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: $x + n = 8$
Match the following List-I with List-II and choose the correct option: List-I (Compounds) | List-II (Shape and Hybridisation) (A) PF\(_{3}\) (I) Tetrahedral and sp\(^3\) (B) SF\(_{6}\) (III) Octahedral and sp\(^3\)d\(^2\) (C) Ni(CO)\(_{4}\) (I) Tetrahedral and sp\(^3\) (D) [PtCl\(_{4}\)]\(^{2-}\) (II) Square planar and dsp\(^2\)
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32