Unit 2: Atomic Structure — Online MCQ Test
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Practice Unit 2: Atomic Structure with a free chapter-wise online MCQ test.
This chapter covers: Nature of electromagnetic radiation photoelectric effect; Spectrum of the hydrogen atom. Bohr model of a hydrogen atom - its postulates derivation of the relations for the energy o....
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Unit 2: Atomic Structure — Important Questions & Answers
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.
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.
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.
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).
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).
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).
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).