Class 11 — Structure of Atom

Chapter Overview & Weightage

Structure of Atom is among the most scoring chapters in CBSE Class 11 Chemistry. Weightage typically $5$ - $7$ marks, distributed across MCQs and short-answer questions. Topics include Bohr’s model, Schrödinger’s equation (qualitative), quantum numbers, orbital shapes, and electronic configuration.

The chapter has both calculation-heavy and concept-heavy questions. About $60\%$ of marks come from numericals based on Bohr’s model and quantum numbers.

Key Concepts You Must Know

Bohr’s model: electrons orbit in fixed energy levels; energy is quantised as $E_n = -13.6/n^2$ eV for hydrogen.
Atomic spectrum: emission/absorption lines correspond to transitions between energy levels. Lyman, Balmer, Paschen series.
Rydberg formula: $1/\lambda = R(1/n_1^2 - 1/n_2^2)$ where $R = 1.097 \times 10^7$ m $^{-1}$ .
Quantum numbers: $n$ (principal), $l$ (azimuthal, $0$ to $n-1$ ), $m_l$ (magnetic, $-l$ to $+l$ ), $m_s$ ( $\pm 1/2$ ).
Aufbau principle, Pauli exclusion, Hund’s rule: filling order, no two electrons same set, max unpaired in degenerate orbitals.
Heisenberg uncertainty: $\Delta x \cdot \Delta p \geq h/(4\pi)$ .
de Broglie wavelength: $\lambda = h/p$ .

Important Formulas

$E_n = -\frac{13.6}{n^2} \text{ eV}$

For hydrogen-like ions: $E_n = -13.6 Z^2/n^2$ eV. Negative means bound.

$\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

For emission, $n_2 > n_1$ . Lyman: $n_1 = 1$ . Balmer: $n_1 = 2$ .

$\lambda = \frac{h}{p} = \frac{h}{m v}$

Apply to electrons, neutrons, or any particle with momentum.

Solved Previous Year Questions

PYQ 1 (CBSE 2023): Calculate the wavelength of light emitted when an electron in hydrogen atom transitions from $n = 4$ to $n = 2$ .

Solution: $1/\lambda = R(1/4 - 1/16) = R \times 3/16 = 1.097 \times 10^7 \times 0.1875$ . So $\lambda \approx 4.86 \times 10^{-7}$ m $= 486$ nm. This is the H-beta line in the Balmer series — a beautiful blue-green colour.

PYQ 2 (CBSE 2022): Write the electronic configuration of $\text{Cr}$ ( $Z = 24$ ). Explain why it is anomalous.

Solution: Expected: $[\text{Ar}] 3d^4 4s^2$ . Actual: $[\text{Ar}] 3d^5 4s^1$ . The half-filled $3d$ subshell ( $d^5$ ) provides extra exchange energy stabilisation, so one $4s$ electron promotes to $3d$ .

PYQ 3 (CBSE 2021): Calculate the de Broglie wavelength of an electron moving at $10^7$ m/s.

Solution: $\lambda = h/(mv) = (6.63 \times 10^{-34})/(9.1 \times 10^{-31} \times 10^7) \approx 7.29 \times 10^{-11}$ m $= 0.73$ Å.

Difficulty Distribution

Easy (45%): Quantum numbers, electronic configurations, definitions.
Medium (40%): Rydberg formula numericals, Bohr energy levels, de Broglie.
Hard (15%): Heisenberg uncertainty applications, anomalous configurations of Cr/Cu, multi-electron transitions.

Expert Strategy

For numericals, keep $hc = 1240$ eV nm handy. Computing $E = hc/\lambda$ in eV is much faster than doing it in joules and converting back.

For electronic configurations, follow the Aufbau order: $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, ...$ . Memorise this — it’s worth easy marks.

Conceptual questions on quantum numbers test boundary cases. For $l = 2$ (d-orbital), $m_l$ can be $-2, -1, 0, +1, +2$ . Five orbitals per d-subshell. Always count carefully.

Common Traps

Trap 1: Forgetting the negative sign in Bohr energy. Energies are negative because the electron is bound. The transition energy is the difference of magnitudes.

Trap 2: Using $Z$ for hydrogen ( $Z = 1$ ). Many forget the $Z^2$ factor when working with He $^+$ or Li $^{2+}$ .

Trap 3: Mixing up Lyman ( $n_1 = 1$ , UV) and Balmer ( $n_1 = 2$ , visible). Lyman is invisible, Balmer is colourful.

Trap 4: Writing Cr as $3d^4 4s^2$ instead of $3d^5 4s^1$ . Cr and Cu are the two anomalous configurations in 3d series — drill them.