Atomic Structure Practice

Take timed practice tests on Atomic Structure for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.

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Atomic Structure Practice Tests

Six 20-question timed sessions plus a 60-question chapter module. Each item is original and reframed for copyright safety.

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JEE Main Session 1

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Preview all 44 questions in JEE Main Session 1 (no login required)

Bohr Model
1. The energy of the electron in the nth orbit of hydrogen atom is:
- A. -13.6/n² eV (Correct)
- B. -13.6n² eV
- C. 13.6/n² eV
- D. -13.6/n eV
Explanation: En = −13.6Z²/n² eV. For hydrogen (Z=1): En = −13.6/n² eV. The negative sign indicates bound state.
Bohr Model
2. The radius of the nth orbit of hydrogen atom is:
- A. 0.529n² Å (Correct)
- B. 0.529/n² Å
- C. 0.529n Å
- D. 0.529 Å only
Explanation: rn = 0.529 × n²/Z Å. For H (Z=1): rn = 0.529n² Å. The Bohr radius (n=1) is 0.529 Å (a₀).
Energy Levels
3. The energy required to remove an electron from the ground state of hydrogen atom is:
- A. 13.6 eV (Correct)
- B. 3.4 eV
- C. 1.51 eV
- D. 0.85 eV
Explanation: Ionisation energy of H = |E₁| = 13.6 eV (to ionise from n=1).
Spectral Lines
4. The spectral lines in the Balmer series of hydrogen correspond to transitions:
- A. n→1
- B. n→2 (Correct)
- C. n→3
- D. n→4
Explanation: Balmer series: transitions from higher levels (n=3,4,5...) to n=2. These fall in visible/UV region.
Rydberg Equation
5. The wavelength of light emitted when hydrogen electron falls from n=4 to n=2 (Rydberg constant R = 1.097×10⁷ m⁻¹) is:
- A. 486 nm (Correct)
- B. 656 nm
- C. 434 nm
- D. 410 nm
Explanation: 1/λ = R(1/4 − 1/16) = R×(3/16). λ = 16/(3R) = 16/(3×1.097×10⁷) ≈ 4.86×10⁻⁷ m = 486 nm (Hβ line, blue-green).
Bohr Model
6. For a hydrogen-like ion He⁺ (Z=2), the energy of n=2 orbit is:
- A. -13.6 eV (Correct)
- B. -6.8 eV
- C. -27.2 eV
- D. -54.4 eV
Explanation: En = −13.6Z²/n² = −13.6×4/4 = −13.6 eV. He⁺ with Z=2, n=2 has the same energy as H n=1.
de Broglie
7. The de Broglie wavelength of a particle with momentum p is:
- A. λ = h/p (Correct)
- B. λ = hp
- C. λ = h/mv²
- D. λ = p/h
Explanation: de Broglie wavelength: λ = h/p = h/(mv). Gives wave character to all matter particles.
Quantum Numbers
8. The maximum number of electrons in the M shell (n=3) is:
- A. 8
- B. 10
- C. 18 (Correct)
- D. 32
Explanation: Max electrons in nth shell = 2n². For n=3: 2×9 = 18. The M shell has 3s, 3p, 3d subshells.
Orbital Quantum Numbers
9. For the 3d subshell, the azimuthal quantum number l is:
- A. 0
- B. 1
- C. 2 (Correct)
- D. 3
Explanation: For d subshell: l = 2. The subshell letter s,p,d,f corresponds to l = 0,1,2,3 respectively.
Magnetic Quantum Number
10. The number of orbitals in a subshell with l = 2 is:
- A. 3
- B. 5 (Correct)
- C. 7
- D. 1
Explanation: Number of orbitals = 2l+1 = 2(2)+1 = 5. The d subshell has 5 orbitals.
Electron Configuration
11. The electronic configuration of Fe (Z=26) is:
- A. [Ar] 3d⁸
- B. [Ar] 3d⁶ 4s² (Correct)
- C. [Ar] 4s² 3d⁶
- D. [Ar] 3d⁵ 4s²4p¹
Explanation: Fe: [Ar] 3d⁶ 4s². Iron has 6 electrons in 3d and 2 in 4s. This is fully compatible with Hund's and Aufbau rules.
