The oxidation state of chrominium in `Cr(CO)_(6)` is

Options

A. `0`
B. `+2`
C. `-2`
D. `+6`

Explanation

To determine the oxidation state of chromium in the complex \( Cr(CO)_6 \), we can follow these steps:

### Step 1: Understand the Complex
The complex \( Cr(CO)_6 \) consists of one chromium atom and six carbon monoxide (CO) ligands.

### Step 2: Assign Oxidation States
- Let \( X \) be the oxidation state of chromium (Cr).
- The oxidation state of carbon monoxide (CO) is neutral, which means it has an oxidation state of 0.

### Step 3: Set Up the Equation
Since the overall charge of the complex is zero, we can set up the following equation based on the sum of the oxidation states:
\[ X + 6 \times (0) = 0 \]
This simplifies to:
\[ X + 0 = 0 \]

### Step 4: Solve for X
From the equation, we can see that:
\[ X = 0 \]

### Conclusion
The oxidation state of chromium in \( Cr(CO)_6 \) is 0.