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Chapter 2: Structure of Atom

2.1 Discovery of Subatomic Particles

Discovery of Electron — Cathode Ray Experiment (J.J. Thomson, 1897)

When a high voltage is applied across a cathode ray discharge tube at very low pressure (~0.001 mm Hg), a stream of particles moves from the cathode to the anode. These are called cathode rays.

Cathode + Anode Cathode Rays (stream of electrons) Cathode Ray Discharge Tube glow To vacuum pump

Properties of Cathode Rays:

  1. Travel in straight lines
  2. Consist of negatively charged particles (electrons)
  3. Independent of the material of the cathode or gas in the tube
  4. Possess kinetic energy and can rotate a paddle wheel

Charge-to-mass ratio of electron (J.J. Thomson):

\[ \frac{e}{m_e} = 1.758820 \times 10^{11};\text{C/kg} \]

Charge of electron (R.A. Millikan — Oil Drop Experiment):

\[ e = 1.6022 \times 10^{-19};\text{C} \]

Mass of electron:

\[ m_e = 9.1094 \times 10^{-31};\text{kg} \]

Discovery of Proton — Anode Rays (Canal Rays)

When a perforated cathode is used in a discharge tube, a stream of positively charged particles moves towards the cathode. These are anode rays or canal rays.

Properties:

  • The charge-to-mass ratio depends on the gas in the tube
  • The lightest particles are obtained with hydrogen → protons

\[ m_p = 1.6726 \times 10^{-27};\text{kg}, \quad e_p = +1.6022 \times 10^{-19};\text{C} \]

Discovery of Neutron (James Chadwick, 1932)

Chadwick bombarded beryllium with α-particles and observed the emission of electrically neutral particles with a mass slightly greater than that of protons — neutrons.

\[ {}^{9}_{4}\text{Be} + {}^{4}_{2}\text{He} \rightarrow {}^{12}_{6}\text{C} + {}^{1}_{0}\text{n} \]

\[ m_n = 1.6750 \times 10^{-27};\text{kg} \]

ParticleSymbolCharge (C)Mass (kg)Mass (u)
Electron\(e^-\)\(-1.6022 \times 10^{-19}\)\(9.109 \times 10^{-31}\)0.00055
Proton\(p^+\)\(+1.6022 \times 10^{-19}\)\(1.6726 \times 10^{-27}\)1.00727
Neutron\(n^0\)0\(1.6750 \times 10^{-27}\)1.00866

2.2 Atomic Number, Mass Number, Isotopes, and Isobars

Atomic number (Z) = Number of protons = Number of electrons (in a neutral atom)

Mass number (A) = Number of protons + Number of neutrons

\[ A = Z + \text{number of neutrons} \]

Notation: \( {}^{A}_{Z}\text{X} \), e.g., \( {}^{12}_{6}\text{C} \), \( {}^{23}_{11}\text{Na} \)

TermDefinitionExample
IsotopesSame Z, different A\({}^{1}_{1}\text{H}\), \({}^{2}_{1}\text{H}\), \({}^{3}_{1}\text{H}\)
IsobarsSame A, different Z\({}^{40}_{18}\text{Ar}\), \({}^{40}_{19}\text{K}\), \({}^{40}_{20}\text{Ca}\)
IsotonesSame number of neutrons\({}^{14}_{6}\text{C}\), \({}^{15}_{7}\text{N}\) (both have 8 neutrons)

2.3 Thomson’s Model of Atom (1904)

J.J. Thomson proposed the “plum pudding” model: the atom is a sphere of positive charge in which electrons are embedded, like plums in a pudding.

+ + + + + + + e⁻ e⁻ e⁻ e⁻ e⁻ Thomson's "Plum Pudding" Model Uniform positive charge with embedded electrons

Limitation: Could not explain the results of Rutherford’s scattering experiment.

2.4 Rutherford’s Nuclear Model (1911)

The α-Particle Scattering Experiment

Rutherford bombarded a thin gold foil (0.0004 cm thick) with α-particles from a radioactive source.