Exception
12. The electronic configuration of Cu (Z=29) is:
- A. [Ar] 3d⁹ 4s²
- B. [Ar] 3d¹⁰ 4s¹ (Correct)
- C. [Ar] 3d⁸ 4s²4p¹
- D. [Ar] 4s² 3d⁹
Explanation: Cu adopts [Ar] 3d¹⁰ 4s¹ (not 3d⁹ 4s²) because a completely filled d subshell is extra stable. This is a key exception.
Ionisation Energy
13. The first ionisation energy of Li is 5.4 eV. The energy needed to remove the electron from Li²⁺ (hydrogen-like, Z=3) from n=1 is:
- A. 13.6 eV
- B. 30.6 eV
- C. 122.4 eV (Correct)
- D. 54.4 eV
Explanation: IE of H-like Li²⁺ = 13.6×Z²/n² = 13.6×9/1 = 122.4 eV.
Uncertainty
14. Heisenberg uncertainty principle states Δx·Δp ≥ h/4π. If the uncertainty in position of an electron is 1 Å, the minimum uncertainty in momentum is approximately:
- A. 5.3×10⁻²⁵ kg m/s (Correct)
- B. 5.3×10⁻²⁷ kg m/s
- C. 1.05×10⁻²⁴ kg m/s
- D. h/2π
Explanation: Δp ≥ h/(4π·Δx) = 6.63×10⁻³⁴/(4π×10⁻¹⁰) ≈ 5.3×10⁻²⁵ kg·m/s.
Spectral series
15. The series limit of the Lyman series of hydrogen corresponds to the electron transitioning from:
- A. n=2 to n=1
- B. n=∞ to n=1 (Correct)
- C. n=3 to n=1
- D. n=∞ to n=2
Explanation: The series limit is the shortest wavelength (highest energy) line in a series. For Lyman, n=∞→n=1. IE = 13.6 eV for this limit.
Spin Quantum Number
16. The spin quantum number (ms) can have values:
- A. Only +1/2
- B. Only −1/2
- C. +1/2 or −1/2 (Correct)
- D. 0, +1, −1
Explanation: ms = ±1/2 for electrons. Positive is often called 'spin up' (↑) and negative 'spin down' (↓).
Pauli Exclusion
17. Pauli exclusion principle states that:
- A. No two electrons have same energy
- B. No two electrons in same atom can have identical quantum numbers (Correct)
- C. Electrons fill orbitals from lowest energy
- D. Half-filled subshells are extra stable
Explanation: Pauli: no two electrons in the same atom can have all four quantum numbers (n, l, ml, ms) identical. This limits each orbital to two electrons (with opposite spins).
Hund Rule
18. Hund's rule of maximum multiplicity states that:
- A. Electrons fill from lowest energy first
- B. Orbitals of equal energy are filled singly before pairing (Correct)
- C. No two electrons have same set of quantum numbers
- D. Paired electrons are more stable
Explanation: Hund's rule: in degenerate orbitals, electrons occupy singly with same spin before pairing. This maximises unpaired electrons (maximum multiplicity).
Aufbau Principle
19. In Aufbau filling, orbitals are filled in order of increasing:
- A. Principal quantum number only
- B. Azimuthal quantum number only
- C. n + l value, and lower n first for same n+l (Correct)
- D. Magnetic quantum number
Explanation: Aufbau filling order: subshells with lower (n+l) fill first. For same n+l, lower n fills first. This gives the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, ...
Unpaired electrons
20. The number of unpaired electrons in Fe²⁺ ([Ar] 3d⁶) is:
- A. 2
- B. 4 (Correct)
- C. 5
- D. 6
Explanation: Fe²⁺: 3d⁶. Filling 5 d-orbitals: 5 electrons go in singly (Hund), 6th pairs with one. Unpaired = 4. Magnetic moment = √(n(n+2)) = √24 ≈ 4.9 BM.
Orbital shape
21. The shape of a p-orbital is:
- A. Spherical
- B. Dumbbell (Correct)
- C. Cloverleaf
- D. Toroidal
Explanation: p-orbitals (l=1) have a dumbbell (bi-lobed) shape along one of the three axes (px, py, pz).
Number of spectral lines
22. When an electron falls from n=4 to the ground state in a hydrogen atom, the maximum number of spectral lines emitted is:
- A. 3
- B. 4
- C. 6 (Correct)
- D. 8
Explanation: Maximum spectral lines when electron falls from level n to lower levels = n(n−1)/2. For n=4: 4×3/2 = 6 lines.
