Animated Solution for Chemistry - Coordination Compounds: The coordination numbers of Co and Al in $[CoCl(en{)}_2]Cl$ and $K_3[Al(C_2{O}_4{)}_3]$, respectively, are (en = ethane-1, 2-diamine)

Imagine a central metal atom as a powerful magnet, and the ligands as iron filings drawn to it. But in the quantum world, these 'filings' don't just stick randomly; they form precise, directed connections called coordinate bonds. The total number of these bonds is what we call the coordination number. It is a fundamental property that dictates the geometry, stability, and reactivity of the complex.

Today, we are going to decode the coordination numbers for two specific complexes: $[CoCl(en{)}_2]Cl$ and $K_3[Al(C_2{O}_4{)}_3]$.

What Counts and What Doesn't

Before we start counting, we need to know where to look. In any coordination compound, the square brackets denote the coordination sphere. Everything inside this bracket is directly bonded to the central metal. Anything outside the bracket belongs to the ionization sphere and acts merely as a counter-ion to balance the charge.

When calculating the coordination number, we strictly ignore the ionization sphere. Our entire focus must be on the ligands trapped inside the square brackets with the metal.

The Trap of Denticity

Let's dive into our first complex: $[CoCl(en{)}_2]Cl$.

Inside the coordination sphere, we have our central Cobalt ($Co$) atom accompanied by one chloride ion ($C{l}^{-}$) and two molecules of ethane-1,2-diamine, commonly abbreviated as $en$.

Now, here is where many students fall into a trap. They simply count the number of ligands ($1+2=3$) and assume the coordination number is 3. But coordination number is about the number of bonds, not the number of molecules. This brings us to the concept of denticity—how many 'teeth' a ligand uses to bite onto the metal.

Chloride is a unidentate ligand. It has only one donor site, so it forms exactly $1$ coordinate bond.

On the other hand, $en$ is a famous bidentate ligand. It possesses two nitrogen atoms, each capable of donating a lone pair of electrons. Therefore, a single $en$ molecule forms $2$ coordinate bonds with the metal.

Let's do the math for Cobalt: - Bonds from Chloride = $1\times 1=1$ - Bonds from $en$ = $2\times 2=4$

Total coordination number of $Co=1+4=5$.

Cobalt is forming a total of five coordinate bonds, giving it a coordination number of 5.

The Power of Oxalate

Now, let's shift our focus to the second complex: $K_3[Al(C_2{O}_4{)}_3]$.

Ignoring the potassium ions outside, we look inside the bracket. We have a central Aluminum ($Al$) atom surrounded by three oxalate ions ($C_2{O}_4^{2-}$).

Just like $en$, the oxalate ion is a powerful bidentate ligand. It uses two oxygen atoms to donate electron pairs, forming two coordinate bonds per molecule.

Let's calculate the coordination number for Aluminum: - Bonds from Oxalate = $3\times 2=6$

Total coordination number of $Al=6$.

Aluminum is forming a total of six coordinate bonds, giving it a coordination number of 6.

By carefully analyzing the denticity of the ligands within the coordination sphere, we have successfully decoded the puzzle. The coordination number for Cobalt is 5, and for Aluminum, it is 6.

This perfectly matches option (d). The key takeaway here is to never just count the ligands. Always respect their denticity, and the coordination number will reveal itself effortlessly!