Question

Calculate the oxidation state of vanadium in the following complex compound $R{b_4}Na[H{V_{10}}{O_{28}}]$ .

Answer 1

Hint: The given compound is a coordination compound. The coordination complex as a whole is neutral. The atoms present in it carry different charges. The various substituents are attached to the metal ion in the complex.

Complete step by step answer:
Let us first understand what are coordination compounds
Coordination compound is a compound formed by a combination of two or more different compounds. These complexes retain their identity in the solid as well as the dissolved state.
In a coordination compound there is a central metal atom or ion. To this metal atom or ion, ligands are attached. Ligands are neutral molecules, cations or anions that are directly linked to the metal atom by co-ordinate or dative bond. Ligands donate lone pairs of electrons to the metal atom. Ligands are lewis bases while the metal atom is lewis acid.
The central metal atom and the ligands that are directly attached to the metal are enclosed in a square bracket. This is called the coordination sphere. It behaves as a single unit because the ligands present in the coordination sphere are held tightly by the metal atom or ion. Any ion present outside this sphere is separated from the complex when the complex is dissolved in water or any other solvent.
The charge on the complex ion is the algebraic sum of oxidation number of metal and charges carried by the ligands.
Now in the given complex, $R{b_4}Na[H{V_{10}}{O_{28}}]$ we need to calculate the oxidation number of vanadium.
For that we should know the charges carried by each ligand and the charge on the complex.
As we can see the charge on the whole complex is zero. Now let us look at each of the ligands.
Rubidium $(Rb)$ is an alkali metal. It carries a charge of $ + 1$ or the oxidation state is $ + 1$ .
Sodium $(Na)$ also is an alkali metal. It has a charge of $ + 1$ .
Now in the coordination sphere, hydrogen and oxygen are attached to the metal ion.
Hydrogen carries a charge of $ + 1$ .
Oxygen being an element of group $16$ and having high electronegativity carries a charge of $ - 2$ .
So, now we can calculate the oxidation state of vanadium.
Charge on complex $ = $ Oxidation number of metal ion $ + $ charges carried by the ligands.
Let the charge on vanadium be $x$ .
Outside the complex ion, sodium and rubidium are present. They in total will carry a charge of $ + 5$ (since there are four rubidium atoms and one sodium atom)
Now to make the whole compound neutral, the coordination sphere will have a charge of $ - 5$ (so as to nullify the charge of sodium and rubidium).
So now charge on complex ions is $ - 5$ .
Substituting the values and multiplying them with the number of ligands present we get
$ - 5 = [( + 1) + 10x + ( - 2 \times 28)]$ (since ten vanadium ions are present)
$ - 5 = [10x - 55]$
$10x = 50$
$x = + 5$
The oxidation state of vanadium is $ + 5$ .

Note: Ligands can be neutral, cationic or anionic.
-Neutral ligands such as $N{H_3},{H_2}O,en,CO$ carry zero charge in such complexes.
-Ligands are also classified on the basis of the number of donor atoms present in a ligand.