NEET Structure of Atom PYQs & Practice Questions

10 free NEET Chemistry questions on Structure of Atom, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Bohr Model - Energy and Radius · Thomson's Model and Atomic Discovery · Rutherford's Nuclear Model · Bohr's Model of Atom · Quantum Numbers · Photoelectric Effect and Dual Nature · Electronic Configuration and Orbitals · de Broglie's Hypothesis · Heisenberg's Uncertainty Principle · Rydberg Formula / Hydrogen Spectrum · Rutherford's Model · Electronic Configuration

Q1NEET 2026Radial Wavefunction and Radial Nodes
Consider schematic plots of orbital wavefunction (ψr) against distance (r) from the nucleus for four orbitals A, B, C and D. Plot A shows a smooth monotonic decay from a positive value to zero with no crossing of the axis. Plot B shows the curve dipping below zero once before returning to a small positive plateau. Plot C shows the curve dipping below zero, crossing back above zero, then dipping below zero a second time before settling to a small positive plateau. Plot D shows a similar multi-crossing pattern. The figure representing two radial nodes in the orbital is:
- A. B
- B. C
- C. D
- D. A
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Answer: A — B
The number of radial nodes equals the number of times the radial wavefunction crosses zero (changes sign) as r increases from 0 to infinity, excluding r=0 and r=infinity. Plot A has zero crossings (0 radial nodes). Plot B crosses zero once (1 radial node). Plot C crosses zero twice (2 radial nodes), matching the question. Plot D would represent 3 radial nodes.
Q2NEET 2025Rydberg Formula / Hydrogen Spectrum
The ratio of the wavelengths of the light absorbed by a hydrogen atom when it undergoes n=2→n=3 and n=4→n=6 transitions, respectively, is
- A. 1/36
- B. 1/16
- C. 1/9
- D. 1/4
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Answer: C — 1/9
Using Rydberg formula 1/λ = RZ²[1/n1²−1/n2²]: for 2→3, 1/λ1 = R[1/4−1/9] = R(5/36). For 4→6, 1/λ2 = R[1/16−1/36] = R(5/144). λ1/λ2 = (5/144)/(5/36) = 36/144 = 1/4.
Q3NEET 2025Bohr Model - Energy and Radius
Energy and radius of first Bohr orbit of He+ and Li2+ are [Given RH = 2.18×10⁻¹⁸J, a0 = 52.9 pm]
- A. En(Li2+)=−19.62×10¹⁸J; rn(Li2+)=17.6pm; En(He+)=−8.72×10⁻¹⁸J; rn(He+)=26.4pm
- B. En(Li2+)=−8.72×10⁻¹⁸J; rn(Li2+)=26.4pm; En(He+)=−19.62×10⁻¹⁸J; rn(He+)=17.6pm
- C. En(Li2+)=−19.62×10⁻¹⁶J; rn(Li2+)=17.6pm; En(He+)=−8.72×10⁻¹⁶J; rn(He+)=26.4pm
- D. En(Li2+)=−8.72×10⁻¹⁶J; rn(Li2+)=17.6pm; En(He+)=−19.62×10⁻¹⁶J; rn(He+)=17.6pm
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Answer: —
E = −E0×Z²/n². For Li2+ (Z=3,n=1): E = −2.18×10⁻¹⁸×9 = −19.62×10⁻¹⁸ J. For He+ (Z=2,n=1): E = −2.18×10⁻¹⁸×4 = −8.72×10⁻¹⁸ J. r = r0×n²/Z. For Li2+: r = 52.9/3 = 17.6 pm. For He+: r = 52.9/2 = 26.45 pm.
Q4NEET 2023Quantum Numbers
The relation between n_m (n_m = the number of permissible values of magnetic quantum number m) for a given value of azimuthal quantum number (l), is:
- A. l = 2n_m + 1
- B. n_m = 2l^2 + 1
- C. n_m = l + 2
- D. l = (n_m - 1)/2
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Answer: C — n_m = l + 2
For a given azimuthal quantum number l, the magnetic quantum number m takes values from -l to +l, giving a total of (2l+1) permissible values. So n_m = 2l+1, which rearranges to l = (n_m - 1)/2.

