Vidaara.orgClass 11 · Chemistry
CodeVID-C11-02-T2-01
Quantum Mechanical Model — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The quantum number that fixes the shape of an orbital is:
- A.$n$
- B.$l$
- C.$m_l$
- D.$m_s$
2.
The maximum number of electrons in a shell with principal quantum number $n$ is:
- A.$2n$
- B.$n^2$
- C.$2n^2$
- D.$2l+1$
3.
The number of orbitals in the $d$ sub-shell is:
- A.1
- B.3
- C.5
- D.7
4.
A $2s$ orbital has a total number of nodes equal to:
- A.0
- B.1
- C.2
- D.3
5.
The de Broglie wavelength is largest for which particle moving at the same speed?
- A.proton
- B.neutron
- C.electron
- D.$\alpha$-particle
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State Heisenberg's uncertainty principle and explain why it rules out fixed Bohr orbits.
7.
Write the allowed sets of quantum numbers for the electrons in the $2p$ sub-shell (give $n,l,m_l$).
8.
Why is the wave nature of a moving car not observable?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Calculate the de Broglie wavelength of a proton ($m=1.67\times10^{-27}\,\text{kg}$) moving at $1.0\times10^{5}\,\text{m s}^{-1}$.
10.
For $n=4$, list the sub-shells, the number of orbitals in each, and the total electron capacity.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Define the four quantum numbers, give their allowed values, and use them to explain how many electrons the $M$ shell ($n=3$) can hold.
Answer Key
Section A — Multiple Choice Questions
- (B) $l$
- (C) $2n^2$
- (C) 5
- (B) 1
- (C) electron
Section B — Short Answer (2 marks)
- $\Delta x\cdot\Delta p\ge\frac{h}{4\pi}$. A fixed orbit needs exact position and momentum together, which the principle forbids.
- $n=2$, $l=1$, $m_l=-1,0,+1$ (three orbitals, each with $m_s=\pm\tfrac{1}{2}$).
- Its mass is huge, so $\lambda=h/mv$ is around $10^{-34}\,\text{m}$ — immeasurably small compared with any object.
Section C — Short Answer (3 marks)
- $\lambda=h/mv=6.626\times10^{-34}/(1.67\times10^{-27}\times1.0\times10^{5})=3.97\times10^{-12}\,\text{m}$.
- $4s(1),4p(3),4d(5),4f(7)$; $16$ orbitals; capacity $2n^2=32$ electrons.
Section D — Long Answer (5 marks)
- $n$ (1,2,3…ize/energy), $l$ (0 to $n-1$, shape), $m_l$ ($-l$ to $+l$, orientation), $m_s$ ($\pm\tfrac{1}{2}$, spin). For $n=3$: $3s,3p,3d$ = $1+3+5=9$ orbitals × 2 = $18$ electrons ($=2n^2$).
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