
Lattice energy is defined as the energy released when \[1\,mol\] of ionic solid is formed from its ions in the gaseous phase. From the following given compound, the lattice energy is maximum for (compare the only magnitude)
(A) $N{a_2}O$
(B) $NaF$
(C) $Mg{F_2}$
(D) $MgC{l_2}$
Answer
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Hint: when the ions combine to form a compound a certain amount of energy is related to lower the energy of the compound so formed. This energy is the lattice energy of the compound. If we want to break the compound again to its constituent elements then the energy equivalent to the lattice energy needs to be supplied to the compound.
Complete step by step answer:
The magnitude of the lattice energy depends on basically two factors that are namely the charge of the ion and the radius of the ions which are involved.This relationship can be represented by the relation
$U = \dfrac{{{q_1}{q_2}}}{d}$ , where $U$ represents the lattice energy and $q$ represents the charge of the ions involved. $d$ is the size of the ions involved.
The size of the ion depends on the electronic charge that it possesses and the outermost orbital in which the electron of the atom or ion that is under consideration. If the effective nuclear charge of the ion is large, then the nuclear attraction of the ion attracts the outermost shell electron and hence reduces the size of the ion. Whereas the addition of extra shells on the ions increases the size of the ion. so the size of the ions is the resultant of the resultant effect of the two above factors.
The size of the cations involved in the question can be seen as
$Mg < Na$ and
The size of the anions involved are,
$F < O < Cl$
Considering the above relation of lattice energy and the size of the specified ions we can conclude with the answer being
$NaF < MgC{l_2}N{a_2}O < Mg{F_2}$
So, the correct answer is Option C .
Note: The lattice energy is generally calculated from the Born-Haber Cycle. The formation of the crystal lattice is an exothermic process since it deals with the energy by bringing the gaseous atoms to the ionic lattice structure. The relation between molar lattice energy and molar lattice enthalpy is given by:
$\Delta U = \Delta H - p\Delta {V_m}$ ,
$\Delta U,\Delta H,\Delta V$ are lattice energy, lattice enthalpy, and change of the volume per mole respectively.
Complete step by step answer:
The magnitude of the lattice energy depends on basically two factors that are namely the charge of the ion and the radius of the ions which are involved.This relationship can be represented by the relation
$U = \dfrac{{{q_1}{q_2}}}{d}$ , where $U$ represents the lattice energy and $q$ represents the charge of the ions involved. $d$ is the size of the ions involved.
The size of the ion depends on the electronic charge that it possesses and the outermost orbital in which the electron of the atom or ion that is under consideration. If the effective nuclear charge of the ion is large, then the nuclear attraction of the ion attracts the outermost shell electron and hence reduces the size of the ion. Whereas the addition of extra shells on the ions increases the size of the ion. so the size of the ions is the resultant of the resultant effect of the two above factors.
The size of the cations involved in the question can be seen as
$Mg < Na$ and
The size of the anions involved are,
$F < O < Cl$
Considering the above relation of lattice energy and the size of the specified ions we can conclude with the answer being
$NaF < MgC{l_2}N{a_2}O < Mg{F_2}$
So, the correct answer is Option C .
Note: The lattice energy is generally calculated from the Born-Haber Cycle. The formation of the crystal lattice is an exothermic process since it deals with the energy by bringing the gaseous atoms to the ionic lattice structure. The relation between molar lattice energy and molar lattice enthalpy is given by:
$\Delta U = \Delta H - p\Delta {V_m}$ ,
$\Delta U,\Delta H,\Delta V$ are lattice energy, lattice enthalpy, and change of the volume per mole respectively.
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