Structure of Atom Class 11 Chemistry — Revision Notes

The atom consists of a nucleus containing protons and neutrons, surrounded by electrons. Understanding atomic structure is essential for understanding chem

Structure of Atom Class 11 Chemistry — Revision Notes

The atom consists of a nucleus containing protons and neutrons, surrounded by electrons. Understanding atomic structure is essential for understanding chem

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TL;DR: The atom consists of a nucleus containing protons and neutrons, surrounded by electrons. Understanding atomic structure is essential for understanding…

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The atom consists of a nucleus containing protons and neutrons, surrounded by electrons. Understanding atomic structure is essential for understanding chem

Subatomic Particles

  • Proton has positive charge and resides in nucleus
  • Neutron has no charge and resides in nucleus
  • Electron has negative charge and revolves around nucleus
  • Proton and neutron have nearly equal mass
  • Electron mass is approximately 1/1836 of proton mass

Nuclear Atom Model

  • Rutherford discovered that atom has dense nucleus with electrons around it
  • Most of atom's mass is concentrated in the nucleus
  • Nucleus is extremely small compared to overall atom size
  • Electrons are attracted to nucleus by electrical force
  • Electrons are in constant motion around nucleus

Bohr's Model of Atom

  • Bohr proposed that electrons occupy specific energy levels or orbits
  • Electrons can jump between orbits by absorbing or emitting energy
  • First orbit holds maximum 2 electrons; second holds maximum 8
  • Energy increases with distance from nucleus
  • Model explains hydrogen atom spectrum but fails for larger atoms

Quantum Model of Atom

  • Electrons exist in regions called orbitals with specific energy and shape
  • Orbital is region where electron is likely to be found
  • Probability of finding electron is shown by electron density
  • Aufbau principle: electrons fill lowest energy orbitals first
  • Pauli exclusion principle: no two electrons have identical quantum numbers

Electronic Configuration

  • Electronic configuration shows arrangement of electrons in atoms
  • Written as 1s2 2s2 2p6 3s2 3p6 for example
  • Valence electrons are outermost electrons determining chemical properties
  • Octet rule: atoms gain, lose, or share electrons to have 8 valence electrons
  • Configuration determines element's position in periodic table

Key Terms

  • Atom: Smallest unit of element that retains properties of element
  • Nucleus: Central part of atom containing protons and neutrons
  • Orbital: Region of space where electron is likely to be found
  • Valence Electron: Electron in outermost shell that participates in chemical bonding

Why Rutherford's model had to be abandoned

Rutherford's alpha-particle scattering experiment was decisive about where the mass and positive charge sit. Most alpha particles passed through the gold foil undeflected, showing that the atom is largely empty space. A small fraction were deflected through small angles, and about one in twenty thousand was deflected through more than ninety degrees or bounced straight back, showing that all the positive charge and nearly all the mass are concentrated in a tiny central nucleus. The nucleus has a radius of about ten to the power minus fifteen metres against an atomic radius of about ten to the power minus ten metres.

The model that followed — electrons revolving around this nucleus like planets round the sun — then ran into a fatal objection from classical electrodynamics. A charged particle moving in a circular path is accelerating, and an accelerating charge must continuously radiate energy. Losing energy, the electron would spiral inwards and collapse into the nucleus in a fraction of a second.

So the model predicted that no atom could exist, which is plainly wrong. It also predicted a continuous spectrum as the electron spiralled through every possible energy, whereas real atoms give sharp line spectra. Both failures point the same way: the energy of an electron in an atom cannot take arbitrary values.

From Bohr's orbits to quantum-mechanical orbitals

Bohr fixed the stability problem by decree. He postulated that the electron revolves only in certain permitted orbits of fixed energy, called stationary states, in which it does not radiate; energy is emitted or absorbed only when the electron jumps between them, with the frequency given by the energy difference divided by Planck's constant. The permitted orbits are those in which the angular momentum is an integral multiple of h over two pi.