Observations:

  1. Most α-particles passed through undeflected → atom is mostly empty space
  2. A small fraction was deflected by small angles → positive charge is concentrated
  3. Very few (~1 in 20,000) bounced back → the positive charge occupies a very small volume (nucleus)
α source Gold foil Most pass through (undeflected) Small angle deflection ~ 1 in 20,000 bounced back ZnS Screen

Rutherford’s Conclusions:

  1. The atom has a tiny, dense, positively charged centre called the nucleus (radius ~ \(10^{-15}\) m)
  2. Nearly all mass is concentrated in the nucleus
  3. Electrons revolve around the nucleus in circular orbits
  4. The atom is mostly empty space (radius ~ \(10^{-10}\) m)

Limitations:

  • Could not explain the stability of atoms (accelerating charged particles should radiate energy and spiral into the nucleus)
  • Could not explain line spectra of atoms

2.5 Bohr’s Model of the Hydrogen Atom (1913)

Niels Bohr proposed a model for hydrogen-like atoms based on quantum ideas:

Postulates

  1. Electrons revolve in fixed circular orbits (called stationary states or shells) without radiating energy.

  2. Quantized angular momentum: The angular momentum of an electron in a stationary state is an integral multiple of \(\frac{h}{2\pi}\):

\[ m_e v r = n \frac{h}{2\pi}, \quad n = 1, 2, 3, \ldots \]

  1. Energy transitions: When an electron jumps from a higher orbit (\(n_2\)) to a lower orbit (\(n_1\)), energy is emitted as a photon:

\[ \Delta E = E_{n_2} - E_{n_1} = h\nu \]

Key Results for Hydrogen-like Species

Radius of \(n\)-th orbit:

\[ r_n = \frac{n^2 a_0}{Z} \]

where \( a_0 = 52.9;\text{pm} \) (Bohr radius), \(Z\) = atomic number.

Energy of \(n\)-th orbit:

\[ E_n = -\frac{13.6,Z^2}{n^2};\text{eV} = -\frac{2.18 \times 10^{-18},Z^2}{n^2};\text{J} \]

Velocity of electron:

\[ v_n = \frac{2.18 \times 10^6 , Z}{n};\text{m/s} \]

Hydrogen Spectrum

When excited hydrogen atoms return to lower energy levels, they emit photons of specific wavelengths, producing line spectra.

\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where \( R_H = 1.097 \times 10^7;\text{m}^{-1} \) (Rydberg constant)

Series\(n_1\)\(n_2\)Region
Lyman12, 3, 4, …Ultraviolet
Balmer23, 4, 5, …Visible
Paschen34, 5, 6, …Infrared
Brackett45, 6, 7, …Infrared
Pfund56, 7, 8, …Far infrared
n = ∞ E = 0 n = 5 −0.54 eV n = 4 −0.85 eV n = 3 −1.51 eV n = 2 −3.40 eV n = 1 −13.6 eV Lyman (UV) Balmer (Visible) Paschen (IR)

Limitations of Bohr’s Model:

  1. Works only for hydrogen-like (one-electron) species
  2. Could not explain the fine structure of spectral lines
  3. Could not explain the Zeeman effect or Stark effect
  4. Does not account for electron-electron repulsion in multi-electron atoms

2.6 Dual Nature of Matter and Radiation

Wave-Particle Duality

Light exhibits both wave and particle nature:

  • Wave nature: Diffraction, interference
  • Particle nature: Photoelectric effect, black-body radiation

Planck’s quantum theory:

\[ E = h\nu = \frac{hc}{\lambda} \]

where \( h = 6.626 \times 10^{-34};\text{J·s} \) (Planck’s constant)

Photoelectric effect (Einstein):

\[ h\nu = h\nu_0 + \frac{1}{2}m_e v^2 \]

where \( \nu_0 \) = threshold frequency

de Broglie Relation

Louis de Broglie (1924) proposed that matter also has a dual nature:

\[ \lambda = \frac{h}{mv} = \frac{h}{p} \]

where \( \lambda \) = wavelength, \( m \) = mass, \( v \) = velocity, \( p \) = momentum.

Note: The wave nature is significant only for microscopic particles (electrons, protons) and negligible for macroscopic objects.

2.7 Heisenberg’s Uncertainty Principle

\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]

or equivalently:

\[ \Delta x \cdot m\Delta v \geq \frac{h}{4\pi} \]

It is impossible to simultaneously determine the exact position and exact momentum of an electron. This principle makes the concept of fixed orbits (Bohr model) meaningless.

