$\mathrm{HF}$ is a weak acid, but in ionic state it leads to the formation of ${\mathrm{HF}}^{-}$ ions. Based on the ion-dipole attractions involved, the amount of energy released when one mole of $\mathrm{HF}$ is neutralized with excess strong alkali is expected to be

We know that in heat of neutralisation we have:

Case I: When a strong acid is neutralised by a strong base.

Heat of neutralisation $=-57.3\mathrm{ }\mathrm{k}\mathrm{J}$

Case II: When a strong acid is neutralised by a weak base or vice-versa

Heat of neutralisation $<-57.3\mathrm{ }\mathrm{k}\mathrm{J}$

Here, $\mathrm{HF}$ is a weak acid which is neutralised by a strong base. So, the heat of neutralisation must be less than $-57.3\mathrm{ }\mathrm{k}\mathrm{J}$.

Hence, option (B) is correct.