Although `Na^(2+)` has a higher charge and, therefore, `NaCl_2` should have a greater lattice enthalpy, sodium prefers to form `NaCl` rather than `NaCl_2` because

Options

A. `Delta_iH_2` of `Na` is very high
B. the lattice enthalpy of `NaCl_2` is far too small to compensate for the energy required to produce the `Na^(2+)` ion
C. `Na^(2+)` does not have the noble gas electron configuration
D. the lattice enthalpy of `NaCl_2` is less than the lattice enthalpy of `NaCl`

Answer

B

Explanation

Sum of `Delta_iH_1` and `Delta_iH_2` for `Na` is `(496 kJ mol^(-1)+4560 kJ mol^(-1))5056 kJ mol^(-1)` while the lattice enthalpy of hypothetical `NaCl_2` is `2527 kJ mol^(-1)` (assuming roughly the same as that for `MgCl_2`). It is far too small an energy yield to compensate for the energy required to produce `Na^(2+)` ion. Using similar reasoning, we can explain that `Na_2Cl` does not exist as its lattice enthalpy is not able to compensate the energy needed to form `Na^(+)` and `Cl^(2-)`.