That worked beautifully for hydrogen and explained its line spectrum quantitatively, but it failed for every atom with more than one electron, could not account for the fine structure of spectral lines, the splitting in a magnetic field or an electric field, and said nothing about chemical bonding.

The deeper problem is that Bohr's model keeps the idea of a definite path. De Broglie's proposal that matter has wave character, and Heisenberg's uncertainty principle, together make a precise orbit meaningless — you cannot know position and momentum simultaneously with arbitrary precision. The quantum mechanical model therefore replaces the orbit with an orbital: a region of space where the probability of finding the electron is high, obtained from the square of the wave function.

The subatomic particles and how each was found

Each particle is associated with a specific experiment, and examiners ask for the pairing.

  • The electron came from cathode ray discharge tube experiments. J. J. Thomson determined the charge to mass ratio of the electron as 1.758820 times ten to the power eleven coulomb per kilogram.
  • Millikan's oil drop experiment established the charge on the electron as 1.6 times ten to the power minus nineteen coulomb in magnitude. Combining this with Thomson's ratio gives the mass of the electron as about 9.1094 times ten to the power minus thirty-one kilogram.
  • The proton was identified from canal rays, or anode rays, observed by Goldstein. It carries a charge equal and opposite to that of the electron and has a mass of about 1.67262 times ten to the power minus twenty-seven kilogram.
  • The neutron was discovered by Chadwick in 1932 by bombarding a thin sheet of beryllium with alpha particles. It is electrically neutral and has a mass of about 1.67493 times ten to the power minus twenty-seven kilogram, very slightly greater than that of the proton.
  • Atomic number Z is the number of protons in the nucleus, and in a neutral atom it equals the number of electrons. Mass number A is the sum of protons and neutrons.
  • Isotopes have the same atomic number but different mass numbers, so they differ in the number of neutrons. Isobars have the same mass number but different atomic numbers.

Quantum theory, the photoelectric effect and atomic spectra

Three experimental results forced the abandonment of classical physics for atoms, and all three appear as short-answer questions.

  • Planck proposed that energy is emitted or absorbed not continuously but in discrete packets called quanta, with the energy of one quantum given by E equals h nu, where h is Planck's constant, 6.626 times ten to the power minus thirty-four joule second.
  • In the photoelectric effect, electrons are ejected from a metal surface when light of sufficient frequency falls on it. There is a threshold frequency below which no electrons are emitted no matter how intense the light, and the kinetic energy of the ejected electrons depends on the frequency, not the intensity. Intensity affects only the number of electrons.
  • Einstein explained this in 1905 by treating light as a stream of photons: the photon energy h nu is used partly to overcome the work function of the metal and the rest appears as the kinetic energy of the electron. The work function equals h times the threshold frequency.
  • The hydrogen emission spectrum consists of discrete lines grouped into series. The Lyman series lies in the ultraviolet and ends at n equal to 1; the Balmer series lies in the visible and ends at n equal to 2; the Paschen, Brackett and Pfund series lie in the infrared and end at n equal to 3, 4 and 5 respectively.
  • The wave number of any line is given by the Rydberg expression, 109677 times the quantity one over n one squared minus one over n two squared, in reciprocal centimetres.

Bohr's model — postulates and results

Learn the three postulates as statements, and the two formulae for hydrogen as results.

  • First postulate: the electron revolves round the nucleus in certain permitted circular orbits of fixed radius and energy, called stationary states or orbits, and while in them it does not radiate energy.
  • Second postulate: energy is emitted or absorbed only when the electron moves from one stationary state to another, and the frequency of the radiation is the energy difference divided by Planck's constant.
  • Third postulate: the angular momentum of the electron is quantised, being an integral multiple of h divided by two pi.
  • For hydrogen, the radius of the nth orbit is 52.9 times n squared picometres, so the orbits get rapidly further apart.
  • The energy of the electron in the nth orbit is minus 2.18 times ten to the power minus eighteen joule divided by n squared, which is minus 13.6 divided by n squared electron volts.
  • The energy is negative because it is measured relative to a free electron at rest at infinite distance, which is assigned zero. A more negative value therefore means a more tightly bound electron.
  • Limitations: the model fails for atoms with more than one electron, cannot explain the fine structure of spectral lines, cannot explain the splitting of lines in magnetic or electric fields, and offers no account of chemical bonding.

