The values of coefficients to balance the following reaction are
`Cr(OH)_(3)+ClO^(-)+OH^(-) rarr CrO_(4)^(2-)+Cl^(-)+H_(2)O`

Options

A. `{:(Cr(OH)_(3),ClO^(-),CrO_(4)^(2-),Cl^(-)),(" 2"," 3"," 3","3"):}`
B. `{:(Cr(OH)_(3),ClO^(-),CrO_(4)^(2-),Cl^(-)),(" 2"," 4"," 3","2"):}`
C. `{:(Cr(OH)_(3),ClO^(-),CrO_(4)^(2-),Cl^(-)),(" 2"," 4"," 4","2"):}`
D. `{:(Cr(OH)_(3),ClO^(-),CrO_(4)^(2-),Cl^(-)),(" 2"," 3"," 2","3"):}`

Explanation

To balance the redox reaction given in the question, we will follow a systematic approach. The reaction is:

\[ \text{Cr(OH)}_3 + \text{ClO}^- + \text{OH}^- \rightarrow \text{CrO}_4^{2-} + \text{Cl}^- + \text{H}_2\text{O} \]

### Step 1: Assign Oxidation States
First, we need to assign oxidation states to each element in the reactants and products.

- In \(\text{Cr(OH)}_3\), chromium (Cr) has an oxidation state of +3.
- In \(\text{ClO}^-\), chlorine (Cl) has an oxidation state of +1.
- In \(\text{OH}^-\), oxygen (O) has an oxidation state of -2 and hydrogen (H) is +1.
- In \(\text{CrO}_4^{2-}\), Cr has an oxidation state of +6.
- In \(\text{Cl}^-\), Cl has an oxidation state of -1.
- In \(\text{H}_2\text{O}\), H is +1 and O is -2.

### Step 2: Determine Changes in Oxidation States
Now, we calculate the changes in oxidation states:

- Chromium changes from +3 to +6, which is an increase of 3.
- Chlorine changes from +1 to -1, which is a decrease of 2.

### Step 3: Balance the Changes in Oxidation States
To balance the increase and decrease in oxidation states, we need to multiply the changes by appropriate coefficients:

- For Cr: 1 Cr changes from +3 to +6 (increase of 3).
- For Cl: 2 Cl changes from +1 to -1 (decrease of 2).

Thus, we will use coefficients of 2 for Cr and 3 for ClO⁻:

\[ 2 \text{Cr(OH)}_3 + 3 \text{ClO}^- \rightarrow 2 \text{CrO}_4^{2-} + 3 \text{Cl}^- + \text{H}_2\text{O} \]

### Step 4: Balance the Oxygen and Hydrogen Atoms
Next, we need to balance the oxygen and hydrogen atoms:

- On the left side, we have:
- From 2 Cr(OH)₃: 2 Cr + 6 O + 2 H
- From 3 ClO⁻: 3 O
- Total: 6 + 3 = 9 O and 2 H

- On the right side, we have:
- From 2 CrO₄²⁻: 8 O
- From 3 Cl⁻: 0 O
- From H₂O: 1 O
- Total: 8 + 1 = 9 O and 0 H

To balance the hydrogen atoms, we need to add water molecules:

- We add 5 H₂O to the right side to balance the hydrogen atoms.

### Step 5: Adjust for Basic Medium
Since the medium is basic, we need to add OH⁻ ions to counteract any H⁺ ions produced:

- For every H₂O added, we need to add an equal number of OH⁻ ions to maintain the balance in basic medium.

### Final Balanced Equation
After balancing all components, we arrive at the final balanced equation:

\[ 2 \text{Cr(OH)}_3 + 3 \text{ClO}^- + 10 \text{OH}^- \rightarrow 2 \text{CrO}_4^{2-} + 3 \text{Cl}^- + 15 \text{H}_2\text{O} \]

### Coefficients
The coefficients for the balanced reaction are:
- \( \text{Cr(OH)}_3 \): 2
- \( \text{ClO}^- \): 3
- \( \text{OH}^- \): 10
- \( \text{CrO}_4^{2-} \): 2
- \( \text{Cl}^- \): 3
- \( \text{H}_2\text{O} \): 15