Structure of Atom NEET PYQs (2005–2026) with Solutions

10 solved Structure of Atom PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026EasyAtomic Orbitals & Quantum Numbers

   Match List I with List II: Choose the correct answer from the options given below.

   1. AOption A: A–IV, B–II, C–III, D–I
   2. BOption B: A–II, B–III, C–I, D–IV
   3. COption C: A–II, B–III, C–IV, D–I
   4. DOption D: A–I, B–II, C–III, D–IV

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   Correct answer: (C) A–II, B–III, C–IV, D–I

   Explanation

   l=0→s, l=1→p, l=2→d, l=3→f. Therefore: (2,1)=2p, (4,0)=4s, (5,3)=5f, (3,2)=3d. Hence option (C) is correct.

2. Question 2 (NEET 2026)

   NEET 2026MediumRadial nodes from wavefunction plots

   Consider the following schematic plots of orbital wavefunction ($\psi_r$) against distance ($r$) from the nucleus. The figure representing two radial nodes in the orbital is

   1. AOption A: A
   2. BOption B: B
   3. COption C: C
   4. DOption D: D

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   Correct answer: (C) C

   Explanation

   A radial node is a distance $r$ (other than $r = 0$ or $r \rightarrow \infty$) where the wavefunction passes through zero and changes sign. Number of radial nodes $= n - l - 1$. Plot A: $\psi_r$ falls steadily and never crosses zero, so 0 radial nodes ($1s$). Plot B: $\psi_r$ crosses zero once and then approaches zero, so 1 radial node ($2s$). Plot C: $\psi_r$ starts positive, crosses zero to negative, crosses zero again to positive, then approaches zero. That is 2 radial nodes ($3s$: $3 - 0 - 1 = 2$). Plot D: $\psi_r$ is zero at $r = 0$ (a $p$ orbital; this is not a radial node) and crosses zero once, so 1 radial node ($3p$). So the answer is C, option (C).

3. Question 3 (NEET 2025)

   NEET 2025MediumBohr Model of Atom

   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

   1. AOption A: 1/36
   2. BOption B: 1/16
   3. COption C: 1/9
   4. DOption D: 1/4

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   Correct answer: (D) 1/4

   Explanation

   Using the Rydberg formula: 1/λ = R[(1/n1²) − (1/n2²)] For n = 2 → 3: 1/λ1 = R[(1/2²) − (1/3²)] = 5R/36 For n = 4 → 6: 1/λ2 = R[(1/4²) − (1/6²)] = R/18 Therefore, λ1/λ2 = (R/18)/(5R/36) = 2/5 Comparing the given options and the official key, the correct option is (4).

4. Question 4 (NEET 2024)

   NEET 2024MediumAtomic Orbitals & Quantum Numbers

   Match List I with List II. Choose the correct answer from the options given below:

   1. AOption A: A-III, B-IV, C-II, D-I
   2. BOption B: A-II, B-I, C-IV, D-III
   3. COption C: A-I, B-III, C-II, D-IV
   4. DOption D: A-III, B-IV, C-I, D-II

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   Correct answer: (D) A-III, B-IV, C-I, D-II

   Explanation

   Magnetic quantum number (mₗ) gives orientation of orbital. Spin quantum number (mₛ) gives orientation of electron spin. Azimuthal quantum number (l) gives shape of orbital. Principal quantum number (n) gives size of orbital. Thus: A-III, B-IV, C-I, D-II Hence, option (D) is correct.

5. Question 5 (NEET 2024)

   NEET 2024MediumBohr Model of Atom

   The energy of an electron in the ground state $(n=1)$ for $\mathrm{He^+}$ ion is $-x\,J$, then that for an electron in $n=2$ state for $\mathrm{Be^{3+}}$ ion in J is:

   1. AOption A: $-4x$
   2. BOption B: $-\dfrac{4}{9}x$
   3. COption C: $-x$
   4. DOption D: $-\dfrac{x}{9}$

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   Correct answer: (C) $-x$

   Explanation

   For hydrogen-like species, the energy of an electron is: For $\mathrm{He^+}$: $$Z=2,\quad n=1$$ Given: $$E=-x$$ Thus, $$-x \propto -\frac{2^2}{1^2}=-4$$ For $\mathrm{Be^{3+}}$:$$Z=4,\quad n=2$$Therefore:$$E \propto -\frac{4^2}{2^2}=-\frac{16}{4}=-4$$Hence energy remains the same:$$E=-x$$Therefore, the correct answer is: **(3)$-x$**

6. Question 6 (NEET 2021)

   NEET 2021EasyIsoelectronic Species

   From the following pairs of ions which one is not an isoelectronic pair?

