JEE Chemistry Atomic Structure MCQs Set 02

Chemistry Objective Questions and Answers: Atomic Structure

Access targeted multiple-choice questions for Atomic Structure designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Navigate directly to the 50 objective questions for Atomic Structure using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: The angular momentum in the second Bohr orbit of the H atom is x. The angular momentum in the first excited state of Li2+ ion is:
(a) 3x
(b) 9x
(c) x/2
(d) x
Show Answer & Explanation

Answer: (d) x

Explanation:
1. Angular momentum = nh/2π depends only on n.
2. The first excited state of Li2+ is n = 2, the same n as the H atom orbit, so it equals x.
3. Answer: (d)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 2: The product of electron velocity and radius for an orbit in an atom is:
(a) Proportional to the mass of the electron
(b) Proportional to the square of the mass of the electron
(c) Inversely proportional to the mass of the electron
(d) Independent of the mass of the electron
Show Answer & Explanation

Answer: (c) Inversely proportional to the mass of the electron

Explanation:
1. Bohr: mvr = nh/2π, so vr = nh/(2πm).
2. The product is inversely proportional to m.
3. Answer: (c)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 3: The velocity of the electron in the third excited state of Be3+ ion will be:
(a) (3/4)(2.188 × 108) m s−1
(b) (3/4)(2.188 × 106) m s−1
(c) 2.188 × 106 km s−1
(d) 2.188 × 103 km s−1
Show Answer & Explanation

Answer: (d) 2.188 × 103 km s−1

Explanation:
1. Third excited state means n = 4; for Be3+, Z = 4.
2. v = 2.188 × 106 × Z/n = 2.188 × 106 × 4/4 = 2.188 × 106 m s−1 = 2.188 × 103 km s−1.
3. Answer: (d)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 4: According to Bohr theory, the radius (r) and velocity (v) of an electron vary with the increasing principal quantum number n as:
(a) r increases, v decreases
(b) r and v both increase
(c) r and v both decrease
(d) r decreases, v increases
Show Answer & Explanation

Answer: (a) r increases, v decreases

Explanation:
1. r ∝ n2/Z increases with n.
2. v ∝ Z/n decreases with n.
3. Answer: (a)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 5: In a hydrogen atom, if the energy of an electron in the ground state is −13.6 eV, then the energy in the 2nd excited state is:
(a) −1.51 eV
(b) −3.4 eV
(c) −6.04 eV
(d) −13.6 eV
Show Answer & Explanation

Answer: (a) −1.51 eV

Explanation:
1. The second excited state is n = 3.
2. E = −13.6/9 = −1.51 eV.
3. Answer: (a)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 6: The potential energy is −27.2 eV in the second orbit of He+. The double of the total energy in the first excited state of the hydrogen atom is:
(a) −13.6 eV
(b) −54.4 eV
(c) −6.8 eV
(d) −27.2 eV
Show Answer & Explanation

Answer: (c) −6.8 eV

Explanation:
1. The first excited state of H is n = 2, with total energy −13.6/4 = −3.4 eV.
2. Double = −6.8 eV.
3. Answer: (c)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 7: Maximum frequency of emission is obtained for the transition:
(a) n = 2 to n = 1
(b) n = 6 to n = 2
(c) n = 1 to n = 2
(d) n = 2 to n = 6
Show Answer & Explanation

Answer: (a) n = 2 to n = 1

Explanation:
1. Only n = 2 → 1 and n = 6 → 2 are emissions; the others are absorptions.
2. The energy gap 2 → 1 (10.2 eV) is larger than 6 → 2 (3.0 eV), so it has the highest frequency.
3. Answer: (a)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 8: Which of the following electron transitions will require the largest amount of energy in a hydrogen atom?
(a) From n = 1 to n = 2
(b) From n = 2 to n = 3
(c) From n = ∞ to n = 1
(d) From n = 3 to n = 5
Show Answer & Explanation

Answer: (a) From n = 1 to n = 2

Explanation:
1. Excitation 1 → 2 needs 10.2 eV, 2 → 3 needs 1.9 eV and 3 → 5 needs 0.97 eV.
2. n = ∞ → 1 is an emission, not an energy requirement.
3. Answer: (a)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 9: If the potential energy of the hydrogen electron is −3.02 eV, in which excited level is the electron present?
(a) 1st
(b) 2nd
(c) 3rd
(d) 4th
Show Answer & Explanation

Answer: (b) 2nd

Explanation:
1. Potential energy = 2 × total energy, so E = −1.51 eV.
2. E = −13.6/n2 gives n = 3, which is the 2nd excited level.
3. Answer: (b)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 10: The radiation of low frequency will be emitted in which transition of the hydrogen atom?
(a) n = 1 to n = 4
(b) n = 2 to n = 5
(c) n = 3 to n = 1
(d) n = 5 to n = 2
Show Answer & Explanation

