluates to 1.097\times10^{7}\ \text{m}^{-1} = R_H. n_1 = 1 gives the Lyman series (UV), n_1 = 2 the Balmer series (visible) and n_1 = 3 the Paschen series (IR).

2 (b). Cr [Ar]3d⁵4s¹, Fe [Ar]3d⁶4s², Cu [Ar]3d¹⁰4s¹, Zn [Ar]3d¹⁰4s². Cr and Cu promote one 4s electron to 3d because a half-filled (d⁵) or completely filled (d¹⁰) subshell is symmetrical and of lower energy.

2 (c). \lambda = h/mv = 6.626\times10^{-34}/(9.11\times10^{-31}\times1.0\times10^{7}) = 6.626\times10^{-34}/(9.11\times10^{-24}) = 7.27\times10^{-11} m = 0.727 Å.

Group D — Model answers for long questions (10 marks each, for deeper revision)

1. State the postulates of Bohr's atomic model. Derive the radius and energy of the nth orbit of hydrogen and explain the hydrogen spectrum, with a numerical. [10]
2. What are quantum numbers? Describe the shapes of s and p orbitals and the rules for filling electrons, and write the electronic configurations of Cr, Cu and Fe³⁺. [10]
3. Explain de Broglie's equation and Heisenberg's uncertainty principle, with numericals, and state the failure of Bohr's model. [10]

Model answer D1: Bohr's model and the hydrogen spectrum (10 marks)

Introduction. Rutherford's model could not explain why an orbiting electron does not spiral into the nucleus, or the line spectrum of hydrogen. In 1913 Niels Bohr combined Rutherford's nuclear atom with Planck's quantum idea.

Postulates.

1. An electron moves in fixed circular orbits (stationary states) around the nucleus without radiating energy.
2. The orbits are allowed only if the angular momentum is quantised: mvr = nh/2\pi (n = 1, 2, 3...).
3. Energy is emitted or absorbed only when an electron jumps between orbits: \Delta E = E_{2} - E_{1} = h\nu.
4. The electrostatic attraction provides the centripetal force.

Radius of the nth orbit. Setting \dfrac{mv^{2}}{r} = \dfrac{ke^{2}}{r^{2}} and using v = nh/2\pi mr gives

Energy of the nth orbit. The total energy is the sum of kinetic and potential: E_{n} = -\dfrac{ke^{2}}{2r_{n}}, which gives

The energy is negative (bound), E_{1} = -13.6\ \text{eV} (ground state), and the ionisation energy of hydrogen is 13.6 eV.

Hydrogen spectrum. When an electron falls from n_{2} to n_{1},

Series	n_{1}	n_{2}	Region
Lyman	1	2, 3, 4...	Ultraviolet
Balmer	2	3, 4, 5...	Visible
Paschen	3	4, 5, 6...	Infrared
Brackett	4	5, 6...	Infrared
Pfund	5	6, 7...	Far infrared

Numerical (H-alpha line). For the Balmer transition n = 3 \to 2: \Delta E = 13.6(1/4 - 1/9) = 1.89\ \text{eV}, and 1/\lambda = 1.097\times10^{7}\times0.1389 = 1.524\times10^{6}\ \text{m}^{-1}, so \lambda = 656\ \text{nm} (the red line).

The first Lyman line (2 \to 1) has \Delta E = 10.2\ \text{eV} and \lambda = 121.6\ \text{nm}.

Merits. It explained the stability of the atom and the hydrogen line spectrum.

Limitations: it fails for atoms with more than one electron; it cannot explain the splitting of lines in a magnetic field (Zeeman) or electric field (Stark); it violates the uncertainty principle by giving exact orbits; and it does not explain chemical bonding.

Conclusion. Bohr's model introduced quantised energy levels. Its predictions for hydrogen are accurate, but a complete theory needed quantum mechanics.

Model answer D2: quantum numbers, orbitals and electron configuration (10 marks)

Introduction. In the quantum mechanical model an electron's state is described by an orbital, a region where the probability of finding the electron is high. Four quantum numbers specify an electron.

The four quantum numbers.

Quantum number	Symbol	Allowed values	Gives
Principal	n	1, 2, 3...	Shell, size and energy
Azimuthal (angular momentum)	l	0 to n-1 (s, p, d, f)	Subshell, shape
Magnetic	m_{l}	-l to +l	Orientation; (2l+1) orbitals
Spin	m_{s}	+\tfrac12 or -\tfrac12	Spin of the electron

A shell n holds 2n^{2} electrons. A subshell s holds 2, p holds 6, d holds 10 and f holds 14 electrons.

