JEE Main 2025 · 29 January · evening · Single correct · +4 / −1
The type of oxide formed by the element among Li, Na, Be, Mg, B and Al that has the least atomic radius is :
- (a) AO
- (b) AO₂
- (c) A₂O
- (d) A₂O₃
Answer and solution
Answer (d) A₂O₃
Solution
Chemistry ma’am · 4 steps
Find the smallest atom first: top row, furthest right. That is boron, and boron gives B₂O₃.
- Size trend. Atomic radius decreases from left to right across a period, because nuclear charge rises while electrons go into the same shell. It increases down a group, because a new shell is added.
- Pick the smallest. Li, Be and B are in period 2, so they are smaller than Na, Mg and Al in period 3. Within period 2, B is furthest to the right, so B has the least atomic radius.
Element Group Usual valency Oxide Type Li, Na 1 1 Li₂O, Na₂O A₂O Be, Mg 2 2 BeO, MgO AO B, Al 13 3 B₂O₃, Al₂O₃ A₂O₃ - Balance the charges. Boron shows valency 3 and oxygen shows valency 2. Two B atoms give 2 × 3 = 6 and three O atoms give 3 × 2 = 6, so the formula is B₂O₃.
- So. Boron, the smallest of these atoms, forms an oxide of type A₂O₃, which is option (d).
The idea to keep. For the smallest atom in a list, go to the highest period and then as far right as you can. The oxide formula then follows from the group valency.
Answer from the NTA final answer key. Solution by PrepWiser.