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

Most Viewed Questions

Preparation Products

Knockout BITSAT 2022

An exhaustive program for the complete preparation of Physics Chemistry Mathematics English & Logical Reasoning of BITSAT exam, Adaptive Time Table, Chapterwise Questions, Concepts Flashcards for Quick and Effective Revision, Unlimited Chapter wise Subject wise and Full mock test for enhancing Speed & Accuracy,.

₹ 2999/- ₹ 1999/-
Buy Now
Engineering Prep Combo

An exhaustive e-learning program to boost your preparation for the leading engineering entrance exams like:-, BITSAT, VITEEE, MET, SRMJEEE, AEEE, MHT CET, AP EAPCET, TS EAMCET, Unlimited Mock Test for each exam..

₹ 9999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2022

An exhaustive E-learning program for the complete preparation of JEE Main and BITSAT.

₹ 35999/- ₹ 16999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions