# Question #eb5ac

Feb 1, 2017

Before we consider electronic energy levels, let us consider electrostatics............

#### Explanation:

We interrogate 2 processes:

$\left(i\right)$ $L i \left(g\right) + {\Delta}_{1} \rightarrow L {i}^{+} \left(g\right) + {e}^{-}$

$\left(i i\right)$ $L {i}^{+} \left(g\right) + {\Delta}_{2} \rightarrow L {i}^{2 +} \left(g\right) + {e}^{-}$

Now just from an electrostatic point of view, we would anticipate that ${\Delta}_{2} > {\Delta}_{1}$ simply on the basis that it should be harder to remove an electron from a positively charged species, $L {i}^{+}$, to make a dication, $L {i}^{2 +}$, than from the neutral metal atom to make a cation. Why? Because the electron is more tightly bound to a formal cation.

So, as physical scientists, let us put in some numbers:

This site reports that for lithium, ${\Delta}_{1} = 520.2 \cdot k J \cdot m o {l}^{-} 1$, and ${\Delta}_{2} = 7298.1 \cdot k J \cdot m o {l}^{-} 1$.

So in fact our facile expectation was correct. However, the substantial magnitude of ${\Delta}_{2}$, which dwarfs ${\Delta}_{1}$, also reflects the electronic stucuture of the lithium atom. The second ionization requires the removal of a $1 {s}^{2}$ electron, rather than than a $2 {s}^{1}$, which was the one first removed. Because this second electron is inner shell, its attraction to the nucleus should be quite substantial, and our data support this expectation.