Question

Consider the following enthalpy changes

$$\begin{gathered} {\mathit{F}}^{\mathbf{-}}\left(g\right)\mathbf{+}\mathit{H}\mathit{F}\left(g\right)\mathbf{\to }\mathit{F}\mathit{H}{\mathit{F}}^{\mathbf{-}}\left(g\right) \\ {\left(C{H}_3\right)}_{\mathbf{2}}\mathit{C}\mathbf{=}\mathit{O}\left(g\right)\mathbf{+}\mathit{H}\mathit{F}\left(g\right)\mathbf{\to }{\left(C{H}_3\right)}_{\mathbf{2}}\mathit{C}\mathbf{=}\mathit{O}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathit{H}\mathit{F}\left(g\right) \\ {\mathit{H}}_{\mathbf{2}}\mathit{O}\left(g\right)\mathbf{+}\mathit{H}\mathit{O}\mathit{H}\left(g\right)\mathbf{\to }{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathit{H}\mathit{O}\mathit{H}\left(inice\right) \end{gathered}$$ $$\begin{gathered} \mathbf{∆}\mathit{H}\mathbf{=}\mathbf{-}\mathbf{155}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l} \\ \mathbf{∆}\mathit{H}\mathbf{=}\mathbf{-}\mathbf{46}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l} \\ \mathbf{∆}\mathit{H}\mathbf{=}\mathbf{-}\mathbf{21}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l} \end{gathered}$$

How do the strengths of hydrogen bonds vary with the electronegativity of the element to which hydrogen is bonded? Where in the preceding series would you expect hydrogen bonds of the following type to fall?

Short answer

In all these reactions, it is visible that in l reactions, energy is provided for breaking the bond, and all three are endothermic reactions.

Fluoride is the most electronegative element. That's why it needs maximum energy to break bonds compared to oxygen.

In the first reaction, the hydrogen bonding is present between the molecules of HF, making the strongest hydrogen bond.

In the second reaction, hydrogen bonding is present between Oxygen and Hydrogen, but fluorine in HF molecules shows its effect. That's why its enthalpy is also higher.

In the third reaction, hydrogen bonding is present between Oxygen and Hydrogen. This reaction is also endothermic, but the bond is weaker compared to bonds of the first and second reactions.

Hydrogen bonding present in N- -H-O is much stronger the that of N- -H-N because the electronegativity of the oxygen is greater than that of nitrogen. It causes effects on hydrogen bonding and increases the strength of the bond.

Step-by-step solution

Electronegativity

As we all know, an electronegative atom such as fluorine, oxygen, or nitrogen is a hydrogen bond acceptor; it doesn't matter whether it is bonded to a hydrogen atom.

Greater electronegativity of the hydrogen bond acceptor will create a stronger hydrogen bond.

Step-2 Hydrogen bond

Hydrogen bonding acts between the molecules, which is a weak bond that usually acts in the presence of hydrogen in molecules.

A hydrogen bond will be the attractive electromagnetic interaction between polar molecules, where hydrogen will be a highly electronegative atom, for instance, nitrogen, oxygen, or fluorine.

Explanation

In all these reactions, it is visible that in l reactions, energy is provided for breaking the bond, and all three are endothermic reactions.

Fluoride is the most electronegative element. That's why it needs maximum energy to break bonds compared to oxygen.

In the first reaction, hydrogen bonding is present between the molecules of HF, which make the strongest hydrogen bond.

In the second reaction, the hydrogen bonding is present between Oxygen and Hydrogen, but fluorine in HF molecules shows its effect, which is why its enthalpy is also higher.

In the third reaction, hydrogen bonding is present between Oxygen and Hydrogen. This reaction is also endothermic, but the bond is weaker compared to bonds of the first and second reactions.

Hydrogen bonding present in N- -H-O is much stronger the that of N- -H-N because the electronegativity of the oxygen is greater than that of nitrogen. It causes effects on hydrogen bonding and increases the strength of the bond.