Because of the stability of the formed rings in case of bipyridine.

see http://actachemscand.dk/pdf/acta_vol_31a_p0813-0824.pdf

A spectrochemical series is a list of ligands ordered on ligand strength and a list of metal ions based on oxidation number, group and its identity. In crystal field theory, ligands modify the difference in energy between the d orbitals (Δ) called the ligand-field splitting parameter for ligands or the crystal-field splitting parameter, which is mainly reflected in differences in color of similar metal-ligand complexes.

A partial spectrochemical series listing of ligands from small Δ to large Δ is given below. (for a table, see the ligand page)

I− < Br− < S2− < SCN− < Cl− < NO3− < N3− < F− < OH− < C2O42− ≈ H2O < NCS− < CH3CN < py (pyridine) < NH3 < en (ethylenediamine) < bipy (2,2'-bipyridine) < phen (1,10-phenanthroline) < NO2− < PPh3 < CN− ≈ CO

Ligands arranged on the left end of this spectrochemical series are generally regarded as weaker ligands and cannot cause forcible pairing of electrons within 3d level and thus form outer orbital octahedral complexes. On the other hand ligands lying at the right end are stronger ligands and form inner orbital octahedral complexes after forcible pairing of electrons within 3d level

Bipyridine acts as a bidentate ligand but pyridine is a monodentate ligand. Bidentate ligands forms ring so that stability of the compounds formed by the chilate ligands is more. That is why bipyridine is placed towards strong field ligands.

Both 2,2'-bipyridine and 1,10-phenanthroline can be viewed as alpha-diimines, and as such they have empty pi* orbitals with the proper orientation to serve as pi acceptors placing them in the strong field portion of the spectrochemical series.

I thing the simplest logical still requiring testing explanation would be conjugation effect meaning that placing electronic density on the pi-orbital of the single ring of py destabilizes it (unoccupied orbitals are of anti-bonding nature) hence the gain in M-N bond strength is counterbalanced by loss of ring stability. While if we move on to 2,2'-bipy that have pi-orbital made of two conjugated pyridine ring such destabilization is of lesser extent. It would be good for checking this concept by calculating how much HOMO-LUMO gap has been reduced. It is sufficient to calculate UV-VIS spectrum of both compounds and look for lowest electronic excitation energy change if dropped significantly then the LUMO orbital become more of non-bonding character hence it can store electronic density without weakening ligand molecule. It's just the of hypothesis but worth checking.

It may kindly be treated as an academic discussion where I (aged 70), very humbly, agree as well as disagree with my young friends as follows:

[1]There is no denying the fact that (py) is much weaker than (biby).But both (py) and (bipy) should be looked at in two different ways- first where they can form complexes involving only MLCT phenomenon and no back bonding and secondly where they form complexes involving both MLCT and back bonding phenomenons. In the former case, the transition metal ions are in the higher oxidation states (OS =+2, 3) while in the latter, they should exhibit low OS (0, +1).

[II] You may call it the over simplification if I submit that the main reason for (bipy) acting as a stronger ligand is the chelate effect which can be seen by comparing the reaction of (bipy) and (py) with Ni (2) wrt their log beta and standard delta G values.

Reaction of (py) and (bipy) with Ni (2).

2py (1bipy) = 3.5(6.9) ;-20(39);

4py (2bipy) =5.6(13.6) ; -32 (- 78);

6py (3bipy)=9.8(19.3);9.8(19.3);-56(-110).

[III] There is no denying the fact that because of the electronegative nitrogen in the (py) ring, the molecule is relatively electron deficient (a very week pi-acceptor) while (bipy) having an other electron withdrawing substituent is all the more electron deficient .But this electron deficiency in (bipy) becomes all the more pronounced due to chelate effect and largely enhances its pi-back acceptor character.

[IV] If we compare the aromaticity index then if benzene is 100, (py) is 82 while (bipy) is only 50.The same conclusion can be drawn from their resonance energies- benzene(150), (py)[117] while for (bipy) no value is reported.The least values of both the aromaticity and resonance energy of (bipy) should make it much stronger a pi-acceptor.

[VI]Lastly, no relative values of energies of the pi and pi star molecular orbitals of (py) and (bipy) could be found in the literature.

