Ne = 2s22p6
F = 2s22p5
O = 2s22p4
N = 2s22p3
C = 2s22p2
B = 2s22p1
Along the period, size decreases and hence ionization energy increases in general, however some exception occurs. N is half filled therefore it is 1. F is more than that of O as O has one extra electron to release easily to get a more stable half filled configuration. Similarly B has only one electron in p orbital, therefore it can release this electron easily.
Hence, the correct arrangement is Ne > F > O> N> C> B
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