How many `sigma-bonds` are present in `N_2 O_3` ?

Answer

To determine the number of sigma bonds in the molecule \( N_2O_3 \), we can follow these steps:

### Step 1: Draw the Lewis Structure of \( N_2O_3 \)
- Start by identifying the total number of valence electrons. Nitrogen has 5 valence electrons and oxygen has 6. Therefore, for \( N_2O_3 \):
- Total valence electrons = \( 2 \times 5 + 3 \times 6 = 10 + 18 = 28 \) electrons.
- Arrange the atoms: The structure consists of two nitrogen atoms and three oxygen atoms. The nitrogen atoms will be in the center, bonded to the oxygen atoms.

### Step 2: Determine the Bonds
- In the Lewis structure, each nitrogen atom is bonded to one oxygen atom with a double bond (which consists of one sigma bond and one pi bond) and to another oxygen atom with a coordinate bond.
- The coordinate bond is formed when one atom provides both electrons for the bond.

### Step 3: Count the Sigma Bonds
- From the structure:
- Each double bond contributes 1 sigma bond. Since there are 2 double bonds, that gives us 2 sigma bonds.
- The coordinate bond between one nitrogen and the third oxygen contributes 1 additional sigma bond.

### Step 4: Total Sigma Bonds
- Therefore, the total number of sigma bonds in \( N_2O_3 \) is:
- 2 (from double bonds) + 1 (from the coordinate bond) = 3 sigma bonds.

### Final Answer
The total number of sigma bonds in \( N_2O_3 \) is **3**.

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