Why do many fluorides tend to be less soluble than the corresponding chlorides?
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Solubility takes place in two steps:
Breaking of lattice of a compound: Energy is required to break the lattice and is called Lattice Energy/Enthalpy.
Bond formation between aqueous ions and solvent: Here, energy is released as bonds are formed between the ions and the water molecules. This energy is called Hydration Energy/Enthalpy.
For a substance to be soluble, there should be a release of energy as only then the system would turn into a more stable one(having less energy).
Taking energy released by the system to be negative, the required condition to be soluble is:
L.E. - H.E. < 0 (both in terms of magnitude)
Now H.E. for fluoride ion is more than that of chloride as both carry the same charge but due to smaller size of fluoride ion, it is able to bond with a larger number of water molecules.
And L.E. for fluoride is also more than that of chloride due to same reasons.
In most of the compounds the difference between L.E. and H.E. is greater for fluoride ion(experimental) than for chloride ion. Hence, fluoride compounds are usually less soluble than their corresponding chloride compounds.
Breaking of lattice of a compound: Energy is required to break the lattice and is called Lattice Energy/Enthalpy.
Bond formation between aqueous ions and solvent: Here, energy is released as bonds are formed between the ions and the water molecules. This energy is called Hydration Energy/Enthalpy.
For a substance to be soluble, there should be a release of energy as only then the system would turn into a more stable one(having less energy).
Taking energy released by the system to be negative, the required condition to be soluble is:
L.E. - H.E. < 0 (both in terms of magnitude)
Now H.E. for fluoride ion is more than that of chloride as both carry the same charge but due to smaller size of fluoride ion, it is able to bond with a larger number of water molecules.
And L.E. for fluoride is also more than that of chloride due to same reasons.
In most of the compounds the difference between L.E. and H.E. is greater for fluoride ion(experimental) than for chloride ion. Hence, fluoride compounds are usually less soluble than their corresponding chloride compounds.
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