The molecular geometry of \(NH_3\) and \(H_2O\) can be shown as:
The central atom \((N)\) in \(NH_3\) has one lone pair and there are three bond pairs. In \(H_2O\), there are two lone pairs and two bond pairs.
The two lone pairs present in the oxygen atom of \(H_2O\) molecule repels the two bond pairs. This repulsion is stronger than the repulsion between the lone pair and the three bond pairs on the nitrogen atom.
Since the repulsions on the bond pairs in \(H_2O\) molecule are greater than that in \(NH_3\), the bond angle in water is less than that of ammonia.
The correct order of bond enthalpy \(\left( kJ mol ^{-1}\right)\) is :
\(O - O\) bond length in \(H _2 O _2\) is X than the \(O - O\) bond length in \(F _2 O _2\)The \(O - H\) bond length in \(H _2 O _2\)is Y than that of the\(O - F\) bond in \(F _2 O _2\)Choose the correct option for \(\underline{X} and \underline{Y}\) from those given below :