On complete reaction of $\ce{FeCl3}$ with oxalic acid in aqueous solution containing $\ce{KOH}$, resulted in the formation of product A. The secondary valency of $\ce{Fe}$ in the product A is:

Now my teacher told me that in this compound the secondary valency of iron would be 6, and told me to remember the formula of all the compounds in GRB Organic Chem.

I have a rough idea that six is the most common secondary valency, but beyond that I fail to see any trend.

Can someone help me out, so that I can predict the formula of such compounds without rote learning?

Only tetravalent and hexavalent compounds are in my course, I also know how to predict whether a compound would be high spin or low spin. My doubt is mainly about whether they will be hexavalent or tetravalent.


1 Answer 1


The first thing to notice here would be the fact that you see an acid along side a base.

$$\ce{H2C2O4 + 2KOH -> K2C2O4.H2O + H2O}$$

Now, this reacts with the ferric chloride present in the solution to give potassium tris(oxalato)ferrate(III). Many if not all iron complexes are hexavalent in nature.

$$\ce{3K2C2O4.H2O(aq) + FeCl3.6H2O(aq) -> K3[Fe(C2O4)3].3H2O(aq) + 3KCl(aq)}$$

This is in fact how you would prepare the said complex as explained in this lab manual

Potassium trisoxalatoferrate(III) trihydrate, $\ce{K3[Fe(C2O4)3].3H2O}$ is a green crystalline salt, soluble in hot water but rather insoluble when cold. It can be prepared by the reaction of $\ce{K2C2O4.H2O}$ with $\ce{FeCl3.6H2O}$

  • $\begingroup$ Is there any such rule for elements other than iron? $\endgroup$
    – Tatai
    Jun 20, 2021 at 8:27
  • 4
    $\begingroup$ @SunainaDas, in inorganic chemistry there is no fixed rule for all sets of compounds. Instead of worrying about rote learning, just focus on seeing the different patterns. This is a similar such pattern. $\endgroup$ Jun 20, 2021 at 8:30

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