Why chlorine acts as a strong field ligand in tetrachloridoplatinate(II)?

In most of the cases I have learnted, halogens act as a weak field ligand. However, I recently learned somewhere on the web that the strength of a ligand also depends upon the size of the atom. A classic example is tetrachloridoplatinate(II): $\ce$. Why is it so, and how can I predict the hybridization correctly in such exceptions, if I come across them for the first time?

• inorganic-chemistry
• coordination-compounds
• hybridization

$\begingroup$ @orthocresol Thank you. I really tried to find it but couldn't find this answer before asking. $\endgroup$

$\begingroup$ No problem, it is difficult to find stuff sometimes. (I didn't think it is 100% a duplicate, and therefore didn't close.) $\endgroup$

2 Answers 2

I would like to add an answer stating a few exceptions which would come handy to you in future:

• $\ce>$ has a $3d^8$ configuration. In this case strong field ligands ($\ce$ or $\ce$ donors) will make the complex square planar. In all other cases, the complex will be tetrahedral.
• In the case of $4d^8$ and $5d^8$ configuration and any ligand the hybridisation will be that of a square planar complex.
• With $\ce>$, nearly all ligands behave as strong field ligands ($dsp^2$ hybridisation). Notable exceptions to this rule being $\ce>$ and $\ce$.
• In case of bulky ligands (for instance $\ce$) present with $3d^8$ configuration, the ligands cannot get adjusted in the same plane leading to a tetrahedral shape ($sp^3$ configuration).

$\begingroup$ The last one is logical. Rest needs to be memorised probably. Thanks a lot. $\endgroup$

$\begingroup$ This answer simplifies too much in the wrong places. Most notably, not all N donors are automatically strong-field ligands and not all strong-field ligands automatically make a nickel(II) complex square planar. $\endgroup$

$\begingroup$ @Jan Agreed, I should've added that it's more or less general, and if it's an exam (perhaps at a high-school level) asking OP this , I'd suggest to follow the simplified rules. $\endgroup$

$\begingroup$ I’m very much in favour of not teaching simplified rules even at high school levels if they break down somewhere. Either teach the pitfall as well or leave out the simplified rules altogether. $\endgroup$

$\begingroup$ @Jan To be honest, I myself wasn't aware of the pitfall until you pointed out in your comment; the points I've posted in my answer are the ones they taught me in high school, I didn't take chem for higher studies. $\endgroup$

Halogens still act as weak-field ligands even in the case of square planar $\ce$ complexes. Weak field does not automatically mean high spin and neither does strong field automatically mean low spin.

Whether a complex adopts a high or a low spin state depends on the central metal’s oxidation state, its position in the periodic table, the ligands, whether they are π-acidic, π-basic or π-neutral, whether the bond is predominantly covalent or dative, the distance between the atoms and sometimes even more subtle factors. Indeed, for some metal-ligand combinations both a high and a low spin complex are known and characterised.

One of the very few rather simple rules when it comes to high or low spin is the fact that any transition metal that is not a $\mathrm$ metal will usually have a low spin state. In very simple terms this is because the orbitals of higher transition metals are more diffuse and therefore able to form bonds which are more covalent; this in turn leads to greater energetic stabilisations and destabilisations. Therefore, ignore the chlorido ligands in your analysis; look solely at platinum and immediately assume low spin.