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Sodium metal in liquid ammonia is

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an ionic conductor

an electronic conductor

a mixed conductor

non - conductor

answer is C.

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Detailed Solution

{M_{left( g right)}} to M_{left( g right)}^ + + {e^ - },,left( {M = Alkalim etal} right)

{M^ + } + xN{H_3} to M{left( {N{H_3}} right)_x}^{+}

{e^ - } + yN{H_3} to e{left( {N{H_3}} right)^ - }

Adding all three equations

{M_{left( g right)}} + left( {x + y} right)N{H_3} to M{left( {N{H_3}} right)_x}^ + + eleft( {N{H_3}} right)_y^ -

The solution is containing free ammoniated cation and free ammoniated electrons.Conductance is due to free ions and free electrons. Hence is a Mixed conductor.

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