Electrovalent bonding Questions in English

(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 $\frac{1}{(r_c + r_a)}$.

(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 $|-1031| = 1031 \ kJ \ mol^{-1}$.

(B) Step $1$: Determine the valency of element $A$. The configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$ shows $2$ electrons in the outermost shell. Thus, $A$ loses $2$ electrons to form $A^{2+}$ ion.
Step $2$: Determine the valency of element $B$. The configuration $1s^2 2s^2 2p^6 3s^2 3p^5$ shows $7$ electrons in the outermost shell. Thus, $B$ gains $1$ electron to form $B^-$ ion.
Step $3$: Determine the formula of the ionic compound. To balance the charges, one $A^{2+}$ ion reacts with two $B^-$ ions. The formula is $AB_2$.

(B) Step $1$: An ionic bond is formed by the complete transfer of electrons from a metal atom to a non-metal atom.
Step $2$: For the metal to lose electrons easily, it must have a low ionization enthalpy.
Step $3$: For the non-metal to accept electrons easily and release energy, it must have a high negative value of electron gain enthalpy (high electron affinity).
Step $4$: Therefore, the combination of low ionization enthalpy of the metal and high negative electron gain enthalpy of the non-metal favors ionic bond formation.