Structure of Atom Class 11: Atomic Models, Quantum Numbers and Solved Numericals

What Structure of Atom is about

Weightage: This chapter is Unit II, Structure of Atom, which the syllabus gives 9 of the 70 theory marks in the school annual exam.Source: CBSE Chemistry (043) syllabus 2026-27, Class XI course structure, Unit II Structure of Atom

Key concepts

Discovery of sub-atomic particles

NCERT section 2.1 Discovery of Sub-atomic Particles (2.1.1 to 2.1.4)

Electron: When a high voltage is applied across a gas at very low pressure in a discharge tube, rays travel from the cathode to the anode. These cathode rays travel in straight lines, cast shadows, are deflected towards the positive plate in an electric field (so they are negatively charged), and their properties do not depend on the gas or the electrode material. They are streams of electrons.

J. J. Thomson measured the charge-to-mass ratio of the electron by balancing electric and magnetic deflections: $e/m_e=1.758820\times 10^{11}$ C kg⁻¹. The deflection is larger when the particle's charge is larger, its mass smaller, or the field stronger.

R. A. Millikan's oil drop experiment gave the charge: $e=-1.602\times 10^{-19}$ C. Combining the two gives the mass of the electron, $m_e=9.1094\times 10^{-31}$ kg.

Proton: Using a perforated cathode, Goldstein observed canal rays (anode rays) moving the other way. They are positive ions; their charge-to-mass ratio depends on the gas, and the lightest one, from hydrogen, is the proton (charge +1.602 × 10⁻¹⁹ C, mass 1.672 × 10⁻²⁷ kg, about 1836 times the electron).

Neutron: In 1932 Chadwick bombarded beryllium with alpha particles and found electrically neutral particles slightly heavier than protons: neutrons (mass 1.675 × 10⁻²⁷ kg).

Worked example

Use $e/m_e=1.76\times 10^{11}$ C kg⁻¹ and $e=1.60\times 10^{-19}$ C to calculate the mass of an electron. Then explain why the charge-to-mass ratio of canal rays changes with the gas in the tube but that of cathode rays does not. (3 marks)

Try it first, then show the solutionSolution

1. $m_e=\frac{e}{e/m_e}=\frac{1.60\times 10^{-19}}{1.76\times 10^{11}}=9.09\times 10^{-31}$ kg. (1)
2. Cathode rays are electrons, which are the same in every atom, so their e/m is fixed. (1)
3. Canal rays are positive ions formed from whatever gas is in the tube; different gases give ions of different masses, so e/m changes. (1)

Answer: m_e ≈ 9.1 × 10⁻³¹ kg; electrons are identical in all atoms, but canal-ray ions depend on the gas.

Thomson and Rutherford models; atomic number, mass number, isotopes and isobars

NCERT section 2.2 Atomic Models (2.2.1 to 2.2.5)

Thomson's model pictured the atom as a sphere of uniformly spread positive charge with electrons embedded in it, like seeds in a watermelon. It explained overall neutrality but not later experiments.

Rutherford's alpha-particle scattering experiment: a thin gold foil (about 100 nm thick) was bombarded with fast alpha particles (He²⁺), and a zinc sulphide screen recorded flashes. Observations:
1. Most alpha particles passed straight through.
2. A small fraction were deflected by small angles.
3. A very few (about 1 in 20 000) bounced right back.

Conclusions: most of the atom is empty space; the positive charge and almost all the mass are packed into a tiny central nucleus (radius about 10⁻¹⁵ m against 10⁻¹⁰ m for the atom); electrons move around the nucleus in circular paths called orbits, held by electrostatic attraction.

• Atomic number (Z) = number of protons = number of electrons in a neutral atom.
• Mass number (A) = protons + neutrons (nucleons). Number of neutrons = A − Z. The symbol is written ${}_Z^{A}X$.
• Isotopes: same Z, different A (¹H protium, ²H deuterium, ³H tritium). Chemical properties are decided by electrons, so isotopes behave alike chemically.
• Isobars: same A, different Z (¹⁴C and ¹⁴N).

Drawbacks of Rutherford's model: an electron moving in a circle is accelerating, and by classical electromagnetic theory it should radiate energy and spiral into the nucleus, so the atom would not be stable. The model also says nothing about how electrons are arranged or their energies, and cannot explain line spectra.

