Why don't electrons pair in [Fe(NH3)6]2+?
(If you know the basics, go to last paragraph and skip the introduction part)

There are two types of ligands. Strong field ligands and weak field ligands. Those which only cause a small extent of crystal field splitting is called weak field ligand and those which cause large splitting is called strong field ligand.

Why does crystal field splitting take place?

Let's take an octahedral complex. The $d_{x^2-y^2}$ and $d_{z^2}$ point along x,y,z axis and the $t_{2g}$ orbitals point between the axis. When the 6 ligands approach along the three axes, the energy of the $e_g$ orbitals increase more than the increase in the energy of $t_{2g}$ orbitals. Hence the d-orbital splits into two. Filling each electron in $t_{2g}$ orbitals decrease energy by $0.4\Delta_0$ and filling electrons in $e_g$ increase energy by $0.6\Delta_0$

Coming back to your question, you know how electrons are filled in the orbitals (according to hund's rule of course). But how does electron pairing take place in your case?

Let the energy required to pair an electron be P.
Let's say there are 4 electrons. The two arrangements are one electron in each sub-orbital (no pairing) and 4 electrons all in $t_{2g}$ itself (instead of 4th electron going to $e_g$, its paired up in $t_{2g}$).
In the first case, the total energy is $-0.6\Delta_0$ and in second case, its $-1.6\Delta_0+P$. So, only strong field ligands can form the second arrangement. $P$ value is constant for a given metal ion.

Finally, ammonia is weak field ligand but in certain cases, it acts as a strong field ligand (example: for Cobalt). For Fe, ammonia is a weak field ligand. Hence no pairing takes place.