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Q5NEET 2022Shapes of Orbitals
Identify the incorrect statement from the following.
- A. All the five 4d orbitals have shapes similar to the respective 3d orbitals
- B. In an atom, all the five 3d orbitals are equal in energy free state.
- C. The shapes of dxy, dyz, and dzx orbitals are similar to each other; and dx2-y2 and dz2 are similar to each other.
- D. All the five 5d orbitals are different in size when compared to the respective 4d orbitals
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Answer: B — In an atom, all the five 3d orbitals are equal in energy free state.
The dxy, dyz, dzx orbitals have similar double-dumbbell shapes to each other, but dx2-y2 and dz2 do NOT have similar shapes to each other - dz2 has a unique shape with a donut-shaped ring around it, different from the double-dumbbell shape of dx2-y2. So statement 3 is incorrect.
Q6NEET 2022Shapes and Sizes of d Orbitals
Identify the incorrect statement from the following: (1) All the five 5d orbitals are different in size when compared to the respective 4d orbitals. (2) All the five 4d orbitals have shapes similar to the respective 3d orbitals. (3) In an atom, all the five 3d orbitals are equal in energy in free state. (4) The shapes of dxy, dyz and dzx orbitals are similar to each other; and dx2-y2 and dz2 are similar to each other.
- A. All the five 5d orbitals are different in size when compared to the respective 4d orbitals.
- B. All the five 4d orbitals have shapes similar to the respective 3d orbitals.
- C. In an atom, all the five 3d orbitals are equal in energy in free state.
- D. The shapes of dxy, dyz and dzx orbitals are similar to each other; and dx2-y2 and dz2 are similar to each other.
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Answer: C — In an atom, all the five 3d orbitals are equal in energy in free state.
While dxy, dyz, and dzx orbitals do have similar shapes to each other (all 4-lobed, lying between axes), dx2-y2 and dz2 do NOT have similar shapes to each other: dx2-y2 has 4 lobes along the axes, while dz2 has a unique dumbbell-with-torus shape. So statement 4 is incorrect.
Q7NEET 2022Bohr Model - Radius Calculation
If radius of second Bohr orbit of the He+ ion is 105.8 pm, what is the radius of third Bohr orbit of Li2+ ion?
- A. 158.7 pm
- B. 15.87 pm
- C. 1.587 pm
- D. 158.7 Angstrom
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Answer: —
Bohr radius r = (0.529 x n^2/Z) Angstrom. For He+ (Z=2, n=2): r = 105.8 pm (given). For Li2+ (Z=3, n=3): r_Li/r_He = (n_Li^2/Z_Li)/(n_He^2/Z_He) = (9/3)/(4/2) = 3/2 = 1.5. r_Li = 105.8 x 1.5 = 158.7 pm.
Q8NEET 2020Subatomic particle composition of an isotope
The number of protons, neutrons and electrons in 175-71-Lu, respectively, are:
- A. 71, 104 and 71
- B. 104, 71 and 71
- C. 71, 71 and 104
- D. 175, 104 and 71
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Answer: —
For an atom with mass number 175 and atomic number 71 (Lutetium): protons = atomic number = 71, electrons = 71 (neutral atom), neutrons = mass number - atomic number = 175-71 = 104. So protons=71, neutrons=104, electrons=71.

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Q9NEET 2019Hydrogen Spectral Series
Which of the following series of transitions in the spectrum of hydrogen atom falls in visible region?
- A. Paschen series
- B. Brackett series
- C. Lyman series
- D. Balmer series
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Answer: C — Lyman series
The Balmer series, resulting from electron transitions to the n=2 energy level, falls in the visible region of the electromagnetic spectrum. Lyman series falls in UV, while Paschen, Brackett, and Pfund series fall in the infrared region.
Q10NEET 2019Energy Ordering of Orbitals (n+l rule)
4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is:
- A. 6p > 5f > 4d > 5p
- B. 5f > 6p > 4d > 5p
- C. 5f > 6p > 5p > 4d
- D. 6p > 5f > 5p > 4d
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Answer: B — 5f > 6p > 4d > 5p
Using the (n+l) rule: 5f has (n+l)=8, 6p has (n+l)=7, 5p has (n+l)=6, 4d has (n+l)=6. Among 5p and 4d with same (n+l)=6, lower n (4d) has lower energy. So decreasing energy order is 5f > 6p > 5p > 4d.

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