1. Na

  2. Mg

  3. Al

  4. K

Answers (1)

E.C. of all the elements.

1) Na = Ne 3s1

2) Mg = Ne 3s2

3) K = [Ar] 4s1

4) Al  = [Ne] 3s2 3p1


Now to compare second ionization energy the conf. will be


1) Na = Ne

2) Mg = Ne 3s1

3) K = [Ar]

4) Al  = [Ne] 3s2


Now Sodium and Potassium have achieved their stable conf. i.e noble gas conf. or now are fully filled and thus more stable than Mg and Al.

Now in K and Na,

Na will have highest second ionisation energy because, Neon is smaller than Argon considering difficult to ionize or remove electrons from Neon as compare to Argon

Option (1) is correct