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AQA A-LEVEL INORGANIC CHEMISTRY | FROM QUESTION TO PERFECTION| EVERY QUESTION SOLVED | EVERY SCORE SECURED!

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AQA A-LEVEL INORGANIC CHEMISTRY | FROM QUESTION TO PERFECTION| EVERY QUESTION SOLVED | EVERY SCORE SECURED!

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AQA A-LEVEL INORGANIC CHEMISTRY | FROM QUESTION
TO PERFECTION| EVERY QUESTION SOLVED | EVERY
SCORE SECURED!

Course Code:
Course Title:
Programme:
Academic Year: 2026/2027.
Duration: 2 Hours.
Total Marks: 70%.


Candidate Instructions:
1) Write your Registration Number on every answer booklet used.
2) Answer ALL questions in Section A and ANY TWO (2) questions in Section B.
3) Read each question carefully before answering.
4) Begin each question on a new page.
5) The marks for each question are indicated in brackets.
6) This paper consists of several printed pages, including this page.
7) Ensure your copy is complete before the examination begins.
8) Unauthorized materials and communication with other candidates are not permitted.




Turn Over.




APPHIA – Crafted with Care and Precision for Academic Excellence.
1

, What is the meaning of the term periodicity? Answer: It is a repeating pattern of properties
across a period e.g. the trend in atomic radius in period 2 is repeated in period 3
State the trend in first ionisation energy across periods 2 and 3 Answer: Overall 1st IE increases
with a slight decrease going from groups 2 to 3 and groups 5 and 6
Define and write the equation for the first ionisation energy of an element. Answer: The
minimum energy required to remove one mole of electrons from one mole of GASEOUS atoms
X(g) → X(g) + e⁻
Explain why the first IE increases across periods 2 and 3. Answer: (1) nuclear charge increases
(2) same number of shells (similar shielding) so (3) attraction between outer electron(s) and
nucleus increases (more energy needed to remove an outer electron)
Explain why there is a decrease in first IE between groups 2&3. Answer: In group 2 electron is
removed from s-subshell, in group 3 it is removed from p-subshell. P is further from the nucleus
than s, so weaker attraction between p electron and nucleus.
Explain why there is a decrease in first IE between groups 5&6. Answer: In group 6, the p
electron being removed is paired up with another electron in the same orbital. These electrons
repel, making it easier to remove.
State and explain the trend in atomic radius across a period, Answer: Atomic radius decreases
across a period due to increasing nuclear charge in atoms with the same number of electron
shells (same shielding). The attraction between nucleus and outer electrons is higher and so
electrons are pulled closer to the nucleus.
Explain why cations are smaller than anions within a period Answer: positive ions have one
less electron shell than negative ions in the same period.
Define the term electronegativity Answer: Eₙ is the power of an atom to attract the shared pair
of electrons in a covalent
State and explain the trend in electronegativity across a period Answer: Eₙ increases across a
period as nuclear charge increases but number of electron shells is the same (shielding same) so
attraction between the nucleus and shared pair of electrons in the bond
Which are the three most electronegative elements on the PT? Answer: F, O, N
State and explain (as fully as possible) the trend in mpt/bpt in Period 3 from Na to Al Answer:
mpt/bpt increases from Na to Al because metallic bond is stronger. This is because the caTions
have a higher charge density (Na⁺, Mg²⁺, Al³⁺)and there are more delocalised electrons, so
attractive force between positive ions and delocalised electrons (=the metallic bond) increases
Why does the conductivity increase from Na to Al in Period 3? Answer: There are more
delocalised electrons in the giant metallic structure



APPHIA – Crafted with Care and Precision for Academic Excellence.
2

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