Vidaara.orgClass 11 · Chemistry
CodeVID-C11-02-T3-01
Electronic Configuration — 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 principle that fixes the order of filling of orbitals is the:
- A.Pauli principle
- B.Aufbau principle
- C.Hund's rule
- D.uncertainty principle
2.
The number of unpaired electrons in nitrogen ($Z=7$) is:
- A.1
- B.2
- C.3
- D.0
3.
The configuration $[\text{Ar}]\,3d^5\,4s^1$ belongs to:
- A.Mn
- B.Fe
- C.Cr
- D.V
4.
Electrons are removed from a transition-metal atom first from the:
- A.$3d$ orbital
- B.$4s$ orbital
- C.$4p$ orbital
- D.$3p$ orbital
5.
A completely filled sub-shell that gives extra stability is:
- A.$p^4$
- B.$d^5$
- C.$d^{10}$
- D.$d^9$
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State Hund's rule and apply it to write the orbital diagram of carbon's $2p$ electrons.
7.
Write the electronic configuration of the chloride ion $\text{Cl}^-$ ($Z$ of Cl $=17$).
8.
Why is the $3d^{10}\,4s^1$ configuration of copper more stable than $3d^9\,4s^2$?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Write the electronic configurations of $\text{Cr}$ ($Z=24$) and $\text{Cr}^{3+}$.
10.
Arrange $3d$, $4s$ and $4p$ in increasing order of energy using the $(n+l)$ rule and justify.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
State the Aufbau principle, Pauli exclusion principle and Hund's rule. Use them to derive the ground-state configuration of iron ($Z=26$) and find its number of unpaired electrons.
Answer Key
Section A — Multiple Choice Questions
- (B) Aufbau principle
- (C) 3
- (C) Cr
- (B) $4s$ orbital
- (C) $d^{10}$
Section B — Short Answer (2 marks)
- Degenerate orbitals fill singly with parallel spins before pairing. Carbon $2p^2$: two of $p_x,p_y,p_z$ singly occupied, parallel spins; one empty.
- Cl is $1s^2\,2s^2\,2p^6\,3s^2\,3p^5$; add one electron: $\text{Cl}^-=1s^2\,2s^2\,2p^6\,3s^2\,3p^6$ (= [Ar]).
- The fully-filled $3d^{10}$ sub-shell is symmetrical and has maximum exchange energy, giving extra stability.
Section C — Short Answer (3 marks)
- $\text{Cr}=[\text{Ar}]\,3d^5\,4s^1$; $\text{Cr}^{3+}=[\text{Ar}]\,3d^3$ (remove $4s^1$ then two $3d$).
- $n+l$: $4s=4$, $3d=5$, $4p=5$. Order $4s<3d<4p$; $3d$ before $4p$ as ties are broken by lower $n$.
Section D — Long Answer (5 marks)
- Aufbau: lowest-energy (lowest $n+l$) orbital fills first. Pauli: ≤2 e per orbital, opposite spins. Hund: singly fill degenerate orbitals with parallel spins first. Fe = $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^6=[\text{Ar}]\,3d^6\,4s^2$; the $3d^6$ has 4 unpaired electrons.
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