Question

The decreasing order of $I{{E}_{2}}$ of K, Ca and Ba is:
(a) $K>Ca>Ba$
(b) $Ca>Ba>K$
(c) $Ba>K>Ca$
(d) $K>Ba>Ca$

Answer 1

Hint: First of all, identify the positions of these elements in the periodic table and then, write the decreasing order of their $I{{E}_{1}}$and after that, you can easily know about the decreasing order of there $I{{E}_{2}}$. By the I.E. We mean the energy released when a gaseous atom loses electrons from its valence shell. Now solve it.

Complete answer:
Now, first of all let’s discuss what is ionization energy. By ionization energy we mean the energy released when the gaseous atom loses electrons from its outermost valence shell.
Ionization increases as we move from left to right because size of the atom decreases and electrons are held more tightly by the nucleus and hence, a large amount of energy is released when they lose the electrons and hence, have high ionization energies.
On the other hand, Ionization decreases as we move from top to bottom because size of the atom increases and the outermost electrons are loosely bound to the nucleus and hence, are easily removable and a small amount of energy is released when they lose the electrons and hence, have low ionization energies. Now considering the statement; First of all, we will identify their positions from the periodic table. The decreasing order of $I{{E}_{1}}$ of K, Ca and Ba is as; $Ca>Ba>K$
I.E. of the calcium will be highest. It is so because :
1.It has a smaller size than the potassium and barium.
2.It has stable half-filled electronic configuration and releases a lot of energy when an electron is removed from its valence shell.
Now, next coming to the barium. Though it has larger size than the potassium but it has stable half-filled electronic configuration in its shell and thus, releases more energy than the potassium atom when the electron is removed from its valence shell and thus, has high I.E. then the potassium and potassium have the least $I{{E}_{1}}$.
So, the decreasing order of $I{{E}_{1}}$is justified.
Now, the decreasing order of $I{{E}_{2}}$ of K, Ca and Ba is as;
$K>Ca>Ba$
Potassium has the highest $I{{E}_{2}}$. It is so because after the lost one electron , it acquires a stable electronic configuration of neon in its valence shell and thus, when an electron is removed from it , a lot of energy is released. On the other hand, in case of calcium and barium ions, after the loose of one electron from their valence shell, they can easily lose the second electron but due to the bigger size of the barium atom than the calcium atom, the electron is removed easily and energy released is less as compared to the energy released by the calcium ion and thus, $I{{E}_{2}}$ of Ca is more than the Ba.
So, the decreasing order of $I{{E}_{2}}$ is justified.

Hence, option (a) is correct.

Note: Don’t get confused in the term’s ionization energy and electron affinity. By ionization energy, we mean the energy released when an atom loses the electron. On the other hand, by electron affinity, we mean the energy released when an atom gains the electron.