Arrange the bond angles of $\mathrm{H_2O}$, $\mathrm{NH_3}$ and $\mathrm{CH_4}$ in decreasing order.

All three central atoms are $sp^3$ hybridised so the base bond angle is $109.5°$. Lone pairs compress that angle (the lone-pair-to-bond-pair repulsion is stronger than bond-pair-to-bond-pair). Approximate bond angles: - $\mathrm{CH_4}$ (0 lone pairs): $109.5°$ - $\mathrm{NH_3}$ (1 lone pair): $107°$ - $\mathrm{H_2O}$ (2 lone pairs): $104.5°$ More lone pairs means more compression, so the order is $\mathrm{CH_4} > \mathrm{NH_3} > \mathrm{H_2O}$.