Question

What is the second ionization energy of sodium?

Answer 1

Second ionization energy:

1. The energy needed to eliminate the loosely attached electron from the outermost orbital of an isolated gaseous atom is called ionization energy.
2. The energy needed to eliminate the loosely attached electron from the outermost orbital of a neutral gaseous sodium atom is called its first ionization energy.
3. The general equation of the first ionization process is shown below:
   $\mathrm{Na}\left(\mathrm{aq}\right)\to {\mathrm{Na}}_{}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}$
4. The amount of energy needed to eliminate the outermost electron from the outermost orbital of ${\mathrm{Na}}^{+}$ion of the element is called the second ionization energy of sodium.
5. The second ionization energy of an atom is always higher than the first because the positive charge binds electrons more strongly.
6. The general equation of the second ionization process is shown below:
   ${\mathrm{Na}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Na}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}$

A detailed description of the second ionization energy of sodium:

1. The second ionization energy of sodium is 4562 ${\mathrm{kJmol}}^{-1}$ whereas the first ionization energy is 495.8 ${\mathrm{kJmol}}^{-1}$.
2. For the second ionization, the outer shell of the sodium is full so it is very difficult to take out one electron.
3. More energy is needed to eliminate one electron from the outermost shell so it has high second ionization energy.
4. The electrostatic force of attraction of electron towards the nucleus is more to eliminate outermost shell electron.