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

Since P, Q and R belong to the same period, their properties change gradually from left to right across the periodic table.

Step 1: Identify element P
P readily loses its one valence electron to form a stable ion.
This is a characteristic of Group 1 elements (alkali metals).
These elements form +1 ions easily.
So, P is a metal (alkali metal).

Step 2: Identify element Q
Q shares its electrons in bonding and does not form ions easily.
This is a property of non-metals that form covalent bonds, typically elements from the middle of the period (like carbon, silicon, etc.).
So, Q is a covalent (non-metallic) element.

Step 3: Identify element R
R has high electronegativity.
Electronegativity increases across a period from left to right.
The elements at the end of the period (like halogens) have the highest electronegativity.
So, R is a highly electronegative non-metal (like a halogen).

Final Answer:
P → Metal (Group 1, alkali metal)
Q → Non-metal forming covalent bonds
R → Highly electronegative non-metal (halogen)

Conclusion:
Across a period:
Metals → Covalent non-metals → Highly electronegative non-metals