Technetium forms a complex with merceptoacetyltriglycine ($\ce{MAG3}$), but I don't know where the negative charge of the technetium complex comes from. The oxidation state of technetium is $\mathrm{+V}$

Image of [Tc(MAG3)]-

$\ce{[Tc(HMPO)]}$ (hexamethylpropylene amine oxime) has a net zero charge, while technetium is also in the oxidation state of $\mathrm{+V}$.

Image of [Tc(HMPO)]

What is the method to find out the charge of the metal-complex?

  • 2
    $\begingroup$ Why, that's obvious: Tc forms 6 bonds (counting the double bond as two) with electronegative elements (non-metals), so it would be +6, but we are told that it is +5, hence the minus. $\endgroup$ – Ivan Neretin Sep 8 '16 at 15:18
  • $\begingroup$ Well, complex could be not fully dissociated, but then additional proton would be connected with one of atoms. $\endgroup$ – Mithoron Sep 8 '16 at 19:21
  • $\begingroup$ @IvanNeretin, Tc usually exist in the from of TcO4- form and can't formed complex with ligand because of it's negative charge until and unless it is reduced using reducing agent, the oxidation state of Tc in TcO4- is (+7). What would be the reduced form of TcO4- when Tc having +5 oxidation state. Would it be TcO+3 or just simple (Tc+5). $\endgroup$ – Khan Sep 8 '16 at 23:07
  • $\begingroup$ @IvanNeretin , Also the coordination number of Tc is 5, with 4 sigma bond and one pi bond. it mean one bond correspond to -1, but my understanding it coordinate covalent bond, one atom shared two electron, so it must be -2 instead of -1. make me correct if I am wrong. $\endgroup$ – Khan Sep 8 '16 at 23:28
  • $\begingroup$ What is the prevalent form of $\ce{Tc^{+5}}$ in water solutions, I don't know, nor is it important for the question at hand. As for this complex, that's right, we have Tc with the coordination number of 5, but with 6 bonds (since one of the bonds is double). Coordinate covalent bond is indeed special in this regard (it contributes 0 to the oxidation state), but that's not important either, as we don't have any. See, these are ordinary covalent bonds, all of them. N makes three bonds, O and S make two. $\endgroup$ – Ivan Neretin Sep 9 '16 at 6:20

Rather than looking at the central atom first, look at the surrounding atoms. Break the bonds to the central metal heterolyticly, giving all electrons to the outer atoms. Then, go from atom to atom and determine its charge. For $\ce{[Tc(MAG3)]-}$ we have:

  • A sulphur only connected to one other carbon ($-1$)
  • three nitrogens, each only connected to two other atoms ($-1$ each)
  • An oxido ligand, which will end up as $\ce{O^2-}$.

Add up the total negative charge (and any other charges that may be present elsewhere in the complex). In your example, you end up with $(-1) + 3 \times (-1) + (-2) = -6$.

Next, consider the central metal’s oxidation state. Use it instead of a charge: technetium(V) would thus have a $5+$ charge. Add this to the ligand charge calculated above: $+5 + (-6) = -1$.

Thus, the complex has an overall charge of $-1$.

You’re second complex is drawn incorrectly, which is why you cannot assign the correct charge to it as is. The correct structure is shown in the image below (taken from Wikipedia, where a full list of authors is available).

Structure of [Tc(HMPO)] from Wikipedia.

Note that the top left nitrogen does not have a hydrogen attached to it, and that two $\ce{C=N}$ double bonds were missing at the lower two nitrogens. Also, instead of writing $\ce{O\bond{...}H\bond{...}O}$, it is better practice to write $\ce{O-H\bond{....}^-O}$.

For calculating, ignore the formal positive charge on technetium and go directly from the oxidation state of $\mathrm{+V}$. Since this is a homework-type question I’ll leave it to you to determine why $\pm 0$ is the correct charge for this complex.

|improve this answer|||||
  • $\begingroup$ @N.ALodhi The second complex was incorrectly drawn. I have now given you the correct structure. Also, this is homework, so I’ll let you do it ;) $\endgroup$ – Jan Sep 11 '16 at 13:09

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