Concept: Counting Lone Pairs in N₂O₃

To find the total number of lone pairs of electrons in N₂O₃ (dinitrogen trioxide), we need to draw its Lewis structure. This involves counting valence electrons, arranging atoms, forming bonds, and distributing remaining electrons as lone pairs.

Step-by-Step Solution:

Step 1: Count the total valence electrons.

Nitrogen (N) has 5 valence electrons. There are 2 N atoms: $2 \times 5 = 10$
Oxygen (O) has 6 valence electrons. There are 3 O atoms: $3 \times 6 = 18$
Total valence electrons = $10 + 18 = 28$

Step 2: Draw the skeletal structure.

The central atom is usually the least electronegative. Nitrogen is less electronegative than oxygen, but in N₂O₃, the structure is O=N–O–N=O (with one N–N bond). A common arrangement is:

O – N – O – N – O (but this is linear, not accurate)

Actual structure: It has a central nitrogen bonded to another nitrogen and two oxygens, and the second nitrogen bonded to the third oxygen. Typically, it is represented with N–N bond and terminal oxygens.

Structure: O=N–O–N=O (with one oxygen bridging between the two nitrogens).

More precisely: The Lewis structure has one N=O double bond on each end, and a central O atom single-bonded to both N atoms (bridging oxygen). So, the atoms are connected as: N–O–N, with each N having one terminal O.

Step 3: Distribute bonds and remaining electrons.

Assume the structure: Terminal O1 double-bonded to N1, N1 single-bonded to central O, central O single-bonded to N2, and N2 double-bonded to terminal O2.

Bonds:
- N1=O1 (double bond: 4 electrons)
- N1–O (single bond: 2 electrons)
- O–N2 (single bond: 2 electrons)
- N2=O2 (double bond: 4 electrons)
Total electrons used in bonds: $4 + 2 + 2 + 4 = 12$ electrons.
Remaining electrons: $28 - 12 = 16$ electrons to be placed as lone pairs.

Step 4: Assign lone pairs to complete octets.

Terminal O1 (double-bonded to N1): Needs 2 lone pairs (4 electrons) to complete octet.
Terminal O2 (double-bonded to N2): Needs 2 lone pairs (4 electrons).
Central O (single-bonded to N1 and N2): Has two bonds, so needs 2 lone pairs (4 electrons) for octet.
N1: Has double bond with O1 (4 electrons) and single bond with O (2 electrons). Total 6 electrons. Needs 1 lone pair (2 electrons) to complete octet? But nitrogen can have expanded octet? Actually, in this case, N1 has formal charge.
N2: Similarly, double bond with O2 (4 electrons) and single bond with O (2 electrons). Total 6 electrons. Needs 1 lone pair.

But wait, total lone pairs assigned so far: O1: 2 pairs, O2: 2 pairs, central O: 2 pairs. That is 6 pairs = 12 electrons. Plus, if we add one pair on N1 and one on N2, that is 2 more pairs (4 electrons). Total electrons used in lone pairs: 12 + 4 = 16, which matches.

So, lone pairs:

Terminal O1: 2 lone pairs
Terminal O2: 2 lone pairs
Central O: 2 lone pairs
N1: 1 lone pair
N2: 1 lone pair

Total lone pairs = 2 + 2 + 2 + 1 + 1 = 8 lone pairs.

Step 5: Verify formal charges.

To ensure correctness, calculate formal charges:

Formal charge = valence electrons - nonbonding electrons - 1/2 bonding electrons

Terminal O1: FC = 6 - 4 - 1/2*4 = 0
Terminal O2: FC = 6 - 4 - 1/2*4 = 0
Central O: FC = 6 - 4 - 1/2*4 = 0
N1: FC = 5 - 2 - 1/2*6 = 5 - 2 - 3 = 0
N2: FC = 5 - 2 - 1/2*6 = 0

All formal charges are zero, so the structure is correct.

Final Answer: The total number of lone pairs of electrons in N₂O₃ is $8$ .

Related Topics:

Lewis Structures
Valence Shell Electron Pair Repulsion (VSEPR) Theory
Formal Charge Calculation
Chemical Bonding
Oxides of Nitrogen

Formulae and Theory:

1. Total Valence Electrons: Sum of valence electrons of all atoms.

2. Formal Charge (FC): $F C = V - L - \frac{1}{2} B$
where V = number of valence electrons in free atom, L = number of nonbonding electrons, B = number of bonding electrons.

3. Octet Rule: Atoms tend to form bonds to achieve 8 electrons in their valence shell.

4. Lone Pairs: Pairs of valence electrons that are not shared between atoms.