A) -5.1 eV
B) -10.2 Ev
C) = 2.55 eV
D) + 10.2 eV
Correct Answer: A
Solution :
\[\text{I}{{\text{E}}_{\text{1}}}\]of Na = - Electron gain enthalpy of \[\text{N}{{\text{a}}^{\text{+}}}\] ion = - 5.leV.
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