Shapes. An s orbital is spherical, and it grows larger with n (1s, 2s have a spherical node in 2s).

A p orbital is dumb-bell shaped with two lobes and a nodal plane through the nucleus; there are three, p_{x}, p_{y} and p_{z}, at right angles to each other and of equal energy.

Rules for filling.

1. Aufbau principle: electrons fill the lowest energy orbital first, in the order of increasing (n + l): 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s...
2. Pauli's exclusion principle: no two electrons in an atom can have the same set of four quantum numbers, so an orbital holds at most two electrons of opposite spin.
3. Hund's rule: in degenerate orbitals one electron enters each orbital with parallel spin before pairing starts.

Electronic configurations.

• Cr (Z = 24): expected 4s² 3d⁴, but actual [Ar] 3d⁵ 4s¹, because a half-filled d subshell is extra stable (symmetry and exchange energy).
• Cu (Z = 29): expected 4s² 3d⁹, but actual [Ar] 3d¹⁰ 4s¹, because a completely filled d subshell is extra stable.
• Fe (Z = 26): [Ar] 3d⁶ 4s². Fe³⁺: the 4s electrons leave first, then one 3d electron, giving [Ar] 3d⁵ (half-filled and so more stable than Fe²⁺). It has 5 unpaired electrons.

Example of quantum numbers. For the last electron of Na (3s¹): n = 3, l = 0, m_{l} = 0, m_{s} = +\tfrac12.

Conclusion. The four quantum numbers give a complete address of every electron. The filling rules explain the periodic table and the exceptions of chromium and copper arise from the extra stability of half-filled and fully filled subshells.

Model answer D3: de Broglie wavelength, uncertainty principle and failure of Bohr's model (10 marks)

Introduction. Light behaves as both wave and particle. In 1924 Louis de Broglie suggested that matter also has wave properties. In 1927 Heisenberg showed that there is a basic limit to how exactly we can know the position and momentum of a small particle.

de Broglie's equation. For a photon, E = h\nu = hc/\lambda and E = mc^{2}, giving \lambda = h/mc. Extending this to a particle of mass m and speed v,

The wave nature was proved by electron diffraction (Davisson and Germer, 1927), and it is the basis of the electron microscope.

Numerical 1. An electron moving at 2.19\times10^{6}\ \text{m s}^{-1} (the speed in the first Bohr orbit): \lambda = \dfrac{6.626\times10^{-34}}{9.109\times10^{-31}\times2.19\times10^{6}} = 3.32\times10^{-10}\ \text{m} = 0.332\ \text{nm}, about the size of an atom, so wave behaviour matters.

Numerical 2. A cricket ball of 0.15\ \text{kg} at 40\ \text{m s}^{-1}: \lambda = 6.626\times10^{-34}/(0.15\times40) = 1.1\times10^{-34}\ \text{m}, which is too small to detect. So the wave nature is noticeable only for tiny particles.

Heisenberg's uncertainty principle. It is impossible to measure both the exact position and the exact momentum of a particle at the same time.

Numerical 3. If an electron's position is known to \Delta x = 10^{-10}\ \text{m} (the size of an atom), then \Delta v \geq \dfrac{h}{4\pi m\,\Delta x} = \dfrac{6.626\times10^{-34}}{4\pi\times9.109\times10^{-31}\times10^{-10}} = 5.8\times10^{5}\ \text{m s}^{-1}.

This uncertainty in speed is large, so it makes no sense to talk about a definite path (orbit) of an electron.

Significance. The principle replaced Bohr's orbits with orbitals and probability (the wave function squared gives the probability density).

Failures of Bohr's model.

1. It applies only to one-electron species such as H, He⁺ and Li²⁺.
2. It does not explain the fine structure of spectral lines.
3. It could not explain the Zeeman and Stark effects.
4. It treats the electron as a particle only and ignores its wave nature.
5. It violates the uncertainty principle.
6. It does not explain the shapes of molecules.

Idea	Equation	Meaning
de Broglie	\lambda = h/mv	Every moving particle has a wavelength
Heisenberg	\Delta x\,\Delta p \geq h/4\pi	Position and momentum cannot both be exact
Bohr	mvr = nh/2\pi	Quantised orbits (replaced by orbitals)

Conclusion. Matter shows wave-particle duality, and the uncertainty principle sets a limit on our knowledge. These ideas led to the quantum mechanical model of the atom.