Worked example

A species has 24 electrons, 26 protons and 30 neutrons. (a) Find its atomic number and mass number. (b) Write its symbol with charge. (c) Give the symbol of an isobar of ⁵⁶Fe with Z = 27. (2 marks)

Try it first, then show the solutionSolution

1. (a) Z = protons = 26 (iron); A = 26 + 30 = 56. (½ + ½)
2. (b) Electrons are two fewer than protons, so the charge is +2: ${}_{26}^{56}F{e}^{2+}$. (½)
3. (c) An isobar has the same A and different Z: ${}_{27}^{56}Co$. (½)

Answer: Z = 26, A = 56; ⁵⁶₂₆Fe²⁺; isobar ⁵⁶₂₇Co

Wave nature of electromagnetic radiation

NCERT section 2.3.1 Wave Nature of Electromagnetic Radiation

James Maxwell showed that light is an electromagnetic wave: oscillating electric and magnetic fields at right angles to each other and to the direction of travel. Such waves need no medium and travel in vacuum at $c=3.0\times 10^8$ m s⁻¹.

The quantities used to describe a wave:

• Wavelength (λ): distance between two crests, in m, nm (10⁻⁹ m) or pm (10⁻¹² m).
• Frequency (ν): number of waves passing a point per second, in hertz (Hz = s⁻¹).
• **Wavenumber ($\bar{\nu }$)**: number of wavelengths per unit length, $\bar{\nu }=1/\lambda$, in m⁻¹ or cm⁻¹.
• They are linked by $c=\nu \lambda$.

The electromagnetic spectrum, in order of increasing frequency (decreasing wavelength): radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays (cosmic rays have the highest frequencies). Visible light runs from about 400 nm (violet) to 750 nm (red); our eyes see only this small part.

Worked example

An FM radio station broadcasts at 98.3 MHz. Calculate the wavelength of the radio waves and their wavenumber. ($c=3.0\times 10^8$ m s⁻¹) (2 marks)

Try it first, then show the solutionSolution

1. $\lambda =\frac{c}{\nu }=\frac{3.0\times 10^8}{98.3\times 10^6}=3.05$ m. (1)
2. $\bar{\nu }=\frac{1}{\lambda }=\frac{1}{3.05}=0.328$ m⁻¹. (1)

Answer: λ ≈ 3.05 m; wavenumber ≈ 0.33 m⁻¹

Particle nature of radiation: Planck's quantum theory and the photoelectric effect

NCERT section 2.3.2 Particle Nature of Electromagnetic Radiation: Planck's Quantum Theory

Some observations cannot be explained by waves:

• Black-body radiation: a perfect absorber and emitter (a black body) emits radiation whose intensity–wavelength curve depends only on temperature. As a body is heated it glows dull red, then orange-yellow, then white-blue: the peak moves to shorter wavelength as temperature rises.
• Photoelectric effect: when light of suitable frequency falls on a metal such as potassium or caesium, electrons are emitted immediately.

Planck's quantum theory: energy is emitted or absorbed not continuously but in small packets called quanta. The energy of one quantum is
$E=h\nu$, where Planck's constant $h=6.626\times 10^{-34}$ J s.
Einstein called a quantum of light a photon.

Photoelectric effect explained (Einstein):
1. Electrons are ejected only if the light has at least a minimum frequency, the threshold frequency ${\nu }_0$, specific to each metal. Below it, no electrons come out, however bright the light.
2. Above ${\nu }_0$, one photon gives all its energy to one electron. Part of it ($h{\nu }_0$, the work function $W_0$) frees the electron; the rest becomes kinetic energy: $h\nu =h{\nu }_0+\frac{1}{2}{m}_e{v}^2$.
3. The number of electrons depends on the intensity (number of photons), but their kinetic energy depends only on the frequency.

So radiation has a dual nature: it acts as a wave in interference and diffraction, and as particles when it exchanges energy with matter.

