Total Number of Lone Pairs of Electrons in Dinitrogen Trioxide

To find the total count of non-bonding electron pairs (lone pairs) present in a molecule of dinitrogen trioxide ($\text{N}_2\text{O}_3$), we first determine its total number of valence electrons and analyze its Lewis structure.

---

Step 1: Total Valence Electron Count

• Valence electrons contributed by Nitrogen ($\text{N}$, Group 15): $\text{2 atoms} \times 5 = 10 \text{ electrons}$
• Valence electrons contributed by Oxygen ($\text{O}$, Group 16): $\text{3 atoms} \times 6 = 18 \text{ electrons}$

$\text{Total valence electrons} = 10 + 18 = 28 \text{ electrons}$

---

Step 2: Lewis Structure of $\text{N}_2\text{O}_3$

Dinitrogen trioxide exists in an asymmetrical planar form containing an $\text{N}-\text{N}$ bond, consisting of a nitrosyl group ($-\text{NO}$) bonded to a nitro group ($-\text{NO}_2$):

$$\text{O}_1 = \text{N}_1 - \text{N}_2 \begin{pmatrix} = \text{O}_2 \\ \smallsetminus \text{O}_3^- \end{pmatrix}$$

---

Step 3: Count of Lone Pairs on Each Atom

1. Nitrosyl Nitrogen ($\text{N}_1$):

   • Forms a double bond with $\text{O}_1$ and a single bond with $\text{N}_2$ ($3$ bonding electrons used).
   • Remaining valence electrons = $5 - 3 = 2$ electrons.
   • Lone pairs = 1

2. Nitro Nitrogen ($\text{N}_2$):

   • Forms a single bond with $\text{N}_1$, a double bond with $\text{O}_2$, and a single bond with $\text{O}_3^-$ ($4$ bonding electrons used).
   • Remaining valence electrons = $5 - 4 = 1$ (lost to form a coordinate bond/formal charge $+1$).
   • Lone pairs = 0

3. Nitrosyl Oxygen ($\text{O}_1$):

   • Forms a double bond with $\text{N}_1$ ($2$ bonding electrons used).
   • Remaining non-bonding electrons = $6 - 2 = 4$ electrons.
   • Lone pairs = 2

4. Double-bonded Nitro Oxygen ($\text{O}_2$):

   • Forms a double bond with $\text{N}_2$ ($2$ bonding electrons used).
   • Remaining non-bonding electrons = $6 - 2 = 4$ electrons.
   • Lone pairs = 2

5. Single-bonded Nitro Oxygen ($\text{O}_3$):

   • Forms a single bond with $\text{N}_2$ and carries a formal charge of $-1$ ($1$ bonding electron used, $6 + 1 = 7$ valence shell electrons total).
   • Remaining non-bonding electrons = $7 - 1 = 6$ electrons.
   • Lone pairs = 3

---

Step 4: Total Lone Pair Calculation

$\text{Total Lone Pairs} = \text{Lone pairs on }(\text{N}_1 + \text{N}_2 + \text{O}_1 + \text{O}_2 + \text{O}_3)$

$\text{Total Lone Pairs} = 1 + 0 + 2 + 2 + 3 = 8$

Verification via electron count: $\text{Total electrons} = (6 \text{ bonds} \times 2) + (8 \text{ lone pairs} \times 2) = 12 + 16 = 28 \text{ electrons}$

Thus, the total count of lone pairs of electrons in a molecule of $\text{N}_2\text{O}_3$ is 8.