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Question

Match List-I with List-II.

List-I
(Electronic configuration of elements )
List-II
(I.E. in kJ/mol)
(a) 1s22s2 (i) 801
(b) 1s22s22p4 (ii) 899
(c) 1s22s22p3 (iii) 1314
(d) 1s22s22p1 (iv) 1402

A
(a) – (ii), (b) – (iii), (c) – (iv), (d) – (i)
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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)
1s22s2Be1s22s22p4O1s22s22p3N1s22s22p1B

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.

Hence,
N1402 kJ/molO1314 kJ/molB801 kJ/molBe899 kJ/mol

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