Since none of these comments really answer the question, how about rephrasing it to: " Why is pyridine a weaker field ligand tan ammonia? Is pyridine not a pi acceptor at all?

In fact bipyridine forms a chelate with a metal ion and thus the complex formed is more stable.Therefore, bipyridine is a stronger ligand than pyridine because it is a bidentate ligand.

Fathi Aqra: Did you not read the message right above your post?

Non of the two ligands is pi acceptor at all. Both of them are sigma donors but ammonia is a stronger sigma donor than pyridine because the pi electrons of the ring is electron withdrawing in order to maitain the stability of the ring which in turn weakens the sigma donation of the nitrogen atom of the ring, so it becomes weaker ligand than ammonia.

Pyridine and some other nitrogen tertiary bases such as 3-methyl pyridine,4- methyl pyridine quinoline, isoquinoline being significantly soft ligands are typical in the sense that they are both weak sigma- donors as well as week pi- acceptors and become weaker than NH3 ligand which happens to be only a strong sigma donor ligand. In away, one strong factor (sigm donation) in NH3 predominates over the two week factors in the case of pyridine and other N-tertiary bases.

The fact that (Py) can act as week pi-acid ligand finds references in literature at least at two places:

[1] Chemistry of Heterocyclic compounds:

Pyridine Metal complexes,

Part-6, Vol 14

Chapter 5.

Pi-Coordination Compounds of Pyridine With Metals, pp- 2213.

[2] The complexes [MPy4AB] (M=Fe, Co, Ni; AB=F, Cl, Br) showing pi- acid chacacter of (Py) has been reported.

[3] Now, a humble query if (Py) does possess vacant pi –star MOs (like C6H6) then why should not it act a pi- acceptor, however weak it may be?

[4] Please find below two orders showing relative order of the strengths of NH3 and (Py) (both reported in the literature).

I− < Br− < S2− < SCN− < Cl− < NO3− < N3− < F− < OH− < C2O42− < H2O < NCS− < CH3CN < py < NH3 < en < bipy < phen < NO2− < PPh3 < CN− < CO.

I- < Br- < SCN- < Cl- < F- ≤ OH- , ONO- < OH2 < NCS- < NCCH3 < NH3, py < NO2- < CN- , NO, CO.

[5] Lastly, right or wrong, all credit to Mdm Laura Gasque, RG and friends who participated to make these questions a healthy academic discussion.

OK, sorry to write so late in an old discussion, but it took me some time to get my hands on my PhD thesis. I studied in it biphosphabenzene, and made an experimental and theoretical comparisons of benzene, pyridine, phosphabenzene, biphenyl, bipyridine and biphosphabenzene.

First of all, about the difference of pyridine and ammonia. Pyridine is both a sigma-donor, its HOMO being the lone pair, and it is indeed a not-so-weak pi-acceptor, its LUMO being a pi* with a strong coefficient on the chelating nitrogen. The spectrochemical series can be linked to the energy level of the corresponding lone pairs, which is also linked to the ionization potential. In the case of NH3 and pyridine, the link is easy since both HOMOs describe the lone pair, ie they have mainly nitrogen sp3 (NH3) or sp2(pyridine) character. One finds a stronger IP for NH3 (10.07eV, CRC Handbook) than for pyridine (9.70eV, ChemPhys, 1975, 10,345). Everything in the gas phase of course. One finds then that NH3 should be a weaker ligand than pyridine, which is not what is noted in the spectrochemical series available in the litterature. All the more so since pyridine is also a pi-acceptor given its low-lying LUMO.

Though this would be worth a thorough theoretical investigation with simple models of transition metal complexes, my guess is that the answer lies in the study of Chen and Chang (JMolStructTheochem, 1998, 431, 127). Indeed one fundamental diference between pyridine and ammonia is that ammonia is a flexible ligand, with its structure influencing strongly the level of the lone pair. While pyridine with its aromatic ring can be expected to be less influenced by its coordination to a metal. In their study, Chen and Chang show a variation of about 0.08eV per degree for the HNH angle opening of ammonia. So with a starting angle at 107.8deg, you only need to open it up to 112deg to raise the lone pair level to pyridine. And actually that is what seems to happen. If you look at the Cambridge database (restricting to R1

The energy level of anti pi orbital of bpy is lower than in py which leads to increase back donation, binding, short bond, meaning high filed.