Animated Solution for Chemistry - Coordination Compounds: The coordination number of Th in $K_4[Th(C_2{O}_4{)}_4(O{H}_2{)}_2]$ is ($C_2{O}_4^{2-}=\text{Oxalato}$)

Imagine a world where metals don't just form simple ionic bonds, but instead, they invite a whole entourage of molecules and ions to surround them in a beautifully structured geometric dance. This is the fascinating realm of coordination chemistry. When we look at a complex like $K_4[Th(C_2{O}_4{)}_4(O{H}_2{)}_2]$, it might seem like a terrifying jumble of letters and numbers at first glance. But I promise you, once you understand the language of these molecules, it reads like a beautiful architectural blueprint.

To truly appreciate this, we have to travel back to the late 19th century when a brilliant chemist named Alfred Werner proposed a revolutionary idea. He suggested that metals have two types of valencies: primary and secondary. The primary valency is just the oxidation state, the charge of the metal. But the secondary valency—that is where the magic happens. The secondary valency is what we now call the coordination number. It dictates how many "teeth" or donor atoms are directly biting onto the central metal atom to form coordinate covalent bonds.

When you are asked to find the coordination number, the very first thing you must do is put on your blinders. You need to ignore everything outside the square brackets. The square brackets represent the coordination sphere, the inner sanctum where the metal and its closest ligands reside.

In our specific problem, the complex is $K_4[Th(C_2{O}_4{)}_4(O{H}_2{)}_2]$. Notice those four potassium ions ($K_4$) sitting outside the brackets? They are part of the ionization sphere. They are just counter ions balancing the overall charge of the complex. They do not form coordinate bonds with the Thorium ($Th$) atom. Therefore, they have absolutely zero impact on the coordination number.

Our entire focus must be on what is happening inside the brackets: $[Th(C_2{O}_4{)}_4(O{H}_2{)}_2]$. Here, Thorium is the central metal atom, acting as a Lewis acid, eagerly waiting to accept electron pairs. The molecules and ions surrounding it—the oxalates and the waters—are the ligands, acting as Lewis bases, ready to donate their electron pairs.

To calculate the coordination number, we can't just count the number of ligand molecules. Why? Because not all ligands are created equal. This brings us to the crucial concept of denticity.

The word denticity comes from the Latin word dentis, meaning tooth. Think of a ligand as a creature biting onto the metal. A monodentate ligand has only one tooth; it can only form one coordinate bond. A bidentate ligand has two teeth; it can form two coordinate bonds simultaneously. There are even hexadentate ligands like EDTA that can bite onto a metal with six teeth at once!

Therefore, the true formula for coordination number is not just the sum of ligands, but the sum of the number of ligands multiplied by their respective denticities.

Oxalate and Water

Let's break down the ligands in our Thorium complex. First, we have four oxalate ions, written as $C_2{O}_4^{2-}$. If you draw the Lewis structure of an oxalate ion, you will see two carbon atoms double-bonded to two oxygen atoms, and single-bonded to two other oxygen atoms that carry a negative charge.

These two negatively charged oxygen atoms are perfectly spaced to act as electron pair donors. Because it has two donor sites that can simultaneously bind to the same metal atom, the oxalate ion is a bidentate ligand. Every single oxalate ion that approaches the Thorium atom will form two coordinate bonds. This creates a stable five-membered ring structure with the metal, a phenomenon known as chelation.

Next, we have two water molecules, written as $O{H}_2$ or $H_{2}O$. Now, you might be thinking, "Wait a minute, the oxygen atom in water has two lone pairs of electrons! Doesn't that make it bidentate?" That is a brilliant question, but there is a catch here.

While oxygen does have two lone pairs, they are located on the same atom and are pointing in directions that make it geometrically impossible for both of them to bind to the same metal atom at the same time. The steric strain would be immense. Therefore, water can only use one of its lone pairs to form a bond. This makes water a strictly monodentate ligand.

Now that we have decoded the denticity of our ligands, the final calculation is a breeze. Let's put it all together.

We have $4$ oxalate ligands. Since each oxalate is bidentate (denticity = 2), they contribute:

We have $2$ water ligands. Since each water is monodentate (denticity = 1), they contribute:

To find the total coordination number, we simply add these contributions together:

And there we have it! The Thorium atom in this complex is surrounded by a staggering 10 donor atoms, making its coordination number 10. This is a relatively high coordination number, which is characteristic of large actinide metals like Thorium that have plenty of space in their electron clouds to accommodate numerous ligands.

The beauty of this problem lies in its simplicity, provided you don't fall into the trap of just counting the molecules. Always remember to check the denticity, and you will never get these questions wrong!