1 mol of an octahedral metal complex with formula MCl$$_3$$ . 2L on reaction with excess of AgNO$$_3$$ gives 1 mol of AgCl. The denticity of Ligand L is _________. (Integer answer)

We have one mole of an octahedral metal complex whose empirical formula is written as $$\mathrm{MCl_3\;.\;2L}$$. In such condensed formulae every chloride written together with the metal may either be (i) inside the coordination sphere as a ligand, or (ii) outside the sphere as a simple counter-ion. The experiment that decides this is the reaction with excess $$\mathrm{AgNO_3}$$, because only those chlorides that are outside the coordination sphere are free to precipitate as $$\mathrm{AgCl}$$.

According to the statement, one mole of the complex gives exactly one mole of $$\mathrm{AgCl}$$:

$$\mathrm{MCl_3\;.\;2L \;+\; AgNO_3\;(excess)\;\longrightarrow\; AgCl\;(s)\;+\;…}$$

This observation means that only one chloride ion is present as a counter-ion. Hence out of the total three chlorides written in the formula, one is outside the coordination sphere while the remaining two are directly bonded to the metal.

So we rewrite the formula more explicitly as

$$[\mathrm{MCl_2(2L)}]\;\mathrm{Cl}$$

where the square brackets contain the coordination sphere. Inside these brackets we can clearly see:

• $$2$$ chloride ligands, each monodentate, therefore offering $$2$$ donor atoms.
• $$2$$ molecules of the ligand $$\mathrm{L}$$, each of unknown denticity that we shall call $$d$$.

Because the complex is octahedral, the coordination number of the central metal ion is $$6$$. Therefore the total number of donor atoms (coordination sites) contributed by all ligands inside the brackets must equal $$6$$. Stating this requirement first:

(Number of donor atoms from Cl) $$\;+\;\text{(Number of donor atoms from 2L)} \;=\;6$$

We substitute the obvious values:

$$2 \;+\; 2d \;=\; 6$$

Now we solve step by step.

First isolate the term containing $$d$$:

$$2d \;=\; 6 \;-\; 2$$

So

$$2d \;=\; 4$$

Dividing both sides by $$2$$ gives

$$d \;=\; \dfrac{4}{2} \;=\; 2$$

Thus each molecule of the ligand $$\mathrm{L}$$ supplies exactly two donor atoms to the metal centre. A ligand that uses two donor atoms is called a bidentate ligand.

Hence, the denticity of ligand $$\mathrm{L}$$ is $$2$$.

So, the answer is $$2$$.