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NEET Atomic Structure MCQs with answers

561 questions on Atomic Structure for NEET 2027. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easy

1. The maximum number of orbitals in the shell with principal quantum number n = 4 is:

  1. 4
  2. 32
  3. 8
  4. 16
Show answer and explanation

Answer: (D) 16

A shell with principal quantum number n has n² orbitals: 4² = 16.

medium

2. What is the total number of electrons with orbital angular momentum quantum number l = 1 in the ground state of a neutral Copper atom (Z = 29)?

  1. 6
  2. 12
  3. 10
  4. 18
Show answer and explanation

Answer: (B) 12

The ground-state configuration of Cu (Z = 29) is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹. Electrons with azimuthal quantum number l = 1 are p-electrons. Cu contains electrons in 2p (6 electrons) and 3p (6 electrons), giving a total of 6 + 6 = 12 electrons.

hard

3. A particle of mass m is confined in a one-dimensional box of length L. Using Heisenberg's uncertainty principle (taking Δx ≈ L), the estimated minimum zero-point kinetic energy of the particle is of the order of:

  1. h² L² / (8m)
  2. h² / (8m L)
  3. h / (4π m L)
  4. h² / (32π² m L²)
Show answer and explanation

Answer: (D) h² / (32π² m L²)

With Δx ≈ L, the minimum uncertainty in momentum is Δp ≈ h / (4πL). Since the momentum cannot be less than its uncertainty, p_min ≈ h / (4πL). The minimum kinetic energy is K_min = p_min² / (2m) = h² / (32π² m L²).

medium

4. For a hydrogen atom, let ΔE₁ be the energy emitted during a transition from n = 2 to n = 1, and ΔE₂ be the energy emitted during a transition from n = 3 to n = 2. What is the ratio ΔE₁ : ΔE₂?

  1. 27 : 5
  2. 5 : 27
  3. 4 : 9
  4. 9 : 4
Show answer and explanation

Answer: (A) 27 : 5

ΔE₁ = 13.6 · (1/1² - 1/2²) = 13.6 · (3/4) eV. ΔE₂ = 13.6 · (1/2² - 1/3²) = 13.6 · (5/36) eV. Ratio ΔE₁ / ΔE₂ = (3/4) / (5/36) = (3/4) · (36/5) = 27/5.

easy

5. The correct ground-state electronic configuration of the dipositive cation of cobalt, Co²⁺ (Z = 27), is:

  1. [Ar] 3d⁵ 4s²
  2. [Ar] 3d⁷
  3. [Ar] 3d⁶ 4s¹
  4. [Ar] 3d⁸ 4s¹
Show answer and explanation

Answer: (B) [Ar] 3d⁷

Neutral cobalt has the configuration [Ar] 3d⁷ 4s². During ionization, electrons are first removed from the outermost 4s orbital before the 3d orbital, giving Co²⁺ the configuration [Ar] 3d⁷.

hard

6. An electron is present in an orbital that has 3 radial nodes and 2 angular nodes. What is the minimum possible principal quantum number (n) of this electron, and which subshell does it belong to?

  1. n = 6, 6d
  2. n = 5, 5d
  3. n = 5, 5p
  4. n = 6, 6f
Show answer and explanation

Answer: (A) n = 6, 6d

The number of angular nodes is l = 2, which corresponds to a d subshell. The number of radial nodes is n - l - 1 = 3. Substituting l = 2 gives n - 2 - 1 = 3, which yields n = 6. Thus, the orbital is 6d.

medium

7. If the kinetic energy of an electron in a certain Bohr orbit of a hydrogen atom is +0.85 eV, what is the orbital angular momentum of the electron in that state?

  1. h / π
  2. 2h / π
  3. 3h / (2π)
  4. 5h / (2π)
Show answer and explanation

Answer: (B) 2h / π

The total energy is E = -KE = -0.85 eV. For hydrogen, Eₙ = -13.6 / n² eV. Setting -13.6 / n² = -0.85 gives n² = 16, so n = 4. The orbital angular momentum is L = nh / (2π) = 4h / (2π) = 2h / π.

medium

8. For the filling of orbitals in a multielectron neutral atom, consider the following orbitals: 4d, 5p, 5f, and 6p. Which of the following lists them in order of increasing energy according to the (n + l) rule?

