The decreasing order of the first ionization energy (in kJ mol^(-1)) of He, Mg and Na is HegtMggtNa. The increasing order of the 2^(nd) ionization energy (in kJ mol^(-1)) of these elements will be

The decreasing order of the first ionization energy (in kJ mol^(-1)) o

1.	The decreasing order of the first ionization energy (in kJ mol^(-1)) of He, Mg and Na is HegtMggtNa. The increasing order of the 2^(nd) ionization energy (in kJ mol^(-1)) of these elements will be
Answer» `NaltMgltHe` `MgltNaltHe` `MgltHeltNa` `NaltHeltMg` Solution :`He : 1s^(2)underset(-1e^(-))overset(DeltaH_(1))rarr1s^(1)` `Na:1s^(2), 2s^(2)2p^(6),3s^(1)underset(-1e^(-))overset(DeltaH_(1))rarr1s^(2), 2s^(2)2p^(6)` `MG: 1s^(2), 2s^(2)2p^(6),3s^(2)underset(-1e^(-))overset(DeltaH_(1))rarr1s^(2),2s^(2)2p^(6),3s^(1)` He has highest value of II IONIZATION energy due to smallest size. After removal of one electron from Na, the `Na^(+)` ion has inert GAS configuration. To remove the electron from this configuration, a very high energy than Mg. So, the order of II ionization energy is : `MgltNaltHe`