chemistry
Chapter 4
1. What is octet rule? Write two drawbacks.
Answer:
Octet rule states that atoms try to get 8 electrons in their valence shell to become stable.
Drawbacks (any two):
- It is not applicable to hydrogen and helium.
- It fails for molecules like BF₃ where octet is incomplete.
Simple explanation:
Atoms want to be “full” like a filled plate.
Example:
NaCl follows octet rule, BF₃ does not.
2. Write the main postulates of VSEPR theory.
Answer:
- Electron pairs repel each other.
- Molecules take shapes to reduce repulsion.
- Lone pair repulsion is greater than bond pair repulsion.
Example:
H₂O is bent because lone pairs push bonds.
3. A molecule AB₂E₂ has 2 bond pairs and 2 lone pairs. What is its shape?
Answer:
Bent shape (V-shape).
Example:
Water (H₂O).
4. Define dipole moment. Why is dipole moment of BF₃ zero?
Answer:
Dipole moment measures the polarity of a molecule.
Dipole moment of BF₃ is zero because it is symmetrical.
Example:
Like equal people pulling a rope from all sides → no movement.
5. Why is dipole moment of NF₃ less than NH₃?
Answer:
In NF₃, bond dipoles and lone pair dipole act in opposite directions, so dipole moment decreases.
Example:
Two forces in opposite directions cancel each other.
6. State Fajan’s rule.
Answer:
Small cation with high charge and large anion increases covalent character in ionic bonds.
Example:
AlCl₃ is more covalent than NaCl.
7. Define hybridisation. Explain BeCl₂.
Answer:
Hybridisation is the mixing of atomic orbitals to form equal orbitals.
BeCl₂:
Hybridisation – sp
Shape – Linear
Example:
Two sticks in a straight line (180°).
8. Write any two features of Molecular Orbital Theory.
Answer:
- Atomic orbitals combine to form molecular orbitals.
- Bond order indicates stability.
Example:
Higher bond order → stronger bond.
9. Predict hybridisation of phosphorus in PCl₅. Why is it reactive?
Answer:
Hybridisation – sp³d
It is reactive because it has large size and weak bonds.
Example:
Loose bricks break easily.
10. HF is liquid while HCl is gas. Why?
Answer:
HF forms hydrogen bonding, HCl does not.
Example:
HF molecules stick together like magnets.
11. Write MO configuration of F₂ and find bond order.
Answer:
Bond order = 1
Explanation:
One bond → stable molecule.
12. Write MO configuration of O₂. Why is it paramagnetic?
Answer:
O₂ has unpaired electrons, so it is paramagnetic.
Bond order = 2
Example:
Unpaired electrons act like tiny magnets.
13. Define hydrogen bond. Write its types with examples.
Answer:
Hydrogen bond is attraction between hydrogen and electronegative atom.
Types:
- Intermolecular – H₂O
- Intramolecular – o-nitrophenol
14. Why does He₂ molecule not exist?
Answer:
Bond order of He₂ = 0, so it is unstable.
Example:
No bond → no molecule.
15. Complete the table
| Molecule | Hybridisation | Shape |
|---|---|---|
| CH₄ | sp³ | Tetrahedral |
| BF₃ | sp² | Trigonal planar |
| SF₆ | sp³d² | Octahedral |
⭐ Exam Tip for Weak Students
- Learn definitions + one reason
- Write shape + hybridisation
- Even short answers can score marks 👍
If you want next:
- ✔️ Only 2-mark answers
- ✔️ MCQs with explanation
- ✔️ One-page revision sheet
Just tell me 😊
1. NO and NO₂ are two oxides of nitrogen. Which law is illustrated?
Law: Law of Multiple Proportions
Statement: When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
Proposed by: John Dalton
2. State Law of Definite (Constant) Proportions.
Statement: A given compound always contains the same elements combined in the same fixed proportion by mass, irrespective of the source.
Proposed by: Joseph Proust
3. State Law of Conservation of Mass.
Statement: Mass is neither created nor destroyed in a chemical reaction; total mass of reactants equals total mass of products.
Proposed by: Antoine Lavoisier
4. Define atomic mass unit (amu).
Answer: One atomic mass unit is defined as 1/12th of the mass of a carbon-12 atom.
5. Define 1 mole.
Answer: One mole is the amount of substance containing 6.022 × 10²³ particles (atoms/molecules/ions).
6. Number of oxygen atoms in 5 moles of glucose (C₆H₁₂O₆)
O atoms in 1 molecule = 6
Moles of O atoms = 5 × 6 = 30 moles
Number of atoms = 30 × 6.022 × 10²³
= 1.8066 × 10²⁵ oxygen atoms
7. Significant figures
(i) 0.0205 → 3
(ii) 3.00 × 10⁻⁵ → 3
(iii) 1005 → 4
8. Define mass per cent.
Answer:
\text{Mass percent} = \frac{\text{Mass of component}}{\text{Total mass of mixture}} \times 100
9. What are empirical and molecular formulae?
- Empirical formula: Simplest whole-number ratio of atoms
- Molecular formula: Actual number of atoms in a molecule
10. Compound data
H = 4.07%, C = 24.27%, Cl = 71.65%
Molecular mass = 98.96
Step 1: Convert % to moles
| Element | % | Atomic mass | Moles |
|---|---|---|---|
| H | 4.07 | 1 | 4.07 |
| C | 24.27 | 12 | 2.02 |
| Cl | 71.65 | 35.5 | 2.02 |
Step 2: Simplest ratio
H : C : Cl = 2 : 1 : 1
Empirical formula: CH₂Cl
Empirical formula mass = 49.48
Molecular mass / Empirical mass = 98.96 / 49.48 = 2
Molecular formula: C₂H₄Cl₂
11. Moles and molecules in 90 g water
Molar mass of H₂O = 18 g/mol
Moles = 90 / 18 = 5 moles
Molecules = 5 × 6.022 × 10²³
= 3.011 × 10²⁴ molecules
12. What is limiting reagent?
Answer: The reactant that is completely consumed first and limits the amount of product formed.
13. Differentiate between molarity and molality
| Molarity | Molality |
|---|---|
| Moles per litre of solution | Moles per kg of solvent |
| Depends on temperature | Independent of temperature |
| Symbol: M | Symbol: m |
14. Molarity of NaOH solution
Given: 4 g NaOH in 250 mL solution
Molar mass of NaOH = 40 g/mol
Moles = 4 / 40 = 0.1 mol
Volume = 250 mL = 0.25 L
\text{Molarity} = \frac{0.1}{0.25} = \boxed{0.4\ M}
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