Chapter 6, Problem 60A

Interpretation Introduction

Interpretation:

It is to be explained why there is a large jump between the second and third ionization energies of magnesium and also the large gap between the third and fourth ionization energies of aluminum.

Concept Introduction :

The first ionization energy is the energy needed for the removal of the outermost electron from the neutral gaseous atom.

  $X\left(g\right)+I{E}_1\to X^{+}\left(g\right)+e^{-}$

While second ionization energy is the energy needed for the removal of the outermost electron from the unipositive gaseous ion.

  $X^{+}\left(g\right)+I{E}_2\to X^{2+}\left(g\right)+e^{-}$

Answer to Problem 60A

After removing two electrons, magnesium attains a noble gas configuration.

Similarly, after removing three electrons, aluminum attains a noble gas configuration.

Explanation of Solution

The atomic mass of magnesium is 12 and its electronic configuration is as shown:

  $\text{Mg}\left(12\right):1s^{2}2s^{2}2p^{6}3s^2$

After removing one electron, the electronic configuration becomes as shown,

  ${\text{Mg}}^{\text{+}}\left(12\right):1s^{2}2s^{2}2p^{6}3s^1$

Second ionization energy is the energy needed to remove the outermost electron from a unipositive isolated cation. After removing the second electron, the electronic configuration becomes as,

  ${\text{Mg}}^{\text{2+}}\left(12\right):1s^{2}2s^{2}2p^6$

After removing the second electron, the ${\text{Mg}}^{\text{2+}}$ attains noble gas configuration and hence difficult to remove any further electron. This is the reason why there is a large gap between the second and third ionization energies of magnesium.

The atomic mass of aluminum is 13 and its electronic configuration is as shown:

  $\text{Al}\left(13\right):1s^{2}2s^{2}2p^{6}3s^{2}3p^1$

After removing one electron, the electronic configuration becomes as:

  ${\text{Al}}^{\text{+}}\left(13\right):1s^{2}2s^{2}2p^{6}3s^2$

Second ionization energy is the energy needed to remove the outermost electron from a unipositive isolated cation. After removing the second electron, the electronic configuration becomes,

  ${\text{Al}}^{\text{2+}}\left(13\right):1s^{2}2s^{2}2p^{6}3s^1$

Third ionization energy is the energy needed to remove the outermost electron from a dipositive isolated cation. After removing the second electron, the electronic configuration becomes,

  ${\text{Al}}^{\text{3+}}\left(13\right):1s^{2}2s^{2}2p^6$

After removing the third electron, the ${\text{Al}}^{\text{3+}}$ attains noble gas configuration and hence is difficult to remove any further electrons. This is the reason why there is a large gap between the third and fourth ionization energies of aluminum.

Conclusion

Electronic configurations of elements decide their ionization energy.