Quantum numbers and the rules for filling orbitals

Four quantum numbers specify an electron completely, and three rules determine the order in which orbitals fill.

  • The principal quantum number n takes values 1, 2, 3 and so on. It identifies the shell and largely determines the size and the energy.
  • The azimuthal quantum number l takes values from 0 to n minus 1. It identifies the subshell and determines the shape: l equal to 0, 1, 2 and 3 correspond to s, p, d and f respectively.
  • The magnetic quantum number m sub l takes values from minus l to plus l including zero, giving 2l plus 1 values. It specifies the orientation of the orbital in space.
  • The spin quantum number m sub s takes only the values plus one half and minus one half.
  • The number of orbitals in a shell is n squared, and the maximum number of electrons it can hold is 2n squared.
  • The Aufbau principle states that orbitals are filled in order of increasing energy. The order is given by the n plus l rule: the subshell with the lower value of n plus l fills first, and when two subshells have the same n plus l value, the one with the lower n fills first. This is why 4s fills before 3d.
  • The Pauli exclusion principle states that no two electrons in an atom can have the same set of all four quantum numbers, so an orbital can hold at most two electrons and they must have opposite spins.
  • Hund's rule of maximum multiplicity states that pairing of electrons in the orbitals of a subshell does not take place until each orbital of that subshell has one electron, and all such singly occupied orbitals have electrons with parallel spin.
  • Chromium and copper are the standard exceptions, having configurations ending 3d⁵4s¹ and 3d¹⁰4s¹ respectively, because exactly half-filled and completely filled subshells have extra stability.

Solved Numericals — Step by Step

Problem 1. Calculate the energy and the radius of the third orbit of the hydrogen atom.

  1. For hydrogen the energy of the nth orbit is Eₙ = −13.6/n² eV.
  2. For n = 3, E₃ = −13.6 ÷ 9 = −1.51 eV.
  3. The radius of the nth orbit is rₙ = 52.9 n² pm.
  4. For n = 3, r₃ = 52.9 × 9 = 476.1 pm.

Answer: E₃ = −1.51 eV and r₃ = 476.1 pm

Problem 2. Calculate the de Broglie wavelength of an electron moving with a velocity of 2.19 × 10⁶ m/s. Take the mass of the electron as 9.11 × 10⁻³¹ kg.

  1. The de Broglie relation is λ = h ÷ mv.
  2. Compute the denominator: mv = 9.11 × 10⁻³¹ × 2.19 × 10⁶ = 1.995 × 10⁻²⁴ kg m/s.
  3. Substitute h = 6.626 × 10⁻³⁴ J s: λ = 6.626 × 10⁻³⁴ ÷ 1.995 × 10⁻²⁴.
  4. λ = 3.32 × 10⁻¹⁰ m, that is 332 pm — comparable to atomic dimensions, which is why electron wave behaviour matters in atoms.

Answer: 3.32 × 10⁻¹⁰ m (332 pm)

Problem 3. How many orbitals and how many electrons are possible in the shell with n = 3? List the subshells.

  1. For n = 3, the azimuthal quantum number l can take values 0, 1 and 2, giving the 3s, 3p and 3d subshells.
  2. The number of orbitals in a subshell is 2l + 1: 3s has 1, 3p has 3 and 3d has 5.
  3. Total orbitals = 1 + 3 + 5 = 9, which agrees with the formula n² = 3² = 9.
  4. Each orbital holds at most 2 electrons, so the maximum is 18, agreeing with 2n² = 2 × 9.

