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 \( P_4O_{10} \)
- \( P_4O_{10} \) is known as phosphorus pentoxide. It consists of 4 phosphorus (P) atoms and 10 oxygen (O) atoms.
- The structure can be visualized as a central phosphorus atom bonded to oxygen atoms, with some of these bonds being double bonds.

### Step 2: Draw the Lewis Structure
- Begin by arranging the phosphorus and oxygen atoms. Each phosphorus atom is typically bonded to oxygen atoms.
- In the structure, each phosphorus atom is bonded to three oxygen atoms, two of which are single bonds and one is a double bond.

### Step 3: Identify the Types of Bonds
- Each single bond between phosphorus and oxygen is a sigma bond.
- Each double bond consists of one sigma bond and one pi bond.

### Step 4: Count the Sigma Bonds
- For each phosphorus atom:
- It forms 3 sigma bonds with oxygen (2 single bonds and 1 from the double bond).
- Since there are 4 phosphorus atoms, we can calculate the total sigma bonds contributed by phosphorus:
- \( 4 \text{ P atoms} \times 3 \text{ sigma bonds per P} = 12 \text{ sigma bonds} \)
- Additionally, we have to count the sigma bonds from the double bonds:
- There are 4 double bonds in total, contributing 4 additional sigma bonds (1 from each double bond).

### Step 5: Total Sigma Bonds Calculation
- Total sigma bonds = Sigma bonds from P-O single bonds + Sigma bonds from P=O double bonds
- Total sigma bonds = \( 12 + 4 = 16 \)

### Conclusion
- The total number of sigma bonds in the structure of \( P_4O_{10} \) is **16**.

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