Question

Which of the following has highest second ionization potential?
(A) O
(B) Ne
(C) Na
(D) Mg​

Answer 1

Answer:

Sodium has 11 electrons which can be written as (2, 8, 1). It's outer orbit has 1 electron When it is ionized for the first time, the outer most electron is removed very easily, so that it can form the electronic configuration of the nearest inert gas Neon (2, 8).

Neon has 8 electrons in its outermost orbit. Thus it is highly stable. Now, when the second electron is removed, a very very high amount of energy is required because it doesn't want to give away one of its outermost electron by disrupting its stability.

Thus, second ionization energy of sodium is extremely high.

Explanation:

answer is Na

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Answer 2

Answer:

O = 1s2 , 2s2, 2p4

Ne = 1s2, 2s2, 2p6

Na = 1s2, 2s2 ,2p6 ,3s1

Mg = 1s2, 2s2, 2p6 , 3s2

to check ionization potential 2 we have to remove 1 electron from all and then check stablity of all

most stable element will have highest second ionization potential

O+ = 1s2, 2s2 ,2p3

Ne+ = 1s2, 2s2 ,2p5

Na+= 1s2 , 2s2, 2p6

Mg+ = 1s2, 2s2 ,2p6,3s1

In this Na + will have highest second IP because it got noble gas configuration