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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.

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3 Answers 3

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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.

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    $\begingroup$ 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. $\endgroup$ Commented Apr 7, 2019 at 15:10
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$\ce{PR3}$ 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.

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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.

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