Question

The element having greatest difference between first and second ionization energies is:
A ) Ca
B ) K
C ) B
D ) Se

Answer 1

Hint: The element having the greatest difference between first and second ionization energies is the one which attains stable noble gas electronic configuration, upon first ionization. Such an element has only one electron in its valence shell.

Complete answer:
Let us write the electronic configurations of each and every element and then see how many electrons in the outermost shell left after first and second ionisation energies.
> The atomic number of Ca is 20. Its electronic configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}\]. It has 2 valence electrons. When it loses one valence electron, the electronic configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}\]. Still the valence shell has one electron. The noble gas electronic configuration is achieved by the loss of a second electron. Hence, there is no greatest difference between first and second ionization energies.
> The atomic number of K is 19. Its electronic configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}\]. It has 1 valence electron. When it loses one valence electron, the electronic configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}\]. The noble gas electronic configuration is achieved by the loss of the first electron. To lose a second electron, a huge amount of energy is needed as stable noble gas electronic configuration is disturbed. Hence, there is the greatest difference between first and second ionization energies.
> The atomic number of Ba is 56. Its electronic configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^2}4{d^{10}}5{p^6}6{s^2}\]. It has 2 valence electrons. When it loses one valence electron, the electronic configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^2}4{d^{10}}5{p^6}6{s^1}\]. Still the valence shell has one electron. The noble gas electronic configuration is achieved by the loss of a second electron. Hence, there is no greatest difference between first and second ionization energies.
> The atomic number of Se is 34. Its electronic configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^2}4{p^4}\]. It has six valence electrons. When it loses one valence electron, the electronic configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^2}4{p^3}\]. Still the valence shell has five electrons. The noble gas electronic configuration is achieved by the loss of a sixth electron. Hence, there is no greatest difference between first and second ionization energies.
Thus, the element having the greatest difference between first and second ionization energies is K.

Hence, the correct option is the option B ).

Note: The ionization energy is the amount of energy needed to remove an electron from an isolated gaseous atom. When the first electron is removed, the ionization energy is the first ionization energy. But when the second electron is removed from mono positive cation, the ionization energy is the second ionization energy.