Energy calculation
23. The energy of the electron in the third Bohr orbit of He⁺ (Z=2) is:
- A. -6.04 eV (Correct)
- B. -13.6 eV
- C. -3.4 eV
- D. -1.51 eV
Explanation: En = −13.6Z²/n² = −13.6×4/9 = −54.4/9 ≈ −6.04 eV.
Photoelectric effect
24. The maximum kinetic energy of photoelectrons in the photoelectric effect is:
- A. hν
- B. hν − φ (Correct)
- C. hν + φ
- D. φ − hν
Explanation: KE_max = hν − φ, where φ is the work function (minimum energy to eject electron). This is Einstein's photoelectric equation.
de Broglie
25. An electron accelerated through potential V volts has de Broglie wavelength λ = 12.27/√V Å. At V = 100 V, λ is:
- A. 1.227 Å (Correct)
- B. 0.1227 Å
- C. 12.27 Å
- D. 122.7 pm
Explanation: λ = 12.27/√100 = 12.27/10 = 1.227 Å = 122.7 pm. This formula is derived from KE = eV and λ = h/√(2meV).
Orbital nodes
26. The number of radial nodes in the 3s orbital is:
- A. 0
- B. 1
- C. 2 (Correct)
- D. 3
Explanation: Radial nodes = n − l − 1 = 3 − 0 − 1 = 2. The 3s orbital has 2 spherical nodal surfaces.
Angular nodes
27. The number of angular nodes in a d-orbital is:
- A. 0
- B. 1
- C. 2 (Correct)
- D. 3
Explanation: Angular nodes = l. For d subshell, l = 2, so 2 angular (planar/conical) nodes.
Spectral series
28. Which series of hydrogen emission spectrum lies in the UV region?
- A. Lyman (Correct)
- B. Balmer
- C. Paschen
- D. Brackett
Explanation: Lyman series (n→1) emits high-energy photons in the UV region. Balmer (n→2) is visible, Paschen and beyond are infrared.
Bohr velocity
29. The velocity of an electron in the nth orbit of hydrogen is proportional to:
- A. n
- B. 1/n (Correct)
- C. n²
- D. 1/n²
Explanation: From Bohr's quantisation: v = Ze²/(nħ) ∝ 1/n. As n increases, orbital radius grows and speed decreases.
Hydrogen-like
30. The shortest wavelength in the Lyman series of Li²⁺ (Z=3) corresponds to:
- A. 1/λ = 9R × 1
- B. 1/λ = R
- C. 1/λ = 9R (Correct)
- D. 1/λ = R/9
Explanation: Series limit (n→∞ to n=1): 1/λ = RZ²(1/1 − 0) = R×9 = 9R. Wavelength is 1/9 of H Lyman limit.
Orbital penetration
31. For the same principal quantum number n, the order of orbital penetration to nucleus is:
- A. s > p > d > f (Correct)
- B. f > d > p > s
- C. s = p = d = f
- D. d > s > p > f
Explanation: s orbitals have greatest penetration (electron density near nucleus), followed by p, d, f. This determines orbital energy in multi-electron atoms: s fills before p, etc.
Wave mechanical
32. Assertion (A): The exact position and momentum of an electron cannot be simultaneously determined with arbitrary precision. Reason (R): Electrons are waves with inherent uncertainty. Which is correct?
- A. Both A and R are true, R explains A (Correct)
- B. Both true but R does not explain A
- C. A true, R false
- D. A false, R true
Explanation: Heisenberg uncertainty arises because electrons have wave-particle duality. The wave nature introduces inherent uncertainty — measuring position disturbs momentum. R correctly explains A.
Electron filling
33. The element with Z=24 (Cr) has configuration [Ar] 3d⁵ 4s¹ instead of [Ar] 3d⁴ 4s². This is because:
- A. 3d⁵ has a completely filled subshell
- B. 3d⁵ 4s¹ has extra stability due to half-filled 3d (Correct)
- C. 4s electron shielding decreases
- D. The energy difference is negligible
Explanation: Half-filled d subshell (3d⁵ = 5 unpaired electrons symmetrically arranged) has extra stability from exchange energy. This overcomes the normal Aufbau order.
