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Why is the atomic radius of Boron smaller than that of Be?

  1. Oct 12, 2009 #1
    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?
  2. jcsd
  3. Oct 12, 2009 #2
    Because the configuration (1s2 2s2) in B is smaller in size due to higher nucleus charge, no?
    Last edited: Oct 12, 2009
  4. Oct 13, 2009 #3
    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?
  5. Oct 13, 2009 #4
    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?
    Last edited: Oct 13, 2009
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