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Hint: There is a general rule that in crystalline form, symmetrical molecules exhibit higher melting temperatures as compared to molecules with similar structure having lower symmetry.
Complete step by step answer: Let us consider the simple aromatic hydrocarbon and number the carbon positions as shown below:
If two substituents occupy the positions 1 and 2, then it is called ortho substitution and the isomer is called the ortho-isomer. If two substituents occupy the positions 1 and 3, then it is called meta substitution and the isomer is called the meta-isomer and if two substituents occupy the positions 1 and 4, then it is called para substitution and the isomer is called the para-isomer.
Let us consider the substituent to be X. Then the isomers will be:
The para-isomer has symmetrical structure and so it can easily fit into a crystal lattice. As a result, the intermolecular forces of attraction are very strong in the para-isomer and so the amount of energy which is required to break the crystal lattice of the para-isomer is also very large. Or in other words, more energy is required to melt the para-isomer and hence it has a very high melting point.
On the other hand, the ortho-isomer and the meta-isomer are not so symmetrical in comparison to the para-isomer and hence they do not form a very closely packed crystal lattice. As a result, the intermolecular forces of attraction are not very strong as compared to those in the para-isomer. Thus, the ortho-isomer and the meta-isomer have very weak forces of attraction and so the amount of energy required to break down the crystal lattices of the ortho-isomer and the meta-isomer is also very less in comparison to that in the para-isomer. This means that the energy required to melt the crystal lattices of the ortho and meta-isomers is smaller than that needed for the para-isomer. So they will have lower melting points than the para-isomer.
Hence, the melting point of the para-isomer is high as compared to the ortho and meta-isomers because the para-isomer is more symmetrical than the ortho and meta-isomers. So, both assertion and reason for the given question are correct and the reason is the correct explanation for assertion. So the correct option is option A.
Note: Molecular packing affects the melting point order. A substance will have high melting point if its molecules are more tightly packed and lower melting point if they are not packed well.
Complete step by step answer: Let us consider the simple aromatic hydrocarbon and number the carbon positions as shown below:
If two substituents occupy the positions 1 and 2, then it is called ortho substitution and the isomer is called the ortho-isomer. If two substituents occupy the positions 1 and 3, then it is called meta substitution and the isomer is called the meta-isomer and if two substituents occupy the positions 1 and 4, then it is called para substitution and the isomer is called the para-isomer.
Let us consider the substituent to be X. Then the isomers will be:
The para-isomer has symmetrical structure and so it can easily fit into a crystal lattice. As a result, the intermolecular forces of attraction are very strong in the para-isomer and so the amount of energy which is required to break the crystal lattice of the para-isomer is also very large. Or in other words, more energy is required to melt the para-isomer and hence it has a very high melting point.
On the other hand, the ortho-isomer and the meta-isomer are not so symmetrical in comparison to the para-isomer and hence they do not form a very closely packed crystal lattice. As a result, the intermolecular forces of attraction are not very strong as compared to those in the para-isomer. Thus, the ortho-isomer and the meta-isomer have very weak forces of attraction and so the amount of energy required to break down the crystal lattices of the ortho-isomer and the meta-isomer is also very less in comparison to that in the para-isomer. This means that the energy required to melt the crystal lattices of the ortho and meta-isomers is smaller than that needed for the para-isomer. So they will have lower melting points than the para-isomer.
Hence, the melting point of the para-isomer is high as compared to the ortho and meta-isomers because the para-isomer is more symmetrical than the ortho and meta-isomers. So, both assertion and reason for the given question are correct and the reason is the correct explanation for assertion. So the correct option is option A.
Note: Molecular packing affects the melting point order. A substance will have high melting point if its molecules are more tightly packed and lower melting point if they are not packed well.
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