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Chemistry · Class 11 · Chapter 2

Structure of Atom

This chapter follows the atom from the discovery of the electron to quantum numbers and electronic configuration. Bohr model numericals, the hydrogen spectrum and quantum numbers carry most of the questions, and electronic configuration is used in every chapter after this one.

In this chapter: electron, proton and neutron, Thomson and Rutherford models, atomic number, mass number, isotopes and isobars, electromagnetic radiation, Planck's quantum theory, photoelectric effect, line spectrum of hydrogen, Bohr model, de Broglie relation, Heisenberg's uncertainty principle, quantum numbers, shapes of orbitals, Aufbau principle, Pauli exclusion principle, Hund's rule and electronic configuration.

Subatomic particles

In the cathode ray discharge tube, rays travel from the cathode to the anode, move in straight lines, are deflected towards the positive plate by an electric field and carry negative charge. Their behaviour does not depend on the gas or the electrode material, which showed that electrons are present in all atoms. J.J. Thomson measured the charge to mass ratio of the electron, e/me = 1.758820 × 1011 C kg−1, and Millikan's oil drop experiment gave its charge.

ParticleCharge (C)Relative chargeMass (kg)
Electron−1.602176 × 10−19−19.10939 × 10−31
Proton+1.602176 × 10−19+11.67262 × 10−27
Neutron001.67493 × 10−27

The proton came from studies of canal rays (positive rays) in modified discharge tubes. The neutron was discovered by Chadwick in 1932 by bombarding a thin sheet of beryllium with α-particles.

Atomic models

Thomson model: the atom is a uniform sphere of positive charge with electrons embedded in it, like seeds in a watermelon (the plum pudding model). It explained the overall neutrality of the atom but not the results of scattering experiments.

Rutherford's α-particle scattering: a thin gold foil was bombarded with α-particles.

  • Most α-particles passed through undeflected, so most of the atom is empty space.
  • A small fraction was deflected by small angles, so the positive charge is concentrated in a small region.
  • A very few (about 1 in 20,000) bounced back through nearly 180°. The positive charge and most of the mass sit in a tiny, dense nucleus, of radius about 10−15 m against about 10−10 m for the atom.

Electrons revolve round the nucleus in orbits. The drawback: by classical electromagnetic theory an electron moving in a circle is accelerating, so it should radiate energy and spiral into the nucleus. The model also says nothing about how electrons are arranged or what their energies are.

Atomic number Z = number of protons (= electrons in a neutral atom). Mass number A = protons + neutrons. Isotopes have the same Z but different A (11H, 21H, 31H); their chemical properties are the same because these depend on the number of electrons. Isobars have the same A but different Z (146C and 147N).

Electromagnetic radiation and quantum theory

Electromagnetic waves travel in vacuum at c = 3.0 × 108 m s−1, and c = νλ. Wavenumber is ν̄ = 1/λ, in m−1 or cm−1. In increasing order of frequency: radio waves, microwaves, infrared, visible (about 400 nm violet to 750 nm red), ultraviolet, X-rays, γ-rays.

The wave picture could not explain black body radiation or the photoelectric effect. Planck proposed that energy is emitted or absorbed in discrete packets called quanta (photons for light):

E = hν = hcλh = 6.626 × 10−34 J s (Planck's constant)

Photoelectric effect

When light of suitable frequency falls on a metal like potassium or caesium, electrons are ejected at once. No electrons come out below a threshold frequency ν0, however intense the light. Above it, the kinetic energy of the electrons increases with frequency, while the number of electrons increases with intensity.

hν = hν0 + ½mev2hν0 is the work function W0 of the metal

Light therefore shows both wave nature (interference, diffraction) and particle nature (photoelectric effect, black body radiation). This is its dual behaviour.

Line spectrum of hydrogen

White light gives a continuous spectrum. Atoms of an element, when excited, emit radiation only at particular wavelengths, giving a line (emission) spectrum; the dark lines seen when light passes through the vapour form the absorption spectrum. Each element has its own line spectrum, which is why spectra are called fingerprints of atoms. For hydrogen, all the lines fit the Rydberg formula:

ν̄ = 109,677 ( 1n12 − 1n22 ) cm−1n2 > n1; 109,677 cm−1 is the Rydberg constant for hydrogen
Seriesn1n2Region
Lyman12, 3, 4 ...Ultraviolet
Balmer23, 4, 5 ...Visible
Paschen34, 5, 6 ...Infrared
Brackett45, 6, 7 ...Infrared
Pfund56, 7, 8 ...Infrared

