Question

The decreasing order of atomic radii (pm) of $\text{Li}, \, \text{Be}, \, \text{B}, \, \text{C}$ is:

• A. Be > Li > B > C
• B. Li > Be > B > C
• C. C > B > Be > Li
• D. Li > C > Be > B

Answer 1

The Correct Option is B

Li, Be, B, and C are elements of the second period. Across a period from left to right, the atomic radii generally decrease. This is because as we move across a period, the number of protons in the nucleus increases, increasing the effective nuclear charge. This stronger positive charge pulls the electrons in the outermost shell closer to the nucleus, resulting in a smaller atomic radius.

Li (Lithium) is in group 1. Be (Beryllium) is in group 2. B (Boron) is in group 13. C (Carbon) is in group 14.

Therefore, the atomic radii decrease in the order: $Li > Be > B > C.$