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Question

Rationalize the following lattice energy values:

CompoundLattice Energy (kJ/mol) CaSe Na2Se CaTe Na2Te2862213027212095\begin{matrix} \text{Compound} & & \text{Lattice Energy (kJ/mol)}\\ \hline \\ \begin{array} { l } { \text { CaSe } } \\ { \mathrm { Na } _ { 2 } \mathrm { Se } } \\ { \text { CaTe } } \\ { \mathrm { Na } _ { 2 } \mathrm { Te } } \end{array} & & \begin{array} { l } { - 2862 } \\ { - 2130 } \\ { - 2721 } \\ { - 2095 } \end{array} \end{matrix}

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The lattice energy represents the amount of energy required to dissociate 1 mole of an ionic solid into the gaseous ions.

Lattice energy, U = kQ1Q2ro\textbf{Lattice energy, U = $\dfrac{k'Q_{1}Q_{2}}{r_{o}}$}

Lattice energy is directly proportional to the charges of the ions and is inversely related to the internuclear distance. It is also inversely proportional to the size of the ions.

Hence, lattice energies are highest for substances with small, highly charged ions.

Here, we have

CaSe, Na2Se, CaTe and Na2Te\textbf{CaSe, Na$_2$Se, CaTe and Na$_2$Te}

Ca2+^{2+} has a greater charge than Na+^{+} because of which CaSe and CaTe will have higher lattice energies compared to Na2_2Se and Na2_2Te.

Also, Se2^{2-} is smaller than Te2^{2-} because of which lattice energy of anions containing Se2^{2-} will be higher as compared to Te2^{2-} anion.

The effect of charge on the lattice energy is greater than the effect of size. Hence, these factors combine to give the following trend in lattice energies:

CaSe > CaTe > Na2Se > Na2Te\textbf{CaSe $>$ CaTe $>$ Na$_2$Se $>$ Na$_2$Te}

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