Structure of Atom ISC Class 11: From Electrons to Orbitals

What Structure of Atom is about

Weightage: Structure of Atom belongs to unit 2 of the ISC Class XI Chemistry syllabus. CISCE weights units 1-6 together (Physical Chemistry) at 41 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Chemistry syllabus, examination year 2028, Class XI unit weightage

Key concepts

Discovery of electron, proton and neutron

Cathode rays, produced in a discharge tube at very low pressure, travel from the cathode in straight lines, cast shadows, turn a light paddle wheel and bend towards a positive plate, showing they are negatively charged particles with mass: electrons. Thomson measured their charge-to-mass ratio, e/m ≈ 1.76 × 10¹¹ C kg⁻¹, the same for every gas and cathode. Millikan's oil-drop experiment gave e = 1.602 × 10⁻¹⁹ C. Anode (canal) rays, seen through a perforated cathode, are positive ions whose e/m depends on the gas; with hydrogen they are the lightest positive particle, the proton. Chadwick bombarded beryllium with α-particles and detected neutral, penetrating particles of mass close to the proton: the neutron (⁹₄Be + ⁴₂He → ¹²₆C + ¹₀n).

Worked example

Give two properties of cathode rays that show they are (i) material particles (ii) negatively charged.

Try it first, then show the solutionSolution

1. (i) They turn a light paddle wheel placed in their path (they carry momentum) and travel in straight lines casting sharp shadows
2. (ii) They bend towards the positive plate in an electric field and are deflected by a magnetic field as negative charges would be

Answer: Paddle wheel and straight-line shadows; deflection towards a positive plate

Atomic number, mass number, isotopes and isobars

Atomic number Z = number of protons (= electrons in a neutral atom). Mass number A = protons + neutrons, so neutrons = A − Z. The symbol is written ᴬ_ZX. Isotopes have the same Z but different A (e.g. ¹H, ²H, ³H; ³⁵Cl, ³⁷Cl): same chemistry, different masses. Isobars have the same A but different Z (⁴⁰Ar and ⁴⁰Ca): different elements. For ions, adjust the electrons: a negative ion has gained electrons.

Worked example

Find the numbers of protons, neutrons and electrons in ³⁵₁₇Cl⁻ and ⁵⁶₂₆Fe³⁺.

Try it first, then show the solutionSolution

1. Cl⁻: p = 17, n = 35 − 17 = 18, e = 17 + 1 = 18
2. Fe³⁺: p = 26, n = 56 − 26 = 30, e = 26 − 3 = 23

Answer: 17, 18, 18; 26, 30, 23

Thomson's and Rutherford's models and their limitations

Thomson's model: a uniform sphere of positive charge with electrons embedded in it (the 'plum pudding'); it explained neutrality but not the scattering results. Rutherford's experiment: a thin gold foil was bombarded with α-particles. Most passed straight through (the atom is mostly empty), a few were deflected through large angles, and about 1 in 20 000 bounced back (a tiny, dense, positive nucleus). Rutherford's model: a nucleus (radius about 10⁻¹⁵ m) holding nearly all the mass, with electrons revolving around it. Limitations: by classical electromagnetism a revolving (accelerating) electron should radiate energy and spiral into the nucleus, so the model cannot explain the stability of atoms; it also says nothing about the arrangement of electrons or line spectra.

Worked example

Which observation of the gold-foil experiment shows that the nucleus is very small, and which shows that it is positively charged and massive?

Try it first, then show the solutionSolution

1. Most α-particles pass through undeflected: the nucleus occupies a tiny fraction of the atom's volume
2. A few are sharply deflected or bounced back: they met a concentrated, massive positive charge that repelled them

Answer: Small: most pass straight through. Positive and massive: large-angle deflections and rebounds

Electromagnetic radiation, Planck's quantum theory and the photoelectric effect

Electromagnetic waves obey c = νλ (c = 3.0 × 10⁸ m s⁻¹); the wavenumber ν̄ = 1/λ. Wave theory failed to explain black-body radiation (the intensity-wavelength curve of a hot body) and the photoelectric effect (no emission below a threshold frequency ν₀, however bright the light; emission is instant; electron energy depends on frequency, not intensity). Planck's quantum theory: radiation is emitted or absorbed in discrete packets (quanta, or photons) of energy E = hν = hc/λ, h = 6.626 × 10⁻³⁴ J s. For photoemission: hν = hν₀ + ½mv², where hν₀ is the work function.

Worked example

A metal has threshold wavelength 600 nm. Find the maximum kinetic energy of electrons emitted by light of wavelength 400 nm.

