The oxidation states of
Cr in `[Cr(H_(2)O)_(6)]Cl_(3). [Cr(C_(6)H_(6))_(2)]` and
`K_(2)[Cr(CN)_(2)(O)_(2)(O_(2))(NH_(3))]` respectively are

Options

A. `+3, +2` and `+4`
B. `+3, 0` and `+6`
C. `+3, 0 ` and `+4`
D. `+3, +4` and `+6`

Explanation

To find the oxidation states of chromium (Cr) in the given coordination compounds, we will analyze each compound step by step.

### Step 1: Determine the oxidation state of Cr in `[Cr(H₂O)₆]Cl₃`

1. **Identify the components**: The complex is `[Cr(H₂O)₆]` and there are 3 chloride ions (Cl⁻).
2. **Determine the charge of Cl**: Each Cl has a charge of -1, so for 3 Cl ions, the total charge is -3.
3. **Calculate the charge on the complex**: Since the total charge from Cl is -3, the charge on the complex `[Cr(H₂O)₆]` must be +3 to balance it out.
4. **Set up the equation**: Let the oxidation state of Cr be \( x \). The water ligands (H₂O) are neutral (0 charge).
\[ x + 0 = +3 \]
5. **Solve for x**:
\[ x = +3 \]

### Step 2: Determine the oxidation state of Cr in `[Cr(C₆H₆)₂]`

1. **Identify the components**: The complex is `[Cr(C₆H₆)₂]`.
2. **Determine the charge of C₆H₆**: Benzene (C₆H₆) is a neutral ligand, contributing 0 charge.
3. **Calculate the charge on the complex**: Since both ligands are neutral, the overall charge of the complex is 0.
4. **Set up the equation**: Let the oxidation state of Cr be \( x \).
\[ x + 0 = 0 \]
5. **Solve for x**:
\[ x = 0 \]

### Step 3: Determine the oxidation state of Cr in `K₂[Cr(CN)₂(O)₂(O₂)(NH₃)]`

1. **Identify the components**: The complex is `[Cr(CN)₂(O)₂(O₂)(NH₃)]` and there are 2 potassium ions (K⁺).
2. **Determine the charge of K**: Each K has a charge of +1, so for 2 K ions, the total charge is +2.
3. **Determine the charges of ligands**:
- CN⁻ has a charge of -1, so for 2 CN, the total is -2.
- O (oxide) has a charge of -2, so for 2 O, the total is -4.
- O₂ (peroxide) has a charge of -2.
- NH₃ is neutral (0 charge).
4. **Calculate the total charge from ligands**:
\[ -2 (from \, CN) + -4 (from \, O) + -2 (from \, O₂) + 0 (from \, NH₃) = -8 \]
5. **Calculate the charge on the complex**: The overall charge of the complex must balance the +2 from K, so:
\[ \text{Charge on complex} = -2 (from K) + 8 (from ligands) = -2 \]
6. **Set up the equation**: Let the oxidation state of Cr be \( x \).
\[ x + (-8) = -2 \]
7. **Solve for x**:
\[ x - 8 = -2 \implies x = +6 \]

### Summary of Oxidation States
- In `[Cr(H₂O)₆]Cl₃`, the oxidation state of Cr is **+3**.
- In `[Cr(C₆H₆)₂]`, the oxidation state of Cr is **0**.
- In `K₂[Cr(CN)₂(O)₂(O₂)(NH₃)]`, the oxidation state of Cr is **+6**.

### Final Answer
The oxidation states of Cr in the respective compounds are:
1. +3
2. 0
3. +6