
The right order of ionization potential of Li, Be, B and C is :
A) C > Be > B > Li
B) C > B > Be > Li
C) C > B > Li > Be
D) B > C > Be > Li
Answer
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Hint: Ionization Potential is the amount of energy required to remove the electron from an atom. It is also known as Ionization Energy. The removal of electrons always takes place from the outermost shell of an atom.
Complete answer:
Now we have to find out the ionization potential of the given elements. All the elements belong to the same period i.e second. When we are moving in a period the electrons to be added in the same shell but nuclear charge and number of electrons will differ.
So according to the above definition, across a period the Ionization potential increases from left to right due to increase in net nuclear charge of elements. So it should be like C > B > Be > Li. But there is an exception in the case of Be and B the ionization potential of Be is higher than that of B. It is because of the fact that electronic configuration of Be is 1${s^2}$ 2${s^2}$ and that of B is 1${s^2}$2${s^2}$2${p^1}$ and as electron in p can be removed easily as compared to the electron from s in Be. This is because of fully filled s orbitals. So less energy will be required in the case of B than in Be. So the Ionization potential of Be will be greater than that of B. So the correct order will be like C > Be > B > Li instead of like C > B > Be > Li.
Therefore the correct answer is (A).
Note: Fully filled or half filled electron configurations are more stable so more energy will be required if we have to remove the electrons from these shells. This type of exceptional behaviour can be seen many times in the periodic table so always write the electronic configurations first and then choose the right answer.
Complete answer:
Now we have to find out the ionization potential of the given elements. All the elements belong to the same period i.e second. When we are moving in a period the electrons to be added in the same shell but nuclear charge and number of electrons will differ.
So according to the above definition, across a period the Ionization potential increases from left to right due to increase in net nuclear charge of elements. So it should be like C > B > Be > Li. But there is an exception in the case of Be and B the ionization potential of Be is higher than that of B. It is because of the fact that electronic configuration of Be is 1${s^2}$ 2${s^2}$ and that of B is 1${s^2}$2${s^2}$2${p^1}$ and as electron in p can be removed easily as compared to the electron from s in Be. This is because of fully filled s orbitals. So less energy will be required in the case of B than in Be. So the Ionization potential of Be will be greater than that of B. So the correct order will be like C > Be > B > Li instead of like C > B > Be > Li.
Therefore the correct answer is (A).
Note: Fully filled or half filled electron configurations are more stable so more energy will be required if we have to remove the electrons from these shells. This type of exceptional behaviour can be seen many times in the periodic table so always write the electronic configurations first and then choose the right answer.
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