Oxidation number of P in `PO_(4)^(3-)`, of S in `SO_(4)^(2-)` and that of `Cr_(2)O_(7)^(2-)` are respectively

Options

A. `+3, +6 and +6`
B. `+5 , +6 and +6`
C. `+3,+6 and +5`
D. `+5 , +3 and +6`

Explanation

To find the oxidation numbers of phosphorus in \( PO_4^{3-} \), sulfur in \( SO_4^{2-} \), and chromium in \( Cr_2O_7^{2-} \), we can follow these steps:

### Step 1: Calculate the oxidation number of phosphorus in \( PO_4^{3-} \)

1. Let the oxidation number of phosphorus be \( x \).
2. Oxygen has an oxidation number of \(-2\).
3. There are 4 oxygen atoms, so the total contribution from oxygen is \( 4 \times (-2) = -8 \).
4. The overall charge of the ion is \(-3\).

Setting up the equation:
\[ x + (-8) = -3 \]
\[ x - 8 = -3 \]
\[ x = -3 + 8 \]
\[ x = +5 \]

### Step 2: Calculate the oxidation number of sulfur in \( SO_4^{2-} \)

1. Let the oxidation number of sulfur be \( y \).
2. Again, oxygen has an oxidation number of \(-2\).
3. There are 4 oxygen atoms, so the total contribution from oxygen is \( 4 \times (-2) = -8 \).
4. The overall charge of the ion is \(-2\).

Setting up the equation:
\[ y + (-8) = -2 \]
\[ y - 8 = -2 \]
\[ y = -2 + 8 \]
\[ y = +6 \]

### Step 3: Calculate the oxidation number of chromium in \( Cr_2O_7^{2-} \)

1. Let the oxidation number of chromium be \( z \).
2. There are 2 chromium atoms, so the contribution from chromium is \( 2z \).
3. Oxygen has an oxidation number of \(-2\).
4. There are 7 oxygen atoms, so the total contribution from oxygen is \( 7 \times (-2) = -14 \).
5. The overall charge of the ion is \(-2\).

Setting up the equation:
\[ 2z + (-14) = -2 \]
\[ 2z - 14 = -2 \]
\[ 2z = -2 + 14 \]
\[ 2z = +12 \]
\[ z = +6 \]

### Final Results

The oxidation numbers are:
- Phosphorus in \( PO_4^{3-} \): \( +5 \)
- Sulfur in \( SO_4^{2-} \): \( +6 \)
- Chromium in \( Cr_2O_7^{2-} \): \( +6 \)

Thus, the final answer is \( +5, +6, +6 \).