The oxidation number of Cr in `[Cr(C_(6)H_(2))_(2]` is
`0` (b) `+2`
(c ) `+3`
(d) `+6` .

Explanation

To determine the oxidation number of chromium (Cr) in the complex `[Cr(C₆H₆)₂]`, we can follow these steps:

### Step 1: Identify the Ligand
The ligand in this complex is C₆H₆ (benzene), which is a neutral ligand. Neutral ligands do not contribute any charge to the overall complex.

### Step 2: Write the General Formula
The general formula for the oxidation state calculation in coordination compounds is:
\[ \text{Oxidation State of Metal} + \text{Sum of Oxidation States of Ligands} = \text{Overall Charge of the Complex} \]

Since the overall charge of the complex is not given, we assume it to be zero.

### Step 3: Assign Oxidation States
Let the oxidation state of chromium be \( x \). Since there are two benzene ligands, we can express the equation as follows:
\[ x + (0 \times 2) = 0 \]
Here, \( 0 \) is the oxidation state of each neutral benzene ligand.

### Step 4: Solve for x
Now, we can simplify the equation:
\[ x + 0 = 0 \]
Thus,
\[ x = 0 \]

### Step 5: Conclusion
The oxidation number of chromium in the complex `[Cr(C₆H₆)₂]` is \( 0 \).

### Final Answer
The correct answer is (a) 0.

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