Answer: 9 orbitals (3s, 3p, 3d) holding a maximum of 18 electrons

Problem 4. Find the number of radial nodes, angular nodes and total nodes in a 3p orbital.

  1. For a 3p orbital, n = 3 and l = 1.
  2. Angular nodes equal l, so there is 1 angular node.
  3. Radial nodes equal n − l − 1 = 3 − 1 − 1 = 1.
  4. Total nodes equal n − 1 = 2, which checks against the sum 1 + 1 = 2.

Answer: 1 radial node, 1 angular node, 2 nodes in total

Problem 5. Calculate the energy released when an electron in a hydrogen atom falls from n = 3 to n = 2.

  1. Energy of each level: E₂ = −13.6/4 = −3.40 eV and E₃ = −13.6/9 = −1.51 eV.
  2. Energy released = E₃ − E₂ = −1.51 − (−3.40) = 1.89 eV.
  3. Equivalently, ΔE = 13.6 × (1/2² − 1/3²) = 13.6 × (1/4 − 1/9) = 13.6 × 5/36 = 1.89 eV.
  4. This transition ends at n = 2, so it belongs to the Balmer series and lies in the visible region, at about 656 nm.

Answer: 1.89 eV, a visible Balmer line at about 656 nm

Problem 6. Write the electronic configuration of chromium (Z = 24) and explain why it is an exception to the expected filling.

  1. Filling in the order given by the n + l rule up to 24 electrons would predict [Ar] 3d⁴ 4s².
  2. The observed configuration is [Ar] 3d⁵ 4s¹.
  3. One 4s electron is promoted to 3d because the 4s and 3d energies are very close in this region.
  4. The promotion produces an exactly half-filled 3d subshell together with a half-filled 4s, and half-filled subshells have extra stability arising from symmetrical distribution and greater exchange energy.

Answer: [Ar] 3d⁵ 4s¹, adopted for the extra stability of the half-filled 3d subshell

Worked Examples

  • 4s fills before 3d — By the n + l rule, 4s has n + l = 4 + 0 = 4 while 3d has 3 + 2 = 5. The lower sum fills first, which is why potassium and calcium put electrons into 4s.
  • 3d and 4p both have n + l = 5 — When the sums are equal, the subshell with the lower n fills first, so 3d fills before 4p.
  • Nitrogen is 1s² 2s² 2p³ with three unpaired electrons — By Hund's rule the three 2p electrons occupy the three separate 2p orbitals with parallel spins rather than pairing up in one.
  • Isotopes of hydrogen — Protium, deuterium and tritium all have Z = 1 but mass numbers 1, 2 and 3, so they differ only in neutron number — the defining feature of isotopes.

Common Mistakes Students Make

  • Saying an electron's energy is negative because the electron has negative charge. The sign comes from choosing zero energy for a free electron at rest at infinity; a bound electron has less energy than that, so the value is negative.
  • Treating an orbital as an orbit. An orbit is a definite path, which the uncertainty principle rules out; an orbital is a region of high probability, obtained from the square of the wave function.
  • Writing l values up to n instead of n − 1. For n = 3, l can be 0, 1 or 2 only, so there is no 3f subshell.
  • Applying Bohr's energy formula to multi-electron atoms. The expression with −13.6/n² is for hydrogen and hydrogen-like species only; Bohr's model fails once there is more than one electron.
  • Ignoring Hund's rule and pairing electrons in the first orbital of a subshell. Each orbital of the subshell must receive one electron with parallel spin before any pairing begins.
  • Writing chromium as [Ar] 3d⁴ 4s². The observed configuration is 3d⁵ 4s¹, because exactly half-filled subshells are extra stable; copper is similarly 3d¹⁰ 4s¹.
  • Claiming intensity of light determines the kinetic energy of photoelectrons. Frequency determines the kinetic energy; intensity determines only the number of electrons ejected.
  • Confusing isotopes with isobars. Isotopes share the atomic number and differ in mass number; isobars share the mass number and differ in atomic number.