Bohr's model of the hydrogen atom

  • The electron moves in circular orbits (stationary states) of fixed energy and does not radiate energy while it stays in one.
  • Energy is absorbed or emitted only when the electron jumps between orbits: ν = ΔE/h = (E2 − E1)/h.
  • Angular momentum is quantised: mevr = nh/2π, n = 1, 2, 3 ...
En = −2.18 × 10−18 Z2n2 Jrn = 52.9 n2Z pmFor hydrogen Z = 1; the same formulas hold for one-electron ions such as He+, Li2+, Be3+

The negative sign means the electron in the atom has lower energy than a free electron at rest (taken as zero at n = ∞). As n increases, the energy becomes less negative and the orbits get farther apart in radius. For a transition,

ΔE = 2.18 × 10−18 Z2 ( 1ni2 − 1nf2 ) JPositive when energy is absorbed (nf > ni)
Hydrogen energy levels and the Lyman, Balmer and Paschen serieswww.iitmedicoguide.comHydrogen atom: En = −13.6/n² eV (not to scale)n = 1n = 2n = 3n = 4n = 5n = ∞−13.6 eV−3.40 eV−1.51 eV−0.85 eV−0.54 eV0Lyman seriesends at n = 1 (ultraviolet)Balmer seriesends at n = 2 (visible)Paschen seriesends at n = 3 (infrared)www.iitmedicoguide.com
Every line of the hydrogen spectrum is a drop from a higher level to a lower one. The series is named after the level where the electron lands: n = 1 for Lyman, n = 2 for Balmer and n = 3 for Paschen.
Worked example: Calculate the wavelength of the line emitted when the electron in a hydrogen atom falls from n = 3 to n = 2.
Solution: ν̄ = 109,677 (1/22 − 1/32) = 109,677 × 5/36 = 15,233 cm−1. λ = 1/ν̄ = 6.565 × 10−5 cm = 656.5 nm, a red line in the visible Balmer series.

Limitations: the Bohr model works only for hydrogen and one-electron ions. It cannot explain the fine structure of lines, the splitting of lines in a magnetic field (Zeeman effect) or an electric field (Stark effect), or how atoms combine to form molecules. It also ignores the dual nature of the electron and the uncertainty principle.

Towards the quantum mechanical model

de Broglie relation

Matter, like radiation, shows dual behaviour. A particle of mass m moving with velocity v has a wavelength

λ = hmv = hp

The wavelength is significant only for microscopic particles like electrons. The electron microscope uses the wave nature of electrons.

Worked example: Find the de Broglie wavelength of an electron moving at 2.0 × 106 m s−1. (me = 9.1 × 10−31 kg)
Solution: λ = h/mv = 6.626 × 10−34 / (9.1 × 10−31 × 2.0 × 106) = 6.626 × 10−34 / 1.82 × 10−24 = 3.64 × 10−10 m = 364 pm.

Heisenberg's uncertainty principle

Δx × Δpx ≥ h4π

The exact position and exact momentum of an electron cannot be known at the same time. This rules out the idea of fixed, well-defined orbits. The effect matters for microscopic particles; for a cricket ball the uncertainties are far too small to notice.

Schrödinger equation and orbitals

In quantum mechanics the electron is described by a wave function ψ, obtained by solving the Schrödinger equation, Ĥψ = Eψ. ψ itself has no physical meaning; ψ2 gives the probability density of finding the electron at a point. An atomic orbital is the wave function for one electron in an atom, and the region where the electron is most likely to be found is drawn as its boundary surface.

Quantum numbers

Quantum numberAllowed valuesWhat it tells you
Principal, n1, 2, 3 ... (shells K, L, M, N)Size and energy of the orbital; a shell has n2 orbitals and up to 2n2 electrons
Azimuthal, l0 to (n − 1): s = 0, p = 1, d = 2, f = 3Shape of the orbital (subshell); a subshell has (2l + 1) orbitals
Magnetic, ml−l to +l, including 0Orientation of the orbital in space
Spin, ms+½ or −½Spin of the electron (clockwise or anticlockwise)

An orbital is fully described by n, l and ml; an electron needs all four.

Shapes and nodes

s orbitals are spherical. The three p orbitals (px, py, pz) have two lobes each, along the three axes, and are equal in energy. There are five d orbitals: dxy, dyz, dzx (lobes between the axes), dx²−y² (lobes on the axes) and dz².

Radial nodes = n − l − 1Angular nodes = lTotal nodes = n − 1

So 2s has one radial node, 3p has one radial and one angular node, and 3d has no radial node and two angular nodes.

