The decreasing order of the second ionization potential of $K, C a$ and $B a$ is :

K > C a > B a

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C a > B a > K

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B a > K > C a

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K > B a > C a

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Solution

The correct option is A $K > C a > B a$
As we move down the group size increases and the electrons experience less effective nuclear charge. Due to this reason, ionization potential decreases down the group. Thus, the second ionization potential of Ca is greater than that of Ba.

The second ionization potential of K is greater than that of Ca as K acquires stable noble gas configuration after losing one electron.

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