The first ionization potential (eV) of Be and B respectively are
(A) 8.29, 9.32                       (B) 9.32, 8.29                   (C) 9.32, 9.32                 (D) 8.29, 8.29
 

36a,38a
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D
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Guys 1st ionisation energy of be is always greater than b so there is only single option i.e. 9.32,8.29 . Hence this is the correct option
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First ionisation potential of Be is greater than boron due to following configuration ₄Be=1s²,2s² ₅B=1s²,2s²2p¹ Order of attraction of electrons towards nucleus 2s>2p, so more amount of energy is required to remove the electron with 2s-orbital in comparison to 2p orbital
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