# π-Acceptor ligands

Which of the following are π-acceptor ligands?

1. $$\ce{PR3}$$
2. $$\ce{Cl-}$$
3. $$\ce{NH3}$$
4. $$\ce{H-}$$

I know that $$\ce{PR3}$$ is a π-acceptor, and that $$\ce{Cl-}$$ is a π-donor ligand from my lectures. I am uncertain however as to whether $$\ce{NH3}$$ and $$\ce{H-}$$ can be π-acceptor ligands. My first thoughts is that $$\ce{H-}$$ would not be a π-acceptor, while I have no idea how to tell if $$\ce{NH3}$$ is a π-donor or π-acceptor.

Like $\ce{PR3}$, $\ce{NH3}$ or $\ce{NR3}$ are π-acceptor ligands because they have an unoccupied σ* orbital, which can accept electrons from the metal's d-orbitals. For both phosphine and ammona, there is a backbonding $n_{M}$$\ce{->}$$σ^*_{N/P}$ interaction. These ligands can act as π-acceptors in much the same way as $\ce{CO}$, except that they have unoccupied σ* orbitals rather than π* orbitals. $\ce{NH3}$ is, however, a weaker π-acceptor than $\ce{PH3}$, but it is still a fairly high field-ligand. By electronegativity, you would expect the nitrogen ligand to hold the metal's d electrons more strongly, but in the case of ammonia vs. phosphine based ligands, the phosphorus ligand has bigger p-orbitals, which have better overlap with the metal's d-orbitals, so the $n_{M}$$\ce{->}$$σ^*_{P}$ interaction is stronger than $n_{M}$$\ce{->}$$σ^*_{N}$.

$\ce{H^-}$ is $1s^2$, so it has no low-energy p-orbitals to π-bond with. It is neither a π-acceptor or a π-donor. It is only a σ-donor.

• By looking at ammonia's placing in the spectrochemical series, it is logical to deduce that the interaction with the N-H antibonding MO is rather weak since it is actually placed together with most of the sigma donors, instead of being placed with the other pi acceptors. I agree that this is likely due to the difference in size relative to the metal d orbitals. – Tan Yong Boon Apr 7 '19 at 15:10

$$\ce{PR3}$$ is π-acceptor (from metal d orbital to phosphorous d orbital electron cloud is transferred).

$$\ce{Cl-}$$ is π-donor (minus charge form σ bond with the metal orbitals, $$\mathrm{e_g}$$, $$\mathrm{t_{1u}}$$ and $$\mathrm{a_{1g}}$$ orbitals in $$O_\mathrm{h}$$ field. The lone pair of electron on $$\ce{Cl}$$ will try to form π-bonding with $$\mathrm{t_{2g}}$$ orbitals of metal, electron cloud is transferred from $$\ce{Cl}$$ to metal d orbitals).

$$\ce{NH3}$$ is σ-donor. The lone pair is used to form sigma bond with metal. No extra electron to give to metal and also no d orbital to accept electron from metal. $$\ce{P}$$ and $$\ce{N}$$ belong to same group but differ here.

$$\ce{H-}$$ is σ-donor; reason same as above.

Alright; let me try.

In Cl-, the “HOMO” of the ion itself are the p orbitals; the two pi-bonding p orbitals act as a pi bonding (weak field) ligand.

In NH3, the HOMO is the 3a1 bonding MO (sigma donor) and the LUMO is the 4a1 antibonding MO. The LUMO+1 are the antibonding 2e orbitals (which involve the p orbitals). As the LUMO can’t act as a pi acceptor, it’s a sigma donor only.

In PR3, the HOMO is again a bonding MO similar to the 3a1 is NH3, BUT the antibonding “e” orbitals involving the p orbitals are lower in energy than the “a1” antibonding. As the LUMO is hence a pi acceptor, it’s strong field.

In H-, the case is similar to NH3, where the p orbitals are too high in energy.