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Structure of Atom – Class 11 Chemistry notes, important questions and MCQs

NCERT Class 11 Chemistry, chapter 2. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Atoms contain electrons, protons and neutrons. Models evolved from Thomson and Rutherford to Bohr and the quantum mechanical model with orbitals and quantum numbers.

Models

Thomson: plum pudding; Rutherford's gold foil: small dense positive nucleus.
Bohr: electrons in fixed orbits; Eₙ = −13.6/n² eV for hydrogen; angular momentum = nh/2π.
de Broglie λ = h/mv; Heisenberg: Δx·Δp ≥ h/4π.

Quantum numbers and configuration

n (shell), l (0 to n − 1: s, p, d, f), mₗ (−l to +l), mₛ (±½).
Aufbau (n + l rule), Pauli exclusion (max 2 electrons per orbital), Hund's rule (fill singly first).
Exceptions: Cr = [Ar] 3d⁵ 4s¹, Cu = [Ar] 3d¹⁰ 4s¹ (half- and fully-filled stability).

Key terms

Orbital: region where the probability of finding an electron is high
Isotopes: atoms with the same atomic number but different mass numbers
Photoelectric effect: emission of electrons from a metal surface when light above a threshold frequency falls on it

Formulas

Eₙ = −13.6/n² eV
λ = h/mv
Number of orbitals in a shell = n²

Important questions with answers

Write the four quantum numbers for the outermost electron of sodium.2 marks

Na: 1s² 2s² 2p⁶ 3s¹. n = 3, l = 0, mₗ = 0, mₛ = +½ (or −½).

State Heisenberg's uncertainty principle and explain why it is not noticeable for large objects.3 marks

It is impossible to know both the exact position and exact momentum of a particle at the same time: Δx·Δp ≥ h/4π. For large objects the mass is huge, so Δv = h/(4πmΔx) is extremely small and the uncertainty cannot be detected.

Calculate the energy and wavelength of the photon emitted when an electron in a hydrogen atom falls from n = 3 to n = 2, and name the series. Also write the electronic configurations of Cu (Z = 29) and explain its stability.5 marks

ΔE = 13.6(1/2² − 1/3²) eV = 13.6 × 5/36 = 1.89 eV = 1.89 × 1.6 × 10⁻¹⁹ ≈ 3.02 × 10⁻¹⁹ J. λ = hc/ΔE = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(3.02 × 10⁻¹⁹) ≈ 6.56 × 10⁻⁷ m = 656 nm, the first line (red) of the Balmer series. Cu: [Ar] 3d¹⁰ 4s¹ instead of 3d⁹ 4s², because a completely filled d subshell has extra stability from symmetry and exchange energy.

Quick revision

Max electrons in a shell = 2n².
Cr and Cu are exceptions.
Hund's rule: singly before pairing.

Exam tips

Number of radial nodes = n − l − 1; angular nodes = l.
Fill orbitals by the (n + l) rule: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d … (lower n + l first; if equal, lower n first).

Practice MCQs

The nucleus was discovered by
1. Thomson
2. Chadwick
3. Bohr
4. Rutherford
Show answer
Answer: (d) Rutherford. Gold foil experiment.
Maximum electrons in an orbital is
1. 1
2. 2
3. 6
4. 8
Show answer
Answer: (b) 2. Pauli principle.
For n = 3, possible values of l are
1. 0, 1
2. 0 only
3. 1, 2, 3
4. 0, 1, 2
Show answer
Answer: (d) 0, 1, 2. 0 to n − 1.
Electronic configuration of Cr (Z = 24) is
1. [Ar] 3d⁴ 4s²
2. [Ar] 4s² 4p⁴
3. [Ar] 3d⁶
4. [Ar] 3d⁵ 4s¹
Show answer
Answer: (d) [Ar] 3d⁵ 4s¹. Half-filled d.
Number of electrons in Fe²⁺ (Z = 26) is
1. 26
2. 23
3. 28
4. 24
Show answer
Answer: (d) 24. Loses 2 electrons.
Energy of electron in the 2nd orbit of hydrogen is
1. −13.6 eV
2. −6.8 eV
3. −1.51 eV
4. −3.4 eV
Show answer
Answer: (d) −3.4 eV. −13.6/4.
Number of radial nodes in a 3p orbital is
1. 0
2. 1
3. 2
4. 3
Show answer
Answer: (b) 1. 3 − 1 − 1.
Wavelength of a photon of energy 3.3 × 10⁻¹⁹ J (h = 6.6 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s) is
1. 300 nm
2. 900 nm
3. 400 nm
4. 600 nm
Show answer
Answer: (d) 600 nm. λ = hc/E.

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Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.