# Why is the second ionization energy greater than the first ionization energy in atoms?

Nov 23, 2016

$M \left(g\right) + {\Delta}_{1} \rightarrow {M}^{+} \left(g\right) + {e}^{-}$ $\text{1st ionization energy}$

${M}^{+} \left(g\right) + {\Delta}_{2} \rightarrow {M}^{2 +} \left(g\right) + {e}^{-}$ $\text{2nd ionization energy}$

#### Explanation:

Let's look at the process purely on the basis of electrostatics. It should take more energy to remove an electron from a positively charged atom, than from a neutral atom, because the valence electrons of a positive species should be more strongly attracted to the nuclear core, and shielding of the nuclear charge by electrons should diminish.

On this basis alone, ${\Delta}_{2} > {\Delta}_{1}$.

In the special case of the alkali metals, the second ionization energy would involve the removal of a non-valence, inner shell electron. And thus, in this instance, ${\Delta}_{2}$ should be PROHIBITIVELY HIGH.....

Are the given data consistent with what we have argued? Why is a second ionization energy not listed for hydrogen?