Number of sigma bonds in `P_(4)O_(10)` is :

Options

A. 6
B. 7
C. 17
D. 16

Explanation

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

### Step 1: Understand the Structure of \( P_4O_{10} \)
- \( P_4O_{10} \) consists of 4 phosphorus (P) atoms and 10 oxygen (O) atoms.
- The structure can be visualized as a central tetrahedral arrangement of phosphorus atoms with oxygen atoms bridging them.

### Step 2: Draw the Structure
- The structure of \( P_4O_{10} \) can be represented as follows:
- Each phosphorus atom is bonded to three oxygen atoms.
- Two of these bonds are typically double bonds (P=O), and one is a single bond (P-O).

### Step 3: Count the Sigma Bonds
- For each P atom:
- Each P-O single bond contributes 1 sigma bond.
- Each P=O double bond contributes 1 sigma bond (the other part of the double bond is a pi bond).

- In \( P_4O_{10} \):
- Each phosphorus atom forms 3 bonds with oxygen.
- Therefore, for 4 phosphorus atoms, the total number of bonds is \( 4 \times 3 = 12 \) bonds.
- However, since there are double bonds, we need to account for them correctly.

### Step 4: Final Count of Sigma Bonds
- Each of the 4 phosphorus atoms forms:
- 1 P=O double bond contributes 1 sigma bond.
- 2 P-O single bonds contribute 2 sigma bonds.

- Total sigma bonds from phosphorus:
- \( 4 \times 1 \) (from double bonds) + \( 4 \times 2 \) (from single bonds) = 4 + 8 = 12 sigma bonds.

- Additionally, there are 4 more sigma bonds from the bridging oxygens, leading to:
- Total sigma bonds = 12 (from P-O and P=O) + 4 (from bridging oxygens) = 16 sigma bonds.

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