The correct order of the atomic radii of C, Cs, Al, and S is:

🧠 Period number is the dominant factor — Cs (Period 6) is huge, C (Period 2) is tiny, Al and S are in between (Period 3) Atomic radius is dominated by which shell the valence electrons sit in. Period 2 has just 2 shells. Period 3 has 3 shells. Period 6 has 6 shells. So Cs >> Al, S > C in size, simply by shell count. Within Period 3, Al and S follow the standard trend (size shrinks left to right).

🗺️ Place each element C: Period 2, Group 14. Smallest of all four — only 2 shells. Approx 77 pm. S: Period 3, Group 16. Three shells, but on the right side of the period. Approx 104 pm. Al: Period 3, Group 13. Three shells, on the left side of the period — bigger than S. Approx 143 pm. Cs: Period 6, Group 1. Six shells, leftmost element. Approx 265 pm. Largest by far.

Increasing order: $\ce{C} < \ce{S} < \ce{Al} < \ce{Cs}$.

⚠️ The trap Option (c) $\ce{S} < \ce{C} < \ce{Al} < \ce{Cs}$ swaps C and S. Students sometimes apply the period-3 trend "S is to the right of Al, so S is smaller than Al" (correct ✓) and assume "Period 2 atoms are even smaller, so C is smaller than S" — but they then place S < C, accidentally flipping the order. Period 2 < Period 3 always; C is the smallest, S is in the middle of this set.

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