Quantum number validity
34. Which set of quantum numbers is NOT possible?
- A. n=2, l=1, ml=0, ms=+1/2
- B. n=3, l=2, ml=−2, ms=−1/2
- C. n=2, l=2, ml=1, ms=+1/2 (Correct)
- D. n=4, l=3, ml=+3, ms=+1/2
Explanation: l can be 0 to (n−1). For n=2, l can be 0 or 1 only. l=2 is not allowed when n=2. This set is invalid.
Frequency of radiation
35. The frequency of radiation absorbed when H electron moves from n=1 to n=3 is (R=3.29×10¹⁵ Hz):
- A. 2.93×10¹⁵ Hz (Correct)
- B. 2.46×10¹⁵ Hz
- C. 3.29×10¹⁵ Hz
- D. 1.46×10¹⁵ Hz
Explanation: ν = cR(1/1 − 1/9) = R×(8/9) = 3.29×10¹⁵ × 8/9 ≈ 2.93×10¹⁵ Hz.
Photoelectric
36. The threshold frequency for a metal is ν₀. If light of frequency 2ν₀ hits it, the maximum KE of emitted electrons is:
- A. hν₀ (Correct)
- B. 2hν₀
- C. h(2ν₀ − ν₀) = hν₀
- D. 3hν₀
Explanation: KE_max = hν − hν₀ = h(2ν₀) − hν₀ = hν₀.
Number of electrons
37. The maximum number of electrons in a subshell with l = 3 (f subshell) is:
- A. 6
- B. 10
- C. 14 (Correct)
- D. 18
Explanation: Number of orbitals in f subshell = 2l+1 = 7. Each holds 2 electrons max → 14 electrons.
Bohr model accuracy
38. Bohr model fails to explain which of these?
- A. Energy levels of hydrogen
- B. Zeeman splitting of spectral lines (Correct)
- C. Ionisation energy of hydrogen
- D. Rydberg formula
Explanation: Bohr model cannot explain the splitting of spectral lines in a magnetic field (Zeeman effect), fine structure, or spectra of multi-electron atoms. It works only for H-like species.
Basic
39. The charge on one mole of electrons is:
- A. 1.6×10⁻¹⁹ C
- B. 96500 C (Correct)
- C. 6.022×10²³ C
- D. None
Explanation: 1 Faraday = charge on 1 mole of electrons = NA × e = 6.022×10²³ × 1.6×10⁻¹⁹ ≈ 96500 C.
Spectral series
40. In Paschen series, transitions are from n > 3 to n = 3. The region of electromagnetic spectrum is:
- A. UV
- B. Visible
- C. Infrared (Correct)
- D. Microwave
Explanation: Paschen series photons have lower energy (n→3) than Balmer (n→2). They fall in the infrared (IR) region.
Orbital
41. How many total nodes does a 3p orbital have?
- A. 1
- B. 2 (Correct)
- C. 3
- D. 0
Explanation: Total nodes = n − 1 = 2. Angular nodes (planes) = l = 1. Radial nodes = n−l−1 = 1. Total = 1+1 = 2.
Hard
42. The ratio of the kinetic energy of an electron in the 2nd Bohr orbit of He⁺ to that in the 3rd orbit of H is:
- A. 4
- B. 9 (Correct)
- C. 16
- D. 12
Explanation: KE ∝ Z²/n². For He⁺ n=2: Z²/n²=4/4=1. For H n=3: Z²/n²=1/9. Ratio = 1/(1/9) = 9. Wait: KE of He⁺ n=2 = 13.6×4/4 = 13.6 eV; KE of H n=3 = 13.6/9 = 1.51 eV. Ratio = 13.6/1.51 ≈ 9. Correct answer: 9.
Wave-particle
43. An electron and a proton have the same de Broglie wavelength. The ratio of their kinetic energies KE_e/KE_p is:
- A. mp/me (Correct)
- B. me/mp
- C. 1
- D. (mp/me)²
Explanation: Same λ → same momentum p. KE = p²/(2m). So KE_e/KE_p = mp/me ≈ 1836.
Configuration
44. Which ion has the configuration [Ar] 3d⁵?
- A. Mn²⁺
- B. Fe³⁺
- C. Cr³⁺
- D. Both A and B (Correct)
Explanation: Mn (Z=25): [Ar] 3d⁵ 4s². Mn²⁺: remove 4s² → [Ar] 3d⁵. Fe (Z=26): [Ar] 3d⁶ 4s². Fe³⁺: remove 4s² and one 3d → [Ar] 3d⁵. Both are [Ar] 3d⁵.

