# Why don't electrons pair in [Fe(NH3)6]2+?

Why doesn't electron pairing take place in $\ce{[Fe(NH3)6]^2+}$ though $\ce{NH3}$ is a strong ligand field producer?

• Because $\ce{NH3}$ isn’t a strong-field ligand. – Jan Sep 16 '16 at 15:25
• Good thing I found this page, for an entire year I thought $\ce{NH3}$ was a strong ligand. – Pritt Balagopal May 19 '17 at 13:12

(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.

• There are not two types of ligands. At the very best, there are two extremes and a very broad range of in-betweens between them. – Jan Sep 16 '16 at 15:45
• @Jan actually, you can classify them based on whether $\Delta$ is greater or less than pairing energy for the electrons. – Pritt Balagopal May 19 '17 at 13:13
• @PrittBalagopal No you cannot. A certain combination of metal and ligand can exist in both a high and low-spin configuration. One of my coordination-chemistry answers has the data which I am too lazy to dig up right now. All I remember is that the ligand is nitrogen-based. – Jan Jun 7 '17 at 20:30

The dichotomy between weak- and strong-field ligands that I have often seen constructed here is false and physically inaccurate.

Complexes can be in a low-spin state or in a high-spin state. The high-spin state is the one according to Hund’s rule where electrons are filled into the five d orbitals one after the other. A low-spin state is any complex in which the Hund rule is not strictly observed, but some orbitals are filled with two electrons while others remain empty. Classic examples for high-spin complexes are, e.g. $\ce{[FeCl6]^3-, [CoCl4]^2-}$ and others. Examples for low-spin complexes are, e.g. $\ce{[Fe(CN)6]^3-, [Cr(CN)6]^3-}$ and others.

While these examples may seem to exemplify your stated dichotomy (with chloride being on the weak-field side and cyanide being on the high-field side), reality is much more complex and the spin state of a complex depends on ligands, metal, oxidation state, counterions and so on. In fact, for 2-picolylamine ($\ce{2{-}pic}$) as a ligand to iron(II), both the low-spin and the high-spin $\ce{[Fe(2{-}pic)3]^2+}$ complexes have been isolated and characterised by crystal structures. The $\ce{Fe-N}$ bond length differs by $19~\mathrm{pm}$ with the high-spin complex obviously having longer bonds.

Complexes have a tendency to be in the low-spin state if

• the metal is not a 3d metal
• the metal’s oxidation state is higher
• metal to ligand π-backbonding can take place.

Especially that final point is important. Ammonia is very bad at π backbonding (and amide is even π-donating). So calling it a high-field ligand is misappreciating physicochemical reality. It is just ammonia’s strong σ bonding ability that causes a slightly larger field split than water. As such, ammonia is best termed a medium-field ligand. (The same is true for most neutral nitrogen donors.)

• Nice thank.! So for third point, N does not have empty orbital in Nh3 is that reason for no pi back bond. Thank generous soul – King Tut Jan 22 '18 at 15:21

$\ce{NH3}$ is placed in the middle of the spectrochemical series. It mostly acts as SFL but sometimes may act as WFL. When a metal ion having higher charge is interacting with ligand NH3 it acts as a SFL. When the metal ion has lower charge it acts as WFL.

EG: $\ce{Co^2+}$ and $\ce{Co^3+}$ In $\ce{Co^2+}$ $\ce{NH3}$ acts as WFL.

SFL:Strong field ligand

WFL: weak..