Question

Which of the next is a correct arrangement of the values of the 1st ionization energy? Ne >F>0<N>C>B Ne <F<O<N<C<B Ne <F<O>Nt<B O Ne>F>O>N>C>B. All calculation

Answer 1

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