The alkaline earth metal nitrate that does not crystallise with molecules is:

We are asked to select that alkaline-earth metal nitrate which does not form crystals containing water of crystallisation. We have the four possible nitrates $$Ba(NO_3)_2,\;Sr(NO_3)_2,\;Mg(NO_3)_2,\;Ca(NO_3)_2.$$

Whether a salt retains water molecules in its crystal lattice depends largely on two opposite energetic factors. First, there is the hydration enthalpy; a small, highly charged cation such as $$Mg^{2+}$$ possesses a strong ability to attract and hold water dipoles, leading to hydrated crystals. Second, there is the lattice enthalpy; a large cation such as $$Ba^{2+}$$ forms an extended ionic lattice of comparatively lower charge density, so the gain in stability on attaching water is insufficient to offset the lattice enthalpy that would be lost.

Let us arrange the cations of the group in order of increasing ionic radius:

$$Mg^{2+}\lt Ca^{2+}\lt Sr^{2+}\lt Ba^{2+}.$$

Because the charge is the same (+2) throughout the series, a smaller radius means a larger charge density and, hence, a stronger hydration enthalpy. Thus we expect

$$\Delta H_{\text{hydr}}(Mg^{2+})\gt \Delta H_{\text{hydr}}(Ca^{2+})\gt \Delta H_{\text{hydr}}(Sr^{2+})\gt \Delta H_{\text{hydr}}(Ba^{2+}).$$

Consequently, the nitrates of $$Mg^{2+},\;Ca^{2+},$$ and $$Sr^{2+}$$ tend to bind water molecules when they crystallise, forming hydrates such as $$Mg(NO_3)_2\!\cdot\!6H_2O,$$ $$Ca(NO_3)_2\!\cdot\!4H_2O,$$ and $$Sr(NO_3)_2\!\cdot\!4H_2O.$$

In contrast, $$Ba^{2+}$$ is the largest cation of the list. Its low charge density gives a hydration enthalpy too small to stabilise water within the lattice. So, when $$Ba(NO_3)_2$$ crystallises, it does so anhydrously:

$$Ba^{2+} + 2\,NO_3^- \longrightarrow Ba(NO_3)_2\;(s)$$ $$\text{(no water attached)}.$$

Thus among the four nitrates provided, only $$Ba(NO_3)_2$$ fails to incorporate any molecules of water in its crystalline form.

Hence, the correct answer is Option A.