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JEE Main Session 1

15Questions

60s/QTimer

+4 / -1Scoring

Preview all 44 questions in JEE Main Session 1 (no login required)

Bohr Model
1. The energy of the electron in the nth orbit of hydrogen atom is:
- A. -13.6/n² eV (Correct)
- B. -13.6n² eV
- C. 13.6/n² eV
- D. -13.6/n eV
Explanation: En = −13.6Z²/n² eV. For hydrogen (Z=1): En = −13.6/n² eV. The negative sign indicates bound state.
Bohr Model
2. The radius of the nth orbit of hydrogen atom is:
- A. 0.529n² Å (Correct)
- B. 0.529/n² Å
- C. 0.529n Å
- D. 0.529 Å only
Explanation: rn = 0.529 × n²/Z Å. For H (Z=1): rn = 0.529n² Å. The Bohr radius (n=1) is 0.529 Å (a₀).
Energy Levels
3. The energy required to remove an electron from the ground state of hydrogen atom is:
- A. 13.6 eV (Correct)
- B. 3.4 eV
- C. 1.51 eV
- D. 0.85 eV
Explanation: Ionisation energy of H = |E₁| = 13.6 eV (to ionise from n=1).
Spectral Lines
4. The spectral lines in the Balmer series of hydrogen correspond to transitions:
- A. n→1
- B. n→2 (Correct)
- C. n→3
- D. n→4
Explanation: Balmer series: transitions from higher levels (n=3,4,5...) to n=2. These fall in visible/UV region.
Rydberg Equation
5. The wavelength of light emitted when hydrogen electron falls from n=4 to n=2 (Rydberg constant R = 1.097×10⁷ m⁻¹) is:
- A. 486 nm (Correct)
- B. 656 nm
- C. 434 nm
- D. 410 nm
Explanation: 1/λ = R(1/4 − 1/16) = R×(3/16). λ = 16/(3R) = 16/(3×1.097×10⁷) ≈ 4.86×10⁻⁷ m = 486 nm (Hβ line, blue-green).
Bohr Model
6. For a hydrogen-like ion He⁺ (Z=2), the energy of n=2 orbit is:
- A. -13.6 eV (Correct)
- B. -6.8 eV
- C. -27.2 eV
- D. -54.4 eV
Explanation: En = −13.6Z²/n² = −13.6×4/4 = −13.6 eV. He⁺ with Z=2, n=2 has the same energy as H n=1.
de Broglie
7. The de Broglie wavelength of a particle with momentum p is:
- A. λ = h/p (Correct)
- B. λ = hp
- C. λ = h/mv²
- D. λ = p/h
Explanation: de Broglie wavelength: λ = h/p = h/(mv). Gives wave character to all matter particles.
Quantum Numbers
8. The maximum number of electrons in the M shell (n=3) is:
- A. 8
- B. 10
- C. 18 (Correct)
- D. 32
Explanation: Max electrons in nth shell = 2n². For n=3: 2×9 = 18. The M shell has 3s, 3p, 3d subshells.
Orbital Quantum Numbers
9. For the 3d subshell, the azimuthal quantum number l is:
- A. 0
- B. 1
- C. 2 (Correct)
- D. 3
Explanation: For d subshell: l = 2. The subshell letter s,p,d,f corresponds to l = 0,1,2,3 respectively.
Magnetic Quantum Number
10. The number of orbitals in a subshell with l = 2 is:
- A. 3
- B. 5 (Correct)
- C. 7
- D. 1
Explanation: Number of orbitals = 2l+1 = 2(2)+1 = 5. The d subshell has 5 orbitals.
Electron Configuration
11. The electronic configuration of Fe (Z=26) is:
- A. [Ar] 3d⁸
- B. [Ar] 3d⁶ 4s² (Correct)
- C. [Ar] 4s² 3d⁶
- D. [Ar] 3d⁵ 4s²4p¹
Explanation: Fe: [Ar] 3d⁶ 4s². Iron has 6 electrons in 3d and 2 in 4s. This is fully compatible with Hund's and Aufbau rules.
