Question

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.

Hint:

This specific problem because Avogadro's Law links volume directly to the number of molecules, regardless of the gas's identity!

Answer 1

Based on the descriptions: P is an Alkali Metal (Group 1), Q is likely a Group 14 element (like Carbon or Silicon), and R is a Halogen (Group 17).
(a) Which element would be most difficult to reduce?
Element P. P is highly electropositive and wants to lose electrons (oxidize). Forcing it to gain electrons (reduction) is extremely difficult.
(b) Smallest atomic radius?
Element R. Atomic radius decreases across a period from left to right due to increased nuclear charge. R is furthest to the right.
(c) Decreasing order of ionization potential:
Ionization potential increases across a period. Therefore, the order of decreasing IP is:
R > Q > P