2.8 Quantum Mechanical Model of the Atom

Schrödinger Wave Equation

\[ \hat{H}\psi = E\psi \]

The full time-independent equation:

\[ \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2} + \frac{8\pi^2 m}{h^2}(E - V)\psi = 0 \]

  • \(\psi\) = wave function
  • \(|\psi|^2\) = probability density of finding the electron
  • \(E\) = total energy
  • \(V\) = potential energy

Quantum Numbers

Each electron in an atom is described by a set of four quantum numbers:

Quantum NumberSymbolValuesDescribes
Principal\(n\)1, 2, 3, …Shell (energy level), size of orbital
Azimuthal\(l\)0 to \(n-1\)Subshell (shape of orbital)
Magnetic\(m_l\)\(-l\) to \(+l\)Orientation of orbital in space
Spin\(m_s\)\(+\frac{1}{2}\) or \(-\frac{1}{2}\)Spin of electron

Subshell notation:

\(l\)0123
Subshellspdf
No. of orbitals1357
Max. electrons261014

2.9 Shapes of Orbitals

s Orbitals (\(l = 0\)) — Spherical

1s 2s (with nodal sphere)
  • Spherically symmetric
  • Node at nucleus for \(n \geq 2\) (radial node)
  • Number of radial nodes = \(n - l - 1\)

p Orbitals (\(l = 1\)) — Dumbbell shaped

+ x p_x + y p_y + z p_z
  • Each p orbital has two lobes with a nodal plane at the nucleus
  • Three p orbitals are oriented along x, y, and z axes (mutually perpendicular)

d Orbitals (\(l = 2\)) — Cloverleaf shapes

There are five d orbitals: \(d_{xy}\), \(d_{yz}\), \(d_{xz}\), \(d_{x^2-y^2}\), \(d_{z^2}\).

  • \(d_{xy}\), \(d_{yz}\), \(d_{xz}\): four lobes between the axes
  • \(d_{x^2-y^2}\): four lobes along x and y axes
  • \(d_{z^2}\): two lobes along z-axis with a doughnut-shaped ring in the xy plane

2.10 Rules for Filling Electrons in Orbitals

1. Aufbau Principle

Electrons fill orbitals in order of increasing energy (\(n + l\) rule). If \(n + l\) values are equal, the orbital with the lower \(n\) fills first.

Energy order:

\[ 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f \ldots \]

2. Pauli’s Exclusion Principle

No two electrons in an atom can have the same set of four quantum numbers. Consequently, each orbital can hold a maximum of 2 electrons with opposite spins.

3. Hund’s Rule of Maximum Multiplicity

Electrons are distributed among orbitals of a subshell in such a way as to give the maximum number of unpaired electrons with parallel spins. Pairing occurs only after all degenerate orbitals are singly occupied.

Hund's Rule Example: Nitrogen (Z = 7) 1s² 2s² 2p³ ✓ Correct: Maximum unpaired spins 2p³ ✗ Wrong: Pairing before singly filling

2.11 Electronic Configuration of Atoms

The electronic configuration is written as: \(nl^x\), where \(n\) = principal quantum number, \(l\) = subshell, \(x\) = number of electrons.

ElementZElectronic Configuration
H1\(1s^1\)
He2\(1s^2\)
Li3\([He],2s^1\)
C6\([He],2s^2,2p^2\)
N7\([He],2s^2,2p^3\)
O8\([He],2s^2,2p^4\)
Ne10\([He],2s^2,2p^6\)
Na11\([Ne],3s^1\)
Ar18\([Ne],3s^2,3p^6\)
K19\([Ar],4s^1\)
Ca20\([Ar],4s^2\)
Fe26\([Ar],3d^6,4s^2\)
Cu29\([Ar],3d^{10},4s^1\) ★
Cr24\([Ar],3d^5,4s^1\) ★
Zn30\([Ar],3d^{10},4s^2\)

Anomalous configurations: Cr and Cu achieve extra stability from half-filled (\(d^5\)) and completely filled (\(d^{10}\)) d-subshells respectively.