Where This Shows Up in Real Life

Flame tests and the colours of fireworks

Heating a metal salt promotes electrons to higher energy levels, and when they fall back the energy difference is emitted as light of a definite frequency. Because the level spacings differ from element to element, each gives a characteristic colour — the basis of the flame test used to identify sodium, potassium, calcium and copper salts in the laboratory, and of the colours chosen in fireworks. The line spectrum is effectively a fingerprint, which is also how elements are identified in the light of distant stars.

Why the periodic table has the shape it does

The lengths of the periods follow directly from the number of orbitals available: two electrons for an s subshell, six for p, ten for d and fourteen for f. Combined with the n plus l filling order this produces periods of 2, 8, 8, 18, 18 and 32 elements, and it explains why the d block appears only from the fourth period and the f block only from the sixth. The block an element belongs to is simply the subshell its last electron enters.

Electron microscopes and de Broglie's relation

An optical microscope cannot resolve detail smaller than the wavelength of visible light. Since the de Broglie wavelength of a particle is Planck's constant divided by its momentum, electrons accelerated to high speed have wavelengths thousands of times shorter than visible light. An electron microscope exploits exactly this to image objects far below the optical limit, which makes the wave nature of matter a working engineering principle rather than an abstraction.

Photocells and the photoelectric effect

A photocell uses a metal with a low work function so that ordinary light exceeds its threshold frequency and ejects electrons, producing a current that stops the moment the light is blocked. This is the operating principle behind automatic doors, burglar alarms and light meters. It also illustrates why intensity alone is not enough: light below the threshold frequency produces no current however bright it is, which is exactly the observation classical wave theory could not explain.

Board-Style Questions with Model Answers

Describe Rutherford's alpha-particle scattering experiment and the conclusions drawn from it. (5 marks)

A beam of fast-moving alpha particles was directed at a very thin gold foil, and the scattered particles were detected on a fluorescent screen. Three observations were made. Most of the alpha particles passed straight through the foil undeflected, from which it was concluded that most of the space inside the atom is empty. A small fraction were deflected through small angles, indicating the presence of a concentrated positive charge that repelled them. A very small number, about one in twenty thousand, were deflected through large angles or bounced straight back, which could only happen if all the positive charge and nearly all the mass of the atom were concentrated in an extremely small volume. Rutherford therefore proposed the nuclear model, in which a tiny dense positively charged nucleus of radius about ten to the power minus fifteen metres lies at the centre of an atom of radius about ten to the power minus ten metres, with electrons revolving around it.

What were the two main drawbacks of Rutherford's model of the atom? (2 marks)

First, it could not explain the stability of the atom. An electron revolving in a circular orbit is continuously accelerating, and according to classical electromagnetic theory an accelerating charged particle must radiate energy continuously; losing energy, the electron would follow a spiral path and fall into the nucleus almost instantly, so no atom could survive. Second, the model said nothing about the distribution of electrons around the nucleus or their energies, and so could not explain why atoms give discrete line spectra rather than the continuous spectrum the spiralling electron would produce.

State the postulates of Bohr's model of the hydrogen atom. (3 marks)

First, the electron revolves round the nucleus in certain selected circular paths of fixed radius and energy, called stationary states or orbits, and while revolving in these orbits it does not radiate energy, which resolves the stability problem. Second, energy is emitted or absorbed only when the electron makes a transition from one stationary state to another, and the frequency of the radiation absorbed or emitted equals the difference in energy between the two states divided by Planck's constant. Third, the angular momentum of the electron in a permitted orbit is quantised, being an integral multiple of h divided by two pi, and only those orbits satisfying this condition are allowed.

