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The correct order of increasing bond angles in the following species is
A
$ClO_2^- < Cl_2O < ClO_2$
B
$Cl_2O < ClO_2 < ClO_2^-$
C
$ClO_2 < Cl_2O < ClO_2^-$
D
$Cl_2O < ClO_2^- < ClO_2$
Detailed Solution
$ClO_2^-$: the central Cl has two bond pairs and two lone pairs ($sp^3$); strong lone pair–lone pair repulsion compresses the angle to about $110^\circ$ (smallest).
$Cl_2O$: the central O has two bond pairs and two lone pairs ($sp^3$), but the two bulky Cl atoms repel each other and open the angle slightly, to about $110.9^\circ$.
$ClO_2$: the central Cl has two bond pairs, one lone pair and one odd (unpaired) electron; a single electron repels much less than a lone pair, so the angle opens up to about $118^\circ$ (largest).
Increasing order of bond angle: $ClO_2^- < Cl_2O < ClO_2$
$Cl_2O$: the central O has two bond pairs and two lone pairs ($sp^3$), but the two bulky Cl atoms repel each other and open the angle slightly, to about $110.9^\circ$.
$ClO_2$: the central Cl has two bond pairs, one lone pair and one odd (unpaired) electron; a single electron repels much less than a lone pair, so the angle opens up to about $118^\circ$ (largest).
Increasing order of bond angle: $ClO_2^- < Cl_2O < ClO_2$