Well, draw out the Lewis structure.
#"Cl"#: #7# valence electrons
#"O"#: #6# valence electrons
Therefore,
#"ClO"_2#: #7 + 2 xx 6 = 21# valence electrons
#"ClO"_2^(-)#: #7 + 2 xx 6 + 1 = 22# valence electrons
So now, put the largest atom in the middle (#"Cl"#) and surround it with the other atoms. Put two bonds on #"Cl"# by default, since it is the central atom. Add three lone pairs on each oxygen and begin to form the appropriate #pi# bonds using those electrons.
Do #"ClO"_2^-# first, then take one electron off of #"Cl"# to get #"ClO"_2#. For #"ClO"_2# you will have #11# valence electrons on #"Cl"#:
#"ClO"_2^-# really has resonance going on, so the hybrid structure has #1.5#-bonds. On the other hand, #"ClO"_2# has two double bonds instead.
FACTOR 1: BONDING ELECTRONS
The bonding electrons in #"ClO"_2# repel each other more, since there are more electrons in a double bond than a #1.5#-bond. That would increase the bond angle of #"ClO"_2# compared to #"ClO"_2^-#.
FACTOR 2: NONBONDING ELECTRONS
There is one more nonbonding electron in #"ClO"_2^-# to repel the bonding electrons, which would decrease the bond angle of #"ClO"_2^-# compared to #"ClO"_2#.
Therefore, #bb("ClO"_2)# would have the larger bond angle. In fact, #"ClO"_2# has #/_OClO = 117.4033^@#, whereas #"ClO"_2^-# has an approximate bond angle of #/_OClO ~~ 110^@ (pm 2^@)#).
