The correct order of increasing bond angles in the following species is:

Analyze $ClO_2$ (Chlorine Dioxide): The central Cl atom has 7 valence electrons. It forms double bonds with two O atoms and has one unpaired electron (odd electron species). The repulsion from a single odd electron is less than that of a full lone pair. The hybridization is effectively $sp^2$ with the odd electron in an orbital. The bond angle is approximately 117.6° (close to 120°).

Analyze $Cl_2O$ (Dichlorine Monoxide): The central atom is Oxygen ($sp^3$ hybridized). It has 2 bond pairs (with Cl) and 2 lone pairs. Normally, 2 lone pairs would compress the angle (as in $H_2O$, 104.5°). However, the Chlorine atoms are large and bulky, leading to significant steric repulsion (van der Waals repulsion) which opens up the bond angle to approximately 110.9° (greater than the tetrahedral angle of 109.5°).

Analyze $ClO_2^-$ (Chlorite Ion): The central atom is Chlorine ($sp^3$ hybridized). It has 2 bond pairs (with O) and 2 lone pairs. The strong lone pair-lone pair repulsion on the central Chlorine atom compresses the bond angle. The angle is experimentally found to be approximately 110–111°.

Conclusion: Comparing the specific values often used in this context ($ClO_2^- \approx 110-111^\circ$ and $Cl_2O \approx 111^\circ$), the order is often debated. However, based on the standard answer key for this AIPMT 2010 question, the order follows $ClO_2^- < Cl_2O < ClO_2$, implying the lone pair repulsion in the chlorite ion compresses the angle slightly more than the net result of steric hindrance in $Cl_2O$.