  1. 4d < 5p < 5f < 6p
  2. 4d < 5p < 6p < 5f
  3. 5p < 4d < 6p < 5f
  4. 4d < 6p < 5p < 5f
Show answer and explanation

Answer: (B) 4d < 5p < 6p < 5f

Calculate (n + l) values: for 4d, n + l = 4 + 2 = 6; for 5p, n + l = 5 + 1 = 6; for 6p, n + l = 6 + 1 = 7; for 5f, n + l = 5 + 3 = 8. Between 4d and 5p (both have n + l = 6), 4d has smaller n so lower energy (4d < 5p). Next is 6p (n + l = 7), followed by 5f (n + l = 8). Hence, the order is 4d < 5p < 6p < 5f.

medium

9. A monochromatic light source of power 33 W emits 1.0 × 10²⁰ photons per second. What is the wavelength of the emitted radiation? (Given: h = 6.6 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m s⁻¹)

  1. 400 nm
  2. 600 nm
  3. 500 nm
  4. 300 nm
Show answer and explanation

Answer: (B) 600 nm

Power P = n × (hc/λ), where n is the number of photons emitted per second. Substituting the given values: 33 = 1.0 × 10²⁰ × (6.6 × 10⁻³⁴ × 3.0 × 10⁸) / λ. This gives 33 = (1.98 × 10⁻⁵) / λ, so λ = 1.98 × 10⁻⁵ / 33 = 6.0 × 10⁻⁷ m = 600 nm.

easy

10. According to Bohr's theory, the relationship between the principal quantum number (n) and the velocity of an electron (v) in a hydrogen-like species with atomic number Z is:

  1. v ∝ n / Z
  2. v ∝ Z / n²
  3. v ∝ Z / n
  4. v ∝ Z² / n
Show answer and explanation

Answer: (C) v ∝ Z / n

The orbital velocity of an electron in Bohr's model is given by v = (2.18 × 10⁶) × (Z / n) m/s. Thus, velocity is directly proportional to Z and inversely proportional to n.

medium

11. In a hydrogen atom, the potential energy of an electron in an excited state is -1.70 eV. What is the value of the principal quantum number (n) for this orbit?

  1. 2
  2. 3
  3. 5
  4. 4
Show answer and explanation

Answer: (D) 4

Total energy E_n = Potential Energy / 2 = -1.70 eV / 2 = -0.85 eV. Using E_n = -13.6 / n² eV, we get -0.85 = -13.6 / n², which gives n² = 13.6 / 0.85 = 16, so n = 4.

medium

12. Which of the following graphs of the radial wave function R(r) versus r correctly represents a 2s orbital of a hydrogen atom?

  1. Starts at zero, crosses the r-axis once, and approaches zero as r increases
  2. Starts at zero, rises to a single maximum, and decays asymptotically to zero
  3. Starts at a positive finite value and decays monotonically to zero without intersecting the r-axis
  4. Starts at a positive finite value, crosses the r-axis once, and approaches zero from negative values as r increases
Show answer and explanation

Answer: (D) Starts at a positive finite value, crosses the r-axis once, and approaches zero from negative values as r increases

For an s orbital, R(0) is finite and non-zero. Since a 2s orbital has (n - l - 1) = 2 - 0 - 1 = 1 radial node, R(r) must cross the horizontal r-axis exactly once, transitioning from positive to negative values before asymptotically reaching zero.

easy

13. The ratio of the radius of the first orbit of H to the radius of the second orbit of Be³⁺ is:

  1. 1 : 4
  2. 1 : 2
  3. 2 : 1
  4. 1 : 1
Show answer and explanation

Answer: (D) 1 : 1

Radius of Bohr orbit is given by r ∝ n² / Z. For H (n = 1, Z = 1), r₁ = 1²/1 = 1. For Be³⁺ (n = 2, Z = 4), r₂ = 2²/4 = 1. Therefore, the ratio r(H, 1) : r(Be³⁺, 2) is 1 : 1.

medium

14. The uncertainty principle has negligible significance for a cricket ball but is important for an electron because:

  1. h is larger for electrons
  2. The electron's mass is extremely small
  3. The cricket ball moves too slowly
  4. The electron is always at rest
Show answer and explanation

Answer: (B) The electron's mass is extremely small

For macroscopic objects the uncertainties are vanishingly small. For an electron, small mass makes Δv very large, ruling out precise orbits.

NCERT Class 11

easy

15. An electron has a kinetic energy E and a de Broglie wavelength λ. If its de Broglie wavelength is changed to 3λ, the new kinetic energy of the electron must be:

  1. 9E
  2. E / 9
  3. E / 3
  4. 3E
Show answer and explanation

Answer: (B) E / 9

The de Broglie wavelength is related to kinetic energy by λ = h / √(2mE), so E = h² / (2mλ²). When the wavelength becomes 3λ, the kinetic energy becomes E' = h² / [2m(3λ)²] = E / 9.

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