The addition of which metal in liquid $NH_3$ leads to the formation of a blue solution:
A. Li
B. Sr
C. Cs
D. Ba
Answer
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Hint: Alkali metals and alkaline earth metals are soluble in ammonia. The metal releases electrons in solution. The solution has an unpaired electron. The s-block contains alkali and alkaline earth metals.
Complete step by step solution: The alkali and alkaline earth metals are dissolved in ammonia, form amides and release hydrogen gas. A general reaction between metals of s-block and ammonia is as follows:
${\text{M}}\,{\text{ + }}\,{\text{N}}{{\text{H}}_{\text{3}}}\, \to {\text{MN}}{{\text{H}}_2}\, + \,{{\text{H}}_2}$
Here, M is the alkali and alkaline earth metals.
As the metal is dissolved in ammonia, the metal forms an ion and releases an electron. The metal ion is surrounded by ammonia molecules as the metal ion is electron-deficient and nitrogen of ammonia has a lone pair of electrons.
The free electron is surrounded by ammonia molecules through hydrogen atoms. This is known as an ammoniated anion. The magnetic property and colour of the solution are due to this ammoniated anion.
The ammonium solution of alkali and alkaline earth metals is of blue colour and paramagnetic due to unpaired electrons.
The first group elements are alkali metals that are, lithium, sodium, potassium, rubidium, cesium and francium.
The second group of elements are alkaline earth metals that are, beryllium, magnesium, calcium, strontium and barium.
The given elements, Li, Sr, Ca and Ba, all are s-block elements so, all these on dissolving in liquid ${\text{N}}{{\text{H}}_{\text{3}}}$ leads to the formation of a blue solution.
Therefore, option (A) Li, (B) Sr, (C) Cs and (D) Ba, are correct.
Note: The solution is paramagnetic and of blue colour at a lower concentration of ammonia. At higher concentration, the solution changes into bronze colour and becomes diamagnetic. These solutions are a good conductor of electricity and have higher conductivity. On increasing the concentration the pairing of electrons takes place, so the magnetic susceptibility decreases and the colour of the solution feds.
Complete step by step solution: The alkali and alkaline earth metals are dissolved in ammonia, form amides and release hydrogen gas. A general reaction between metals of s-block and ammonia is as follows:
${\text{M}}\,{\text{ + }}\,{\text{N}}{{\text{H}}_{\text{3}}}\, \to {\text{MN}}{{\text{H}}_2}\, + \,{{\text{H}}_2}$
Here, M is the alkali and alkaline earth metals.
As the metal is dissolved in ammonia, the metal forms an ion and releases an electron. The metal ion is surrounded by ammonia molecules as the metal ion is electron-deficient and nitrogen of ammonia has a lone pair of electrons.
The free electron is surrounded by ammonia molecules through hydrogen atoms. This is known as an ammoniated anion. The magnetic property and colour of the solution are due to this ammoniated anion.
The ammonium solution of alkali and alkaline earth metals is of blue colour and paramagnetic due to unpaired electrons.
The first group elements are alkali metals that are, lithium, sodium, potassium, rubidium, cesium and francium.
The second group of elements are alkaline earth metals that are, beryllium, magnesium, calcium, strontium and barium.
The given elements, Li, Sr, Ca and Ba, all are s-block elements so, all these on dissolving in liquid ${\text{N}}{{\text{H}}_{\text{3}}}$ leads to the formation of a blue solution.
Therefore, option (A) Li, (B) Sr, (C) Cs and (D) Ba, are correct.
Note: The solution is paramagnetic and of blue colour at a lower concentration of ammonia. At higher concentration, the solution changes into bronze colour and becomes diamagnetic. These solutions are a good conductor of electricity and have higher conductivity. On increasing the concentration the pairing of electrons takes place, so the magnetic susceptibility decreases and the colour of the solution feds.
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