Worked example

The work function of potassium is 2.25 eV. Light of wavelength 400 nm falls on it. (a) Calculate the energy of one photon in eV. (b) Find the maximum kinetic energy of the emitted electrons. (c) What is the threshold wavelength? ($h=6.626\times 10^{-34}$ J s, $c=3.0\times 10^8$ m s⁻¹, 1 eV = 1.602 × 10⁻¹⁹ J) (3 marks)

Try it first, then show the solutionSolution

1. (a) $E=\frac{hc}{\lambda }=\frac{6.626\times 10^{-34}\times 3.0\times 10^8}{400\times 10^{-9}}=4.97\times 10^{-19}$ J = 4.97 × 10⁻¹⁹ ÷ 1.602 × 10⁻¹⁹ = 3.10 eV. (1)
2. (b) KE = 3.10 − 2.25 = 0.85 eV (= 1.36 × 10⁻¹⁹ J). (1)
3. (c) ${\lambda }_0=\frac{hc}{W_0}=\frac{1.988\times 10^{-25}}{2.25\times 1.602\times 10^{-19}}=5.51\times 10^{-7}$ m = 551 nm. (1)

Answer: (a) 3.10 eV (b) 0.85 eV (c) about 551 nm

Atomic spectra and Bohr's model of the hydrogen atom

NCERT section 2.3.3 Evidence for Quantized Electronic Energy Levels: Atomic Spectra; 2.4 Bohr's Model for Hydrogen Atom (2.4.1, 2.4.2)

White light passed through a prism gives a continuous spectrum. Atoms excited by heat or electricity emit light only at certain wavelengths, giving a line spectrum (emission spectrum); the same atoms absorb exactly those wavelengths from white light (absorption spectrum). Each element has its own line spectrum, like a fingerprint; rubidium, caesium, thallium, indium, gallium and scandium were discovered this way.

The lines of hydrogen fall into series given by the Rydberg equation:
$\bar{\nu }=109677\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ cm⁻¹, with $n_2>n_1$.
Lyman ($n_1$ = 1, ultraviolet), Balmer ($n_1$ = 2, visible), Paschen ($n_1$ = 3), Brackett ($n_1$ = 4) and Pfund ($n_1$ = 5) (all three infrared).

Bohr's postulates (1913):
1. The electron moves round the nucleus in fixed circular orbits (stationary states) without losing energy.
2. Each orbit has a fixed energy; energy increases outwards.
3. Energy is absorbed or emitted only when the electron jumps between orbits: $\Delta E=E_2-E_1=h\nu$ (Bohr's frequency rule).
4. Angular momentum is quantised: $m_{e}vr=n\frac{h}{2\pi }$.

Results for hydrogen (and hydrogen-like ions such as He⁺, Li²⁺ with atomic number Z):

• Radius: $r_n=52.9\frac{n^2}{Z}$ pm.
• Energy: $E_n=-2.18\times 10^{-18}\left(\frac{Z^2}{n^2}\right)$ J.

The energy is negative because the bound electron has less energy than a free electron at rest (taken as zero at $n=\infty$). The ground state ($n$ = 1) is the most stable.

For a jump from a higher level n_i to a lower level n_f, the energy given out is $|\Delta E|=2.18\times 10^{-18}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$ J for $n_i>n_f$. The number of lines possible when an electron falls from level n to the ground state is $\frac{n(n-1)}{2}$.

Limitations: Bohr's model fails for atoms with more than one electron, cannot explain the fine structure of lines or the splitting of lines in electric (Stark) and magnetic (Zeeman) fields, cannot explain chemical bonding, and contradicts de Broglie's waves and the uncertainty principle.

Worked example

Calculate the energy and wavelength of the photon emitted when the electron in a hydrogen atom drops from n = 6 to n = 2. To which series and region does the line belong? ($h=6.626\times 10^{-34}$ J s, $c=3.0\times 10^8$ m s⁻¹) (3 marks)

Try it first, then show the solutionSolution

1. $\Delta E=2.18\times 10^{-18}\left(\frac{1}{2^2}-\frac{1}{6^2}\right)=2.18\times 10^{-18}\times \frac{8}{36}=4.84\times 10^{-19}$ J. (1)
2. $\lambda =\frac{hc}{\Delta E}=\frac{6.626\times 10^{-34}\times 3.0\times 10^8}{4.84\times 10^{-19}}=4.10\times 10^{-7}$ m = 410 nm. (1)
3. The lower level is n = 2, so it is a Balmer line, at the violet end of the visible region. (1)

Answer: ΔE ≈ 4.84 × 10⁻¹⁹ J; λ ≈ 410 nm; Balmer series, visible (violet) region

Dual behaviour of matter and Heisenberg's uncertainty principle

NCERT section 2.5 Towards Quantum Mechanical Model of the Atom (2.5.1, 2.5.2)

de Broglie (1924) proposed that matter, like radiation, has both particle and wave properties. A particle of mass m moving with velocity v has a wavelength
$\lambda =\frac{h}{mv}=\frac{h}{p}$
where p is the momentum. This was confirmed when electron beams were found to be diffracted by crystals; the electron microscope uses the wave nature of electrons to magnify about 15 million times.