Try it first, then show the solutionSolution

1. KE = hc(1/λ − 1/λ₀) = 6.626 × 10⁻³⁴ × 3.0 × 10⁸ × (1/(400 × 10⁻⁹) − 1/(600 × 10⁻⁹))
2. = 1.988 × 10⁻²⁵ × (2.500 × 10⁶ − 1.667 × 10⁶)
3. = 1.988 × 10⁻²⁵ × 8.33 × 10⁵ = 1.66 × 10⁻¹⁹ J

Answer: 1.66 × 10⁻¹⁹ J (about 1.03 eV)

Spectra and the hydrogen spectrum

An emission spectrum is the light given out by excited atoms; an absorption spectrum shows dark lines where light is absorbed from a continuous source. Atoms give line spectra (sharp separate lines, a fingerprint of the element); molecules give band spectra (groups of closely spaced lines); hot solids give continuous spectra. Hydrogen's lines fit the Rydberg formula 1/λ = R_H(1/n₁² − 1/n₂²), R_H = 1.097 × 10⁷ m⁻¹, n₂ > n₁. Series: Lyman (n₁ = 1, ultraviolet), Balmer (n₁ = 2, visible), Paschen (3), Brackett (4), Pfund (5) (infrared). Electrons falling from level n to the ground state can produce n(n − 1)/2 different lines.

Worked example

Find the wavelength of the Balmer line for the transition n = 3 → n = 2.

Try it first, then show the solutionSolution

1. 1/λ = 1.097 × 10⁷ × (1/4 − 1/9) = 1.097 × 10⁷ × 5/36 = 1.524 × 10⁶ m⁻¹
2. λ = 6.56 × 10⁻⁷ m

Answer: 656 nm (red line)

Bohr's model: postulates, results and limitations

Postulates: (1) the electron moves round the nucleus in certain fixed circular orbits (stationary states) without radiating energy; (2) only orbits with angular momentum mvr = nh/2π are allowed (n = 1, 2, 3, …); (3) energy is emitted or absorbed only when the electron jumps between orbits, with ΔE = E₂ − E₁ = hν. Results for hydrogen-like species (derivations not required): rₙ = 0.529 n²/Z Å; vₙ = 2.18 × 10⁶ Z/n m s⁻¹; Eₙ = −2.18 × 10⁻¹⁸ Z²/n² J = −13.6 Z²/n² eV. The negative energy means the electron is bound; E = 0 at n = ∞ (ionisation). Limitations: it fails for atoms with more than one electron, cannot explain the fine splitting of lines or their splitting in magnetic (Zeeman) and electric (Stark) fields, cannot explain molecule formation, and contradicts de Broglie and Heisenberg by assuming exact orbits.

Worked example

For He⁺ (Z = 2), find the radius, speed and energy of the electron in the n = 2 orbit.

Try it first, then show the solutionSolution

1. r = 0.529 × 2²/2 = 1.058 Å
2. v = 2.18 × 10⁶ × 2/2 = 2.18 × 10⁶ m s⁻¹
3. E = −13.6 × 2²/2² = −13.6 eV

Answer: 1.06 Å; 2.18 × 10⁶ m s⁻¹; −13.6 eV

Dual nature: de Broglie and Heisenberg

Light shows both wave behaviour (interference) and particle behaviour (photoelectric effect). de Broglie proposed that matter also has waves: λ = h/mv = h/p. The wavelength is measurable only for very small masses (electrons are diffracted by crystals); for everyday objects it is far too small to detect. Heisenberg's uncertainty principle: the position and momentum of a particle cannot both be known exactly at the same time: Δx × Δp ≥ h/4π (so Δx × Δv ≥ h/4πm). This destroys the idea of a fixed orbit and leads to the probability picture of orbitals.

Worked example

(i) Find the de Broglie wavelength of an electron (m = 9.109 × 10⁻³¹ kg) moving at 1.0 × 10⁶ m s⁻¹. (ii) If an electron's position is known to within 1.0 × 10⁻¹⁰ m, find the minimum uncertainty in its speed.

