
Which of the following does not occur when sodium metal is dissolved in a moderate amount in liquid ammonia at low temperature?
Blur color of the solution
$N{a^ + }$ ions formed in the solution
Liquid ammonia becomes a good conductor of electricity
Liquid ammonia remains diamagnetic
Answer
545.1k+ views
Hint:We know that sodium is a group $1$ element and is represented by the symbol $Na$. Since it is a group $1$ element, it is an alkali metal. Liquid ammonia dissolves all alkali metals to give solvated electrons. Solvated electrons are the smallest possible anions.
Complete solution:
We know that sodium is a group $1$ element. It is represented by the symbol $Na$ and has an atomic number $11$. We also know that all group $1$ elements are alkali metals. Alkali metals are those metals which when dissolved in water give an alkaline solution. Liquid ammonia dissolves all alkali metals. When sodium is dissolved in liquid ammonia following reaction occurs:
$Na + (x + y)N{H_3} \to {[Na{(N{H_3})_x}]^ + } + {[e{(NH)_y}]^ - }$
From the above reaction, we can see that sodium reacts with ammonia to form solvated or ammoniated electrons. Solvated electrons are free electrons and are the smallest possible anions. In the above reaction, we can see that electrons released from sodium metal get solvated, and are surrounded by multiple ammonia molecules. This solvated electron gives a blue color to the solution when present in low concentration and gives copper color when present in higher concentration. And since this electron is free, it conducts electricity in the solution. Thus liquid ammonia conducts electricity. Now, if we again observe the reaction, we see that $Na$ forms a cation that is coordinated to ammonia molecules. Now since the solution has many solvated or ammoniated electrons that are free, the solution becomes paramagnetic.
So, the correct answer will be option number D.
Note:It is important to note that this solution of sodium metal in liquid ammonia is called Birch reduction. It selectively reduces benzene molecules. In Birch reduction, we add sodium, ammonia, and alcohol as catalysts to selectively reduce benzene molecules.
Complete solution:
We know that sodium is a group $1$ element. It is represented by the symbol $Na$ and has an atomic number $11$. We also know that all group $1$ elements are alkali metals. Alkali metals are those metals which when dissolved in water give an alkaline solution. Liquid ammonia dissolves all alkali metals. When sodium is dissolved in liquid ammonia following reaction occurs:
$Na + (x + y)N{H_3} \to {[Na{(N{H_3})_x}]^ + } + {[e{(NH)_y}]^ - }$
From the above reaction, we can see that sodium reacts with ammonia to form solvated or ammoniated electrons. Solvated electrons are free electrons and are the smallest possible anions. In the above reaction, we can see that electrons released from sodium metal get solvated, and are surrounded by multiple ammonia molecules. This solvated electron gives a blue color to the solution when present in low concentration and gives copper color when present in higher concentration. And since this electron is free, it conducts electricity in the solution. Thus liquid ammonia conducts electricity. Now, if we again observe the reaction, we see that $Na$ forms a cation that is coordinated to ammonia molecules. Now since the solution has many solvated or ammoniated electrons that are free, the solution becomes paramagnetic.
So, the correct answer will be option number D.
Note:It is important to note that this solution of sodium metal in liquid ammonia is called Birch reduction. It selectively reduces benzene molecules. In Birch reduction, we add sodium, ammonia, and alcohol as catalysts to selectively reduce benzene molecules.
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