Question

Find the coordination number and oxidation number of X in the given compound.
$\left[\text{X}{\left(\text{N}{\text{H}}_{\text{3}}\right)}_{\text{5}}\left(\text{S}{\text{O}}_{\text{4}}\right)\right]\text{Cl}$ .

Answer 1

Hint: The coordination number of the central atom or the central ion can be determined by the number of sigma bonds formed by the ligands with the central atom or the central ion. The oxidation number of the central metal atom is the charge that it carries as calculated by assigning appropriate charges to the ligands and equating the sum of the charges on the ligands and central atom to the charge on the coordination sphere.

Complete step by step answer:
The coordination number of the central metal atom/ ion is used to refer to the total number of unidentate ligands along with double the number of bidentate ligands if any, attached to the central metal atom/ ion through coordinate bonds.
In other words, coordination number gives us the number of atoms of the ligands which are directly bonded to the central metal atom/ ion.
For example, in the complex ion ${\left[\text{Ag}{\left(\text{CN}\right)}_{\text{2}}\right]}^{\text{ - }}$ , the coordination number of Ag is 2. But in the complex ion ${\left[\text{Fe}{\left({\text{C}}_2{\text{O}}_4\right)}_3\right]}^{\text{2 - }}$ , the coordination number of Fe is 6 as ${\text{C}}_{\text{2}}{{\text{O}}_{\text{4}}}^{\text{2 - }}$ is a bidentate ligand.
Now, in the given compound $\left[\text{X}{\left(\text{N}{\text{H}}_{\text{3}}\right)}_{\text{5}}\left(\text{S}{\text{O}}_{\text{4}}\right)\right]\text{Cl}$ , ammonia and sulphate ions are both unidentate ligands and there are 5 ammonia ligands and 1 sulphate ligand. So, there are a total of 6 unidentate ligands attached to the central metal atom X. So, the coordination number of X in $\left[\text{X}{\left(\text{N}{\text{H}}_{\text{3}}\right)}_{\text{5}}\left(\text{S}{\text{O}}_{\text{4}}\right)\right]\text{Cl}$ is 6.
The simple ion whether cation or anion behaves like a free ion and carries usual charge as one radical. The charge on the complex ion is calculated on the basis that it is equal to the charge on the central metal atom plus the total charge carried by the ligands. The total positive charge must be equal to the total negative charge since the complex as a whole is electrically neutral.
Now, in the given compound $\left[\text{X}{\left(\text{N}{\text{H}}_{\text{3}}\right)}_{\text{5}}\left(\text{S}{\text{O}}_{\text{4}}\right)\right]\text{Cl}$ , chloride ion is the simple anion. Since chloride ions have a charge of -1, the complex ion will have a charge of +1.
Let z be the charge on the central atom and the charge on sulphate ion is -2 and ammonia is a neutral ligand. So, we have
$$\begin{gathered} +1=\text{z}+\left(5\times 0\right)+\left(-2\right) \\ \Rightarrow \text{z}=+1+2 \\ \Rightarrow \text{z}=+3 \end{gathered}$$
Therefore, oxidation number of X is +3.

Note:
Within the coordination sphere, the symbols of the ligands which follow the symbol of the metal atom are arranged alphabetically according to their names. This alphabetical arrangement is irrespective of the charge present on the ligands.
For example, for the coordination compound tetraammineaquachlorocobalt (III) chloride, the formula is written as $\left[\text{Co}{\left(\text{N}{\text{H}}_{\text{3}}\right)}_{\text{4}}\left({\text{H}}_{\text{2}}\text{O}\right)\text{Cl}\right]\text{C}{\text{l}}_2$ where the three ligands present are ammine, aqua and chloro.