`N to N^(-)` `F to F^(-)` `Cl to Cl^(-)` `H to H^(-)`
Solution :Nitrogen has stable `2p^(3)` CONFIGURAITON and also due to HIGH `e^(-)` charge density at outermost orbital it requires energy to add one EXTRA `e^(-)` in its outer most shell i.e., its first electron gain enthalpy is positive.