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Electrovalent bonding Questions in English

Class 11 Chemistry · Chemical Bonding and Molecular Structure · Electrovalent bonding

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151
MediumMCQ
$AB$ is an ionic solid. The ionic radii of $A^{+}$ and $B^{-}$ are respectively $r_c$ and $r_a$. Lattice energy of $AB$ is proportional to:
A
$\frac{r_c}{r_a}$
B
$(r_c + r_a)$
C
$\frac{r_a}{r_c}$
D
$\frac{1}{(r_c + r_a)}$

Solution

(D) The lattice energy $(U)$ of an ionic crystal is defined as the energy released when gaseous ions combine to form one mole of the solid ionic crystal.
According to the Born-Landé equation,the lattice energy is inversely proportional to the inter-ionic distance $(r_0)$.
Since the inter-ionic distance $r_0$ is the sum of the ionic radii of the cation $(r_c)$ and the anion $(r_a)$,we have $r_0 = r_c + r_a$.
Therefore,the lattice energy $U$ is proportional to $\frac{1}{(r_c + r_a)}$.
152
DifficultMCQ
If the enthalpy of sublimation of $Li$ is $155 \ kJ \ mol^{-1}$,enthalpy of dissociation of $F_2$ is $150 \ kJ \ mol^{-1}$,ionization enthalpy of $Li$ is $520 \ kJ \ mol^{-1}$,electron gain enthalpy of $F$ is $-313 \ kJ \ mol^{-1}$,and standard enthalpy of formation of $LiF$ is $-594 \ kJ \ mol^{-1}$,then the magnitude of the lattice enthalpy of $LiF$ is . . . . . . $kJ \ mol^{-1}$ (nearest integer).
A
$1000$
B
$1031$
C
$1150$
D
$950$

Solution

(B) According to the Born-Haber cycle for the formation of $LiF(s)$:
$\Delta H_f^{\circ} = \Delta H_{sub}(Li) + \frac{1}{2} \Delta H_{diss}(F_2) + IE(Li) + EGE(F) + U$
Where $U$ is the lattice enthalpy.
Substituting the given values:
$-594 = 155 + \frac{150}{2} + 520 + (-313) + U$
$-594 = 155 + 75 + 520 - 313 + U$
$-594 = 437 + U$
$U = -594 - 437 = -1031 \ kJ \ mol^{-1}$
The magnitude of the lattice enthalpy is $|-1031| = 1031 \ kJ \ mol^{-1}$.

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