Try it first, then show the solutionSolution

1. (i) λ = 6.626 × 10⁻³⁴/(9.109 × 10⁻³¹ × 1.0 × 10⁶) = 7.27 × 10⁻¹⁰ m
2. (ii) Δv ≥ h/(4πmΔx) = 6.626 × 10⁻³⁴/(4π × 9.109 × 10⁻³¹ × 1.0 × 10⁻¹⁰)
3. = 5.79 × 10⁵ m s⁻¹

Answer: 7.27 × 10⁻¹⁰ m (0.727 nm); 5.79 × 10⁵ m s⁻¹

Quantum mechanical model, quantum numbers, orbital shapes and nodes

Schrödinger's wave equation (no derivation) gives wave functions ψ for the electron. ψ itself has no physical meaning; |ψ|² gives the probability density of finding the electron at a point. An orbital is the region where this probability is high. Quantum numbers: n (principal: shell, size and energy; 1, 2, 3, …); l (azimuthal: subshell and shape; 0 to n − 1 for s, p, d, f); m_l (magnetic: orientation; −l to +l, so 2l + 1 orbitals); m_s (spin: +½ or −½). Shapes: s spherical; p dumb-bell (pₓ, p_y, p_z along the axes); d mostly four-lobed (dₓᵧ, d_yz, d_zx between axes; d_{x²−y²} along x and y; d_z² two lobes along z with a ring). Nodes (zero probability): radial nodes = n − l − 1, angular nodes (nodal planes) = l, total n − 1. Radial probability curves plot the chance of finding the electron at distance r; the number of peaks is n − l.

Worked example

For a 3p orbital, give n, l, the possible m_l values, and the numbers of radial and angular nodes.

Try it first, then show the solutionSolution

1. n = 3, l = 1; m_l = −1, 0, +1
2. Radial nodes = 3 − 1 − 1 = 1; angular nodes = l = 1

Answer: n = 3, l = 1, m_l = −1, 0, +1; 1 radial and 1 angular node

Filling orbitals and electronic configurations

Aufbau principle: orbitals fill in order of increasing energy, given by the (n + l) rule: lower n + l fills first; for equal n + l, lower n fills first. So 4s (n + l = 4) fills before 3d (5), and 3d (n + l = 5, n = 3) before 4p (also 5, but n = 4). Pauli's exclusion principle: no two electrons in an atom have all four quantum numbers the same, so an orbital holds at most two electrons with opposite spins. Hund's rule of maximum multiplicity: in degenerate orbitals, electrons occupy them singly with parallel spins before pairing. Extra stability of half-filled and completely filled subshells (symmetry and exchange energy) gives Cr [Ar]3d⁵4s¹ and Cu [Ar]3d¹⁰4s¹. For ions of transition metals, electrons leave the 4s orbital before 3d.

Worked example

Write the configurations of Fe (Z = 26) and Fe³⁺, and the number of unpaired electrons in Fe³⁺.

Try it first, then show the solutionSolution

1. Fe: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶, i.e. [Ar]3d⁶4s²
2. Fe³⁺: remove two 4s electrons and one 3d electron: [Ar]3d⁵
3. 3d⁵ with Hund's rule: five singly occupied orbitals

Answer: [Ar]3d⁶4s²; [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 and Electronic Configuration

Structure of Atom formula sheet

Wave relations and photon energy

$c=\nu \lambda ;\bar{\nu }=\frac{1}{\lambda };E=h\nu =\frac{hc}{\lambda }$

h = 6.626 × 10⁻³⁴ J s.

Photoelectric equation

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

hν₀ = work function.

Rydberg formula

$\frac{1}{\lambda }=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),R_H=1.097\times 10^7{\text{m}}^{-1}$

Multiply R_H by Z² for hydrogen-like ions.

Bohr radius and speed

$r_n=0.529\frac{n^2}{Z}\text{A˚};v_n=2.18\times 10^6\frac{Z}{n}{\text{m s}}^{-1}$

Hydrogen-like species only.

Bohr energy

$E_n=-2.18\times 10^{-18}\frac{Z^2}{n^2}\text{J}=-13.6\frac{Z^2}{n^2}\text{eV}$

1 eV = 1.602 × 10⁻¹⁹ J.

de Broglie wavelength

$\lambda =\frac{h}{mv}$

Matter waves.

Uncertainty principle

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

Δp = mΔv.

Nodes

$\text{radial}=n-l-1;\text{angular}=l;\text{total}=n-1$

Orbitals in a subshell: 2l + 1.

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.

• Isotopes share protons, isobars share bars of mass

  isoTOPEs: same Place (Z) in the table; isoBARs: same mass number (barometer of mass).

  Use it when: Isotopes vs isobars.

• Lyman Ultra, Balmer Visible

  Lyman to 1 (UV), Balmer to 2 (visible), then Paschen, Brackett, Pfund (IR): 'Lazy Boys Prefer Bright Parties'.

  Use it when: Spectral series.

• Radial n − l − 1, angular l

  Total nodes are n − 1; l of them are planes.

  Use it when: Nodes.