Stability of Half-filled and Completely Filled Subshells

Two factors contribute:

  1. Symmetrical distribution of electrons → lower energy
  2. Exchange energy — electrons with parallel spins in degenerate orbitals can exchange positions, releasing energy. More exchanges → greater stability.

Practice Questions

Multiple Choice Questions (MCQs)

1. The number of angular nodes for a 4d orbital is:

 (a) 1

 (b) 2

 (c) 3

 (d) 4


2. Which of the following sets of quantum numbers is NOT possible?

 (a) \(n = 2,\ l = 1,\ m_l = 0,\ m_s = +\frac{1}{2}\)

 (b) \(n = 3,\ l = 2,\ m_l = -2,\ m_s = -\frac{1}{2}\)

 (c) \(n = 2,\ l = 2,\ m_l = 0,\ m_s = +\frac{1}{2}\)

 (d) \(n = 4,\ l = 0,\ m_l = 0,\ m_s = -\frac{1}{2}\)


3. If uncertainty in position of an electron is zero, the uncertainty in its momentum would be:

 (a) zero

 (b) \(\geq \frac{h}{4\pi}\)

 (c) \(\lt \frac{h}{4\pi}\)

 (d) infinite


4. The wavelength of a ball of mass 100 g moving with a velocity of 100 m/s is (\(h = 6.6 \times 10^{-34}\) J·s):

 (a) \(6.6 \times 10^{-35}\) m

 (b) \(6.6 \times 10^{-34}\) m

 (c) \(6.6 \times 10^{-33}\) m

 (d) \(6.6 \times 10^{-32}\) m


5. The electronic configuration of Cu (Z = 29) is:

 (a) \([Ar],3d^9,4s^2\)

 (b) \([Ar],3d^{10},4s^1\)

 (c) \([Ar],3d^{10},4s^2\)

 (d) \([Ar],3d^8,4s^2,4p^1\)

Short Answer Questions (2–3 Marks)

6. Calculate the wavelength of an electron moving with a velocity of \(2.05 \times 10^7\) m/s.


7. Write the electronic configuration of Fe²⁺ and Fe³⁺ ions. Which one is more stable and why?


8. What is the maximum number of electrons that can have the quantum numbers \(n = 3\), \(l = 2\)?


9. State Heisenberg’s uncertainty principle. Why is it significant for microscopic particles but not for macroscopic objects?


10. Write the four quantum numbers for the last electron of sodium (Z = 11).

Long Answer Questions (5 Marks)

11. (a) Explain the Bohr model of hydrogen atom. Derive the expression for the radius and energy of the \(n\)-th orbit.

 (b) Calculate the wavelength of the first line in the Balmer series of hydrogen spectrum.


12. (a) Draw the shapes of the five d orbitals. How do \(d_{z^2}\) and \(d_{x^2-y^2}\) differ from the other three d orbitals?

 (b) Explain why Cr has the electronic configuration \([Ar],3d^5,4s^1\) and not \([Ar],3d^4,4s^2\).


13. (a) State and explain the photoelectric effect. How did it provide evidence for the particle nature of light?

 (b) A photon of wavelength 4 × 10⁻⁷ m strikes a metal surface, the work function of the metal being 2.13 eV. Calculate the kinetic energy and velocity of the emitted photoelectron.

Assertion-Reason Questions

14. Assertion (A): The energy of 2s orbital is less than that of 2p orbital in multi-electron atoms.

Reason (R): 2s electrons have greater penetration power than 2p electrons.


15. Assertion (A): The total number of nodes for 3p orbital is 2.

Reason (R): Total nodes = \(n - 1\), angular nodes = \(l\), radial nodes = \(n - l - 1\).


Answer Key

QAnswer
1(b) — Angular nodes = \(l\) = 2
2(c) — For \(n=2\), max \(l = 1\), so \(l=2\) is not possible
3(d) — If \(\Delta x = 0\), then \(\Delta p \to \infty\) by uncertainty principle
4(a) — \(\lambda = h/mv = 6.6 \times 10^{-34}/(0.1 \times 100) = 6.6 \times 10^{-35}\) m
5(b) — \([Ar],3d^{10},4s^1\) due to stability of completely filled d subshell
14(a) — Both A and R true; R is the correct explanation of A
15(a) — Both A and R true; R is the correct explanation (total = 3-1 = 2; angular = 1; radial = 1)