State and explain the Pauli exclusion principle and Hund's rule of maximum multiplicity. (3 marks)

The Pauli exclusion principle states that no two electrons in an atom can have the same set of all four quantum numbers. Since an orbital is specified by n, l and m sub l, two electrons occupying the same orbital must differ in the spin quantum number, which can take only two values. It follows that an orbital can accommodate a maximum of two electrons and that these must have opposite spins. Hund's rule of maximum multiplicity states that pairing of electrons in the orbitals belonging to the same subshell does not take place until each orbital of that subshell has been singly occupied, and that all the singly occupied orbitals have electrons with parallel spins. This is why nitrogen, with three 2p electrons, has them in three different 2p orbitals with parallel spins and therefore has three unpaired electrons.

Explain why the electronic configuration of chromium is [Ar] 3d⁵ 4s¹ and not [Ar] 3d⁴ 4s². (3 marks)

In the region of the periodic table where chromium lies, the energies of the 3d and 4s subshells are very close together, so the energy cost of moving an electron from 4s to 3d is small. The configuration [Ar] 3d⁵ 4s¹ gives an exactly half-filled 3d subshell along with a half-filled 4s orbital. Exactly half-filled and completely filled subshells possess extra stability, which arises from the symmetrical distribution of electron density and from the greater exchange energy available when a larger number of electrons have parallel spins. The gain in stability outweighs the small promotion energy, so the observed configuration is the half-filled one. Copper is an analogous case, adopting [Ar] 3d¹⁰ 4s¹ for the stability of a completely filled 3d subshell.

State Heisenberg's uncertainty principle and explain its significance for the concept of an orbit. (3 marks)

The principle states that it is impossible to determine simultaneously and with arbitrary precision both the position and the momentum of a microscopic particle such as an electron; the product of the uncertainties in position and momentum is at least h divided by four pi. The significance is that the very idea of a definite orbit becomes meaningless for an electron, because specifying an orbit would require knowing both the exact position and the exact momentum of the electron at each instant. This is why Bohr's picture of fixed circular paths had to be abandoned, and why the quantum mechanical model speaks instead of an orbital, meaning a region of space around the nucleus where the probability of finding the electron is high.

Assertion–Reason Questions (CBSE Format)

For each question choose: (a) both A and R are true and R is the correct explanation of A; (b) both are true but R is not the correct explanation; (c) A is true but R is false; (d) A is false but R is true.

1. Assertion (A): The concept of a definite orbit for an electron is not acceptable.
Reason (R): The position and momentum of an electron cannot be determined simultaneously with arbitrary precision.

Answer: (a) — Both statements are true and the reason explains the assertion. Specifying an orbit requires exact position and momentum at every instant, which Heisenberg's uncertainty principle forbids, so the orbit is replaced by the orbital.

2. Assertion (A): The 4s orbital is filled before the 3d orbital.
Reason (R): An orbital can accommodate a maximum of two electrons, and these must have opposite spins.

Answer: (b) — Both statements are true, but the reason does not explain the assertion. The capacity of an orbital follows from the Pauli exclusion principle and governs how many electrons fit, not the order in which subshells fill. The order follows the n plus l rule, and 4s has the lower value.

3. Assertion (A): The energy of an electron in a hydrogen atom is negative.
Reason (R): The electron carries a negative charge.

Answer: (c) — The assertion is true but the reason is false as an explanation. The negative sign arises from taking the energy of a free electron at rest at infinite separation as zero; a bound electron has lower energy than this reference, not from the sign of its charge.

4. Assertion (A): In the photoelectric effect, increasing the intensity of the incident light increases the kinetic energy of the emitted electrons.
Reason (R): Intensity determines the number of photons striking the surface per second.

Answer: (d) — The assertion is false and the reason is true. Intensity governs how many electrons are ejected, while the kinetic energy of each electron is determined by the frequency of the light through the relation involving the work function.

