KVPY SX SB Exam 2015 Question Paper

Question : 51 of 120

Marks: +1, -0

The electronic configuration of an element with the largest difference between the 1

^{\st}

and 2

^{\nd}

ionization energies is

Solution:

Na = 1s^2 2s^2 2p^6 3s^1

Na^{+} → 1s^2 2s^22p^6 [Ne]

[Ne] is inert gas, so, electron removal is very difficult so I.P. is very high.

Na^{+2} –––––▸ 1s^2 2s^2 2p

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