Structure of Atom

10 NEET-UG practice questions on Structure of Atom, part of Chemistry. The first 10 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Easy

   How many unpaired electrons does a gaseous Fe³⁺ ion (Z = 26) have in its ground state?

   1. A5
   2. B4
   3. C6
   4. D3

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   Answer: A · 5

   Iron is [Ar] 3d⁶ 4s², and the 4s electrons leave first, so Fe³⁺ is [Ar] 3d⁵. By Hund's rule, the five 3d electrons occupy five orbitals singly.

   • B: This is the count for Fe²⁺, which is 3d⁶.
   • C: This counts the six 3d electrons of the neutral atom as unpaired.
   • D: This removes all three electrons from 3d, leaving 3d³ 4s².

   Remember this

   Transition metal ions lose 4s electrons before 3d, and Hund's rule then fixes the number of unpaired electrons.

2. Question 2 · Hard

   The first Balmer line of hydrogen comes from the transition n = 3 → n = 2. Which transition in Li²⁺ (Z = 3) emits light of the same wavelength? (Use 1/λ ∝ Z²(1/n₁² − 1/n₂²).)

   1. An = 6 → n = 4
   2. Bn = 27 → n = 18
   3. Cn = 9 → n = 6
   4. Dn = 3 → n = 2

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   Answer: C · n = 9 → n = 6

   For the same wavelength, Z²(1/n₁² − 1/n₂²) must match, and Z² = 9 is cancelled when each n is multiplied by Z = 3. So 9(1/36 − 1/81) = 1/4 − 1/9, and the transition is n = 9 → n = 6.

   • A: This scales n by 2, which is the He⁺ answer with Z = 2.
   • B: This scales n by Z² = 9 instead of by Z = 3.
   • D: This leaves n unchanged and ignores the Z² factor.

   Remember this

   A hydrogen line reappears in a one-electron ion when both n values are multiplied by Z.

3. Question 3 · Easy

   A 3p orbital has how many radial nodes?

   1. A3
   2. B1
   3. C2
   4. D0

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   Answer: B · 1

   Radial nodes = n − l − 1 = 3 − 1 − 1 = 1. A p orbital also has one angular node, which gives two nodes in all.

   • A: This takes the principal quantum number as the node count.
   • C: This is the total number of nodes, radial plus angular.
   • D: This assumes a p orbital has only an angular node.

   Remember this

   Radial nodes = n − l − 1, angular nodes = l, total nodes = n − 1.

4. Question 4 · Medium

   Particles X and Y have masses in the ratio 1 : 4. The uncertainties in their velocities are in the ratio 2 : 1. In what ratio are the minimum uncertainties in their positions, Δx(X) : Δx(Y)?

   1. A1 : 2
   2. B8 : 1
   3. C2 : 1
   4. D1 : 8

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   Answer: C · 2 : 1

   From Heisenberg, Δx × m × Δv ≥ h/4π, so the minimum Δx is proportional to 1/(mΔv). The values are 1/(1 × 2) = 1/2 for X and 1/(4 × 1) = 1/4 for Y, a ratio of 2 : 1.

   • A: This compares the products mΔv directly, without taking the inverse.
   • B: This inverts the mass ratio but not the velocity uncertainty ratio.
   • D: This inverts the velocity uncertainty ratio but not the mass ratio.

   Remember this

   Minimum Δx varies inversely with the product of mass and Δv.

5. Question 5 · Medium

   Arrange the following orbitals of a multi-electron atom in increasing order of energy. A. 4p B. 3d C. 4s D. 3p

   1. AD, C, B, A
   2. BD, C, A, B
   3. CD, B, C, A
   4. DC, D, B, A

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   Answer: A · D, C, B, A

   The n + l values are 4p = 5, 3d = 5, 4s = 4 and 3p = 4. For equal n + l, the orbital with lower n has lower energy, so 3p < 4s < 3d < 4p.

   • B: This places 4p below 3d, ignoring the lower n rule at equal n + l.
   • C: This places 3d below 4s, using n alone.
   • D: This places 4s below 3p, ignoring the lower n rule at equal n + l.

   Remember this

   The (n + l) rule: lower sum fills first; for the same sum, lower n fills first.

6. Question 6 · Medium

   What is the de Broglie wavelength of a ball of mass 0.1 kg moving at 33 m/s? (Take h = 6.6 × 10⁻³⁴ J s)

   1. A2 × 10⁻³⁷ m
   2. B2 × 10⁻³⁴ m
   3. C6.6 × 10⁻³³ m
   4. D2 × 10⁻³³ m

   Answer and explanation in LearnFree account

7. Question 7 · Medium

   Which sets of quantum numbers are allowed for an electron? A. n = 3, l = 2, m = −2, s = +1/2 B. n = 2, l = 2, m = 0, s = −1/2 C. n = 4, l = 0, m = 0, s = −1/2 D. n = 3, l = 1, m = +2, s = +1/2

   1. AA and C only
   2. BA and D only
   3. CB and C only
   4. DA, C and D only

   Answer and explanation in LearnFree account

8. Question 8 · Hard

   The energy levels of a one-electron species are drawn in the figure; the lowest level is its ground state. Using the Bohr model, with Bohr radius 52.9 pm and hydrogen ground-state energy −13.6 eV, find the radius of the orbit for level X.

   

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   1. A238 pm
   2. B159 pm
   3. C79 pm
   4. D476 pm

   Answer and explanation in LearnFree account

9. Question 9 · Easy

   Four emission transitions of the hydrogen atom are marked in the figure. Match List-I with List-II. List-I (transition) A. Arrow A B. Arrow B C. Arrow C D. Arrow D List-II (spectral series) I. Lyman II. Balmer III. Paschen IV. Brackett

   

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   1. AA-III, B-II, C-IV, D-I
   2. BA-II, B-I, C-III, D-IV
   3. CA-III, B-I, C-IV, D-II
   4. DA-IV, B-I, C-III, D-II

   Answer and explanation in LearnFree account

10. Question 10 · Medium

    A hydrogen orbital with azimuthal quantum number l = 1 has the radial probability curve shown. Which orbital is it?

    

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    1. A2p
    2. B4p
    3. C3p
    4. D5p

    Answer and explanation in LearnFree account