Quick Self-Check

Q1. The neutron was discovered by:

  • A. Millikan
  • B. Goldstein
  • C. J. J. Thomson
  • D. James Chadwick

Answer: D. James Chadwick — Chadwick discovered the neutron in 1932 by bombarding a thin sheet of beryllium with alpha particles. Thomson found the charge to mass ratio of the electron and Millikan its charge.

Q2. For n = 4, the permitted values of the azimuthal quantum number l are:

  • A. 1, 2, 3, 4
  • B. 0, 1, 2, 3, 4
  • C. 0, 1, 2, 3
  • D. −4 to +4

Answer: C. 0, 1, 2, 3 — The azimuthal quantum number runs from 0 to n − 1, so for n = 4 it takes the values 0, 1, 2 and 3, corresponding to the 4s, 4p, 4d and 4f subshells.

Q3. The maximum number of electrons that can be accommodated in a shell with n = 4 is:

  • A. 8
  • B. 16
  • C. 18
  • D. 32

Answer: D. 32 — The maximum is 2n², which for n = 4 gives 2 × 16 = 32. The number of orbitals is n² = 16, each holding two electrons.

Q4. The number of radial nodes in a 3s orbital is:

  • A. 0
  • B. 2
  • C. 1
  • D. 3

Answer: B. 2 — Radial nodes equal n − l − 1. For 3s, n = 3 and l = 0, so there are 3 − 0 − 1 = 2 radial nodes. There are no angular nodes because l = 0.

Q5. Which series of the hydrogen spectrum lies in the visible region?

  • A. Balmer
  • B. Lyman
  • C. Paschen
  • D. Pfund

Answer: A. Balmer — The Balmer series, whose transitions end at n = 2, lies in the visible region. The Lyman series is ultraviolet and the Paschen, Brackett and Pfund series are infrared.

Key Terms Glossary

Atomic number (Z)
The number of protons in the nucleus of an atom; equal to the number of electrons in a neutral atom.
Mass number (A)
The total number of protons and neutrons in the nucleus.
Isotopes
Atoms of the same element having the same atomic number but different mass numbers.
Isobars
Atoms having the same mass number but different atomic numbers.
Quantum
The smallest discrete packet of energy that can be emitted or absorbed, of magnitude hν.
Planck's constant
The constant h, of value 6.626 × 10⁻³⁴ J s, relating the energy of a quantum to its frequency.
Work function
The minimum energy required to eject an electron from a metal surface, equal to h times the threshold frequency.
Threshold frequency
The minimum frequency of incident light below which no photoelectric emission occurs, however intense the light.
Stationary state
One of the permitted orbits in Bohr's model in which an electron may revolve without radiating energy.
Orbital
A region of space around the nucleus where the probability of finding an electron is high; the square of the wave function gives the probability density.
Principal quantum number
n, which identifies the shell and largely determines the size and energy of the orbital.
Azimuthal quantum number
l, running from 0 to n − 1, which identifies the subshell and determines the shape of the orbital.
Magnetic quantum number
m sub l, running from −l to +l, which specifies the orientation of the orbital in space.
Aufbau principle
The rule that orbitals are filled in order of increasing energy, given in practice by the n + l rule.
Node
A region where the probability of finding the electron is zero; radial nodes number n − l − 1 and angular nodes number l.

