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I was going through a list of monodentate ligands, and I came across examples like $\ce{H2O}$ (which has 2 lone pairs), $\ce{Br-}$, $\ce{I-}$ (both of them have more than one lone pair). Similarly, thiourea has 4 lone pairs in all (2 from nitrogen and sulphur each).

When I searched about it, I realized that the oxygen in water can't simultaneously form bonds using both the lone pairs (due to some obvious geometrical reasons, it said).

However, I couldn't figure out any such (obvious) reasons in thiourea.

Can anyone please help?

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    $\begingroup$ There are plenty $\mu_2$-thiourea complexes with Cd, Pb, Ag, Cu. Just saying. $\endgroup$
    – andselisk ♦
    Dec 7 '17 at 9:13
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    $\begingroup$ He said that thiourea can act as a bridging ligand. It is also wrong to say that water can only form one co-ordinate bond, the reasons why it often doesn't are not at all obvious. Could you comment, at which level you are studying and which source said that it is obvious. $\endgroup$
    – Martin - マーチン ♦
    Dec 7 '17 at 11:17
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This is not really a standalone answer, rather a pointer or an exeption. There are few complexes where deprotonated thiourea $\ce{S=C(NH2)NH}^-$ acts as a bidentate ligand. One example is (aminomethanethioamido)-hydrido-tetrakis(trimethylphosphine)-technetium(III) $\ce{[Tc(P(CH3)3)4(SCN2H3)]+}$ complex[1]:

enter image description here

The fact that thiourea ligand is deprotonated and is coordinated asymmetrically within equatorial plane in $\eta^2-\ce{N,S}$ bonding mode has been proven by NMR and SCXRD studies. From the article [1, p. 4642] (emphasis mine; $\ce{tu}$ = thiourea $\ce{S=C(NH2)2}$):

The $\ce{Tc-N}$ distance, 2.190(13) Å, is long when compared to known $\ce{Tc(III)}$ complexes. The range for $\ce{Tc-N}$ bonds was found to be 2.041 Å ... to 2.124 Å ... . This reflects delocalization of negative charge within the bidentate deprotonated thiourea ligand. This elongation is also seen in the $\ce{Tc-S}$ bond, 2.543(4) Å, which is significantly longer than $\ce{Tc-S}$ bonds in the parent compound $\ce{[Tc(\textit{S}-tu)6][Cl]3}$ ... .

As Martin pointed out, the reasons for the observed denticity are not always transparent. Simply put, it seems like in this case in order to be a bidentate ligand, thiourea molecule needs to increase its nucleophilicity and remove steric hindrance caused by $\ce{H}$-atoms lying within the $\sigma_\mathrm{h}$ plane. Both can be achieved by eliminating a single proton from the thiourea molecule.

Reference

  1. Watson, P. L.; Albanese, J. A.; Calabrese, J. C.; Ovenall, D. W.; Smith, R. G. Inorg. Chem. 1991, 30 (24), 4638–4643. DOI: 10.1021/ic00024a035.
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  • $\begingroup$ Okay. So as far as it has all the H atoms, it acts as a monodentate ligand. If we somehow remove an H atom, it can act as a bidentate ligand, right? Is there any possibility that it can act as a tridentate ligand (because it has 4 lone pairs) ? $\endgroup$
    – user166465
    Dec 7 '17 at 13:07
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    $\begingroup$ @user166465 Well, I haven't discovered any examples of existing compounds (certainly not in crystal structure database), plus how you would present it geometrically? Thiourea ligand is basically a triangle which you are trying to attach to a single point by all three vertices, which leaves two nucleophilic centers (metal and central carbon – refer to the picture) in the vicinity, which would greatly destabilize the complex. To answer quantitatively, one need to do quantum chemistry calculations, which I' unfortunately not capable of. $\endgroup$
    – andselisk ♦
    Dec 7 '17 at 13:12

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