Chapter 11.4, Problem 6RQ

Interpretation Introduction

Interpretation: The property in previous problem needs to be explained by using ionization energy trends.

Concept introduction:The ionization energy of an atom is the energy required to remove an electron from an individual atom in the gas phase.

Answer to Problem 6RQ

  ${\text{IP}}_2>{\text{IP}}_{\text{1}}$ due to full filled stability.

Explanation of Solution

Let us take the example of Boron,

  ${}_{\text{5}}{\text{Br}}^{\text{11}}={\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{1}}$

Here, p-orbital have 1 electron. Removing the 1^st electron from the orbit is ${\text{IP}}_1$

  ${}_{\text{5}}{\text{Br}}^{\text{+}}={\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^0$

Here, p-orbital have 0 electron. Removing the 1^st electron from the orbit is ${\text{IP}}_2$

So, ${\text{IP}}_{\text{2}}>{\text{IP}}_{\text{1}}$ .

Conclusion

Electronic configuration is used to find the ionization energy.