Chapter Exam

Unit 2: Atomic Structure — Online MCQ Test

Q: Identify a molecule which does not exist.

Answer: D. He2 — According to Molecular Orbital Theory, the bond order of He2 is zero (1/2(2-2) = 0), so it does not exist.

Q: The number of protons, neutrons and electrons in Lu (Lutetium), respectively, are:

Answer: D. 71, 104 and 71 — For Lutetium (Lu, Z=71), mass number is approximately 175. Protons = Z = 71; Electrons = 71; Neutrons = 175 - 71 = 104.

Q: Two electrons occupying the same orbital are distinguished by:

Answer: C. Spin quantum number — According to the Pauli exclusion principle, no two electrons in an atom can have the same four quantum numbers. Since they are in the same orbital (n, l, and ml are the same), they must differ in their spin quantum number (ms).

Q: Given the value of Rydberg constant is 10^7 m^-1, the wave number of the last line of the Balmer series in hydrogen spectrum will be:

Answer: B. 0.25 × 10^7 m^-1 — Wave number nu_bar = R * (1/n1^2 - 1/n2^2). For the last line of Balmer series, n1 = 2 and n2 = infinity. Thus, nu_bar = R * (1/4 - 0) = R/4 = 10^7 / 4 = 0.25 × 10^7 m^-1.

Q: Match List I with List II. List I (Quantum Number) A. ml B. ms C. l D. n List II (Information provided) I. shape of orbital II. size of orbital III. orientation of orbital IV. orientation of spin of electron Choose the correct answer from the options given below:

Answer: D. A-III, B-IV, C-I, D-II — ml gives orientation of orbital, ms gives spin orientation, l gives shape of orbital, and n gives size of orbital.

Q: Two electrons occupying the same orbital are distinguished by:

Answer: D. Spin quantum number — Two electrons occupying the same orbital have the same values for principal (n), azimuthal (l), and magnetic (m_l) quantum numbers, but differ by their spin quantum number (s).

Q: Which of the following series of transitions in the spectrum of hydrogen atom falls in visible region?

Answer: D. Balmer series — In the hydrogen spectrum, the Balmer series corresponds to transitions ending at n=2, which fall within the visible region.

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Unit 2: Atomic Structure — Important Questions & Answers (FAQ)

Frequently asked questions from NEET NEET (All) CHEMISTRY — Unit 2: Atomic Structure, with answers and explanations. These are sample questions; the exam has its own separate question set.

Identify a molecule which does not exist.

A. H2
B. C2
C. O2
D. He2 ✓

Answer: D. He2
According to Molecular Orbital Theory, the bond order of He2 is zero (1/2(2-2) = 0), so it does not exist.

The number of protons, neutrons and electrons in Lu (Lutetium), respectively, are:

A. 104, 71 and 71
B. 71, 71 and 104
C. 175, 104 and 71
D. 71, 104 and 71 ✓

Answer: D. 71, 104 and 71
For Lutetium (Lu, Z=71), mass number is approximately 175. Protons = Z = 71; Electrons = 71; Neutrons = 175 - 71 = 104.

Energy and radius of first Bohr orbit of He+ and Li2+ are [Given RH = 2.18 × 10⁻¹⁸ J, a₀ = 52.9 pm]

A. En(Li²⁺) = –19.62 × 10⁻¹⁸ J; rn(Li²⁺) = 17.6 pm; En(He⁺) = –8.72 × 10⁻¹⁸ J; rn(He⁺) = 26.4 pm ✓
B. En(Li²⁺) = –8.72 × 10⁻¹⁸ J; rn(Li²⁺) = 26.4 pm; En(He⁺) = –19.62 × 10⁻¹⁸ J; rn(He⁺) = 17.6 pm
C. En(Li²⁺) = –19.62 × 10⁻¹⁶ J; rn(Li²⁺) = 17.6 pm; En(He⁺) = –8.72 × 10⁻¹⁶ J; rn(He⁺) = 26.4 pm
D. En(Li²⁺) = –8.72 × 10⁻¹⁶ J; rn(Li²⁺) = 17.6 pm; En(He⁺) = –19.62 × 10⁻¹⁶ J; rn(He⁺) = 17.6 pm

