A metal M readily forms water soluble su...

A metal `M` readily forms water soluble sulphate `MSO_(4)`, water insoluble hydroxide `M (OH)_(2)` and oxide `MO` which becomes inert on heating. The hydroxide is soluble in `NaOH`. The `M` is

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M is Be which forms water soluble `BeSO_(4),` water insoluble `Be(OH)_(2) and BeO. Be(OH)_(2)` is soluble in NaoH giving sodium beryllate `Na_(2) BeO_(2).`

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The solubility of a crystalline compound in water is influenced by two factors. These are lattice enthalpy and hydration enthalpy. Lattice enthalpy is needed to separate the ions from the crystal lattice. At the same time hydration enthalpy is released when the ion are dissolved in water. The resultant of these two opposing tendencies guides the solubility of the compound in water. If lattice enthalpy has a greater magnitude, the compound has low solubility. In case hydration enthalpy is more, then the compound is highly soluble in water. A metal M readily forms a water soluble sulphate MSO_(4) water insoluble hydroxide M(OH)_(2) and oxide MO which becomes inert upon heating. The hydroxide is soluble in NaOH. The metal M is

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