Animated Solution for Chemistry - s and p-Block Elements: Choose the correct statement from the following.

Welcome to a fascinating exploration of s-block thermodynamics! This problem challenges us to interpret a graph detailing the standard enthalpy of formation ($\Delta H_f$) for various alkali metal halides. The key to unlocking this graph lies in paying close attention to the axes. The y-axis represents $-\Delta H_f$. This means that an upward slope indicates that $-\Delta H_f$ is increasing, which mathematically translates to the actual enthalpy of formation ($\Delta H_f$) becoming more negative.

Let's trace the blue line representing the fluorides. Notice how it starts high at Lithium and slopes downwards towards Cesium? This downward slope means the value of $-\Delta H_f$ is decreasing. In physical terms, the actual enthalpy of formation is becoming less negative as we descend the group. This happens because the lattice enthalpy drops sharply as the metal ion gets larger, making the formation of the heavier fluorides less exothermic.

Now, contrast this with the chlorides, bromides, and iodides. Look at their lines—they are sloping upwards! For these larger halogens, the enthalpy of formation actually becomes more negative down the group. This completely contradicts Option (a), which claims bromides become less negative. Furthermore, looking at the peak of the fluoride line, we see that $LiF$ has the highest $-\Delta H_f$, meaning it has the most negative enthalpy of formation, instantly disproving Option (d).

Lattice vs. Hydration

Let's shift gears and talk about solubility. Imagine the thermodynamic battle that happens when a salt dissolves in water: the lattice enthalpy tries to keep the crystal together, while the hydration enthalpy tries to pull it apart and surround the ions with water molecules.

Look at Lithium Fluoride ($LiF$). Both the $L{i}^{+}$ ion and the $F^{-}$ ion are exceptionally tiny. They pack together incredibly tightly in the solid state, creating a massive lattice enthalpy. Because this lattice enthalpy is so overwhelmingly high, the energy released by hydrating these ions just isn't enough to break the crystal apart. This makes $LiF$ the least soluble among all alkali metal halides. Therefore, Option (c) is a perfectly correct statement!

Just to be thorough, let's look at Option (b). It claims Cesium Iodide ($CsI$) has low solubility due to high lattice enthalpy. Think about the physical reality—Cesium is a huge ion, and Iodide is also a huge ion. They do not pack well together, so their lattice enthalpy is actually very low!

So why is $CsI$ relatively insoluble? It's because its hydration enthalpy is even lower than its already low lattice enthalpy. Large ions have very low charge densities, meaning water molecules aren't strongly attracted to them. Option (b) is a classic trap designed to test if you truly understand the interplay of these two thermodynamic forces.