For everyday objects the wavelength is so tiny that it cannot be detected; the wave nature matters only for very light particles such as electrons.

Using the kinetic energy: $\frac{1}{2}m{v}^2=KE$ gives $v=\sqrt{\frac{2KE}{m}}$, then $\lambda =\frac{h}{mv}$.

Heisenberg's uncertainty principle (1927): it is impossible to find both the exact position and the exact momentum (or velocity) of an electron at the same time:
$\Delta x\times \Delta p_x\ge \frac{h}{4\pi }$, or $\Delta x\times \Delta v_x\ge \frac{h}{4\pi m}$.
To locate an electron precisely we must hit it with light of very short wavelength, whose high-energy photons change its momentum.

Significance: the idea of a definite path (orbit) for an electron has no meaning. We can talk only of the probability of finding the electron in a region. For large objects the uncertainty is negligible, so classical mechanics works for them. This is why Bohr's model, with fixed circular orbits, had to be replaced.

Worked example

(a) Calculate the de Broglie wavelength of an electron moving at 1.0 × 10⁶ m s⁻¹. (b) If the position of this electron is known to within 40 pm, find the minimum uncertainty in its velocity, and comment. ($h=6.626\times 10^{-34}$ J s, $m_e=9.11\times 10^{-31}$ kg) (3 marks)

Try it first, then show the solutionSolution

1. (a) $\lambda =\frac{h}{mv}=\frac{6.626\times 10^{-34}}{9.11\times 10^{-31}\times 1.0\times 10^6}=7.27\times 10^{-10}$ m = 727 pm. (1)
2. (b) $\Delta v=\frac{h}{4\pi m\Delta x}=\frac{6.626\times 10^{-34}}{4\times 3.14\times 9.11\times 10^{-31}\times 40\times 10^{-12}}=1.45\times 10^6$ m s⁻¹. (1)
3. The uncertainty in velocity is larger than the velocity itself, so an exact path for the electron cannot be defined. (1)

Answer: (a) ≈ 727 pm (b) Δv ≈ 1.45 × 10⁶ m s⁻¹, larger than v itself

Quantum mechanical model: orbitals, quantum numbers and shapes

NCERT section 2.6 Quantum Mechanical Model of Atom (2.6.1 Orbitals and Quantum Numbers; 2.6.2 Shapes of Atomic Orbitals)

In 1926 Schrödinger wrote an equation whose solutions, wave functions (ψ), describe electrons in atoms. ψ itself has no physical meaning, but ψ² gives the probability density of finding the electron at a point. An orbital is the region where the probability of finding the electron is high, described by one wave function. Each orbital holds at most two electrons.

Four quantum numbers describe an electron:

• Principal (n) = 1, 2, 3 …: the shell; decides size and (for hydrogen) energy. Shell n has n² orbitals and holds 2n² electrons.
• Azimuthal or subsidiary (l) = 0 to (n − 1): the subshell and the shape. l = 0, 1, 2, 3 are s, p, d, f.
• Magnetic (mₗ) = −l to +l, giving (2l + 1) orbitals per subshell: one s, three p, five d, seven f. It gives the orientation in space.
• Spin (mₛ) = +½ or −½: the two spin orientations of an electron.

Shapes:

• s orbitals are spherical. The 1s orbital has no node; ns has (n − 1) radial nodes.
• p orbitals are dumb-bell shaped, with two lobes along an axis (pₓ, p_y, p_z) and a nodal plane through the nucleus.
• d orbitals: four have four lobes (d_xy, d_yz, d_xz, d_x²−y²) and d_z² has two lobes along z with a doughnut ring.

Nodes: radial nodes = n − l − 1; angular nodes = l; total nodes = n − 1.