• Lower n + l first

  Tie on n + l? Lower n first.

  Use it when: Aufbau order.

• Last in, first out? Not for 4s

  4s fills before 3d but also empties before 3d in ions.

  Use it when: Configurations of transition-metal ions.

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 D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 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 in ions

  Students reverse the filling order.

  Wrong working

  Fe²⁺ = [Ar]3d⁴4s².

  Right working

  Fe²⁺ = [Ar]3d⁶ (both 4s electrons are lost first).

  ✓Do this: Write the neutral atom first, then remove from the highest n.

  Usually costs: 1 mark.

• ✗Mistake: Pairing electrons too early

  Hund's rule is ignored.

  Wrong working

  Nitrogen 2p: ↑↓ ↑ _.

  Right working

  2p: ↑ ↑ ↑, three unpaired electrons.

  ✓Do this: Fill each degenerate orbital singly first.

  Usually costs: 1 mark.

• ✗Mistake: Wavelengths in nm left unconverted

  Substituting 589 instead of 589 × 10⁻⁹ m gives absurd energies.

  Wrong working

  E = 6.626 × 10⁻³⁴ × 3 × 10⁸/589.

  Right working

  Use 589 × 10⁻⁹ m: E = 3.37 × 10⁻¹⁹ J.

  ✓Do this: Convert every quantity to SI before substituting.

  Usually costs: 1-2 marks.

• ✗Mistake: Rydberg with n₁ and n₂ swapped

  A negative 1/λ results.

  Wrong working

  1/λ = R(1/16 − 1/4) for 4 → 2.

  Right working

  n₁ is the lower level: R(1/4 − 1/16).

  ✓Do this: n₁ < n₂ always; the lower level goes first.

  Usually costs: 1 mark.

• ✗Mistake: Confusing isotopes with isobars

  Both start with 'iso'.

  Wrong working

  ⁴⁰Ar and ⁴⁰Ca are isotopes.

  Right working

  Same A, different Z: isobars.

  ✓Do this: Isotopes are the same element.

  Usually costs: 1 mark.

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 maximum number of electrons in a subshell with l = 2 is:
   (a) 2
   (b) 6
   (c) 10
   (d) 14

   Hint

   2(2l + 1).

   Answer

   (c)

2. Q2 · 1 mark · MCQExam level

   Which pair are isobars?
   (a) ¹²C and ¹⁴C
   (b) ⁴⁰Ar and ⁴⁰Ca
   (c) ¹H and ²H
   (d) ²³Na and ²⁴Mg

   Hint

   Same mass number.

   Answer

   (b)

3. Q3 · 1 mark · Very short answerExam level

   What is the shape of an orbital with l = 1?

   Hint

   p orbital.

   Answer

   Dumb-bell

4. Q4 · 1 mark · Very short answerExam level

   How many angular nodes does a 3d orbital have?

   Hint

   Equal to l.

   Answer

   2

5. Q5 · 2 marks · Short answerExam level

   Write the electronic configuration of Mn²⁺ (Z = 25) and state the number of unpaired electrons.

   Hint

   Remove the 4s electrons.

   Answer

   [Ar]3d⁵; 5 unpaired

6. Q6 · 2 marks · Short answerHarder exam level

   Find the frequency and wavenumber of light of wavelength 400 nm.

   Hint

   c = νλ.

   Answer

   ν = 3.0 × 10⁸/4.00 × 10⁻⁷ = 7.5 × 10¹⁴ Hz; ν̄ = 2.5 × 10⁶ m⁻¹

7. Q7 · 2 marks · Short answerHarder exam level

   Find the energy (in eV) and radius of the n = 2 orbit of Li²⁺ (Z = 3).

   Hint

   Use Z²/n² and n²/Z.

   Answer

   E = −13.6 × 9/4 = −30.6 eV; r = 0.529 × 4/3 = 0.705 Å

8. Q8 · 3 marks · Short answerHarder exam level

   Find the wavelength of the first line of the Paschen series (n = 4 → 3) of hydrogen.

   Hint

   1/9 − 1/16.

   Answer

   1/λ = 1.097 × 10⁷ × 7/144 = 5.33 × 10⁵ m⁻¹; λ = 1.88 × 10⁻⁶ m (1875 nm, infrared)

9. Q9 · 3 marks · Short answerHarder exam level

   The position of a proton (m = 1.673 × 10⁻²⁷ kg) is known to within 1.0 × 10⁻¹² m. Find the minimum uncertainty in its speed.

   Hint

   Δv ≥ h/(4πmΔx).

