Does ionization energy have anything to do with how reactive an element is?
This comes about as a consequence of the effective nuclear charge on the atom or Z. eff, and the size of the atom. Z effective if you are not familiar is a concept usually touched upon in chem 2 as a periodic trend. It is a measure of the charge felt by the outer most electron. It is calculated by accounting for the electrons on the inner shells essentially canceling out or shielding the outer electron from feeling all of the charge from the positive nucleus. As you could expect this make the electron easier or harder to take away causing ionization depending on the Zeff. A small atomic radius causes the energy barrier to be higher to remove an electron, because the electron is closer to the nucleus. If the atom is larger it will be easier to remove the outer electron because it's further away from the nucleus. As you move along the periods of the periodic table from left to right and low to high, the effective nuclear charge increases. As you move along the periodic table from top down and right to left, the atomic radius increases. 

We find that fluorine is at the top right of the periodic table, where the effective nuclear charge is high, and the radius of the atom is small. This means that it has a small condensed positive charge permeating through the inner shells. This causes the reactivity of fluorine to other atoms due to it's electron affinity from the unshielded nucleus. It is hard to ionize compared to sodium because of its smaller than sodium, and it's z eff is higher. These both cause the electron to be attracted to the nucleus more, causing a higher energy barrier to take it away.