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The lattice energy of NaCl is ‘X’. If the ionic size of A$^{2 + }$ is equal to that of Na+ and B$^{2 - }$ is equal to Cl- , then lattice energy associated with the crystal AB is
(a) X
(b) $2$X
(c) $4$X
(d) $8$X

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Last updated date: 16th Jul 2024
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Answer
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Hint: The energy associated with the crystal lattice of a compound is called the lattice energy. It is directly proportional to the product of the ion charges and inversely proportional to the internuclear distance. In order to determine the lattice energy associated with the crystal AB, substitute the charge of A and B in the equation and then reorder the equation in terms of the lattice energy of NaCl.

Complete step-by-step solution:
The given values are:
The lattice energy of NaCl is X
That is $U_{NaCl}$ =X
$U_{NaCl}$$ = k\dfrac{{{q_1}{q_2}}}{d} = X$ …………… (1) Charge of Na\[ = + 1\] , Cl\[ = - 1\]
To find that lattice energy associated with the crystal AB
$U_{AB}$$ = k\dfrac{{(2{q_1})*(2{q_2})}}{d}$ Charge of A\[ = + 2\] , B\[ = - 2\]
Simplifying we get, $U_{AB}$$ = k\dfrac{{4*{q_1}{q_2}}}{d} = 4*(k\dfrac{{{q_1}{q_2}}}{d})$
From equation (1) $U_{NaCl}$$ = k\dfrac{{{q_1}{q_2}}}{d} = X$
Substitute the lattice energy value of NaCl in the lattice energy formula of AB $U_{NaCl}$ in $U_{AB}$
From this we get, lattice energy of AB, $U_{AB}$$ = 4*(k\dfrac{{{q_1}{q_2}}}{d}) = 4*{U_{NaCl}} = 4*X = 4X$
Option (c) is the correct answer.

Note: Lattice energy is the energy which is released when two oppositely charged ions attract each other to form an ionic solid. The two main factors that affect the lattice energy of ionic compounds are the magnitude of charge and the distance between the ions. In these types of questions the values of distance between the ion and charge value of the ion need not be substituted in the equation.