Worked example

(a) Give the designation of the orbital with n = 4, l = 2, and list its possible mₗ values. (b) How many radial and angular nodes does a 3p orbital have? (c) How many electrons in an atom can have n = 3? (3 marks)

Try it first, then show the solutionSolution

1. (a) l = 2 is d, so the orbital is 4d; mₗ = −2, −1, 0, +1, +2 (five 4d orbitals). (1)
2. (b) 3p: n = 3, l = 1. Radial nodes = 3 − 1 − 1 = 1; angular nodes = l = 1. (1)
3. (c) Shell n = 3 has n² = 9 orbitals, each holding 2 electrons: 18 electrons. (1)

Answer: (a) 4d; mₗ = −2 to +2 (b) one radial and one angular node (c) 18

Energies of orbitals, filling rules and electronic configuration

NCERT section 2.6.3 to 2.6.6 Energies of Orbitals, Filling of Orbitals, Electronic Configuration, Stability of Half-filled and Completely Filled Orbitals

In hydrogen, the energy depends only on n: 1s < 2s = 2p < 3s = 3p = 3d …. Orbitals of the same energy are degenerate.

In atoms with many electrons, energy depends on both n and l because inner electrons shield the outer ones from the nucleus. The attraction actually felt is the effective nuclear charge. s electrons penetrate closer to the nucleus than p, p closer than d, so for the same n: s < p < d < f.

(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. This gives the filling order:
1s, 2s, 2p, 3s, 3p, then 4s before 3d, then 4p, 5s, 4d and so on (6s comes before 4f and 5d)

Three filling rules:
1. Aufbau principle: fill orbitals in order of increasing energy.
2. Pauli exclusion principle: an atom cannot contain two electrons whose four quantum numbers are all identical, so an orbital holds at most two electrons with opposite spins.
3. Hund's rule of maximum multiplicity: in degenerate orbitals (p, d, f), electrons first occupy each orbital singly with parallel spins; pairing starts only when each orbital has one electron.

Examples: N = 1s² 2s² 2p³ (three unpaired electrons); Fe = [Ar] 3d⁶ 4s².

Exceptions: chromium and copper. Cr is [Ar] 3d⁵ 4s¹, not 3d⁴ 4s², and Cu is [Ar] 3d¹⁰ 4s¹, not 3d⁹ 4s². Half-filled and completely filled subshells are extra stable because (a) their electron distribution is symmetrical, so shielding is smaller and the electrons are held more tightly, and (b) they have the maximum exchange energy: electrons of parallel spin in degenerate orbitals can exchange positions, and each exchange releases energy.

Worked example

Write the electronic configurations of (a) Cr (Z = 24) and (b) Fe³⁺ (Z of Fe = 26), and give the number of unpaired electrons in each. Explain the configuration of Cr. (3 marks)

Try it first, then show the solutionSolution

1. (a) Cr: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹; six unpaired electrons (five 3d and one 4s). (1)
2. (b) Fe is [Ar] 3d⁶ 4s². Electrons are removed first from 4s (the outermost shell), then 3d: Fe³⁺ = [Ar] 3d⁵; five unpaired electrons. (1)
3. Cr takes 3d⁵ 4s¹ because a half-filled d subshell has symmetrical distribution and maximum exchange energy, which makes it more stable than 3d⁴ 4s². (1)

Answer: (a) Cr [Ar] 3d⁵ 4s¹, 6 unpaired (b) Fe³⁺ [Ar] 3d⁵, 5 unpaired

Chemistry in real life: Structure of Atom

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Bohr Model and the Hydrogen Spectrum
• Quantum Numbers, Orbitals and Electronic Configuration

Structure of Atom formula sheet

Mass number and neutrons

$A=Z+N,N=A-Z$

Z = protons; for an ion, electrons = Z − charge.

Wave relations

$c=\nu \lambda ,\bar{\nu }=\frac{1}{\lambda }$

c = 3.0 × 10⁸ m s⁻¹.

Energy of a photon

$E=h\nu =\frac{hc}{\lambda }$

h = 6.626 × 10⁻³⁴ J s. Multiply by $N_A$ for one mole of photons.

Photoelectric equation

$h\nu =h{\nu }_0+\frac{1}{2}{m}_e{v}^2$

$h{\nu }_0$ is the work function $W_0$.

Bohr energy and radius (hydrogen-like)

$E_n=-2.18\times 10^{-18}\frac{Z^2}{n^2}\text{J},r_n=52.9\frac{n^2}{Z}\text{pm}$

Z = 1 for H, 2 for He⁺, 3 for Li²⁺.

Energy of a transition

$\Delta E=2.18\times 10^{-18}{Z}^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\text{J}$

Positive for emission written this way ($n_i>n_f$).

Rydberg equation

$\bar{\nu }=109677\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right){\text{cm}}^{-1}$

$n_1$ = 1, 2, 3, 4, 5 for Lyman, Balmer, Paschen, Brackett, Pfund.

