1should be B<C<O<N
as N has half filled 2p orbital making it more stable and thus having higher IP.
1.arrange in order of Increasing ionic size..
Na+, Mg2+, Al3+, F-
2.arrange in order of Increasing first ionization enthalpy
B , C , N, O
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4 Answers
1Al3+ < Mg2+ < Na+ < F-
all the species contain ten electrons but the nuclear charge is highest for Al(13) followed by Mg,Na,F. hence the effective nuclear charge is higher in case of Al followed by the given order & hence the corresponding reduction in size is explained.
11B<C<O<N
in all cases the electron must be removed from the 2p subshell. so they must be arranged in the reverse order of their nuclear charge ie in the reverse oder of the attraction they encounter. but in case of oxygen there is slight gain in stability due to half filled p subshell while nitrogen loses stability due to loss of half filled character.