Question

According to Born-Haber's cycle, the enthalpy of formation of ionic compounds can be determined. The formation of NaCl involves following steps –
(i) $Na\left(s\right)\underset{\text{Energy of subtimation}}{\overset{S}{\to }}Na\left(g\right)$

(ii) $Na\left(g\right)\underset{\text{Ionisation potenital}}{\overset{I}{\to }}N{a}^{+}\left(g\right)+e^{-}$

(iii) $C{l}_2\left(g\right)\underset{\text{Bond dissociation energy}}{\overset{D}{\to }}2Cl\left(g\right)$

(iv) $Cl\left(g\right)+e^{-}\underset{\text{Electron - affinity}}{\overset{E}{\to }}C{l}^{-}\left(g\right)$

(v) $C{l}^{-}\left(g\right)+N{a}^{+}\left(g\right)\underset{\text{Lattice energy}}{\overset{U}{\to }}NaCl\left(s\right)$
The enthalpy of formation of NaCl will be -

• A. $\Delta H_f=S+I-\frac{D}{2}-E-U$
• B. $\Delta H_f=S+I+\frac{D}{2}-E-U$
• C. $\Delta H_f=-S-I-D+\frac{E}{2}+U$
• D. $\Delta H_f=-S-I-D-\frac{E}{2}-U$

Answer 1

The correct option is B $\Delta H_f=S+I+\frac{D}{2}-E-U$

(i) Energy needs to be supplied to convert an atom from one phase to another and therefore the energy of sublimation will be +ve.
Contribution = +S

(ii) Energy is required to remove an electron from an isolated gaseous atom and therefore energy is supplied in case of ionisation potential and thus it is +ve
Contribution = +I

(iii) Energy needs to be supplied to break a bond and covert a diatomic molecule into individual atoms. Therefore the bond dissociation energy is +ve. Since only one atom is required for the formation of the end product, only half of the dissociation energy will be taken into account.
Contribution =+D/2

(iv) Energy is released when an electronegative atom attracts an electron and attains a stable noble gas configuration. Since energy is released in the process, Electron affinity is -ve.
Contribution = -E

(v) Energy is released when the cations and anions occupy the lattice and therefore the lattice nergy is -ve.
Contribution = -U

Summing up all the contributions we have,
$\Delta H_f=S+I+\frac{D}{2}-E-U$