Answer :
Answer:
H2O> NH3> CH4
Explanation:
According to valence shell electron pair repulsion theory (VSEPR), bond angles and repulsion of electron pairs depends on the nature of electron pairs on the central atom of the molecule. Lone pairs cause more repulsion (and distortion of bond angles) than bond pairs). Lone pair- lone pair repulsion is greater than lone pair bond pair repulsion.
Water contains two lone pairs on oxygen hence it experiences the greatest repulsion. Ammonia has only one lone pair on nitrogen hence there is lesser repulsion between lone pairs and bond pairs. Methane possess only bond pairs of electrons hence it has the least repulsion.
Based on bond angles, the order of repulsion from strongest to weakest is; H2O > NH3 > CH4
We are given the following;
Methane; CH4
Ammonia;NH3
Water; H2O
The bond angle of the compounds above are;
Bond angle of CH4 = 109.5°
Bond angle of NH3 = 107°
Bond angle of H2O = 104.5°
Now, according to Valence Shell Electron Pair Repulsion Theory (VSEPR Theory), as the bond angle decreases due to the presence of lone pairs, it leads to more repulsion on the bond pairs and as a result of that, it makes the bond pairs to come closer.
Thus, it means the higher the bond angle the lesser the repulsion and vice versa.
Thus, the order of repulsion from strongest to weakest is;
H2O > NH3 > CH4
Read more about bond angles and repulsions at; https://brainly.com/question/14225705