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View Full Version : Why is the atomic radius of Boron smaller than that of Be?


mahela007
Oct12-09, 12:05 PM
The 1st ionization energy decreases from Be to B. This would suggest that the outermost electron is held more loosely in B. Also, considering the electron configurations of Be (1s2 2s2) and B (1s2 2s2 2p1) one would think that in B, the shielding effect of the 2s electrons on the p electron would also act to increase the atomic radius...
but alas :wink: that isn't how it is... Now.. the universal question: WHY?

Bob_for_short
Oct12-09, 01:18 PM
Because the configuration (1s2 2s2) in B is smaller in size due to higher nucleus charge, no?

mahela007
Oct13-09, 10:23 AM
Wait.. i think I made a mistake in my initial post.
What I meant to ask was why is the atomic radius of B smaller when it has an extra P orbital which is also shielded by full 1s and 2s orbitals?

Bob_for_short
Oct13-09, 10:37 AM
Wait.. i think I made a mistake in my initial post.
What I meant to ask was why is the atomic radius of B smaller when it has an extra P orbital which is also shielded by full 1s and 2s orbitals?

Ion B+ radius is smaller than that of neutral Be despite the same electron configuration 1s2 2s2 just because the charge of nucleus in B is higher. The rest is a collective electronic effect, I think.

Besides, a p-state is not spherical. Which axe (dimension) is considered as the B-size?