explain : solubility of NaF in water is less than that of RbF?
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In Sodium Fluoride the configuration of Sodium is up to 3s1 and in Rubidium Fluoride the configuration of Rubidium is up to 5s1 that means it has the interruption of d-orbital and also has vacant d-orbital. Due to the vacant d-orbital it can accommodate the water molecules rather sodium cannot. Hence NaF has less solubility than RbF in water.
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