Number of spectral lines

$\text{lines}=\frac{n(n-1)}{2}$

From level n down to the ground state.

de Broglie wavelength

$\lambda =\frac{h}{mv}=\frac{h}{p}=\frac{h}{\sqrt{2mKE}}$

Use SI units: m in kg, v in m s⁻¹.

Uncertainty principle

$\Delta x\cdot \Delta p\ge \frac{h}{4\pi },\Delta x\cdot \Delta v\ge \frac{h}{4\pi m}$

h/4π = 5.27 × 10⁻³⁵ J s.

Orbitals, electrons and nodes

$\text{orbitals in shell}=n^2,\text{electrons}=2n^2,\text{radial nodes}=n-l-1,\text{angular nodes}=l$

A subshell has 2l + 1 orbitals.

Structure of Atom: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Structure of Atom

The question patterns to expect, and the method that works for each.

Tricks to remember Structure of Atom

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Raging Martians Invaded Venus Using X-ray Guns

  Radio, Microwave, Infrared, Visible, Ultraviolet, X-rays, Gamma: increasing frequency and energy.

  Use it when: Ordering the electromagnetic spectrum.

• Lazy Boys Play Basketball Peacefully

  Lyman (1), Balmer (2), Paschen (3), Brackett (4), Pfund (5): the number is the lower level n₁.

  Use it when: Naming the series of a hydrogen line.

• Bus seat rule (Hund)

  On a bus, people sit alone on each double seat before sharing. Electrons fill degenerate orbitals singly first, with parallel spins.

  Use it when: Writing orbital diagrams and counting unpaired electrons.

• Diagonal arrows for filling order

  Write 1s; 2s 2p; 3s 3p 3d; 4s 4p 4d 4f … in rows and draw diagonal arrows from top right to bottom left: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s …

  Use it when: Aufbau order without memorising a list.

• n − l − 1 for radial nodes

  Radial = n − l − 1, angular = l, total = n − 1. 'Total is one less than the shell.'

  Use it when: Node questions.

Structure of Atom: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Structure of Atom

The slips examiners see most often, and what to write instead.

• ✗Mistake: Removing 3d electrons before 4s when forming ions

  Students reverse the filling order.

  Wrong working

  Fe²⁺ = [Ar] 3d⁴ 4s²

  Right working

  Fe²⁺ = [Ar] 3d⁶ (the 4s electrons leave first)

  ✓Do this: Electrons leave from the highest n first.

  Usually costs: 1 mark.

• ✗Mistake: Not converting nm or Å to metres

  Students substitute 400 for 400 nm.

  Wrong working

  E = 6.626 × 10⁻³⁴ × 3 × 10⁸ ÷ 400

  Right working

  E = 6.626 × 10⁻³⁴ × 3 × 10⁸ ÷ (400 × 10⁻⁹) = 4.97 × 10⁻¹⁹ J

  ✓Do this: Write every quantity in SI units before substituting.

  Usually costs: The answer mark.

• ✗Mistake: Writing Cr as 3d⁴ 4s²

  Students apply the aufbau order blindly.

  ✓Do this: Learn Cr (3d⁵ 4s¹) and Cu (3d¹⁰ 4s¹) with the two reasons: symmetry and exchange energy.

  Usually costs: 1 mark.

• ✗Mistake: Saying brighter light gives faster photoelectrons

  Intensity and frequency are confused.

  ✓Do this: Intensity → number of electrons; frequency → kinetic energy of each.

  Usually costs: 1 mark in MCQ or assertion-reason.

• ✗Mistake: Forgetting Z² for He⁺ and Li²⁺

  The hydrogen formula is used unchanged.

  Wrong working

  E₁(He⁺) = −2.18 × 10⁻¹⁸ J

  Right working

  E₁(He⁺) = −2.18 × 10⁻¹⁸ × 2² = −8.72 × 10⁻¹⁸ J

  ✓Do this: Write Z²/n² in the formula every time.

  Usually costs: 1-2 marks.

• ✗Mistake: Giving mass of the electron in grams in λ = h/mv

  Units are mixed.

  ✓Do this: h is in J s = kg m² s⁻¹, so mass must be in kg and velocity in m s⁻¹.

  Usually costs: The answer mark.

NCERT exercise guide: Structure of Atom

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Structure of Atom

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The number of radial nodes in a 3s orbital is:
   (a) 0 (b) 1 (c) 2 (d) 3

   Hint

   n − l − 1.

