The atomic radius generally decreases as we move across a period from left to right in the periodic table. This happens because, as we move across a period, the number of protons increases, leading to a greater nuclear charge. The increased nuclear charge pulls the electrons closer to the nucleus, reducing the size of the atom.
Step 1: Atomic radius across the period.
- In the case of the elements in the question, we are dealing with period 2 elements: B (Boron), C (Carbon), N (Nitrogen), O (Oxygen), and F (Fluorine).
- As we move from left to right across the period, the atomic radius decreases.
Step 2: Atomic radius of fluorine.
Fluorine (F) is the rightmost element in the period, and therefore, it has the least atomic radius. This is because it has the highest effective nuclear charge in this period, which pulls its electrons closer to the nucleus.
Step 3: Conclusion.
Therefore, the element with the least atomic radius is fluorine (F), which corresponds to option (E).
Final Answer: (E) F