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^@)).
