Class 11 Chemistry · Original EZT practice
Structure of Atom (Class 11 Chapter 2): summary and practice
Structure of Atom builds the modern picture piece by piece: electron, proton and neutron, the Thomson and Rutherford models, light as wave and particle, Bohr's quantised orbits, and the quantum numbers that label every orbital. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Discovery of sub-atomic particles
Cathode rays revealed the electron, Thomson measured its charge-to-mass ratio, Millikan fixed its charge, and canal rays plus Chadwick's work added the proton and neutron. Section 2.1 runs this discovery story in order.
2. Atomic models
Thomson's plum-pudding gave way to Rutherford's nuclear model after the gold-foil experiment: a tiny dense positive nucleus with electrons around it. The model explained scattering but could not explain why orbiting electrons do not spiral in - its stated drawback.
3. Light as wave and particle
Waves diffract and interfere; the photoelectric effect needs particles. Planck's quantum theory sets photon energy E = h nu, and atomic spectra - sharp lines rather than a continuum - hint that electron energies are quantised too.
4. Bohr's model and the hydrogen spectrum
Bohr fixed electrons into orbits of energy En = -13.6/n^2 eV: light is emitted or absorbed only when an electron jumps between them, which explains every hydrogen line series. Its limits - multi-electron atoms, fine spectra, the uncertainty principle - pushed the theory further.
5. The quantum mechanical model
Matter has dual behaviour and Heisenberg's uncertainty principle bans exact paths, so the model speaks of orbitals, not orbits. Four quantum numbers n, l, m and s label each electron, and the Aufbau principle, Pauli exclusion and Hund's rule fill the orbitals in order.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Energy of a Bohr orbit
Choose the method: En = -13.6 / n^2 eV for hydrogen.
EZT practice question
Find the energy of the n = 2 orbit of a hydrogen atom.
Show the worked solution
- En = -13.6 / n^2.
- = -13.6 / 4.
Answer: -3.4 eV.
Check: Higher orbits sit closer to zero - n = 3 gives -1.51 eV, less tightly bound than n = 2.
Watch for this: Bound-state energies are NEGATIVE - a positive answer means the electron has escaped.
Now try it yourself
Energy of the n = 4 orbit?
Check your answer
-0.85 eV.
-13.6 / 16 = -0.85 eV.
Energy of a spectral jump
Choose the method: Photon energy = difference between the two levels; wavelength = 1240 / energy(eV) nm.
EZT practice question
Find the energy and wavelength of the photon released when hydrogen's electron falls from n = 2 to n = 1 (the first Lyman line).
Show the worked solution
- Delta E = -3.4 - (-13.6) = 10.2 eV released.
- lambda = 1240 / 10.2.
Answer: 10.2 eV, about 122 nm - deep ultraviolet.
Check: Lyman lines land in the UV because every jump ends at the tightly bound ground state.
Watch for this: Subtract the energies in the right order and take the magnitude - the emitted photon carries positive energy.
Now try it yourself
Balmer line n = 3 to n = 2. Energy and wavelength?
Check your answer
1.89 eV, about 656 nm (red).
-1.51 - (-3.4) = 1.89 eV; 1240 / 1.89 = 656 nm.
Frequency from wavelength
Choose the method: nu = c / lambda.
EZT practice question
Find the frequency of light of wavelength 600 nm.
Show the worked solution
- nu = c / lambda.
- = 3 x 10^8 / (600 x 10^-9).
Answer: 5.0 x 10^14 Hz.
Check: Visible light sits in the 10^14 Hz range - 600 nm is orange-red, near the low-frequency end.
Watch for this: Convert nanometres to metres first - 600 nm is 6 x 10^-7 m.
Now try it yourself
Frequency of 700 nm light?
Check your answer
4.3 x 10^14 Hz (approximately).
3 x 10^8 / 7 x 10^-7 = 4.3 x 10^14 Hz.
Photoelectric kinetic energy
Choose the method: KE = photon energy - work function.
