The Lewis structure of the most important resonance form of the ion ClO2– (chlorite ion) can be drawn and explained as follows:
1. Determine the total number of valence electrons:
– Chlorine (Cl) has 7 valence electrons.
– Each Oxygen (O) has 6 valence electrons.
– The ion has an extra electron due to the negative charge.
– Total valence electrons = 7 (Cl) + 2 × 6 (O) + 1 (charge) = 20 electrons.
2. Draw the skeletal structure:
– Place Chlorine (Cl) in the center and Oxygen atoms (O) on either side.
– Connect the atoms with single bonds.
3. Distribute the remaining electrons:
– After forming single bonds, 16 electrons remain.
– Place lone pairs on the Oxygen atoms to satisfy the octet rule.
– Each Oxygen atom will have 3 lone pairs (6 electrons).
4. Formal Charges:
– Chlorine (Cl): Formal charge = 7 – (2 + 4) = +1
– Oxygen (O) with single bond: Formal charge = 6 – (6 + 1) = -1
– Oxygen (O) with double bond: Formal charge = 6 – (4 + 2) = 0
5. Oxidation Numbers:
– Chlorine (Cl): Oxidation number = +3
– Oxygen (O): Oxidation number = -2
6. Resonance Structure:
– The double bond can be formed between Chlorine and either Oxygen atom, leading to resonance structures.
– The most important resonance form has the double bond with the Oxygen atom that has a formal charge of 0.
Here is the Lewis structure:
[O]-
|
[Cl]=[O]
In this structure, the Chlorine atom is double-bonded to one Oxygen atom and single-bonded to the other Oxygen atom. The formal charges and oxidation numbers are as calculated above.