The total number of lone pair of electrons in `N_(2)O_(3)` is

Answer

To determine the total number of lone pairs of electrons in \( N_2O_3 \), we will follow these steps:

### Step 1: Determine the Lewis Structure of \( N_2O_3 \)
- Start by identifying the number of valence electrons for each atom in the molecule.
- Nitrogen (N) has 5 valence electrons, and there are 2 nitrogen atoms: \( 2 \times 5 = 10 \) electrons.
- Oxygen (O) has 6 valence electrons, and there are 3 oxygen atoms: \( 3 \times 6 = 18 \) electrons.
- Total valence electrons = \( 10 + 18 = 28 \) electrons.

### Step 2: Draw the Bonds
- Connect the two nitrogen atoms with a single bond.
- Connect one nitrogen atom to two oxygen atoms with double bonds.
- Connect the other nitrogen atom to one oxygen atom with a dative bond (where one atom provides both electrons).

### Step 3: Assign Lone Pairs to Each Atom
- For the nitrogen atom with three bonds (the one connected to two oxygens with double bonds), it will have 1 lone pair.
- For the nitrogen atom connected to one oxygen with a dative bond, it has 0 lone pairs since it is involved in bonding.
- For the oxygen atoms:
- The oxygen atom involved in the dative bond will have 2 lone pairs.
- The two oxygen atoms involved in double bonds will each have 2 lone pairs.

### Step 4: Count the Total Lone Pairs
- Nitrogen 1: 1 lone pair
- Nitrogen 2: 0 lone pairs
- Oxygen 1 (dative bond): 2 lone pairs
- Oxygen 2 (double bond): 2 lone pairs
- Oxygen 3 (double bond): 2 lone pairs

Now, summing these up:
- Total lone pairs = \( 1 + 0 + 2 + 2 + 2 = 7 \) lone pairs.

### Conclusion
The total number of lone pairs of electrons in \( N_2O_3 \) is **7**.

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