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CodeVID-C11-04-T2-01
VSEPR & Valence Bond Theory - Assignment
Chapter: Chemical Bonding and Molecular Structure
Topic: VSEPR & Valence Bond Theory
Maximum Marks: 30
Time: 60 minutes
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 geometry predicted by VSEPR for 4 bond pairs and no lone pairs is:
  • A.linear
  • B.trigonal planar
  • C.tetrahedral
  • D.octahedral
2.
The strongest repulsion in VSEPR is between:
  • A.bond pair-bond pair
  • B.lone pair-bond pair
  • C.lone pair-lone pair
  • D.all are equal
3.
The hybridisation of beryllium in BeCl2 is:
  • A.sp
  • B.sp2
  • C.sp3
  • D.sp3d
4.
A double bond consists of:
  • A.two sigma bonds
  • B.one sigma and one pi bond
  • C.two pi bonds
  • D.one sigma bond only
5.
Resonance in a molecule results in:
  • A.unequal bond lengths
  • B.higher energy
  • C.equal bond lengths and lower energy
  • D.no change
Section B - Short Answer (2 marks) 3 × 2 = 6 marks
6.
State the basic postulate of VSEPR theory.
7.
Distinguish between a sigma and a pi bond.
8.
Give the hybridisation and shape of PCl5.
Section C - Short Answer (3 marks) 2 × 3 = 6 marks
9.
Explain, with hybridisation, why H2O has a bond angle of 104.5 degrees rather than 109.5 degrees.
10.
Describe the sigma and pi bonding in the nitrogen molecule N2.
Section D - Long Answer (5 marks) 1 × 5 = 5 marks
11.
Using VSEPR and hybridisation, predict and explain the shapes of BF3, CH4, NH3 and SF6.

Answer Key

Section A - Multiple Choice Questions
  1. (C) tetrahedral
  2. (C) lone pair-lone pair
  3. (A) sp
  4. (B) one sigma and one pi bond
  5. (C) equal bond lengths and lower energy
Section B - Short Answer (2 marks)
  1. Electron pairs in the valence shell of the central atom repel one another and arrange themselves as far apart as possible to minimise repulsion, fixing the molecular shape.
  2. A sigma bond forms by axial (end-on) overlap and is strong; a pi bond forms by sideways overlap of p-orbitals and is weaker.
  3. sp3d hybridisation; trigonal bipyramidal shape.
Section C - Short Answer (3 marks)
  1. Oxygen is sp3 hybridised with 2 bond pairs and 2 lone pairs. The two lone pairs repel more strongly than bond pairs, compressing the H-O-H angle from 109.5 to 104.5 degrees.
  2. Each N is sp hybridised; one sp-sp overlap gives a sigma bond and the two unhybridised p orbitals on each atom overlap sideways to give two pi bonds, so N2 has a triple bond (1 sigma + 2 pi).
Section D - Long Answer (5 marks)
  1. BF3: 3 bp, sp2, trigonal planar 120. CH4: 4 bp, sp3, tetrahedral 109.5. NH3: 3 bp + 1 lp, sp3, pyramidal 107 (lone pair repulsion). SF6: 6 bp, sp3d2, octahedral 90. Lone pairs and pair count govern the geometry.
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