   Answer

   (c) 2

2. Q2 · 1 mark · MCQExam level

   Which of the following has the electronic configuration [Ar] 3d⁵?
   (a) Mn²⁺ (b) Fe²⁺ (c) Cr (d) Co³⁺

   Hint

   Count electrons after removing 4s first.

   Answer

   (a) Mn²⁺ (Mn is [Ar] 3d⁵ 4s²; losing 4s² gives 3d⁵). Fe³⁺ would also be 3d⁵, but Fe²⁺ is 3d⁶.

3. Q3 · 2 marks · Very short answerExam level

   Give two differences between an orbit and an orbital. (2 marks)

   Hint

   Path versus probability.

   Answer

   An orbit is a definite circular path (Bohr) while an orbital is a 3-D region of high probability of finding an electron (1). An orbit is two-dimensional and violates the uncertainty principle; an orbital can hold at most two electrons and has a shape (s, p, d) (1).

4. Q4 · 2 marks · Very short answerExam level

   Calculate the frequency and wavenumber of radiation of wavelength 600 nm. (2 marks)

   Hint

   c = νλ.

   Answer

   ν = 3.0 × 10⁸ ÷ 600 × 10⁻⁹ = 5.0 × 10¹⁴ Hz (1); wavenumber = 1/λ = 1.67 × 10⁶ m⁻¹ (1).

5. Q5 · 3 marks · Short answerHarder exam level

   The threshold wavelength of a metal is 540 nm. Ultraviolet light of wavelength 360 nm falls on it. Calculate (a) the threshold frequency, (b) the kinetic energy of the fastest emitted electrons and (c) their speed. ($h=6.626\times 10^{-34}$ J s, $c=3.0\times 10^8$ m s⁻¹, $m_e=9.11\times 10^{-31}$ kg) (3 marks)

   Hint

   KE = hc(1/λ − 1/λ₀).

   Answer

   (a) ν₀ = c/λ₀ = 3.0 × 10⁸ ÷ 540 × 10⁻⁹ = 5.56 × 10¹⁴ Hz (1). (b) KE = 1.988 × 10⁻²⁵ × (1/360 × 10⁻⁹ − 1/540 × 10⁻⁹) = 1.988 × 10⁻²⁵ × 9.26 × 10⁵ = 1.84 × 10⁻¹⁹ J (1). (c) v = √(2 × 1.84 × 10⁻¹⁹ ÷ 9.11 × 10⁻³¹) = √(4.04 × 10¹¹) = 6.36 × 10⁵ m s⁻¹ (1).

6. Q6 · 2 marks · Very short answerHarder exam level

   What is the radius of the third Bohr orbit of He⁺, and how does it compare with the first orbit of hydrogen? (2 marks)

   Hint

   r = 52.9 n²/Z pm.

   Answer

   r = 52.9 × 9 ÷ 2 = 238 pm (1); this is 4.5 times the first orbit of hydrogen (52.9 pm) (1).

7. Q7 · 2 marks · Very short answerHarder exam level

   Arrange the following orbitals in increasing order of energy in a multi-electron atom, giving the (n + l) value for each: 4s, 3d, 4p, 5s, 4d. (2 marks)

   Hint

   Lower n + l first; ties go to lower n.

   Answer

   (n + l): 4s = 4, 3d = 5, 4p = 5, 5s = 5, 4d = 6 (1). Order: 4s < 3d < 4p < 5s < 4d (1).

8. Q8 · 3 marks · Short answerHarder exam level

   Suppose a student shines ultraviolet light of wavelength 250 nm on a metal and finds that the fastest electrons have a kinetic energy of 2.0 eV. (a) Find the work function in eV. (b) What would happen with light of 600 nm? (1 eV = 1.602 × 10⁻¹⁹ J) (3 marks)

   Hint

   Find the photon energy in eV first.

   Answer

   (a) E = 1.988 × 10⁻²⁵ ÷ 250 × 10⁻⁹ = 7.95 × 10⁻¹⁹ J = 4.96 eV (1); W₀ = 4.96 − 2.0 = 2.96 eV (1). (b) A 600 nm photon has 2.07 eV, less than 2.96 eV, so no electrons are emitted (1).

9. Q9 · 3 marks · Short answerHarder exam level

   Write the four quantum numbers for the last electron of sodium (Z = 11), and explain why two electrons in the same orbital must differ in spin. (3 marks)

   Hint

   Na = [Ne] 3s¹.