Exception
12. The electronic configuration of Cu (Z=29) is:
- A. [Ar] 3d⁹ 4s²
- B. [Ar] 3d¹⁰ 4s¹ (Correct)
- C. [Ar] 3d⁸ 4s²4p¹
- D. [Ar] 4s² 3d⁹
Explanation: Cu adopts [Ar] 3d¹⁰ 4s¹ (not 3d⁹ 4s²) because a completely filled d subshell is extra stable. This is a key exception.
Ionisation Energy
13. The first ionisation energy of Li is 5.4 eV. The energy needed to remove the electron from Li²⁺ (hydrogen-like, Z=3) from n=1 is:
- A. 13.6 eV
- B. 30.6 eV
- C. 122.4 eV (Correct)
- D. 54.4 eV
Explanation: IE of H-like Li²⁺ = 13.6×Z²/n² = 13.6×9/1 = 122.4 eV.
Uncertainty
14. Heisenberg uncertainty principle states Δx·Δp ≥ h/4π. If the uncertainty in position of an electron is 1 Å, the minimum uncertainty in momentum is approximately:
- A. 5.3×10⁻²⁵ kg m/s (Correct)
- B. 5.3×10⁻²⁷ kg m/s
- C. 1.05×10⁻²⁴ kg m/s
- D. h/2π
Explanation: Δp ≥ h/(4π·Δx) = 6.63×10⁻³⁴/(4π×10⁻¹⁰) ≈ 5.3×10⁻²⁵ kg·m/s.
Spectral series
15. The series limit of the Lyman series of hydrogen corresponds to the electron transitioning from:
- A. n=2 to n=1
- B. n=∞ to n=1 (Correct)
- C. n=3 to n=1
- D. n=∞ to n=2
Explanation: The series limit is the shortest wavelength (highest energy) line in a series. For Lyman, n=∞→n=1. IE = 13.6 eV for this limit.
Spin Quantum Number
16. The spin quantum number (ms) can have values:
- A. Only +1/2
- B. Only −1/2
- C. +1/2 or −1/2 (Correct)
- D. 0, +1, −1
Explanation: ms = ±1/2 for electrons. Positive is often called 'spin up' (↑) and negative 'spin down' (↓).
Pauli Exclusion
17. Pauli exclusion principle states that:
- A. No two electrons have same energy
- B. No two electrons in same atom can have identical quantum numbers (Correct)
- C. Electrons fill orbitals from lowest energy
- D. Half-filled subshells are extra stable
Explanation: Pauli: no two electrons in the same atom can have all four quantum numbers (n, l, ml, ms) identical. This limits each orbital to two electrons (with opposite spins).
Hund Rule
18. Hund's rule of maximum multiplicity states that:
- A. Electrons fill from lowest energy first
- B. Orbitals of equal energy are filled singly before pairing (Correct)
- C. No two electrons have same set of quantum numbers
- D. Paired electrons are more stable
Explanation: Hund's rule: in degenerate orbitals, electrons occupy singly with same spin before pairing. This maximises unpaired electrons (maximum multiplicity).
Aufbau Principle
19. In Aufbau filling, orbitals are filled in order of increasing:
- A. Principal quantum number only
- B. Azimuthal quantum number only
- C. n + l value, and lower n first for same n+l (Correct)
- D. Magnetic quantum number
Explanation: Aufbau filling order: subshells with lower (n+l) fill first. For same n+l, lower n fills first. This gives the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, ...
Unpaired electrons
20. The number of unpaired electrons in Fe²⁺ ([Ar] 3d⁶) is:
- A. 2
- B. 4 (Correct)
- C. 5
- D. 6
Explanation: Fe²⁺: 3d⁶. Filling 5 d-orbitals: 5 electrons go in singly (Hund), 6th pairs with one. Unpaired = 4. Magnetic moment = √(n(n+2)) = √24 ≈ 4.9 BM.
Orbital shape
21. The shape of a p-orbital is:
- A. Spherical
- B. Dumbbell (Correct)
- C. Cloverleaf
- D. Toroidal
Explanation: p-orbitals (l=1) have a dumbbell (bi-lobed) shape along one of the three axes (px, py, pz).
Number of spectral lines
22. When an electron falls from n=4 to the ground state in a hydrogen atom, the maximum number of spectral lines emitted is:
- A. 3
- B. 4
- C. 6 (Correct)
- D. 8
Explanation: Maximum spectral lines when electron falls from level n to lower levels = n(n−1)/2. For n=4: 4×3/2 = 6 lines.
