Question

Which of the following alkali metal halides has the lowest lattice energy?
(A) $LiF$
(B) $NaCl$
(C) $KBr$
(D) $CsI$

Answer 1

Alkali metals react with halogen and form alkali metal halides. These alkali metal halides are ionic compounds arranged in a crystal lattice. The energy required to dissociate one mole of the compound to its constituent ions. It is directly proportional to the size of the ions in a given compound.

Complete answer:
The elements that belong to group $1$ in the periodic table are known as alkali earth metals. These are metals as they can easily lose electrons. Halogens are the group of chemical elements that belong to group $17$ and can try to attract or gain electrons. These are non-metals.
Both these metals and non-metals combine with each other to form an ionic compound known as alkali metal halides.
The ions in the given compounds are arranged in the definite order known as a crystal lattice. The energy required to dissociate the compound into ions in the gaseous state is known as lattice energy.
The lattice energy is inversely proportional to the size of ions in the compound.
In the given ions, $CsI$ has cesium ion and iodide ion with the largest size. Thus, $CsI$ has the lowest lattice energy.

Note:
In the given compounds, all the cations belong to the same group i.e.., alkali earth metals, and all the anions belong to halogens. The atomic size increases while moving from top to bottom. Thus, in the cation, cesium has more size and in anions, iodide has more size.