   Answer

   n = 3, l = 0, mₗ = 0, mₛ = +½ (or −½) (2). Two electrons in one orbital already share n, l and mₗ; by Pauli's principle they must differ in mₛ (1).

10. Q10 · 3 marks · Short answerHarder exam level

    Why is Bohr's model said to contradict the uncertainty principle? Give two other limitations of the model. (3 marks)

    Hint

    A fixed orbit means exact position and momentum.

    Answer

    A fixed circular orbit of definite radius means the electron's position and velocity are known exactly at every instant, which Heisenberg's principle forbids (1). Limitations (any two): fails for multi-electron atoms; cannot explain fine structure of lines; cannot explain Zeeman and Stark effects; cannot explain chemical bonding (2).

11. Q11 · 3 marks · Short answerHOTS / competency

    Suppose a teacher says: 'In chromium, one 4s electron moves into 3d because 3d⁵ 4s¹ has more exchange energy.' Count the exchange pairs in the 3d subshell for 3d⁴ (four parallel spins) and 3d⁵ (five parallel spins), and use the result to support her statement. Also give the second reason for the stability of half-filled subshells. (3 marks)

    Hint

    Number of pairs from k electrons = k(k − 1)/2.

    Answer

    3d⁴: 4 × 3 ÷ 2 = 6 exchange pairs; 3d⁵: 5 × 4 ÷ 2 = 10 exchange pairs (1). More exchanges release more exchange energy, so 3d⁵ 4s¹ is lower in energy than 3d⁴ 4s² (1). Second reason: the symmetrical distribution of electrons in a half-filled subshell reduces shielding, so electrons are held more strongly (1).

12. Q12 · 5 marks · Long answerHOTS / competency

    A hydrogen atom in its ground state absorbs a photon of wavelength 97.3 nm. (a) To which level is the electron raised? (b) When it returns, list the possible emitted lines that fall in the visible region. ($hc=1.988\times 10^{-25}$ J m) (5 marks)

    Hint

    Set 2.18 × 10⁻¹⁸ (1 − 1/n²) equal to the photon energy.

    Answer

    (a) Photon energy = 1.988 × 10⁻²⁵ ÷ 97.3 × 10⁻⁹ = 2.043 × 10⁻¹⁸ J (1). 1 − 1/n² = 2.043 ÷ 2.18 = 0.937, so 1/n² = 0.063 and n² ≈ 16, n = 4 (2). (b) Visible lines are Balmer lines ending on n = 2 from n ≤ 4: 4 → 2 (486 nm) and 3 → 2 (656 nm) (2).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Who measured the charge on the electron and how?Show answer

   R. A. Millikan, by the oil drop experiment.

2. 2.What did the few alpha particles that bounced back show?Show answer

   That the positive charge and nearly all the mass are concentrated in a tiny nucleus.

3. 3.Define isobars with an example.Show answer

   Atoms with the same mass number but different atomic numbers, e.g. ¹⁴C and ¹⁴N.

4. 4.What is the threshold frequency?Show answer

   The minimum frequency of light that can eject an electron from a given metal surface.

5. 5.In which region of the spectrum is the Lyman series?Show answer

   Ultraviolet.

6. 6.Write Bohr's quantisation condition.Show answer

   Angular momentum mₑvr = nh/2π.

7. 7.State Pauli's exclusion principle.Show answer

   An atom cannot contain two electrons with all four quantum numbers identical.

8. 8.How many orbitals are there in the n = 4 shell?Show answer

   16 (n²).

9. 9.What is the shape of a p orbital?Show answer

   Dumb-bell, with two lobes and a nodal plane through the nucleus.

10. 10.Give the configuration of copper.Show answer

    [Ar] 3d¹⁰ 4s¹.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you list the observations and conclusions of the gold foil experiment?
• ☐Can you convert between λ, ν and wavenumber?
• ☐Can you find the KE of a photoelectron from λ and the work function?
• ☐Can you calculate the energy and radius of any Bohr orbit for H, He⁺ or Li²⁺?
• ☐Can you find the wavelength of a hydrogen line and name its series?
• ☐Can you calculate a de Broglie wavelength from velocity or kinetic energy?
• ☐Can you spot an impossible set of quantum numbers?
• ☐Can you write configurations of Z = 1 to 30, including Cr, Cu and common ions?

Structure of Atom: frequently asked questions