Energy calculation
23. The energy of the electron in the third Bohr orbit of He⁺ (Z=2) is:
- A. -6.04 eV (Correct)
- B. -13.6 eV
- C. -3.4 eV
- D. -1.51 eV
Explanation: En = −13.6Z²/n² = −13.6×4/9 = −54.4/9 ≈ −6.04 eV.
Photoelectric effect
24. The maximum kinetic energy of photoelectrons in the photoelectric effect is:
- A. hν
- B. hν − φ (Correct)
- C. hν + φ
- D. φ − hν
Explanation: KE_max = hν − φ, where φ is the work function (minimum energy to eject electron). This is Einstein's photoelectric equation.
de Broglie
25. An electron accelerated through potential V volts has de Broglie wavelength λ = 12.27/√V Å. At V = 100 V, λ is:
- A. 1.227 Å (Correct)
- B. 0.1227 Å
- C. 12.27 Å
- D. 122.7 pm
Explanation: λ = 12.27/√100 = 12.27/10 = 1.227 Å = 122.7 pm. This formula is derived from KE = eV and λ = h/√(2meV).
Orbital nodes
26. The number of radial nodes in the 3s orbital is:
- A. 0
- B. 1
- C. 2 (Correct)
- D. 3
Explanation: Radial nodes = n − l − 1 = 3 − 0 − 1 = 2. The 3s orbital has 2 spherical nodal surfaces.
Angular nodes
27. The number of angular nodes in a d-orbital is:
- A. 0
- B. 1
- C. 2 (Correct)
- D. 3
Explanation: Angular nodes = l. For d subshell, l = 2, so 2 angular (planar/conical) nodes.
Spectral series
28. Which series of hydrogen emission spectrum lies in the UV region?
- A. Lyman (Correct)
- B. Balmer
- C. Paschen
- D. Brackett
Explanation: Lyman series (n→1) emits high-energy photons in the UV region. Balmer (n→2) is visible, Paschen and beyond are infrared.
Bohr velocity
29. The velocity of an electron in the nth orbit of hydrogen is proportional to:
- A. n
- B. 1/n (Correct)
- C. n²
- D. 1/n²
Explanation: From Bohr's quantisation: v = Ze²/(nħ) ∝ 1/n. As n increases, orbital radius grows and speed decreases.
Hydrogen-like
30. The shortest wavelength in the Lyman series of Li²⁺ (Z=3) corresponds to:
- A. 1/λ = 9R × 1
- B. 1/λ = R
- C. 1/λ = 9R (Correct)
- D. 1/λ = R/9
Explanation: Series limit (n→∞ to n=1): 1/λ = RZ²(1/1 − 0) = R×9 = 9R. Wavelength is 1/9 of H Lyman limit.
Orbital penetration
31. For the same principal quantum number n, the order of orbital penetration to nucleus is:
- A. s > p > d > f (Correct)
- B. f > d > p > s
- C. s = p = d = f
- D. d > s > p > f
Explanation: s orbitals have greatest penetration (electron density near nucleus), followed by p, d, f. This determines orbital energy in multi-electron atoms: s fills before p, etc.
Wave mechanical
32. Assertion (A): The exact position and momentum of an electron cannot be simultaneously determined with arbitrary precision. Reason (R): Electrons are waves with inherent uncertainty. Which is correct?
- A. Both A and R are true, R explains A (Correct)
- B. Both true but R does not explain A
- C. A true, R false
- D. A false, R true
Explanation: Heisenberg uncertainty arises because electrons have wave-particle duality. The wave nature introduces inherent uncertainty — measuring position disturbs momentum. R correctly explains A.
Electron filling
33. The element with Z=24 (Cr) has configuration [Ar] 3d⁵ 4s¹ instead of [Ar] 3d⁴ 4s². This is because:
- A. 3d⁵ has a completely filled subshell
- B. 3d⁵ 4s¹ has extra stability due to half-filled 3d (Correct)
- C. 4s electron shielding decreases
- D. The energy difference is negligible
Explanation: Half-filled d subshell (3d⁵ = 5 unpaired electrons symmetrically arranged) has extra stability from exchange energy. This overcomes the normal Aufbau order.
