How many `N-O (sigma)` bonds are present in `N_2 O_5` ?

Answer

To determine the number of N-O (sigma) bonds present in \( N_2O_5 \), we can follow these steps:

### Step 1: Understand the Structure of \( N_2O_5 \)
The molecular structure of \( N_2O_5 \) consists of two nitrogen atoms and five oxygen atoms. It can be represented as:

- Two nitrogen atoms (N)
- Five oxygen atoms (O)

### Step 2: Draw the Lewis Structure
To visualize the bonding, we can draw the Lewis structure of \( N_2O_5 \):

1. Each nitrogen atom forms a double bond with two oxygen atoms.
2. Each nitrogen atom also forms a coordinate bond with one oxygen atom.

### Step 3: Identify the Types of Bonds
In the Lewis structure of \( N_2O_5 \):
- Each double bond between nitrogen and oxygen consists of one sigma bond and one pi bond.
- Each coordinate bond (where one atom donates both electrons) is also considered a sigma bond.

### Step 4: Count the Sigma Bonds
Now, let’s count the sigma bonds:
- From the two nitrogen atoms, we have:
- 2 double bonds (each contributes 1 sigma bond) = 2 sigma bonds
- 2 coordinate bonds (each contributes 1 sigma bond) = 2 sigma bonds
- Additionally, there are 2 more sigma bonds from the remaining double bonds between nitrogen and oxygen.

Thus, we can summarize:
- Total sigma bonds from double bonds = 2 (from N=O)
- Total sigma bonds from coordinate bonds = 2 (from N→O)
- Total sigma bonds from remaining N-O bonds = 2

### Step 5: Calculate Total Sigma Bonds
Adding all the sigma bonds together:
- Total N-O sigma bonds = 2 (from double bonds) + 2 (from coordinate bonds) + 2 (from remaining N-O bonds) = 6 sigma bonds.

### Final Answer
The total number of N-O (sigma) bonds in \( N_2O_5 \) is **6**.

---