According to VSEPR theory, the central atoms in $\text{CH}_4$, $\text{NH}_3$, and $\text{H}_2\text{O}$ all undergo $sp^3$ hybridisation. However, they possess different numbers of lone pairs: carbon in $\text{CH}_4$ has 0, nitrogen in $\text{NH}_3$ has 1, and oxygen in $\text{H}_2\text{O}$ has 2 lone pairs. Because lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion, which is in turn stronger than bond pair-bond pair repulsion, the bond angle decreases as the number of lone pairs increases.

• In $\text{CH}_4$, the bond angle is the ideal tetrahedral angle of $109.5^\circ$.
• In $\text{NH}_3$, the presence of one lone pair compresses the H-N-H angle to $107^\circ$.
• In $\text{H}_2\text{O}$, the presence of two lone pairs compresses the H-O-H angle further to $104.5^\circ$. As all these angles ($109.5^\circ$, $107^\circ$, and $104.5^\circ$) are greater than $90^\circ$, option 4 is correct. The correct order of bond angles is $\text{CH}_4 > \text{NH}_3 > \text{H}_2\text{O}$. Therefore, the statement claiming that the H-O-H bond angle in $\text{H}_2\text{O}$ is larger than the H-C-H bond angle in $\text{CH}_4$ is factually incorrect.