Quantum number validity
34. Which set of quantum numbers is NOT possible?
- A. n=2, l=1, ml=0, ms=+1/2
- B. n=3, l=2, ml=−2, ms=−1/2
- C. n=2, l=2, ml=1, ms=+1/2 (Correct)
- D. n=4, l=3, ml=+3, ms=+1/2
Explanation: l can be 0 to (n−1). For n=2, l can be 0 or 1 only. l=2 is not allowed when n=2. This set is invalid.
Frequency of radiation
35. The frequency of radiation absorbed when H electron moves from n=1 to n=3 is (R=3.29×10¹⁵ Hz):
- A. 2.93×10¹⁵ Hz (Correct)
- B. 2.46×10¹⁵ Hz
- C. 3.29×10¹⁵ Hz
- D. 1.46×10¹⁵ Hz
Explanation: ν = cR(1/1 − 1/9) = R×(8/9) = 3.29×10¹⁵ × 8/9 ≈ 2.93×10¹⁵ Hz.
Photoelectric
36. The threshold frequency for a metal is ν₀. If light of frequency 2ν₀ hits it, the maximum KE of emitted electrons is:
- A. hν₀ (Correct)
- B. 2hν₀
- C. h(2ν₀ − ν₀) = hν₀
- D. 3hν₀
Explanation: KE_max = hν − hν₀ = h(2ν₀) − hν₀ = hν₀.
Number of electrons
37. The maximum number of electrons in a subshell with l = 3 (f subshell) is:
- A. 6
- B. 10
- C. 14 (Correct)
- D. 18
Explanation: Number of orbitals in f subshell = 2l+1 = 7. Each holds 2 electrons max → 14 electrons.
Bohr model accuracy
38. Bohr model fails to explain which of these?
- A. Energy levels of hydrogen
- B. Zeeman splitting of spectral lines (Correct)
- C. Ionisation energy of hydrogen
- D. Rydberg formula
Explanation: Bohr model cannot explain the splitting of spectral lines in a magnetic field (Zeeman effect), fine structure, or spectra of multi-electron atoms. It works only for H-like species.
Basic
39. The charge on one mole of electrons is:
- A. 1.6×10⁻¹⁹ C
- B. 96500 C (Correct)
- C. 6.022×10²³ C
- D. None
Explanation: 1 Faraday = charge on 1 mole of electrons = NA × e = 6.022×10²³ × 1.6×10⁻¹⁹ ≈ 96500 C.
Spectral series
40. In Paschen series, transitions are from n > 3 to n = 3. The region of electromagnetic spectrum is:
- A. UV
- B. Visible
- C. Infrared (Correct)
- D. Microwave
Explanation: Paschen series photons have lower energy (n→3) than Balmer (n→2). They fall in the infrared (IR) region.
Orbital
41. How many total nodes does a 3p orbital have?
- A. 1
- B. 2 (Correct)
- C. 3
- D. 0
Explanation: Total nodes = n − 1 = 2. Angular nodes (planes) = l = 1. Radial nodes = n−l−1 = 1. Total = 1+1 = 2.
Hard
42. The ratio of the kinetic energy of an electron in the 2nd Bohr orbit of He⁺ to that in the 3rd orbit of H is:
- A. 4
- B. 9 (Correct)
- C. 16
- D. 12
Explanation: KE ∝ Z²/n². For He⁺ n=2: Z²/n²=4/4=1. For H n=3: Z²/n²=1/9. Ratio = 1/(1/9) = 9. Wait: KE of He⁺ n=2 = 13.6×4/4 = 13.6 eV; KE of H n=3 = 13.6/9 = 1.51 eV. Ratio = 13.6/1.51 ≈ 9. Correct answer: 9.
Wave-particle
43. An electron and a proton have the same de Broglie wavelength. The ratio of their kinetic energies KE_e/KE_p is:
- A. mp/me (Correct)
- B. me/mp
- C. 1
- D. (mp/me)²
Explanation: Same λ → same momentum p. KE = p²/(2m). So KE_e/KE_p = mp/me ≈ 1836.
Configuration
44. Which ion has the configuration [Ar] 3d⁵?
- A. Mn²⁺
- B. Fe³⁺
- C. Cr³⁺
- D. Both A and B (Correct)
Explanation: Mn (Z=25): [Ar] 3d⁵ 4s². Mn²⁺: remove 4s² → [Ar] 3d⁵. Fe (Z=26): [Ar] 3d⁶ 4s². Fe³⁺: remove 4s² and one 3d → [Ar] 3d⁵. Both are [Ar] 3d⁵.