Elements P, Q, and R are in the same period of the modern periodic table.
- P readily loses its one valence electron to form a stable ion.
- Q shares its electrons in bonding but does not form ions easily.
- R has high electronegativity.
Answer the following questions based on the above information:
(a) Which element would be most difficult to reduce among P, Q and R?
(b) Which element is expected to have the smallest atomic radius among P, Q and R?
(c) Arrange P, Q and R in order of decreasing ionisation potential.
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P loses one valence electron, so P is a Group 1 metal (most electropositive). R has high electronegativity, so R is a non-metal on the right side of the periodic table. Q shares electrons (covalent), so Q is in Group 14 (between P and R).
(a) P would be most difficult to reduce.
Reason — Reduction means gain of electrons. Element P readily loses its valence electron and is the most electropositive (strongest reducing agent itself). Hence, its tendency to gain electrons (be reduced) is the lowest, so it is most difficult to reduce.
(b) R has the smallest atomic radius.
Reason — Across a period from left to right, the atomic radius decreases due to increasing nuclear charge. R is on the rightmost side, so it has the smallest atomic radius.
(c) Decreasing ionisation potential: R > Q > P
Reason — Across a period, ionisation potential increases from left to right because atomic size decreases and the effective nuclear charge increases.