To determine the bond angles, we apply VSEPR theory and consider steric hindrance as well as electronegativity:

• $\text{ClO}_2$: The central chlorine atom is $sp^2$ hybridised and possesses one lone pair along with an odd electron. Because the repulsion caused by an odd electron is less than that of a lone pair or a bond pair, the bond angle remains relatively large, approximately $118^\circ$.
• $\text{Cl}_2\text{O}$: The central oxygen atom is $sp^3$ hybridised with two bond pairs and two lone pairs. Normally, the angle would be around $104.5^\circ$ (like water), but due to the large size of the terminal chlorine atoms, severe steric hindrance (repulsion between Cl atoms) pushes the bonds apart, widening the angle to approximately $111^\circ$.
• $\text{ClO}_2^-$: The central chlorine atom is $sp^3$ hybridised with two bond pairs and two lone pairs. The terminal oxygen atoms are smaller than chlorine, leading to less steric hindrance than in $\text{Cl}_2\text{O}$. Additionally, oxygen is more electronegative than chlorine, drawing the bond pairs away from the central atom, which decreases bond pair-bond pair repulsion. This results in a slightly smaller bond angle than $\text{Cl}_2\text{O}$.

Therefore, the correct increasing order of bond angles is $\text{ClO}_2^- < \text{Cl}_2\text{O} < \text{ClO}_2$.