Shapes of s, p and d orbitals (boundary surface diagrams)www.iitmedicoguide.comxyxyxzxyxyxz1s2px2pz3dxy3dx²−y²3dz²spherical, l = 0dumb-bell, l = 1lobes on z axislobes between axeslobes on x and y axestwo lobes and a ringtwo colours = opposite signs (phases) of the wave functionwww.iitmedicoguide.com
Boundary surface diagrams of s, p and d orbitals. The dyz and dzx orbitals look like dxy, lying in their own planes, and all five 3d orbitals have the same energy in an isolated atom.

Filling electrons into orbitals

In hydrogen, orbital energy depends only on n (2s = 2p). In multi-electron atoms it depends on n and l, because inner electrons shield the outer ones and s electrons penetrate closer to the nucleus than p, and p more than d. The (n + l) rule: the lower the value of (n + l), the lower the energy; if two orbitals have the same (n + l), the one with lower n has lower energy.

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s
  • Aufbau principle: orbitals are filled in order of increasing energy.
  • Pauli 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, with opposite spins.
  • Hund's rule of maximum multiplicity: pairing in orbitals of the same subshell (p, d or f) does not start until each orbital has one electron with parallel spin. Nitrogen is 1s2 2s2 2px1 2py1 2pz1.

Exceptions: Cr (Z = 24) is [Ar] 3d5 4s1, not 3d4 4s2, and Cu (Z = 29) is [Ar] 3d10 4s1, not 3d9 4s2. Half-filled and completely filled subshells are extra stable because of their symmetrical distribution and larger exchange energy.

Common mistakes: (1) In the Rydberg formula n1 is the lower level; the series is fixed by where the electron ends. (2) Forgetting Z2 in energy and Z in radius for He+ or Li2+. (3) Writing l = n for a subshell; l goes only up to n − 1, so there is no 1p or 2d. (4) Saying intensity of light increases the kinetic energy of photoelectrons; it increases only their number. (5) Filling 3d before 4s, and missing the Cr and Cu exceptions.

JEE and NEET focus

  • Bohr model: energy, radius and velocity trends with n and Z; energy and wavelength of transitions.
  • Hydrogen spectrum series, the number of spectral lines from level n (n(n − 1)/2), and the region of each series.
  • Photoelectric effect equation, work function and threshold frequency calculations.
  • de Broglie wavelength and uncertainty principle numericals, keeping SI units throughout.
  • Allowed sets of quantum numbers, number of orbitals and electrons in a shell, radial and angular nodes.
  • Electronic configurations of the first 30 elements with the Cr and Cu exceptions, and unpaired electrons in atoms and ions.

Practice questions

The number of radial nodes in a 3p orbital is:

  1. 0
  2. 1
  3. 2
  4. 3
Show answer
B. n − l − 1 = 3 − 1 − 1 = 1.

Which set of quantum numbers is not possible?

  1. n = 2, l = 1, ml = 0
  2. n = 3, l = 2, ml = −2
  3. n = 2, l = 2, ml = 1
  4. n = 4, l = 0, ml = 0
Show answer
C. l can be at most n − 1, so l = 2 is not allowed for n = 2.

The radius of the first Bohr orbit of Li2+ is about:

  1. 52.9 pm
  2. 158.7 pm
  3. 26.5 pm
  4. 17.6 pm
Show answer
D. r = 52.9 × n2/Z = 52.9 × 1/3 = 17.6 pm.

The energy of the electron in the second orbit of He+ is:

  1. −2.18 × 10−18 J
  2. −5.45 × 10−19 J
  3. −8.72 × 10−18 J
  4. −1.09 × 10−18 J
Show answer
A. E = −2.18 × 10−18 × Z2/n2 = −2.18 × 10−18 × 4/4.

The ground state electronic configuration of chromium (Z = 24) is:

  1. [Ar] 3d4 4s2
  2. [Ar] 3d5 4s1
  3. [Ar] 3d6
  4. [Ar] 4s2 4p4
Show answer
B. A half-filled 3d subshell is more stable.

In the photoelectric effect, the kinetic energy of the emitted electrons depends on:

  1. Intensity of light
  2. Time of exposure
  3. Frequency of light
  4. Area of the metal plate
Show answer
C. ½mv2 = h(ν − ν0); intensity changes only the number of electrons.

The lines of the Balmer series of hydrogen lie in the:

  1. Ultraviolet region
  2. Visible region
  3. Infrared region
  4. Microwave region
Show answer
B. Transitions ending at n = 2 fall in the visible region.

The maximum number of electrons that can be accommodated in the shell with n = 3 is:

  1. 8
  2. 9
  3. 32
  4. 18
Show answer
D. 2n2 = 18 (9 orbitals, two electrons each).
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