A metal M readily forms its sulphate `MSO_(4)` which is water soluble. It forms oxide `MO` which becomes inert on heating. It forms insoluble hydroxide which is soluble in `NaOH`. The metal `M` is:

A metal M readily forms its sulphate MSO_(4) which is water soluble. It forms oxide Mo which becomes inert on heating. It forms insouble hydroxide which is soluble NaOH. The metal is

A metal M readily forms its sulphate MSO_(4) which is water soluble. It forms its oxide MO which becomes inert on heating. It forms its insoluble hydroxide M(OH)_(2) which is soluble in NaOH solution. Then M is

A metal M readily forms its sulphate MSO_(4) which is watersoluble. It forms its oxide MO which becomes inert on heating. It forms its insoulbe hydroxide M(OH)_(2) which is soluble in NaOH solution. Then M is

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

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 metal is :

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