The increasing order of the atomic radii of the following elements is: (a) C (b) O (c) F (d) Cl (e) Br

🧠 Period number dominates atomic radius — Period 4 atom is bigger than Period 3, which is bigger than Period 2 The five elements span three periods. Br (P4) > Cl (P3) > any Period 2 element (C, O, F). Among the Period 2 atoms, radius decreases left-to-right: C > O > F. 🗺️ Map letters and rank (a) C — Period 2, Group 14. (b) O — Period 2, Group 16. (c) F — Period 2, Group 17. (d) Cl — Period 3, Group 17. (e) Br — Period 4, Group 17.

Approximate atomic radii (pm): F = 64, O = 66, C = 77, Cl = 99, Br = 114.

Increasing order: F < O < C < Cl < Br → (c) < (b) < (a) < (d) < (e).

⚠️ The trap Option (a) C < O < F < Cl < Br applies "size decreases across a period" backwards — it puts F as bigger than C, but C is to the LEFT of F in Period 2, so C should be bigger than F. Always remember: in the same period, leftmost = biggest. The question's letter labels are (a)=C, (c)=F, so within Period 2 the order from smallest is F → O → C, i.e., (c) → (b) → (a).

$\boxed{\text{Answer: (c)}}$