EZT practice question
Light of 400 nm strikes a metal with work function 2.0 eV. Find the kinetic energy of the emitted electrons.
Show the worked solution
- Photon energy = 1240 / 400 = 3.1 eV.
- KE = 3.1 - 2.0.
Answer: 1.1 eV.
Check: If the photon energy fell below 2.0 eV, no electron would leave at all - however bright the light.
Watch for this: Intensity sets how MANY electrons leave; frequency sets whether any leave and how fast.
Now try it yourself
Same metal, 500 nm light. Kinetic energy?
Check your answer
0.48 eV.
Photon energy 1240 / 500 = 2.48 eV; 2.48 - 2.0 = 0.48 eV.
Count orbitals from quantum numbers
Choose the method: l runs 0 to n - 1; each l holds 2l + 1 orbitals.
EZT practice question
How many orbitals does the n = 3 shell contain?
Show the worked solution
- l = 0, 1, 2 - the s, p and d subshells.
- Orbitals: 1 + 3 + 5.
Answer: 9 orbitals.
Check: The shell total is always n^2: 3^2 = 9, and it holds 2n^2 = 18 electrons at most.
Watch for this: Do not mix n with l - n names the shell, l names the subshell shape within it.
Now try it yourself
Orbitals in the n = 2 shell?
Check your answer
4.
l = 0 and 1 give 1 + 3 = 4 orbitals, matching n^2 = 4.
Fill orbitals in order
Choose the method: Aufbau order, Pauli pairs, Hund spreads electrons singly first.
EZT practice question
Write the ground-state electronic configuration of nitrogen (Z = 7).
Show the worked solution
- Fill in Aufbau order: 1s, 2s, 2p.
- 1s2 2s2 uses 4 electrons; 3 remain for 2p.
- Hund's rule spreads them one per p orbital.
Answer: 1s2 2s2 2p3, with the three 2p electrons singly occupying the three p orbitals.
Check: Half-filled subshells like p3 carry extra stability - section 2.6.6 calls this out explicitly.
Watch for this: Do not pair electrons in one p orbital while another p orbital sits empty - Hund's rule forbids it.
Now try it yourself
Configuration of oxygen (Z = 8)?
Check your answer
1s2 2s2 2p4.
One more electron pairs up in a 2p orbital after the three singles.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Bohr orbit energy?
Show answer and reason
En = -13.6 / n^2 eV.
Negative because the electron is bound.
2. Photon energy?
Show answer and reason
E = h nu, and c = nu lambda.
1240 / lambda(nm) gives eV directly.
3. Four quantum numbers?
Show answer and reason
n, l, m, s.
n shell, l shape, m orientation, s spin.
4. Filling rules?
Show answer and reason
Aufbau, Pauli, Hund.
Lowest energy first, no twins, spread before pairing.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Find the energy of the n = 4 Bohr orbit of hydrogen.
- Find the frequency of 700 nm light.
- Write the electronic configuration of oxygen (Z = 8).
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Questions about this practice page
- Why did Rutherford's model fail?
- An orbiting electron is an accelerating charge, so classical electromagnetism says it should radiate energy continuously and spiral into the nucleus in a fraction of a second. Atoms are stable, so the model had to be wrong somewhere - Bohr's quantised orbits were the fix.
- What does a quantum number actually mean?
- Each quantum number pins down one property of an electron's state: n the shell and rough size, l the orbital's shape, m its orientation, and s the spin. Together the four act like a unique address - Pauli's rule says no two electrons share the same full address.
- Why are half-filled and fully filled subshells extra stable?
- Symmetrically spread electrons - one in each orbital, all spinning the same way - exchange places freely and lower the atom's energy. That is why p3, d5, f7 and the filled d10 show up as especially stable configurations.
- Why does the Bohr model fail for multi-electron atoms?
- Its energy formula assumes a single electron around the nucleus. With more electrons, electron-electron repulsion reshapes the energies, and finer spectral details plus the uncertainty principle need the full quantum mechanical model of section 2.6.
- Are these the NCERT exercise questions for Structure of Atom (Class 11)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.