Calculate the oxidation number of phosphorous in `Mg_2P_2O_7`.

Answer

To calculate the oxidation number of phosphorus in the compound \( \text{Mg}_2\text{P}_2\text{O}_7 \), we can follow these steps:

### Step 1: Identify the oxidation states of known elements
- Magnesium (Mg) typically has an oxidation state of +2.
- Oxygen (O) typically has an oxidation state of -2.

### Step 2: Set up the equation
Let the oxidation number of phosphorus (P) be \( x \). Since there are 2 phosphorus atoms in the compound, their contribution to the total oxidation state will be \( 2x \).

### Step 3: Write the equation based on the overall charge
The overall charge of the compound \( \text{Mg}_2\text{P}_2\text{O}_7 \) is 0 (neutral compound). Therefore, we can set up the equation as follows:

\[ 2(\text{oxidation state of Mg}) + 2(\text{oxidation state of P}) + 7(\text{oxidation state of O}) = 0 \]

Substituting the known oxidation states:

\[ 2(+2) + 2(x) + 7(-2) = 0 \]

### Step 4: Simplify the equation
Now, simplify the equation:

\[ 4 + 2x - 14 = 0 \]

Combine like terms:

\[ 2x - 10 = 0 \]

### Step 5: Solve for \( x \)
Now, isolate \( x \):

\[ 2x = 10 \]
\[ x = 5 \]

### Conclusion
The oxidation number of phosphorus in \( \text{Mg}_2\text{P}_2\text{O}_7 \) is +5.

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