JEE Main 2024 — Coordination Compounds Question with Solution

To solve this problem, we need to determine the oxidation state of cobalt ($$x$$) and the number of ammonia molecules ($$n$$) in the complex $${\mathrm{CoCl}}_3\cdot {\mathrm{nNH}}_3$$, given that it produces 2 moles of $$\mathrm{AgCl}$$ upon reaction with excess $${\mathrm{AgNO}}_3$$.

First, let's write the reaction between the complex and $${\mathrm{AgNO}}_3$$.

The precipitate formation indicates that some chloride ions are free (not coordinated to the metal). Since 2 moles of $$\mathrm{AgCl}$$ are formed, it indicates that there are 2 chloride ions that are free to react with $${\mathrm{AgNO}}_3$$.

Therefore, we can conclude that in the complex, 1 chloride ion is coordinated to the cobalt, and 2 chloride ions are free. This can be represented as:

$$[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_{\mathrm{n}}\mathrm{Cl}]\mathrm{C}{\mathrm{l}}_2$$

Now, let's determine the oxidation state of cobalt (Co). The charge on the entire complex should be zero, so we can write the charge balance equation as follows:

The sum of the charges is:

$$x+(0\cdot n)+(-1\cdot 1)+(-1\cdot 2)=0$$

Solving this, we get:

$$x-1-2=0$$

$$x-3=0$$

$$x=3$$

So, the oxidation state of Co is 3.

Now, we need to find the value of $$n$$. Since the coordination number of Co in an octahedral complex is typically 6 and we have 1 chloride ion coordinated to Co, the number of ammonia molecules coordinated to Co would be:

$$n=6-1=5$$

Finally, we calculate $$x+n$$:

$$x+n=3+5=8$$

So, the value of "$$x+n$$" is 8.

Therefore, the correct option is:

Option D: 8