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B
(a) – (iv), (b) – (i), (c) – (ii), (d) – (iii)
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C
(a) – (i), (b) – (iv), (c) – (iii), (d) – (ii)
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D
(a) – (i), (b) – (iii), (c) – (iv), (d) – (ii)
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Solution
The correct option is A (a) – (ii), (b) – (iii), (c) – (iv), (d) – (i) 1s22s2→Be1s22s22p4→O1s22s22p3→N1s22s22p1→B
The ionization enthalpy order is B<Be<O<N.Be has more I.E. compared to B due to extra stabe fully filled 2s orbital & N has more I.E. compared to O due to extra stable half filled 2p orbital.