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Here’s how I would do it.
You can find the procedure .
For ##ClO_2^- Cl## is the less electronegative atom. So ##Cl## is the central atom
The skeleton structure is ##O-Cl-O.##
The trial structure is
You have 20 in your trial structure.
The you have available are: ##1 Cl + 2 O + 1 e = 17 + 26 + 1 = 20##.
Hence the trial structure has the correct number of electrons.
The formal charge on each atom is:
##Cl = 7 – 4 – (4) = +1; O = 6 6 – (2) = -1##
Every atom has a formal charge.
We can reduce the number of formal charges by moving a lone pair of electrons from ##O## to form a ##Cl=O## double bond.
This gives us two new structures in which one ##O## atom has ##FC = -1## and the other atoms have ##FC = 0##:
The actual structure is nine of these. Rather it is a resonance hybrid of them both