A Level Chemistry
CIE A Level Chemistry Solved Past Paper Oct/Nov 2019 P22

CIE A Level Chemistry Solved Past Paper Oct/Nov 2019 P22

1 aiii) Ionisation energy is energy required to remove an electron from a gaseous atom’s outermost shell. When nucleur attraction increases, ionisation energy increases and nucleur attraction increases when number of electrons increase and shielding effect remains same. This is only possible where electrons keep adding by one (atomic number increases by one) as we go along Period III.

iv) When the removal of electrons is done from a shell and now when its turn for an inner shell’s electrons, electron from that inner shell has a lot higher ionisation energy as that of electron from the outer shell because of more force of attraction between electrons in the inner shell and nucleus. Always remember removal of electrons starts from outermost shell or higher energy level, this Q has a huge difference in 4th ionisation energy from the 3rd one, this shows that the atom has 3 electrons in its outermost shell.

bii) Bonding between two non-metals, Si and N, and its insolubility in water shows that its a structure with covalent bonding in. Its high melting point evidences that bonding within atoms is very strong, its poor conductivity also shows that structure has got no delocalised or mobile electrons which carry charges.

di) Increase in pressure will have no effect on equilibrium mixture in this Q since there is 1 mole of SO2 and 1 mole of O3 in total 2 moles of reactants reacting and 1 mole of SO3 and 1 mole of O2 or 2 moles of products are forming.

ii) Le Chatelier’s Principle says if you change reaction conditions like temperature, concentration or pressure, equilibrium shifts or favours that side of equation which cancels out that change. Likewise if you increase the temperature in this Q, equilibrium shifts to the right side favouring formation of SO3 and O2 because forward reaction is endothermic (temperature of surroundings decreases in this case).

2 ci) Pressure,p in calculations should be in Pascals, volume, V needs to be in m^3, R has a constant value of 8.314 and temperature, T should be in Kelvin.

iii) Each Oxygen atom carries a negative charge means each Oxygen atom has gained an electron and now number of protons in an Oxygen atom are not equal to number of electrons instead each Oxygen has now become an anion.

3 ciii) NaCl + H2SO4 ——–> HCl + NaHSO4

NaI + H2SO4 ——> HI + NaHSO4

HI + H2SO4 ——–> I2 + S + H2O or HI + H2SO4 —–> I2 + H2S + H2O

When NaCl reacts with H2SO4, HCl forms which possesses more oxidizing power than H2SO4 does. That is the reason this reaction stops there unlike HI which decomposes easily into its elements because HI is not a better oxidizing agent than H2SO4 rather HI is a better reducing agent.

4a) C-Cl of G breaks heterolytically where Cl takes away both its and C’s electron which was being shared in covalent bond. This results in Cl having a partial negative charge and Carbon having a partial positive charge. Theory of unlike charges attracting comes into action here, Carbon in G carrying a partial positive charge attracts :OH- of NaOH which also broke heterolytically and Cl- of G attracts Na+ from NaOH.

bi) Reaction with steam is addition of water molecule to unsaturated Carbon structure but the problem is that -OH from H2O has 2 positions to choose from. If -OH attaches to the C atom which is already attached to 2 methyl groups, that would make product have a primary and a tertiary alcohol. If -OH attaches to C next to the C attached to 2 methyl groups, the product will end up having a primary and a secondary alcohol. Third product is the optical isomer of my second product since it has got a middle C atom attached to 4 different atoms or group of atoms. My first product if reacts with acidified potassium dichromate(VI), primary alcohol oxidizes to carboxylic acid, no change for tertiary alcohol and when that is added to 2,4-DNPH, no precipitate forms. If my second and third products had reacted with acidified potassium dichromate(VI), primary alcohol would oxidise to carboxylic acid: secondary alcohol would oxidise to ketone and that when is added to 2,4-DNPH, orange precipitate forms. Hence J is my first product.

eiv) Reaction I product is low in yield because Cl attached to that C atom which forms primary carbocation that is less stable. If Cl attaches to the C atom which is already attached to 2 methyl groups, a tertiary carbocation would form that is most stable carbocation due to more alkyl groups near C-C-Cl bond (alkyl groups are electron donating groups that make structure stable).

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