Question:medium
The decreasing order of atomic radii (pm) of \( \text{Li}, \, \text{Be}, \, \text{B}, \, \text{C} \) is:
The decreasing order of atomic radii (pm) of \( \text{Li}, \, \text{Be}, \, \text{B}, \, \text{C} \) is:
Updated On: Nov 26, 2025
- Be > Li > B > C
- Li > Be > B > C
- C > B > Be > Li
- Li > C > Be > B
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The Correct Option is B
Solution and Explanation
The elements Lithium (Li), Beryllium (Be), Boron (B), and Carbon (C) belong to the second period of the periodic table. As one moves from left to right across a period, atomic radii typically diminish. This trend is attributable to an increasing number of protons in the nucleus, which enhances the effective nuclear charge. This augmented positive charge exerts a stronger pull on the electrons in the outermost shell, drawing them nearer to the nucleus and consequently reducing the atomic radius.
Lithium (Li) is located in Group 1. Beryllium (Be) is in Group 2. Boron (B) is in Group 13. Carbon (C) is in Group 14.
Consequently, the atomic radii decrease in the following sequence: \(Li > Be > B > C.\)
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