To determine which statement is false, we need to analyze the bond angles in $CH_4, NH_3,$ and $H_2O. \\$1. $CH_4$ (methane) has a tetrahedral geometry with bond angles of $109.5^{\circ}. \\$2. $NH_3$ (ammonia) has a trigonal pyramidal geometry due to the lone pair on nitrogen,$\\$ resulting in bond angles of approximately $107^{\circ}. \\$3. $H_2O$ (water) has a bent geometry due to two lone pairs on oxygen,$\\$ resulting in bond angles of approximately $104.5^{\circ}. \\$Now, let's evaluate each option:$\\$Option 1: The H – C – H bond angle in $CH_4,$ the H – N – H bond angle in $NH_3,$ and the H – O – H bond angle in $H_2O$ are all greater than $90^{\circ}. \\$ - This is true because all the mentioned bond angles are greater than $90^{\circ}. \\$Option 2: The H – O – H bond angle in $H_2O$ is larger than the H – C – H bond angle in $CH_4. \\$ - This is false because $104.5^{\circ}$ (H – O – H) is smaller than $109.5^{\circ}$ (H – C – H).$\\$Option 3: The H – O – H bond angle in $H_2O$ is smaller than the H – N – H bond angle in $NH_3. \\$ - This is true because $104.5^{\circ}$ (H – O – H) is smaller than $107^{\circ}$ (H – N – H).$\\$Option 4: The H – C – H bond angle in $CH_4$ is larger than the H – N – H bond angle in $NH_3. \\$ - This is true because $109.5^{\circ}$ (H – C – H) is larger than $107^{\circ}$ (H – N – H).$\\$Therefore, the false statement is Option 2.