According to VSEPR model, what makes the shapes of ${\mathrm{CH}}_4, {\mathrm{NH}}_3 \mathrm{and} {\mathrm{H}}_2\mathrm{O} \mathrm{different}?$

In ${\mathrm{CH}}_4$ there is no lone pair. In ${\mathrm{NH}}_3,$ one lone pair and In ${\mathrm{H}}_2\mathrm{O}$there are two lone pairs on the respective central atoms .

In ${\mathrm{CH}}_4, {\mathrm{NH}}_3 \mathrm{and} {\mathrm{H}}_2\mathrm{O} \mathrm{the} \mathrm{angle} \mathrm{at} \mathrm{central} \mathrm{atom} \mathrm{are} \mathrm{H} —\mathrm{C}—\mathrm{H}={109}^{\mathrm{o}}28\text{'}, \mathrm{H}—\mathrm{N}—\mathrm{H}={107}^{\mathrm{o}} \mathrm{and} \mathrm{H}—\mathrm{O}—\mathrm{H}=104.5^{\mathrm{o}}.$