Answer: (d) n = 5 to n = 2

Explanation:
1. Emission is a fall to a lower level: (c) 3 → 1 (12.1 eV) and (d) 5 → 2 (2.86 eV). Options (a) and (b) are absorptions.
2. The smaller energy gap gives the lower frequency, so 5 → 2.
3. Answer: (d)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 11: Which of the following is a correct relationship?
(a) E1 of H = ½E2 of He+ = ⅓E3 of Li2+ = ¼E4 of Be3+
(b) E1(H) = E2(He+) = E3(Li2+) = E4(Be3+)
(c) E1(H) = 2E2(He+) = 3E3(Li2+) = 4E4(Be3+)
(d) No relation
Show Answer & Explanation

Answer: (b) E1(H) = E2(He+) = E3(Li2+) = E4(Be3+)

Explanation:
1. En = −13.6 Z2/n2 and Z/n = 1 for each of the four cases.
2. So all four energies equal −13.6 eV.
3. Answer: (b)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 12: The first excitation potential of the H atom is:
(a) 10.2 eV
(b) 3.4 eV
(c) 0
(d) −3.4 eV
Show Answer & Explanation

Answer: (a) 10.2 eV

Explanation:
1. First excitation is n = 1 → 2: E = 13.6(1 − 1/4) = 10.2 eV.
2. Answer: (a)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 13: The ionisation energy of an excited hydrogen atom (electron in n ≥ 2) in eV will be:
(a) 13.6
(b) Less than 13.6
(c) Greater than 13.6
(d) Exactly 3.4
Show Answer & Explanation

Answer: (b) Less than 13.6

Explanation:
1. Ionisation energy = 13.6/n2, which is 3.4 eV for n = 2 and smaller for higher n.
2. So it is less than 13.6 eV.
3. Answer: (b)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 14: The ionisation energy of an H-like atom is 13.6 eV. The energy required to excite the atom from the ground state to the next higher state is:
(a) 3.4 eV
(b) 10.2 eV
(c) 12.1 eV
(d) 1.5 eV
Show Answer & Explanation

Answer: (b) 10.2 eV

Explanation:
1. E = 13.6(1 − 1/4) = 10.2 eV.
2. Answer: (b)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 15: The third line of the Balmer series in the spectrum of the H atom is produced by which transition?
(a) 5 to 2
(b) 5 to 1
(c) 4 to 2
(d) 4 to 1
Show Answer & Explanation

Answer: (a) 5 to 2

Explanation:
1. Balmer lines end at n = 2: first 3 → 2, second 4 → 2, third 5 → 2.
2. Answer: (a)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 16: The wavelength of the photon obtained by an electron transition between the same two levels in the H atom and in singly ionised He are λ1 and λ2 respectively. Then:
(a) λ2 = λ1
(b) λ2 = 2λ1
(c) λ2 = λ1/2
(d) λ2 = λ1/4
Show Answer & Explanation

Answer: (d) λ2 = λ1/4

Explanation:
1. 1/λ ∝ Z2; Z = 2 for He+ gives four times the energy.
2. So λ2 = λ1/4.
3. Answer: (d)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 17: The first Lyman transition in the hydrogen spectrum has ΔE = 10.2 eV. The same energy change is observed in the second Balmer transition of:
(a) Li2+
(b) Li+
(c) He+
(d) Be3+
Show Answer & Explanation

Answer: (c) He+

Explanation:
1. The second Balmer transition is n = 4 → 2: ΔE = 13.6 Z2(1/4 − 1/16) = 2.55 Z2 eV.
2. Setting this equal to 10.2 gives Z2 = 4, so Z = 2: He+.
3. Answer: (c)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 18: In the H atom, an electron makes a transition from the 6th orbit to the 2nd orbit in multiple steps. The total number of spectral lines (excluding those of the Balmer series) will be:
(a) 6
(b) 10
(c) 4
(d) 0
Show Answer & Explanation

Answer: (a) 6

Explanation:
1. Lines among levels 6, 5, 4, 3, 2: 10 in total.
2. Lines of the Balmer series (ending on n = 2) are 4: 6→2, 5→2, 4→2, 3→2.
3. Remaining = 6.
4. Answer: (a)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 19: A certain electronic transition from an excited state to the ground state of the H atom, in one or more steps, gives rise to three lines in the ultraviolet region of the spectrum. How many lines does this transition produce in the infrared region of the spectrum?
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. Three Lyman (UV) lines mean the electron started from n = 4 (4→1, 3→1, 2→1).
2. Other lines possible from n = 4: 4→2 and 3→2 (visible, Balmer) and 4→3 (infrared, Paschen).
3. So one line is in the infrared.
4. Answer: (a)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

Question 20: If 9.9 eV of energy is supplied to an H atom (in its ground state), the number of spectral lines emitted is equal to:
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. The least excitation energy is 10.2 eV (n = 1 → 2).
2. 9.9 eV is insufficient, so no transition occurs and no lines are emitted.
3. Answer: (a)

Teacher's Note:
1. Use the Bohr formulas for hydrogen-like species.
2. Check units and the value of Z.

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