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

Question

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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: $Li > Be > B > C.$

Answer

Be > Li > B > C

Answer

Li > Be > B > C

Answer

C > B > Be > Li

Answer

Li > C > Be > B

List I		List II
A	Down’s syndrome	I	11^th chormosome
B	α-Thalassemia	II	‘X’ chromosome
C	β-Thalassemia	III	21^st chromosome
D	Klinefelter’s syndrome	IV	16^th chromosome

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

The Correct Option is B

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