QJEEM24Ch7133: JEE Main 2024 Chemistry - Class 12 JEEM24Ch7133 Solution: An octahedral complex with the formula upon reaction with…

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:

$\mathrm{[Co(NH_3)_{n}Cl]Cl_2}$

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