An octahedral complex with the formula $\mathrm{CoCl}_3 \cdot \mathrm{nNH}_3$ upon reaction with excess of $\mathrm{AgNO}_3$ solution gives 2 moles of $\mathrm{AgCl}$. Consider the oxidation state of $\mathrm{Co}$ in the complex is '$x$'. The value of "$x+n$" is __________.

<p>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 $\mathrm{AgNO}_3$.</p> <p>First, let's write the reaction between the complex and $\mathrm{AgNO}_3$.</p> <p>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 $\mathrm{AgNO}_3$.</p> <p>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:</p> <p>$\mathrm{[Co(NH_3)_{n}Cl]Cl_2}$</p> <p>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:</p> <p>The sum of the charges is:</p> <p>$x + (0 \cdot n) + (-1 \cdot 1) + (-1 \cdot 2) = 0$</p> <p>Solving this, we get:</p> <p>$x - 1 - 2 = 0$</p> <p>$x - 3 = 0$</p> <p>$x = 3$</p> <p>So, the oxidation state of Co is 3.</p> <p>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:</p> <p>$n = 6 - 1 = 5$</p> <p>Finally, we calculate $x + n$:</p> <p>$x + n = 3 + 5 = 8$</p> <p>So, the value of "$x + n$" is 8.</p> <p>Therefore, the correct option is:</p> <p>Option D: 8</p>