Determine the oxidation number of the element as indicated
(ix) Cr in `[Cr(NH_(3))_(6)]Cl_(3)`

Answer

To determine the oxidation number of chromium (Cr) in the complex compound \([Cr(NH_3)_6]Cl_3\), we can follow these steps:

### Step 1: Understand the Structure of the Compound
The compound \([Cr(NH_3)_6]Cl_3\) consists of a complex ion \([Cr(NH_3)_6]^{3+}\) and three chloride ions (\(Cl^-\)). The ammonia (\(NH_3\)) ligands are neutral and do not contribute to the overall charge of the complex.

### Step 2: Assign the Charge of the Chloride Ions
Since there are three chloride ions, each with a charge of \(-1\), the total charge contributed by the chloride ions is:
\[ 3 \times (-1) = -3 \]

### Step 3: Set Up the Equation for the Oxidation Number
Let the oxidation number of chromium be \(X\). The overall charge of the complex ion \([Cr(NH_3)_6]^{3+}\) must balance the charge contributed by the chloride ions. Thus, we can write the equation:
\[ X + 0 = +3 \]
Here, the \(0\) represents the total oxidation number contribution from the six ammonia ligands, since each \(NH_3\) has an oxidation number of \(0\).

### Step 4: Solve for the Oxidation Number
From the equation:
\[ X = +3 \]

### Conclusion
The oxidation number of chromium (Cr) in the complex \([Cr(NH_3)_6]Cl_3\) is \(+3\).

---