Structure of Atom is a Class 11 Chemistry chapter that carries about 3.2% of the Chemistry questions in NEET. Practise it with 42 questions covering 2 topics, each with an explanation.
Sample questions on Structure of Atom with answers
8 questions from the practice bank, spread across the chapter's topics. The full set is in the app.
Question 1: Bohr's Model of the Atom
The wavenumber of the first line of the Balmer series of hydrogen is (R = 1.097 × 10⁷ m⁻¹):
- A 8.2 × 10⁶ m⁻¹
- B 1.097 × 10⁷ m⁻¹
- C 1.52 × 10⁶ m⁻¹
- D 2.74 × 10⁶ m⁻¹
Answer: C. 1.52 × 10⁶ m⁻¹
Why: ν̄ = R(1/4 − 1/9) = R × 5/36.
Question 2: Quantum Numbers
The number of radial nodes in a 3p orbital is:
- A 0
- B 2
- C 1
- D 3
Answer: C. 1
Why: Radial nodes = n − l − 1 = 3 − 1 − 1 = 1.
Question 3: Bohr's Model of the Atom
The radius of the second Bohr orbit of hydrogen is about (a₀ = 52.9 pm):
- A 106 pm
- B 476 pm
- C 212 pm
- D 53 pm
Answer: C. 212 pm
Why: rₙ = n² a₀ = 4 × 52.9 pm.
Question 4: Quantum Numbers
According to the Aufbau principle, electrons are filled into orbitals in order of:
- A decreasing energy
- B increasing energy
- C random order
- D alphabetical order of orbital names
Answer: B. increasing energy
Why: The Aufbau principle states that electrons occupy the lowest-energy orbitals available first, filling orbitals in order of increasing energy.
Question 5: Bohr's Model of the Atom
The wavelength of electromagnetic radiation of frequency 5 × 10¹⁴ Hz is (c = 3 × 10⁸ m/s):
- A 6 × 10⁻⁷ m
- B 6 × 10⁻⁹ m
- C 1.5 × 10²³ m
- D 1.67 × 10⁶ m
Answer: A. 6 × 10⁻⁷ m
Why: λ = c/ν = 3 × 10⁸/5 × 10¹⁴ = 6 × 10⁻⁷ m (600 nm).
Question 6: Quantum Numbers
The total number of electrons that can occupy all the d orbitals of a given shell is:
- A 10
- B 5
- C 14
- D 6
Answer: A. 10
Why: There are five d orbitals and each holds two electrons.
Question 7: Bohr's Model of the Atom
The energy of the electron in the n = 2 orbit of the hydrogen atom is (E₁ = −2.18 × 10⁻¹⁸ J):
- A −1.09 × 10⁻¹⁸ J
- B −8.72 × 10⁻¹⁸ J
- C −2.18 × 10⁻¹⁸ J
- D −5.45 × 10⁻¹⁹ J
Answer: D. −5.45 × 10⁻¹⁹ J
Why: Eₙ = E₁/n² = −2.18 × 10⁻¹⁸/4.
Question 8: Quantum Numbers
In a multi-electron atom, the 4s orbital is filled before the 3d orbital because:
- A 4s is nearer the nucleus
- B the 4s has a lower energy by the (n + l) rule
- C 3d has a higher principal quantum number
- D 4s can hold more electrons
Answer: B. the 4s has a lower energy by the (n + l) rule
Why: The (n + l) values are 4 for 4s and 5 for 3d.