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The Born-Haber cycle for NaCl is shown below:
As can be seen from the diagram:
$$\Delta H_{lat}[NaCl_{(s)}] = -EA_1[Cl_{(g)}] - IE_1[NA_{(g)}]
-\Delta H_{sub}[Na_{(s)}] -\Delta H_{at}[\frac{1}{2}Cl_{2\ (g)}] +
\Delta H_f [NaCl_{(s)}]$$
$$= 349 -496 -108-121 -411$$
$$= -787\ kJ/mol$$
For NaCl$_2$, the following Born-Haber cycle can be
constructed:
The lattice enthalpy is already given in the question, but we can
still use the cycle to calculate the enthalpy of formation of
NaCl$_2$.
$$\Delta H_f [NaCl_{2\ (s)}] = \Delta H_{sub}[Na_{(s)}] + 2\Delta
H_{at}[\frac{1}{2}Cl_{2\ (s)}] + IE_1[Na_{(g)}] + IE_2[Na_{(g)}] +
2EA_1[Cl_{(g)}] + \Delta H_{lat}[NaCl_{2\ (s)}]$$
$$= 108 + 2(121) + 496 + 4563 +2(-349) +(-3360)$$
$$= +1351\ kJ/mol$$
As this calculation reveals, the enthalpy of formation of NaCl$_2$
is actually endothermic (positive) rather than endothermic
(negative), and so the drawn Born-Haber cycle is actually
incorrect. The cycle should reveal that the lattice is higher in
energy than the elements that it is formed from.
It is clear to see that NaCl$_2$ actually requires energy to be
formed from its constituent elements, whereas NaCl releases energy
upon its formation. This is the main reason why NaCl is formed
preferentially over NaCl$_2$, since it is lower in energy.
The main contribution that makes the formation of NaCl$_2$ so
unfavourable is the large second ionisation energy of sodium. The
excessive amount of energy that needs to be put in to remove an
electron from an already positively-charged ion means that the
overall formation of the lattice is not favourable. Because
subsequent ionisation energies are even higher (which are not
compensated for by the increased attraction in the lattice), it is
likely that the formation of other NaCl$_n$ lattices are even more
unlikely.
Simply by plugging the correct ionic radii and born exponents into
the Born-Lande equation should yield the results shown below.
Alongside are the lattice enthalpies calculated experimentally, for
comparison:
It can be seen that the Born-Lande equation gives a reasonable
approximation for the smaller of the halides, but that as the
halide size increases, the discrepancy between theoretical and
experimental lattice enthalpies changes. It should be noted
foremost that the theoretical value is always lower than the
experimental, since it relies solely on electrostatic interactions:
there are of course additional covalent forces which strengthen the
lattice. The increasing discrepancy for increasing size of halide
indicates that the larger halides have an increased amount of
covalent bonding, which is not taken into account by the Born-Lande
equation. Covalent bonding is the sharing of electrons, and it can
be rationalised that increased covalent character is seen for the
larger halides, since these halides are more diffuse and so have
more polarisable electron densities.