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

3.4 Defects of Bohr's theory

Defect	What Bohr could not explain
Multi-electron atoms	The theory works only for one-electron species (H, He⁺, Li²⁺); it fails even for helium
Fine structure	Under a high-resolution spectroscope each line is really several closely spaced lines
Zeeman effect	Splitting of spectral lines in a magnetic field
Stark effect	Splitting of spectral lines in an electric field
Wave nature	It treats the electron purely as a particle, ignoring de Broglie's wave character
Uncertainty principle	Fixing both the radius of an orbit and the momentum in it is forbidden by Heisenberg
Shape of orbits	It assumes circular orbits only; Sommerfeld had to add elliptical ones
Bonding	It cannot explain the shapes of molecules or how chemical bonds form

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.

3.8 Quantum numbers

Quantum number	Symbol	Allowed values	Tells us
Principal	n	1, 2, 3, …	Shell (K, L, M, N); size and main energy of the orbital
Azimuthal / angular momentum	l	0 to (n-1)	Subshell (s, p, d, f) and the shape of the orbital
Magnetic	m	-l to +l, including 0	Orientation of the orbital in space
Spin	s	+\tfrac{1}{2} or -\tfrac{1}{2}	Direction of spin of the electron

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).

3.9 Orbitals and shape of s and p orbitals

	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.

3.11 Electronic configurations of atoms and ions (up to atomic number 30)

Z	Element	Configuration	Z	Element	Configuration
1	H	1s¹	16	S	[Ne]3s²3p⁴
2	He	1s²	17	Cl	[Ne]3s²3p⁵
3	Li	[He]2s¹	18	Ar	[Ne]3s²3p⁶
4	Be	[He]2s²	19	K	[Ar]4s¹
5	B	[He]2s²2p¹	20	Ca	[Ar]4s²
6	C	[He]2s²2p²	21	Sc	[Ar]3d¹4s²
7	N	[He]2s²2p³	22	Ti	[Ar]3d²4s²
8	O	[He]2s²2p⁴	23	V	[Ar]3d³4s²
9	F	[He]2s²2p⁵	24	Cr	[Ar]3d⁵4s¹
10	Ne	[He]2s²2p⁶	25	Mn	[Ar]3d⁵4s²
11	Na	[Ne]3s¹	26	Fe	[Ar]3d⁶4s²
12	Mg	[Ne]3s²	27	Co	[Ar]3d⁷4s²
13	Al	[Ne]3s²3p¹	28	Ni	[Ar]3d⁸4s²
14	Si	[Ne]3s²3p²	29	Cu	[Ar]3d¹⁰4s¹
15	P	[Ne]3s²3p³	30	Zn	[Ar]3d¹⁰4s²

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

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.