0
$\begingroup$

Ionic compounds are in the form of crystal lattices, and when they are dissolved in a polar solvent, they are bonded with the corresponding cations/anions of the polar solvent. For example, $Na^+$ bonds with $OH^-$ and $Cl^-$ bonds with $H^+$ when NaOH is dissolved in $H_2O$.

The reason why $NaCl_{(aq)}$ conducts electricity is because $Na^+$ and $Cl^-$ do not form crystal lattices with $OH^-$ and $H^+$ respectively. But why don't they form crystal lattices?

$\endgroup$
2
$\begingroup$

They actually form: ions can be dissolved up till a certain concentration, and beyond that, they form ionic lattices just as you predict.

The easiest way to describe the scenario as an equilibrium between ions-in-water vs ions-in-lattice:

  • If the ion-water interaction is very strong, the ions can be very stable in water, even more than in an ionic lattice, so ions will be in the solution.
  • If the ion-water interaction is weak compared to the solid, then the solid phase is much more stable, and only a very low concentration of ions can be in the solution.

Looking at only interaction energies is a very simplistic view as it does not consider entropic factors, yet can describe at least part of the story. Entropy is generally a huge driving force for any kind of dissolution and allows some ions to be in solution even if the solid state energetically more stable.

$\endgroup$
1
$\begingroup$

Your statement 'For example, Na+ bonds with OH− and Cl− bonds with H+ when NaOH is dissolved in H2O' is not correct if the ions are in solution.

In an ionic solid each ion is surrounded by other ions of the same charge and those of opposite charge, in NaCl each Na$^+$ ion is surrounded by 6 Cl$^-$ and vice versa. The interaction between the ions is essentially Coulomb in nature with energy that varies with distance $r$ as $E \sim 1/r$. This is a long range interaction and so choosing a 'victim' ion this can interact with all its many neighbours, not just nearest neighbours, thus many interactions have to be summed up. The sum is called a Madeling Sum and the cohesive energy produced is written in the simple form $E\sim se^2/r_0$ where $s$ is the Madelung sum (-1.748 for NaCl), $r_0$ the closest separation of any two ions and $e$ the electronic charge. However the ions could just approach one another and the ionic solid could just collapse. Why does this not happen? What prevents this is the Coulomb repulsion between electrons (and between nuclei) as any two ions (or atoms) approach one another and the Pauli Exclusion principle. This ensures that only two electrons can fill an energy level and so energy is needed to promote an electron from one energy level into an unfilled one on the other atom as they approach. Both these effects are repulsive in nature and vary as $\sim 1/r^{12}$ so are very short range.

Having established that there is a binding energy holding a solid together when an ionic solid is added to a polar solvent we expect it to dissolve and free ions to be produced. Overall if this is to happen the free energy change $\Delta G$ must be negative. This is composed of two parts the enthalpy change (potential energy) $\Delta H$ and the entropy change $\Delta S$ as $\Delta G = \Delta H -T\Delta S$, where $T$ is the temperature. If the free energy is to be negative then there are four possibilities depending on the sign of the change $\Delta H$ and $\Delta S$. The one situation that cannot occur if the solid is to dissolve is $\Delta H \gt 0, \; \Delta S \lt 0$ as this always makes $\Delta G$ positive. It will always dissolve if $\Delta H \lt 0, \; \Delta S \gt 0$ as $\Delta G $ is negative. The other cases depend on particular values.

We expect that $\Delta H$ will be positive on dissolution as the crystalline solid exists and it will take energy to make it dissolve. (This will cool the solution if heat is not supplied to keep the temperature constant). As there is a larger volume for the ions to occupy in solution than in the solid the entropy change is expected to be positive. Simply having more places to occupy will increase the entropy. However, the solid does not have zero entropy, the ions can move albeit by a small amount about their equilibrium positions in the crystal and waves of vibrational motion called phonons exist and so there is already certain amount of entropy. This is lost on dissolution and has to be exceeded if the entropy change is to be positive. Also it is necessary to consider the entropy change in the solvent. As the ions are charged, in a polar solvent the dipoles are attracted to the charge, and partially order around the it. This leads to a decrease in entropy of the solvent as there are ions plus an 'atmosphere' of a few solvent molecules around each one. All these terms need to be considered, but as, say, NaCl does dissolve then the entropy change must be positive.

When the ion is dissolved, why do pairs of ions not form? This is because the solvent is polar and when the dielectric constant of the solvent is considered the energy varies as $E\sim 1/(\epsilon r)$ where $\epsilon$ is the dielectric constant, for water this 78 and 37 for acetonitrile, but for an ionic crystal it is only about 2, similar to that for non-polar hexane. This large dielectric constant means that the interaction of one ion to another is $\epsilon$ times smaller compared to that at the same distance in the crystal or in a non-polar solvent. In other words, in a polar solvent the ions are effectively independent. The interaction is small and comparable to the thermal motion jostling the molecules and so the number of times they form pairs is vanishingly small. This means that the solution will conduct electricity.

$\endgroup$
  • $\begingroup$ "What prevents this is the Coulomb repulsion between electrons (and between nuclei) as any two ions (or atoms) approach one another..." Are you talking about the coloumbic repulsion between two like ions (Cl- - Cl- or Na+ - Na+) as they try approaching the victim ion? "...and the Pauli Exclusion principle" More energy is needed to elevate electrons to the 4s shell in Cl- and 3p shell in Na+: Is that what you're saying? $\endgroup$ – Sashank Sriram Nov 9 '18 at 13:50
  • $\begingroup$ "We expect that ΔH will be positive on dissolution as the crystalline solid exists and it will take energy to make it dissolve." So energy is absorbed to carry out the ionisation. "This will cool the solution if heat is not supplied to keep the temperature constant." Does this mean that the absorbed heat energy will get used up during ionisation? $\endgroup$ – Sashank Sriram Nov 9 '18 at 13:57
  • $\begingroup$ And from your explanation of entropy, do you mean to say that for NaCl, the entropy change due to spacing out significantly exceeds the (absolute value of) the entropy change due to other factors such as restricting the little freedom of ions in crystal structure, electrostatic forces restricting freedom of ions, etc.? And does the relative permittivity of free space of the polar solvent also contribute to a positive change in entropy as electrostatic forces that constrain freedom are reduced? $\endgroup$ – Sashank Sriram Nov 9 '18 at 14:00
  • 1
    $\begingroup$ (a) re Pauli, yes more energy would be needed, (b) the atoms are always ionised in crystal and solution so no ionisation is needed. (c) yes if dimers are not formed as dielectric is large then dielectric indirectly contributes to entropy $\endgroup$ – porphyrin Nov 9 '18 at 16:45

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.