Write down the Lewis structure of i) Na2S2O3 ii) O3

i) Lewis Structure of $\mathrm{Na}{}_2\mathrm{S}{}_2\mathrm{O}{}_3$

Step 1: Determine the total number of valence electrons

• Na: 1 valence electron each (2 atoms) → $2\times 1=2$
• S: 6 valence electrons each (2 atoms) → $2\times 6=12$
• O: 6 valence electrons each (3 atoms) → $3\times 6=18$
• Total valence electrons = $2+12+18=32$

sodium_thiosulfatesodium_thiosulfate

Step 2: Draw the skeletal structure

• The central atoms are S–S with O atoms bonded to them.
• Na⁺ ions are not included in the Lewis structure (they are ionic).

Step 3: Distribute electrons

• Each S forms a single bond with the other S and with O atoms.
• Remaining electrons are placed as lone pairs and double bonds to satisfy the octet rule.
• The structure shows S–S bonded with O atoms around them, with one S having a double bond to an O and a single bond to another O (which carries a negative charge).

Final Lewis structure (simplified for $\mathrm{S}{}_2\mathrm{O}{}_3{}^{2-}$):

   O
   ||
O–S–S–O⁻
   |
   O⁻

• Na⁺ ions are not shown in the Lewis structure (they balance the 2⁻ charge).

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ii) Lewis Structure of $\mathrm{O}{}_3$ (Ozone)

Step 1: Total valence electrons

• O: 6 valence electrons each (3 atoms) → $3\times 6=18$

Step 2: Skeletal structure

• Central O atom bonded to two terminal O atoms.

Step 3: Distribute electrons

• Each terminal O forms a single bond with the central O.
• Remaining electrons are placed as lone pairs and one double bond to satisfy the octet rule.
• The central O has one single bond and one double bond with terminal O atoms.

Final Lewis structure:

   O⁻
   //
O–O⁺=O

• The central O has 7 electrons (expanded octet), and one terminal O carries a partial negative charge, while the other has a partial positive charge (resonance structures exist).

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Resonance form 1 (partial charges)Resonance form 1 (partial charges)