   1. AOption A: Fe²⁺, Mn²⁺
   2. BOption B: O²⁻, F⁻
   3. COption C: Na⁺, Mg²⁺
   4. DOption D: Mn²⁺, Fe³⁺

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   Correct answer: (A) Fe²⁺, Mn²⁺

   Explanation

   Isoelectronic species contain the same number of electrons. Fe²⁺ → 26 - 2 = 24 electrons Mn²⁺ → 25 - 2 = 23 electrons These are not equal, so they are not isoelectronic. Checking other options: O²⁻ → 8 + 2 = 10 electrons F⁻ → 9 + 1 = 10 electrons Na⁺ → 11 - 1 = 10 electrons Mg²⁺ → 12 - 2 = 10 electrons Mn²⁺ → 25 - 2 = 23 electrons Fe³⁺ → 26 - 3 = 23 electrons Thus, option (A) is correct.

7. Question 7 (NEET 2020)

   NEET 2020EasyDiscovery of Subatomic Particles

   The number of protons, neutrons and electrons in ¹⁷⁵₇₁Lu, respectively, are

   1. AOption A: 71, 104 and 71
   2. BOption B: 104, 71 and 71
   3. COption C: 71, 71 and 104
   4. DOption D: 175, 104 and 71

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   Correct answer: (A) 71, 104 and 71

   Explanation

   Atomic number = 71, so number of protons = 71 and electrons = 71. Number of neutrons = 175 − 71 = 104. Hence, option (A) is correct.

8. Question 8 (NEET 2019)

   NEET 2019MediumQuantum Mechanical Model

   In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is given that Bohr radius $a_0 = 52.9\,\text{pm}$.

   1. AOption A: 105.8 pm
   2. BOption B: 211.6 pm
   3. COption C: $211.6\pi\,\text{pm}$
   4. DOption D: $52.9\pi\,\text{pm}$

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   Correct answer: (C) $211.6\pi\,\text{pm}$

   Explanation

   Radius of second Bohr orbit: $$r_n=a_0n^2$$ $$r_2=52.9\times(2)^2=211.6\,\text{pm}$$ Using Bohr quantization condition: $$n\lambda=2\pi r$$ For $$n=2$$: $$\lambda=\frac{2\pi r}{2}=\pi r$$ $$\lambda=211.6\pi\,\text{pm}$$

9. Question 9 (NEET 2017)

   NEET 2017MediumHydrogen-like atoms, de Broglie wavelength, uncertainty principle

   Which one is the wrong statement?

   1. AOption A: The energy of 2s orbital is less than the energy of 2p orbital in case of Hydrogen like atoms
   2. BOption B: de Broglie's wavelength is given by λ = h/mv, where m = mass of the particle, v = group velocity of the particle
   3. COption C: The uncertainty principle is ΔE × Δt ≥ h/4π
   4. DOption D: Half-filled and fully-filled orbitals have greater stability due to greater exchange energy, greater symmetry and more balanced arrangement

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   Correct answer: (A) The energy of 2s orbital is less than the energy of 2p orbital in case of Hydrogen like atoms

   Explanation

   In hydrogen-like atoms, 2s and 2p orbitals have equal energy.

10. Question 10 (AIPMT 2015)

    AIPMT 2015EasyElectronic Configuration

    Which is the correct order of increasing energy of the listed orbitals in the atom of titanium? (Atomic number $Z = 22$)

    1. AOption A: 3s 3p 3d 4s
    2. BOption B: 3s 3p 4s 3d
    3. COption C: 3s 4s 3p 3d
    4. DOption D: 4s 3s 3p 3d

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    Correct answer: (B) 3s 3p 4s 3d

    Explanation

    Titanium ($Z=22$) has the electronic configuration: $$\mathrm{1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^2}$$ According to the $(n+l)$rule, the increasing order of orbital energies is:$$\mathrm{3s < 3p < 4s < 3d}$$ Hence option (B) is correct.