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Vidaara.orgClass 11 · Chemistry
CodeVID-C11-04-CH-01
Chemical Bonding and Molecular Structure - Full Chapter Test
Chapter: Chemical Bonding and Molecular Structure
Topic: All Topics
Maximum Marks: 40
Time: 90 minutes
Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

  • This is a full-length test covering the whole chapter - every topic is included.
  • 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 6 × 1 = 6 marks
1.
Which molecule has an incomplete octet on its central atom?
  • A.CH4
  • B.BF3
  • C.NH3
  • D.H2O
2.
The geometry of CH4 is:
  • A.linear
  • B.trigonal planar
  • C.tetrahedral
  • D.octahedral
3.
The bond order of O2 is:
  • A.1
  • B.2
  • C.3
  • D.0
4.
Hydrogen bonding is found in:
  • A.CH4
  • B.HF
  • C.CO2
  • D.Cl2
5.
The hybridisation of carbon in ethene (C2H4) is:
  • A.sp
  • B.sp2
  • C.sp3
  • D.sp3d
6.
Which molecule is paramagnetic?
  • A.N2
  • B.H2
  • C.O2
  • D.F2
Section B - Short Answer (2 marks) 4 × 2 = 8 marks
7.
Write the formula for formal charge and state its use.
8.
State the bond-order formula of MO theory and find the bond order of N2.
9.
Why is CO2 non-polar despite having polar bonds?
10.
Give the hybridisation and shape of SF6.
Section C - Short Answer (3 marks) 2 × 3 = 6 marks
11.
Explain, using VSEPR, why the bond angle decreases from CH4 to NH3 to H2O.
12.
State Fajans' rules and apply them to compare the covalent character of AlCl3 and AlF3.
Section D - Long Answer (5 marks) 2 × 5 = 10 marks
13.
Draw the MO energy diagram of O2, calculate its bond order, explain its magnetic behaviour, and contrast it with N2.
14.
Discuss hydrogen bonding: define it, distinguish intermolecular from intramolecular bonding, and give two consequences with examples.

Answer Key

Section A - Multiple Choice Questions
  1. (B) BF3
  2. (C) tetrahedral
  3. (B) 2
  4. (B) HF
  5. (B) sp2
  6. (C) O2
Section B - Short Answer (2 marks)
  1. FC = valence electrons - lone-pair electrons - (1/2) bonding electrons; the structure with formal charges nearest zero is the most stable.
  2. Bond order = (Nb - Na)/2; for N2, (10 - 4)/2 = 3.
  3. It is linear, so the two equal and opposite bond dipoles cancel, giving zero net dipole moment.
  4. sp3d2 hybridisation; octahedral shape.
Section C - Short Answer (3 marks)
  1. All are sp3 with 4 electron pairs. CH4 (0 lp) is 109.5; NH3 (1 lp) is 107; H2O (2 lp) is 104.5. Lone pair-bond pair repulsion exceeds bond pair-bond pair repulsion, so more lone pairs compress the angle.
  2. Covalency rises with small cation, large anion and high charge. Cl- is larger and more polarisable than F-, so AlCl3 has greater covalent character than AlF3.
Section D - Long Answer (5 marks)
  1. O2 (16 e) fills sigma2s, sigma*2s, sigma2p, two pi2p, then two pi*2p singly. Bond order = (10-6)/2 = 2; two unpaired pi* electrons make O2 paramagnetic. N2 (14 e) has all electrons paired with bond order 3, so it is diamagnetic.
  2. A hydrogen bond is the attraction between an H atom bonded to F/O/N and a lone pair on a neighbouring electronegative atom (about 10-40 kJ/mol). Intermolecular H-bonding is between molecules (raises boiling point of water/HF); intramolecular is within one molecule (ortho-nitrophenol, lowering boiling point). Consequences: high boiling point of water and the low density of ice (open cage structure).
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