The decreasing order of atomic radii (pm) of \( \text{Li}, \, \text{Be}, \, \text{B}, \, \text{C} \) is:
- Be > Li > B > C
- Li > Be > B > C
- C > B > Be > Li
- Li > C > Be > B
The Correct Option is B
Solution and Explanation
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: \(Li > Be > B > C.\)
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