
Sodium metal on dissolution in liquid ammonia gives a deep blue solution due to the formation of:
A.Sodium-ion ammonia complex
B.Sodamide
C.Sodium-ammonia complex
D.Ammoniated electrons
Answer
565.2k+ views
Hint: To answer this question, you should recall the concept of properties of s-block elements and the effect of their addition in liquid ammonia. Alkali metals in the presence of liquid ammonia release electrons.
Complete step by step answer:
The s block elements are known as alkali metals and have only one electron in their s-orbital. In liquid ammonia, you can generate solvated electrons. Here, when you dissolve sodium metal in very cold liquid ammonia solvent, the sodium metal loses its electron, which results in the ion becoming surrounded by the positive end as ammonia is polar of several ammonia molecules.
The solvated electron is of special importance in radiation chemistry. Alkali metals show the property that when dissolved in liquid ammonia they give deep blue solutions which are conducting in nature. The resulting blue colour of the solution is due to ammoniated electrons which absorb energy in the visible region of light.
Hence, the correct answer to this question is Option D.
Note:
The liquid ammonia required for the production of ammoniated electrons requires Ammonia which is prepared using Haber’s process: ${N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g) + 22.0{\text{kcal}}$ Make sure you remember the conditions which maximize the yield of ammonia:
Using Le Chatelier’s principle, as the above reaction is exothermic, low temperature will shift the equilibrium to the right leading to a greater yield of ammonia. A temperature of \[450^\circ C\] will maximise the preparation of ammonia.
As there is a decrease in gaseous moles, high pressure on the reaction at equilibrium favours the shift of the equilibrium to the right. A pressure of \[{\text{200 atm}}\] will maximize the preparation of ammonia.
To increase the rate of reaction, a catalyst is used and quickly attain equilibrium.
The reactants nitrogen and hydrogen gas should be pure to increase the yield of ammonia.
Complete step by step answer:
The s block elements are known as alkali metals and have only one electron in their s-orbital. In liquid ammonia, you can generate solvated electrons. Here, when you dissolve sodium metal in very cold liquid ammonia solvent, the sodium metal loses its electron, which results in the ion becoming surrounded by the positive end as ammonia is polar of several ammonia molecules.
The solvated electron is of special importance in radiation chemistry. Alkali metals show the property that when dissolved in liquid ammonia they give deep blue solutions which are conducting in nature. The resulting blue colour of the solution is due to ammoniated electrons which absorb energy in the visible region of light.
Hence, the correct answer to this question is Option D.
Note:
The liquid ammonia required for the production of ammoniated electrons requires Ammonia which is prepared using Haber’s process: ${N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g) + 22.0{\text{kcal}}$ Make sure you remember the conditions which maximize the yield of ammonia:
Using Le Chatelier’s principle, as the above reaction is exothermic, low temperature will shift the equilibrium to the right leading to a greater yield of ammonia. A temperature of \[450^\circ C\] will maximise the preparation of ammonia.
As there is a decrease in gaseous moles, high pressure on the reaction at equilibrium favours the shift of the equilibrium to the right. A pressure of \[{\text{200 atm}}\] will maximize the preparation of ammonia.
To increase the rate of reaction, a catalyst is used and quickly attain equilibrium.
The reactants nitrogen and hydrogen gas should be pure to increase the yield of ammonia.
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