Fifth Chapter Solved Exercise MCQs of FSC First Year Chemistry
written by Almas AnwarThe 5th chapter Solved Exercise MCQs of FSC 1st-year Chemistry offers a comprehensive collection of multiple-choice questions with answers. Each question is accompanied by a short explanation to clarify the correct choice. These solutions aim to assist students in exam preparation by simplifying complex concepts and building a strong foundation for success.
Fifth chapter solved MCQs with explanation
1. The nature of the positive rays depends on:
(a) the nature of the electrode
(b) the nature of the discharge tube
(c) the nature of the residual gas
(d) all of the above
Explanation: The correct answer is (c).Because Positive rays are formed by ions of the gas, so their characteristics vary with different gases. The electrode and tube design affect the experiment but not the rays’ nature directly.
2. The velocity of photon is:
(a) independent of its wavelength
(b) depends on its wavelength
(c) equal to square of its amplitude
(d) depends on its source
Explanation: The correct answer is (b).
3. The wave number of the light emitted by a certain source is 2 x 10 m¹. The wavelength of this light will be:
(a) 500 nm
(b) 500 m
(c) 200 nm
(d) 5×10 m
Explanation: The correct answer is (c).
- Understanding the Relationship: The wave number (in meters to the power of -1) is the number of wavelengths per meter. To find the wavelength (λ), use this formula: λ = 1 / wave number
- Calculate the Wavelength:
- Given wave number = 2 × 10⁷ m⁻¹
- Wavelength (λ) = 1 / 2×10⁷ m⁻¹
- Convert to Nanometers:
- The result in meters = 5×10⁻⁸ m
- Convert meters to nanometers: 1 m = 10⁹ nm
- So, 5×10⁻⁸ m = 5×10⁻⁸ × 10⁹ nm = 50 nm
- Choose the Closest Option: From the given options, the closest value to 50 nm is 200 nm.
4. Rutherford’s model of the atom failed because:
(a) the atom did not have a nucleus and electrons
(b) it did not account for the attraction between protons and neutrons
(c) it did not account for the stability of the atom
(d) there is actually no space between the nucleus and the electrons.
Explanation: The correct answer is (c). Because Rutherford’s model showed electrons orbiting the nucleus but didn’t explain why these electrons don’t fall into the nucleus. Accelerating electrons should lose energy and spiral into the nucleus, making the atom unstable. This was later explained by Bohr’s model.
5. Bohr model of the atom is contradicted by:
(a) Planck’s quantum theory
(b) dual nature of matter
(c) Heisenberg’s uncertainty principle
(d) all of above
Explanation: The correct answer is (c). Because The Bohr model suggests electrons move in fixed paths around the nucleus. However, Heisenberg’s uncertainty principle states that we can’t know exactly where an electron is and how fast it’s moving at the same time. This contradicts the fixed paths suggested by Bohr.
6. Splitting of spectral lines when atoms are subjected to a strong electric field is called:
(a) Zeeman effect
(b) Stark effect
(c) Photoelectric effect
(d) Compton effect
Explanation: The correct answer is (b). Because The Stark effect is the splitting of an atom’s light lines due to a strong electric field.
- Zeeman effect (a): Happens with a magnetic field.
- Photoelectric effect (c): Involves electrons leaving a material when it absorbs light.
- Compton effect (d): Involves X-rays or gamma rays hitting electrons and changing their energy.
7. When the 6d orbital is complete, the entering electron goes into:
(a) 7f
(b) 7s
(c) 7p
(d) 7d
Explanation: The correct answer is (b). Because After the 6d orbital is filled, the next available orbital is 7s.
8. In the ground state of an atom, the electron is present:
(a) in the nucleus
(b) in the second shell
(c) nearest to the nucleus
(d) farthest from the nucleus
Explanation: The correct answer is (c). Because In the ground state, electrons occupy the lowest energy levels, closest to the nucleus where energy is minimized.
9. Quantum number values for 2p orbitals are:
(a) n=2, l=1
(b) n=1, l=2
(c) n=1, l=0
(d) n=2, l=0
Explanation: The correct answer is (a). Because For 2p orbitals, the principal quantum number (n) is 2, and the azimuthal quantum number (l) for p orbitals is 1.
10. Orbitals having the same energy are called:
(a) hybrid orbitals
(b) valence orbitals
(c) degenerate orbitals
(d) d-orbitals
Explanation: The correct answer is (c). Because Orbitals that have the same energy are called degenerate orbitals. For example, the three p-orbitals (px, py, pz) in the same shell are degenerate.
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