The total number of sigma bonds in the structure of `P_4O_10` is

Answer

To determine the total number of sigma bonds in the structure of \( P_4O_{10} \), we can follow these steps:

### Step 1: Understand the structure of \( P_4O_{10} \)
- \( P_4O_{10} \) consists of four phosphorus (P) atoms and ten oxygen (O) atoms. The structure can be visualized as a tetrahedral arrangement of phosphorus atoms with oxygen atoms forming bonds.

### Step 2: Identify the types of bonds
- In \( P_4O_{10} \), there are both single and double bonds. Each double bond consists of one sigma bond and one pi bond.

### Step 3: Count the sigma bonds
1. **Phosphorus-Phosphorus (P-P) bonds**:
- There are 3 P-P bonds in the structure. Each P-P bond is a sigma bond.
- Total from P-P bonds = 3 sigma bonds.

2. **Phosphorus-Oxygen (P-O) bonds**:
- Each phosphorus atom is bonded to oxygen atoms. In \( P_4O_{10} \), there are 10 P-O bonds.
- Each P-O bond is a sigma bond.
- Total from P-O bonds = 10 sigma bonds.

3. **Double bonds**:
- In \( P_4O_{10} \), there are 4 P=O double bonds. Each double bond contributes 1 sigma bond.
- Total from P=O bonds = 4 sigma bonds.

### Step 4: Calculate the total number of sigma bonds
- Total sigma bonds = (sigma from P-P) + (sigma from P-O) + (sigma from P=O)
- Total sigma bonds = 3 (from P-P) + 10 (from P-O) + 4 (from P=O) = 17 sigma bonds.

### Conclusion
The total number of sigma bonds in the structure of \( P_4O_{10} \) is **17**.

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