Answer: A. En(Li²⁺) = –19.62 × 10⁻¹⁸ J; rn(Li²⁺) = 17.6 pm; En(He⁺) = –8.72 × 10⁻¹⁸ J; rn(He⁺) = 26.4 pm
Using formulas En = -13.6 × (Z²/n²) eV (converted to Joules) and rn = a₀(n²/Z). For He+ (Z=2), E = -13.6 * 4 * 1.6 * 10^-19 = -8.72 * 10^-18 J. For Li2+ (Z=3), E = -13.6 * 9 * 1.6 * 10^-19 = -19.62 * 10^-18 J.

Two electrons occupying the same orbital are distinguished by:

A. Magnetic quantum number
B. Azimuthal quantum number
C. Spin quantum number ✓
D. Principal quantum number

Answer: C. Spin quantum number
According to the Pauli exclusion principle, no two electrons in an atom can have the same four quantum numbers. Since they are in the same orbital (n, l, and ml are the same), they must differ in their spin quantum number (ms).

Given the value of Rydberg constant is 10^7 m^-1, the wave number of the last line of the Balmer series in hydrogen spectrum will be:

A. 0.5 × 10^7 m^-1
B. 0.25 × 10^7 m^-1 ✓
C. 2.5 × 10^7 m^-1
D. 0.025 × 10^4 m^-1

Answer: B. 0.25 × 10^7 m^-1
Wave number nu_bar = R * (1/n1^2 - 1/n2^2). For the last line of Balmer series, n1 = 2 and n2 = infinity. Thus, nu_bar = R * (1/4 - 0) = R/4 = 10^7 / 4 = 0.25 × 10^7 m^-1.

Match List I with List II. List I (Quantum Number) A. ml B. ms C. l D. n List II (Information provided) I. shape of orbital II. size of orbital III. orientation of orbital IV. orientation of spin of electron Choose the correct answer from the options given below:

A. A-III, B-IV, C-II, D-I
B. A-II, B-I, C-IV, D-III
C. A-I, B-III, C-II, D-IV
D. A-III, B-IV, C-I, D-II ✓

Answer: D. A-III, B-IV, C-I, D-II
ml gives orientation of orbital, ms gives spin orientation, l gives shape of orbital, and n gives size of orbital.

Which one is the wrong statement?

A. The uncertainty principle is Δx * Δp >= h / 4π
B. Half-filled and fully filled orbitals have greater stability due to greater exchange energy, greater symmetry and more balanced arrangement
C. The energy of 2s orbital is less than the energy of 2p orbital in case of Hydrogen like atoms ✓
D. de-Broglie's wavelength is given by λ = h/mv, where m = mass of the particle, v = group velocity of the particle

Answer: C. The energy of 2s orbital is less than the energy of 2p orbital in case of Hydrogen like atoms
In hydrogen and hydrogen-like (single electron) atoms, the energy depends only on the principal quantum number (n), so the energy of 2s and 2p orbitals is the same.

Which one is the wrong statement?

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

Answer: A. The energy of 2s orbital is less than the energy of 2p orbital in case of Hydrogen like atoms
For hydrogen and hydrogen-like (single electron) atoms, the energy depends only on the principal quantum number (n). Therefore, 2s and 2p orbitals have the same energy.

Two electrons occupying the same orbital are distinguished by:

A. Principal quantum number
B. Magnetic quantum number
C. Azimuthal quantum number
D. Spin quantum number ✓

Answer: D. Spin quantum number
Two electrons occupying the same orbital have the same values for principal (n), azimuthal (l), and magnetic (m_l) quantum numbers, but differ by their spin quantum number (s).

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 ✓

Answer: D. Balmer series
In the hydrogen spectrum, the Balmer series corresponds to transitions ending at n=2, which fall within the visible region.

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