
The decreasing order of the second ionization potential of K, Ca and Ba is:
(A) K > Ca > Ba
(B) Ca > Ba > K
(C) Ba > K > Ca
(D) K > Ba > Ca
Answer
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Hint: In order to solve this question electronic configuration of K, Ca and Ba should be considered. The second ionization energy refers to the energy required to remove an electron after an electron has already been removed from the isolated gaseous atom.
Complete step by step answer:
> Ionization energy in simple terms can be described as a measure of the difficulty in removing an electron from an atom or ion or the tendency of an atom or ion to surrender an electron. The loss of electrons usually happens in the ground state of the chemical species.
So, in order to solve this question let’s see the electronic configuration of the compounds,
- Electronic configuration of Potassium: \[\left[ {Ar} \right]4{s^1}\]
The second electron will be released from the '3p' subshell. As this sub-shell is fully filled, so more energy will be required to remove an electron. Hence, this will have the highest second ionization potential of all the elements.
- Electronic configuration of Calcium: \[\left[ {Ar} \right]4{s^2}\]
The second electron will be released from the '4s' subshell. As this subshell is half filled, so less amount of energy will be required to remove an electron than the electron from a fully filled subshell.
- Electronic configuration of Barium: \[\left[ {Xe} \right]6{s^2}\]
The second electron will be released from the '6s' subshell. As this subshell is half filled, so less amount of energy will be required to remove an electron than the electron from a fully filled subshell. As this element lies below Calcium in the same group, so it will have less second ionization potential than calcium.
Hence, the decreasing order of the second ionization potential for the given elements is K > Ca > Ba.
Note: Decreasing order of second ionization potential is K> Ca > Ba and do not confuse it with K< Ca < Ba this is the ascending order of second ionization potential. Comparison signs should be taken care of while marking the answer.
Complete step by step answer:
> Ionization energy in simple terms can be described as a measure of the difficulty in removing an electron from an atom or ion or the tendency of an atom or ion to surrender an electron. The loss of electrons usually happens in the ground state of the chemical species.
So, in order to solve this question let’s see the electronic configuration of the compounds,
- Electronic configuration of Potassium: \[\left[ {Ar} \right]4{s^1}\]
The second electron will be released from the '3p' subshell. As this sub-shell is fully filled, so more energy will be required to remove an electron. Hence, this will have the highest second ionization potential of all the elements.
- Electronic configuration of Calcium: \[\left[ {Ar} \right]4{s^2}\]
The second electron will be released from the '4s' subshell. As this subshell is half filled, so less amount of energy will be required to remove an electron than the electron from a fully filled subshell.
- Electronic configuration of Barium: \[\left[ {Xe} \right]6{s^2}\]
The second electron will be released from the '6s' subshell. As this subshell is half filled, so less amount of energy will be required to remove an electron than the electron from a fully filled subshell. As this element lies below Calcium in the same group, so it will have less second ionization potential than calcium.
Hence, the decreasing order of the second ionization potential for the given elements is K > Ca > Ba.
Note: Decreasing order of second ionization potential is K> Ca > Ba and do not confuse it with K< Ca < Ba this is the ascending order of second ionization potential. Comparison signs should be taken care of while marking the answer.
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