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a
IE1 of N>IE1 of O
b
IE2 of N>IE2 of O
c
IE2 of Li>IE2 of Ne
d
IE1 of Al>IE1 of Ca
answer is A.
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Detailed Solution
(1) True. IE1 of N>IE1 of O, due to half-filled configuration in N.(2) False. N2s22p3 (More stable) High IE1⟶N⊕2s22p2 (Less stable) −e−Low IE2⟶N2+2s12p1O2s22p4 (Less stable) ⟶ (Low IE1O⊕2s22p3(Morestable)−e−High IE2→O2+2s22p2 Hence, IE2 of O>IE2 of N.