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CBSE · Class 11 · Chemistry

Structure of Atom

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
Who discovered the electron, using cathode rays?
  1. James Chadwick
  2. Niels Bohr
  3. Ernest Rutherford
  4. J. J. Thomson
2.
The number of electrons in 2412Mg2+ is:
  1. 10
  2. 12
  3. 24
  4. 11
3.
The energy of one photon of light of wavelength 400 nm is: (h = 6.626 × 10−34 J s, c = 3.0 × 108 m/s)
  1. 4.97 × 10−18 J
  2. 4.97 × 10−16 J
  3. 4.97 × 10−20 J
  4. 4.97 × 10−19 J
4.
Which set of quantum numbers is not possible for an electron?
  1. n = 1, l = 2, mₗ = +1, mₛ = +½
  2. n = 1, l = 0, mₗ = 0, mₛ = −½
  3. n = 4, l = 2, mₗ = 0, mₛ = −½
  4. n = 3, l = 1, mₗ = +1, mₛ = −½
5.
An electron in a hydrogen atom falls from n = 5 to n = 2. The line emitted belongs to the:
  1. Balmer series (visible)
  2. Paschen series (infrared)
  3. Lyman series (ultraviolet)
  4. Brackett series (infrared)
6.
What is the total number of nodes in a 1s orbital?
  1. 2
  2. 1
  3. 0
  4. 4
7.
The ground-state electronic configuration of bromine (Z = 35) is:
  1. [Ar] 3d10 4s2 4p5
  2. [Ar] 3d10 4s2 4p6
  3. [Ar] 3d10 4s2 4p4 5s1
  4. [Ar] 3d10 4s2 4p4
8.
Assertion (A): An atom of copper (Z = 29) has 2 unpaired electrons in its ground state.
Reason (R): Electrons occupy the orbitals of a subshell singly, with parallel spins, before pairing (Hund’s rule).
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
State Hund’s rule of maximum multiplicity.
10.
Find the frequency and the wavenumber of radiation of wavelength 500 nm. (c = 3.0 × 108 m/s)
11.
How much energy is given out when the electron of a hydrogen atom falls from n = 4 to n = 1? (E_n = −2.18 × 10−18/n2 J)
12.
Find the wavenumber and wavelength of the line emitted when the electron of a hydrogen atom falls from n = 3 to n = 2. (R = 109,677 cm−1) (2 decimal places where needed.)
13.
Find the de Broglie wavelength of a ball of mass 200 g moving at 40 m/s. (h = 6.626 × 10−34 J s)
14.
How many photons of wavelength 650 nm make up 20 J of energy? (h = 6.626 × 10−34 J s, c = 3.0 × 108 m/s)
15.
For the 4f subshell, give the values of n and l, the possible values of mₗ, the number of orbitals and the maximum number of electrons.
16.
The threshold frequency of a metal is 5.0 × 1014 Hz. Light of frequency 13.0 × 1014 Hz falls on it. Find the work function of the metal and the kinetic energy of the emitted electrons. (h = 6.626 × 10−34 J s)
Section C — Word problems4 × 4 = 16 marks
17.
Describe Rutherford’s α-particle scattering experiment. What did it show, and what was the main drawback of his model?
18.
The electron of a hydrogen atom falls from n = 3 to n = 2. Find (a) the energy given out, (b) the wavelength of the light in nm, (c) the series and region of the spectrum. (E_n = −2.18 × 10−18/n2 J; h = 6.626 × 10−34 J s; c = 3.0 × 108 m/s)
19.
For selenium (Z = 34): (a) write the ground-state electronic configuration in full, (b) give the number of unpaired electrons, (c) how many electrons have l = 1? (d) how many electrons have n = 3? (e) which subshell receives the last electron?
20.
An ion with mass number 79 carries a charge of −1. It contains 22.22% more neutrons than electrons. (a) Find the numbers of electrons, protons and neutrons. (b) Identify the ion and write its symbol.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 11 · Chemistry

Answer key

Structure of Atom · AW-11C-SOA-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) J. J. Thomson—
2(a) 10Z = 12, A = 24: protons 12, neutrons 24 − 12 = 12, electrons 12 − 2 = 10
3(d) 4.97 × 10−19 JE = hc/λ, with λ in metres
4(a) n = 1, l = 2, mₗ = +1, mₛ = +½l must be less than n (here l = 2, n = 1)
5(a) Balmer series (visible)the series is named by the lower level, n = 2
6(c) 0total nodes = n − 1
7(a) [Ar] 3d10 4s2 4p5fill by the (n + l) rule: 1s 2s 2p 3s 3p 4s 3d 4p
8(d) A is false, but R is true.A is false: [Ar] 3d10 4s1: 1 unpaired. R is true.
9Electrons enter the orbitals of a subshell singly, with parallel spins, before any pairing begins.—
10ν = 6.00 × 1014 Hz; ν̄ = 2.00 × 106 m−1ν = c/λ; ν̄ = 1/λ (λ in metres)
112.04 × 10−18 JΔE = 2.18 × 10−18 (1/12 − 1/42)
12ν̄ = 15,232.92 cm−1; λ = 656.47 nm (Balmer series)ν̄ = R(1/22 − 1/32); λ = 1/ν̄ (1 cm = 107 nm)
138.28 × 10−35 mλ = h/(mv), with m in kg; far too small to observe
146.54 × 1019number = total energy ÷ hc/λ
15n = 4, l = 3; mₗ = −3, −2, −1, 0, +1, +2, +3; 7 orbitals; 14 electronsl = 3 for f; mₗ from −l to +l
16W0 = 3.313 × 10−19 J; KE = 5.301 × 10−19 JW0 = hν0; KE = h(ν − ν0)
17A thin gold foil was bombarded with α-particles. Most passed straight through, a few were deflected through small angles, and very few (about 1 in 20,000) bounced back. So most of the atom is empty space, with its positive charge and almost all its mass in a tiny nucleus; electrons move around it. Drawback: an electron moving in a circle should radiate energy and spiral into the nucleus, so the model cannot explain the stability of the atom (or line spectra).—
18(a) 3.03 × 10−19 J; (b) 656.52 nm; (c) Balmer series, visibleΔE = 2.18 × 10−18 (1/n12 − 1/n22); λ = hc/ΔE
19(a) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4; (b) 2; (c) 16; (d) 18; (e) 4pAufbau order; Hund’s rule for unpaired electrons
20(a) electrons 36, protons 35, neutrons 44; (b) bromine: 7935Br−if electrons = x, neutrons = x(1 + 22.22/100) and protons = x − 1; protons + neutrons = 79