Chapter 2: Quantum Mechanical Model of Atom — Online MCQ Test
CHEMISTRY · CLASS 11th · Tamil Nadu State Board
Practice Chapter 2: Quantum Mechanical Model of Atom with a free chapter-wise online MCQ test.
This chapter covers: Focusing on atomic structure this chapter details Bohr model limitations de Broglie dual nature principle and Heisenberg uncertainty principle. Students explore quantum numbers Sch....
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Chapter 2: Quantum Mechanical Model of Atom — Important Questions & Answers
Which of the following quantum numbers describes the shape of the orbital?
- A. Principal quantum number
- B. Azimuthal quantum number
- C. Magnetic quantum number
- D. Spin quantum number
Answer: B. Azimuthal quantum number
The azimuthal quantum number (l) determines the angular momentum and the shape of the orbital.
The azimuthal quantum number (l) determines the angular momentum and the shape of the orbital.
What is the maximum number of electrons that can be accommodated in the d-subshell?
- A. 6
- B. 10
- C. 14
- D. 2
Answer: B. 10
A d-subshell has 5 orbitals, each holding 2 electrons, totaling 10 electrons.
A d-subshell has 5 orbitals, each holding 2 electrons, totaling 10 electrons.
What is the Heisenberg uncertainty principle equation?
- A. delta x * delta p >= h / 4*pi
- B. delta x * delta p = h / 2*pi
- C. E = mc^2
- D. lambda = h/p
Answer: A. delta x * delta p >= h / 4*pi
The principle states the product of uncertainty in position and momentum is greater than or equal to h/4pi.
The principle states the product of uncertainty in position and momentum is greater than or equal to h/4pi.
Calculate the de Broglie wavelength of an electron moving with velocity 2 x 10^6 m/s (mass = 9.1 x 10^-31 kg).
- A. 3.64 x 10^-10 m
- B. 1.25 x 10^-9 m
- C. 4.12 x 10^-11 m
- D. 6.63 x 10^-34 m
Answer: A. 3.64 x 10^-10 m
Using lambda = h/mv, where h=6.626x10^-34, m=9.1x10^-31, and v=2x10^6, the result is approx 3.64 x 10^-10m.
Using lambda = h/mv, where h=6.626x10^-34, m=9.1x10^-31, and v=2x10^6, the result is approx 3.64 x 10^-10m.
The energy of an electron in the nth orbit of a hydrogen-like ion is given by En = -13.6 Z^2 / n^2 eV. What is the energy for the ground state of He+?
- A. -13.6 eV
- B. -54.4 eV
- C. -27.2 eV
- D. -6.8 eV
Answer: B. -54.4 eV
For He+, Z=2, n=1. E = -13.6 * (2^2) / 1^2 = -54.4 eV.
For He+, Z=2, n=1. E = -13.6 * (2^2) / 1^2 = -54.4 eV.