Question

Lattice energy of NaCl is ‘X’. if the ionic size of ${{A}^{+2}}$is equal to that of$N{{a}^{+}}$and ${{B}^{-2}}$is equal to$C{{l}^{-}}$, then lattice energy associated with the crystal AB is?

Answer 1

Hint: lattice energy of any compound is the energy that is required to completely separate the respective ions in an ionic compound that form their respective gaseous states. It is derived by the charges on both ions upon the distance between the two charged atoms in the ionic compound.

Complete answer:
Lattice energy can be defined as the energy required to completely separate the ions of an ionic compound into their gas states. The ions thus formed are cation (positive ions) and anions (negative ions).
Sodium chloride NaCl is an ionic compound that dissociates as:
$NaCl(s)\to N{{a}^{+}}(g)+C{{l}^{-}}(g)$
The formula of lattice energy can be derived as ${{E}_{lattice}}=k\dfrac{{{Q}_{1}}{{Q}_{2}}}{d}$ , where Q are the charges (positive and negative) and d is the distance between them, k is the constant of proportionality.
So, for NaCl it can be${{E}_{lattice}}NaCl=\dfrac{{{Q}_{1}}{{Q}_{2}}}{d}=X$that is X as given in the question.
Now, we have been given that NaCl is termed as AB and the ionic charges are${{A}^{+2}}$is equal to that of$N{{a}^{+}}$and ${{B}^{-2}}$is equal to$C{{l}^{-}}$, so the formula will be
${{E}_{lattice}}AB=\dfrac{(2{{Q}_{1}})(2{{Q}_{2}})}{d}=$
${{E}_{lattice}}AB=4\left( \dfrac{{{Q}_{1}}{{Q}_{2}}}{d} \right)$
As, $\dfrac{{{Q}_{1}}{{Q}_{2}}}{d}=X$,
Therefore, the lattice energy will be 4X.
Hence, lattice energy associated with the crystal AB is 4X.

Note:
As the charges on the ions, that is the cations and the anions, increase, the lattice energy of the compound also increases. While the distance between the charges of the ions or the ionic radii increases the lattice energy decreases. The formation of an ionic compound is exothermic while the dissociation or the reverse of it is endothermic.