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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 for Tamil Nadu State Board CLASS 11th CHEMISTRY. 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.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter 2: Quantum Mechanical Model of Atom — Important Questions & Answers (FAQ)

Frequently asked questions from Tamil Nadu State Board CLASS 11th CHEMISTRY — Chapter 2: Quantum Mechanical Model of Atom, with answers and explanations. These are sample questions; the exam has its own separate question set.

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.
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.
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.
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.
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.

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