The reaction of $[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2$ with excess AgNO$_3$ is as follows:
\[[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2 + \text{excess AgNO}_3 \rightarrow 2\text{AgCl (2 moles)}\]
In the complex $[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2$:
The inner coordination sphere contains 1 Cl ligand.
The outer coordination sphere contains 2 Cl$^-$ ions, which react with AgNO$_3$ to give 2 moles of AgCl.
Let $x$ be the oxidation state of Co. The total charge on the complex is neutral. Therefore:
\[x + 0 \text{ (from 5 NH}_3\text{)} + (-1 \text{ from 1 Cl}) + (-2 \text{ from 2 Cl}^-) = 0\]
\[x - 1 - 2 = 0\]
\[x = +3\]
Here, $n = 5$ (the number of NH$_3$ ligands). Thus:
\[x + n = 3 + 5 = 8\]
Final Answer: $x + n = 8$
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)