Why does hydrogen have a lower ionization energy than fluorine? I found here that the ionization energy of hydrogen is $\pu{1312kJ/mol}$ and for fluorine, it is $\pu{1681kJ/mol}$. Now clearly, from the data, we can see that hydrogen has a lower ionization energy when compared to fluorine.

I did not expect this because I saw here that fluorine is twice as big as hydrogen (I am referring to the empirical values of the atomic radii). This means that it would be significantly easier to remove an electron from a fluorine atom than from a hydrogen atom. That's why I expected hydrogen to have a higher ionization energy. But I found that it's the other way round. Why is it so? What is wrong with my reasoning? Is this an exception?
Fluorine is bigger, but there are $9$ protons in the nucleus, compared to $1$ in case of hydrogen. So, the outermost electron in fluorine experiences more attraction than what hydrogen's single electron experiences, in spite of larger size of fluorine.

However, the outermost electron in fluorine doesn't experience attraction from all the $9$ protons (nuclear charge). Due to improper shielding, the effective nuclear charge experienced by one $2p$ electron of fluorine becomes -
$\mathrm{ [9 - {(6 \times 0.35) + (2 \times0.85)}]= 5.2}$. 

In comparison, the single electron in hydrogen experiences effective nuclear charge of $1.0$ 

Note: For hydrogen, there is no other electron to shield nuclear charge. So, effective nuclear charge = actual nuclear charge.

Source: To know how to calculate effective nuclear charge visit this Wikihow link.