Oxidation number of Cr in `K_(2)Cr_(2)O_(7),CrO_(5)` and `Cr_(2)(SO_(4))_(3)` respectively is

Options

A. `+6,+10,+3`
B. `+6,+4,+6`
C. `+6,+6,+3`
D. `+6,+3,+3`

Explanation

To find the oxidation number of chromium (Cr) in the compounds \( K_2Cr_2O_7 \), \( CrO_5 \), and \( Cr_2(SO_4)_3 \), we will use the oxidation number method. Let's go through each compound step by step.

### Step 1: Finding the oxidation number of Cr in \( K_2Cr_2O_7 \)

1. **Identify the known oxidation states**:
- Potassium (K) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

2. **Set up the equation**:
- The formula for potassium dichromate is \( K_2Cr_2O_7 \).
- There are 2 potassium atoms contributing \( 2 \times +1 = +2 \).
- There are 7 oxygen atoms contributing \( 7 \times -2 = -14 \).
- Let the oxidation state of chromium be \( x \). Since there are 2 chromium atoms, this contributes \( 2x \).

3. **Write the equation**:
\[ 2x + 2 - 14 = 0 \]

4. **Solve for \( x \)**:
\[ 2x - 12 = 0 \implies 2x = 12 \implies x = +6 \]

### Step 2: Finding the oxidation number of Cr in \( CrO_5 \)

1. **Identify the known oxidation states**:
- Oxygen (O) has an oxidation state of -2.

2. **Set up the equation**:
- The formula for chromium pentoxide is \( CrO_5 \).
- There are 5 oxygen atoms contributing \( 5 \times -2 = -10 \).
- Let the oxidation state of chromium be \( x \).

3. **Write the equation**:
\[ x + (-10) = 0 \]

4. **Solve for \( x \)**:
\[ x - 10 = 0 \implies x = +6 \]

### Step 3: Finding the oxidation number of Cr in \( Cr_2(SO_4)_3 \)

1. **Identify the known oxidation states**:
- Sulfate (\( SO_4^{2-} \)) has an oxidation state of -2 for each sulfate ion.
- Each sulfate contributes \( -2 \) and there are 3 sulfate ions.

2. **Set up the equation**:
- The formula for chromium sulfate is \( Cr_2(SO_4)_3 \).
- Let the oxidation state of chromium be \( x \).
- The contribution from 2 chromium atoms is \( 2x \).
- The contribution from 3 sulfate ions is \( 3 \times -2 = -6 \).

3. **Write the equation**:
\[ 2x - 6 = 0 \]

4. **Solve for \( x \)**:
\[ 2x = 6 \implies x = +3 \]

### Summary of Results:
- The oxidation number of Cr in \( K_2Cr_2O_7 \) is **+6**.
- The oxidation number of Cr in \( CrO_5 \) is **+6**.
- The oxidation number of Cr in \( Cr_2(SO_4)_3 \) is **+3**.