Chemistry - Why is the melting point of p-dichlorobenzene higher than those of o-dichlorobenzene and m-dichlorobenzene?
Solution 1:
Generally, the melting point of para-isomer is quite higher than that of ortho- or meta-isomers. This is due to the fact that it has a symmetrical structure and therefore, its molecules can easily pack closely in the crystal lattice. As a result, intermolecular forces of attraction are stronger and therefore, greater energy is required to break its lattice and its melts at higher temperature. This question is most likely to be answered easily.
If you consider solubility. These are insoluble in water but soluble in organic solvents. Generally, para-isomers are highly soluble in organic solvents than ortho-isomers. This is interesting question.
Every compound (more specifically solids) does not dissolve in a given compound. In general, a liquid dissolves in a liquid if the intermolecular interactions are similar. This is in accordance with the rule "like dissolves like," ionic or polar compounds dissolve more readily in polar solvents. Non-polar compounds (covalent or organic) are soluble in non-polar solvents.
If you consider p-dichlorobenzene, it has zero dipole moment and thus is more non-polar than o-dichlorobenzene which has dipole moment of about $\pu{2.54 D}$. So, non-polar p-dichlorobenzene dissolves more readily than o-dichlorobenzene in non-polar (organic) solvents.
Solution 2:
I had some comments to make, but they're too large to actually fit into the comments, so I'm making another answer. I'll take the opportunity to give a bit of background, which may be going a little too far for you right now, but which may come in useful in the future. I'm sorry all my answers tend to be overly lengthy and dry!
There are many factors that combine to create a substance's observed physical properties. When talking about solubility, melting points, and boiling points, it's essential to analyse the intermolecular forces between molecules.
First let's start with solubility. After defining the concept of polarity for a substance, (based on the strength of the electric dipole present in a molecule) one quickly discovers that, in general, "like dissolves like." This is a very approximate rule of thumb that relies on a qualitative description of intermolecular forces. For a substance $A$ to dissolve in another substance $B$, then it is necessary that intermolecular $A\dotsb A$ forces and $B\dotsb B$ forces be broken in order to create $A\dotsb B$ forces, and this process must be favourable. We say "like dissolves like" because if $A$ and $B$ are similar, then the $A$-$B$ interactions formed are likely about as strong as $A\dotsb A$ and $B\dotsb B$, and so the molecules of each substance will tend to mingle well. If $A$ and $B$ are very different, then it's less likely that $A\dotsb B$ interactions will predominate over $A\dotsb A$ and $B\dotsb B$, and hence the substances stay apart.
Intramolecular interactions are all about electrostatic interactions at long range. From Coloumb's law of electrostatics, it's clear that there are stronger attractions and repulsions the higher the charges involved. Even in neutral molecules, there can often be parts of the molecules with a slightly concentration of negative charge and another part with a slightly concentration of positive charge. We call these oppositely-charged areas in a same object poles, and when there are two oppositely charged areas connected, they form a dipole. The dipole is characterized by something called the dipole moment, which is a vector (a quantity that has direction as well as magnitude). Depending on how spatially concentrated the excess negative and positive charges are, the dipole is said to vary in strength. Weak dipoles will have little charge separation, while strong dipoles have lots of charge separation. The strength of the dipole is what dictates a substance's polarity, which is a quantification of just how polar it is.
As I mentioned, most intermolecular forces are electrostatic in nature. Smaller charges and dipoles tend to cause weaker intermolecular interactions, and larger charges and dipoles tend to cause stronger intermolecular interactions. If you mix substances $A$ and $B$, both with weak dipoles (low polarity) or both with strong dipoles (high polarity), then $A\dotsb B$ interactions are similar to $A\dotsb A$ and $B\dotsb B$, so they tend to dissolve. If, however, you mix $A$ and $B$ such that $A$ has low polarity and $B$ has high polarity, then the electrostatic forces between $B\dotsb B$ will be stronger than $B\dotsb A$, and there is no drive for $B$ to mix with $A$.
All of this talk is to justify why polarity is something you'd expect to be important in determining whether two substances mix. In your specific case, you're studying mixing water (a high polarity substance) with dichlorobenzenes. It is possible to approximately calculate the dipoles present in molecules by comparing the placement of atoms in the molecule and the atoms' electronegativities (electronegativity quantifies how badly an element wants electrons). For example, carbon and hydrogen atoms have similar electronegativities, so when they bond, they form approximately non-polar bonds. When chlorine and carbon bond, the former has a significantly higher electronegativity, so it will tug on the bond's electrons a bit, such that the chlorine acquires a small negative charge, and the carbon acquires a small positive charge, and we say the bond thus formed is polar.
In dichlorobenzenes (DCBs), this means there are dipole moments in the $\ce{C-Cl}$ bonds. In ortho-DCB, the dipoles moments of each bond are pointing in a similar direction (two equal vectors with a $60^\circ$ angle between them), so their effect adds up and the substance has some polarity. In meta-DCB, the dipoles are kind of facing away from each other (two equal vectors with a $120^\circ$ angle between them), so their effects cancel about a bit and the molecule is only left with a slight polarity. For para-DCB, the vectors are exactly opposite and equal (two equal vectors with a $180^\circ$ angle between them), so they cancel out entirely; this means para-DCB is a non-polar molecule, even though it has some polar bonds.
Since water is a highly polar substance, if we assume that the dichlorobenzenes with higher polarity will be more soluble, then we expect that the solubilities increase in the order p-DCB < m-DCB < o-DCB. Experimental data show that their solubilities in water are approximately $\pu{80 mg/L}$, $\pu{125 mg/L}$, and $\pu{156 mg/L}$ for p-DCB, m-DCB, and o-DCB, respectively. Just what we expected! Note that even o-DCB, the most polar of the three, has a very low solubility in water. This is because the polar regions in the DCB molecules are quite small relative to the non-polar regions.
So we got through solubility. Now with the same background in intermolecular forces, explaining the boiling point behaviour is relatively simple. When a substance is heated close to its boiling point, you are giving the molecules almost enough energy to break the intermolecular $A\dotsb A$ forces. The stronger the $A\dotsb A$ forces, the more energy is needed to overcome them. All other things being equal, $A\dotsb A$ interactions are stronger for high-polarity molecules than low-polarity molecules, as there are bigger charges and bigger electrostatic interaction in high-polarity molecules.
Again, by looking at polarity alone, we expect the boiling point trend to be p-DCB < m-DCB < o-DCB. Experimental data shows that the boiling points are $\pu{174 ^\circ C}$, $\pu{173 ^\circ C}$, and $\pu{180 ^\circ C}$ for p-DCB, m-DCB, and o-DCB, respectively. The expected and experimental trends are not exactly the same. This is likely because the polarity difference between p-DCB and m-DCB is small enough that other factors need to be taken into account.
Explaining melting points is quite a bit harder than boiling points. Several things other than polarity can strongly influence melting temperatures. I have a previous post on this matter with some general information. In the case of the dichlorobenzenes, it seems that molecular symmetry and solid packing is the dominant effect. Chem.SE user Uncle Al has provided a list of compounds showing how strong this factor can be.
Solution 3:
I think this reasoning will suffice (not sure though):
p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, it fits more closely than o-and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene. As a result, p-dichlorobenzene has a higher melting point than o-and m-isomers.
Any thoughts?