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 (Spectral Lines of Hydrogen for transitions from)		List II (Wavelength (nm))
A.	n₂ = 3 to n₁ = 2	I.	410.2
B.	n₂ = 4 to n₁ = 2	II.	434.1
C.	n₂ = 5 to n₁ = 2	III.	656.3
D.	n₂ = 6 to n₁ = 2	IV.	486.1

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

The Correct Option is B

Solution and Explanation

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