3
$\begingroup$

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?

Improve this question
$\endgroup$
2
  • 4
    $\begingroup$ There are plenty $\mu_2$-thiourea complexes with Cd, Pb, Ag, Cu. Just saying. $\endgroup$
    – andselisk ♦
    Dec 7, 2017 at 9:13
  • 4
    $\begingroup$ 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, 2017 at 11:17

1 Answer 1

Reset to default
5
$\begingroup$

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.
Improve this answer
$\endgroup$
2
  • $\begingroup$ 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, 2017 at 13:07
  • 1
    $\begingroup$ @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, 2017 at 13:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.