One-Minute Revision

  • Electron: cathode rays, Thomson's e/m and Millikan's charge. Proton: canal rays, Goldstein. Neutron: Chadwick, 1932, beryllium plus alpha particles.
  • Rutherford: most particles undeflected means empty space; about 1 in 20,000 bounced back means a tiny dense nucleus.
  • Rutherford's model failed on stability — an accelerating electron would radiate and spiral in — and on line spectra.
  • Bohr: fixed stationary orbits, transitions give ΔE = hν, angular momentum quantised as nh/2π.
  • Hydrogen: rₙ = 52.9 n² pm and Eₙ = −13.6/n² eV. The sign comes from the zero chosen at infinity, not the electron's charge.
  • Bohr fails for multi-electron atoms, fine structure, magnetic and electric splitting, and bonding.
  • Series: Lyman UV (n₁ = 1), Balmer visible (n₁ = 2), Paschen, Brackett, Pfund infrared (n₁ = 3, 4, 5).
  • Photoelectric effect: frequency sets the kinetic energy, intensity sets the number of electrons.
  • de Broglie λ = h/mv; Heisenberg forbids simultaneous exact position and momentum, which kills the orbit.
  • l runs 0 to n − 1; mₗ runs −l to +l giving 2l + 1 orbitals; a shell has n² orbitals and 2n² electrons.
  • n + l rule: lower sum first, and on a tie the lower n first — hence 4s before 3d, and 3d before 4p.
  • Pauli: no two electrons share all four quantum numbers, so two per orbital with opposite spins. Hund: singly occupy first, spins parallel.
  • Cr is 3d⁵4s¹ and Cu is 3d¹⁰4s¹ — half-filled and fully filled subshells are extra stable.
  • Nodes: radial = n − l − 1, angular = l, total = n − 1.

Syllabus reference: NCERT Class 11 Chemistry — Structure of Atom. It supplies the quantum numbers and electronic configurations used in Classification of Elements and Periodicity and in Chemical Bonding.

Frequently Asked Questions

Why does Bohr's model work for hydrogen but not for larger atoms?

Bohr's model assumes one electron, so it accurately predicts hydrogen's spectrum. For larger atoms with multiple electrons, electron-electron repulsion complicates the energy levels, making the model inaccurate.

What is an orbital and how is it different from an orbit?

An orbit is a fixed circular path where an electron moves, as proposed by Bohr. An orbital is a region where an electron is likely to be found, and has a specific shape and energy level.

Why is the energy of an electron in an atom taken as negative?

The zero of energy is fixed by convention at a free electron at rest infinitely far from the nucleus, where there is no interaction between them. An electron bound in an atom is attracted by the nucleus and has less energy than this reference state, so its energy comes out negative. The sign therefore expresses that work must be done to remove the electron, and a more negative value means a more tightly bound electron. It has nothing to do with the electron carrying a negative charge, which is a common misreading.

What is the difference between an orbit and an orbital?

An orbit is a definite circular path of fixed radius, as in Bohr's model, in which the electron is imagined to move like a planet. An orbital is a three-dimensional region around the nucleus in which the probability of finding the electron is high, obtained from the square of the wave function in the quantum mechanical model. The change was forced by Heisenberg's uncertainty principle, which makes a precise path meaningless, and by de Broglie's demonstration that the electron has wave character. An orbit can hold any number of electrons in principle, whereas an orbital holds at most two, with opposite spins.

Why does the 4s subshell fill before 3d even though its principal quantum number is higher?

Because the order of filling depends on the total energy of the subshell rather than on n alone, and this is captured by the n plus l rule. For 4s the sum is 4 plus 0, which is 4, while for 3d it is 3 plus 2, which is 5. The lower sum corresponds to the lower energy, so 4s is filled first, which is why potassium and calcium place their outermost electrons in 4s before the d block begins. When two subshells have the same n plus l value, such as 3d and 4p which both give 5, the one with the lower n is filled first.

Why could classical physics not explain the photoelectric effect?

Classical wave theory predicts that the energy delivered to an electron should depend on the intensity of the light and that a weak beam should eventually eject electrons if given enough time. Experiment shows the opposite: below a threshold frequency no electrons are emitted however intense the light or however long it shines, and above it emission is immediate with the kinetic energy depending on frequency rather than intensity. Einstein resolved this by treating light as a stream of photons each carrying energy hν, so one photon interacts with one electron and must individually carry enough energy to overcome the work function.

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