   Answer

   6.626 × 10⁻³⁴/(4π × 1.673 × 10⁻²⁷ × 1.0 × 10⁻¹²) = 3.15 × 10⁴ m s⁻¹

10. Q10 · 3 marks · Short answerHarder exam level

    Give the possible quantum numbers for an electron in a 5d orbital, and the numbers of radial and angular nodes of 5d.

    Hint

    n = 5, l = 2.

    Answer

    n = 5, l = 2, m_l = −2, −1, 0, +1 or +2, m_s = +½ or −½. Radial nodes 5 − 2 − 1 = 2; angular nodes 2; total 4 = n − 1.

11. Q11 · 3 marks · Short answerHarder exam level

    A laser pointer emits 10 mW of light of wavelength 632.8 nm. How many photons does it emit per second?

    Hint

    Power ÷ energy per photon.

    Answer

    E = hc/λ = 3.14 × 10⁻¹⁹ J; 0.010/3.14 × 10⁻¹⁹ = 3.2 × 10¹⁶ photons s⁻¹

12. Q12 · 2 marks · Short answerHarder exam level

    Compare the numbers of unpaired electrons in Fe²⁺ and Fe³⁺ and explain which ion is more stable.

    Hint

    3d⁶ vs 3d⁵.

    Answer

    Fe²⁺ [Ar]3d⁶: 4 unpaired; Fe³⁺ [Ar]3d⁵: 5 unpaired. Fe³⁺ is more stable because its half-filled 3d subshell is symmetric with maximum exchange energy.

13. Q13 · 5 marks · Long answerHarder exam level

    (a) Find the radius of the third Bohr orbit of hydrogen. [1]
    (b) Find the speed of the electron in this orbit. [1]
    (c) Find the energy needed to excite the electron from this orbit to n = 5, per atom and per mole. [2]
    (d) State one limitation of Bohr's model. [1]

    Hint

    Z = 1.

    Answer

    (a) 0.529 × 9 = 4.76 Å. (b) 2.18 × 10⁶/3 = 7.27 × 10⁵ m s⁻¹. (c) 2.18 × 10⁻¹⁸ × (1/9 − 1/25) = 2.18 × 10⁻¹⁸ × 16/225 = 1.55 × 10⁻¹⁹ J; × 6.022 × 10²³ = 9.34 × 10⁴ J = 93.4 kJ mol⁻¹. (d) It cannot explain the spectra of atoms with more than one electron (or the Zeeman/Stark effects).

14. Q14 · 3 marks · Short answerHOTS / competency

    An electron and a proton have the same de Broglie wavelength. Find the ratio of their speeds (m_p/m_e = 1836).

    Hint

    Equal λ means equal momentum.

    Answer

    λ = h/mv equal ⇒ m_e v_e = m_p v_p ⇒ v_e/v_p = m_p/m_e = 1836: the electron is 1836 times faster.

15. Q15 · 3 marks · Short answerHOTS / competency

    Find the ionisation energy of He⁺ from its ground state in eV, and the wavelength of a photon that can just ionise it.

    Hint

    13.6 Z²; λ(nm) ≈ 1240/E(eV).

    Answer

    13.6 × 4 = 54.4 eV; λ = 1240/54.4 ≈ 22.8 nm (far ultraviolet)

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 discovered the neutron?Show answer

   Chadwick, by bombarding beryllium with α-particles.

2. 2.Main conclusion of Rutherford's experiment?Show answer

   A tiny, dense, positively charged nucleus.

3. 3.Planck's equation?Show answer

   E = hν.

4. 4.Energy of the nth Bohr orbit?Show answer

   −13.6 Z²/n² eV.

5. 5.de Broglie relation?Show answer

   λ = h/mv.

6. 6.Uncertainty principle?Show answer

   Δx Δp ≥ h/4π.

7. 7.Radial nodes in an orbital?Show answer

   n − l − 1.

8. 8.Configuration of Cu?Show answer

   [Ar]3d¹⁰4s¹.

Quick revision checklist

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

• ☐Can you describe Chadwick's experiment with its nuclear equation?
• ☐Can you list three observations of Rutherford's experiment and their conclusions?
• ☐Can you solve a photoelectric numerical for maximum speed?
• ☐Can you find the wavelength of any hydrogen line?
• ☐Can you find radius, speed and energy for a hydrogen-like ion?
• ☐Can you compute a de Broglie wavelength and an uncertainty?
• ☐Can you state allowed quantum numbers and count nodes?
• ☐Can you write configurations of Cr, Cu and transition-metal ions?

Structure of Atom: frequently asked questions