Incorrect paper 1

Question 1

Answer 1

C

The correct answer is C: Br⁻(aq) can reduce Cl₂(aq) because:

1. Understanding Oxidising and Reducing Agents
   
   • A reducing agent donates electrons and gets oxidised.
   • An oxidising agent accepts electrons and gets reduced.
   • Halogens (X₂) act as oxidising agents, while halide ions (X⁻) act as reducing agents.
2. Oxidising Strength of Halogens (X₂)
   
   • F₂ > Cl₂ > Br₂ > I₂ (fluorine is the strongest oxidising agent).
   • This means Cl₂ is strong enough to oxidise Br⁻ into Br₂ but not the other way around.
3. Reducing Strength of Halide Ions (X⁻)
   
   • I⁻ > Br⁻ > Cl⁻ > F⁻ (iodide is the best reducing agent).
   • This means Br⁻ can reduce Cl₂ to Cl⁻ because Cl₂ is a stronger oxidising agent.
4. Why is C Correct?
   
   • Br⁻ donates electrons to Cl₂, reducing Cl₂ to Cl⁻.
   • Br⁻ itself gets oxidised to Br₂.

Equation:

This matches option C, making it the correct answer.

Question 2

Answer 2

A

Question 3

Answer 3

A

Question 4

Answer 4

Question 5

Answer 5

Question 6

Answer 6

B

Question 7

Answer 7

Since Na+ is smaller than K+, the distance between Na+ and O2- is shorter, leading to a stronger attraction and a more exothermic lattice enthalpy.
- Rember atoms and ions aren’t the same thing so you can’t talk about atomic radius ion context of an ion.
- And we know that in this context we are looking at ions as the question states lattice enthalpy which is : The enthalpy change when **one mole of ionic compound is formed from gauseous ions under standard states and conditions. Remember IONS IONS IONS !!!

Question 8

Answer 8

Question 9

Answer 9

Question 10

Answer 10

C

Question 11

Answer 11

C

Question 12

Answer 12

Question 13

Answer 13

Question 14

Answer 14

Question 15

Answer 15

Question 16

Calculate the bond enthalpy of the F–F bond.

Answer 16

Or use hess law

Question 17

Answer 17

Kp gets smaller so denominator must be getting bigger therefore we know that equilibrium shifts left.
- Answer to (i):

The forward reaction is exothermic.

Reason:
From the table, K_p decreases significantly as temperature increases (from 2.19 \times 10^{12} at 298K to 2.03 \times 10^{-1} at 1000K).

According to Le Chatelier’s Principle, if the forward reaction were endothermic, increasing temperature would increase K_p . However, since K_p decreases with temperature, this indicates that the equilibrium shifts towards the reactants at higher temperatures. This happens when the forward reaction releases heat (exothermic), meaning increasing temperature favors the reverse (endothermic) reaction.

Final Answer: The forward reaction is exothermic.

Question 18

Answer 18

Think logically

Question 19

Answer 19

Question 20

Answer 20

Forces
London forces increase
OR induced dipole(-dipole) interactions increase v
Reason
(Number of) electrons increases
Link to energy and particles
More energy to break intermolecular forces to break London forces
OR
to break induced dipole(-dipole) interactions v

Question 21

Answer 21

Question 22

Explain the guidance ms
Q : Boiling point down group 7 increases, explain why

Answer 22

• Chlorine (Cl₂) molecules only have London dispersion forces between them because they are nonpolar. The only intermolecular force they experience is due to temporary induced dipoles (London dispersion forces).
• Permanent dipole-dipole forces occur in polar molecules, where there is a permanent separation of charge due to differences in electronegativity (e.g., HCl).
• Hydrogen bonding is a special, stronger type of dipole-dipole interaction that occurs when hydrogen is directly bonded to N, O, or F (e.g., in water, H₂O).

Question 23

Answer 23

Question 24

You always get this wrong

Answer 24

Question 25

This was a bit tricky to wrap your head around

Answer 25

Key : use units to help you ! The key idea is that: **Mass (g) per 1 molecule** multiplied by **Number of molecules in 1 mole (Avogadro’s number)** gives you the molar mass (g/mol).

Question 26

Answer 26

Always do the cycle !!!

Question 27

Answer 27

Bromine exists as liquid in standard state

Question 28

Answer 28

Question 29

Answer 29

Question 30

Answer 30

B

Question 31

Answer 31

Common error: 3 marks 84.9 °C (not divided mol of AgNO3 by 2)

Question 32

Answer 32

Question 33

In relation to flashcard 32

Question 34

Answer 34

Question 35

Answer 35

Question 36

Answer 36

Question 37

Answer 37

Question 38

Answer 38

Question 39

Answer 39

Question 40

Answer 40

Question 41

Answer 41

Question 42

Answer 42

Question 43

Answer 43

Question 44

Answer 44

Question 45

Answer 45

Question 48

How does ionisation change across the first 3 periods

Answer 48

As we go from left to right nuclear charge increases and shielding stays relatively the same so on the right (noble gases) you will have the elements with the highest ionisation energies because it requires a lot of energy to remove outer shell electron.

Question 49

Answer 49

Question 50

Difference between empirical and molecular formula

Answer 50

Question 51

Explain in what scenarios you would compare intermolecular forces with intermolecular forces

Answer 51

Question 52

Answer 52

Question 53

Answer 53

Question 54

Answer 54