When $KMn{{O}_{4}}$ is added to $KOH$, which of the following colours is observed?
A. Pale pink
B. Brown
C. Black
D. Green
Answer
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Hint: \[KMnO_4\] is water-soluble and consists of two ions: Permanganate ion and potassium ion. It is a dark purple odourless solid in its physical state. When potassium permanganate crystals are dissolved in water the solution formed is purple. It is considered as a strong oxidizing agent(the oxidation states of atoms increase as the elements become more electronegative. Therefore, permanganate is a good oxidizing agent) and does not produce toxic products. It is usually prepared from other minerals such as manganese oxide
This compound has an ionic bond between the potassium cation and the permanganate anion.
The solution under examination loses its pink colour once all the permanganate ions are used up in the reaction. It suggests the end of the reaction and therefore a self-indicator is called potassium permanganate, as it serves as an indicator apart from being one of the reactants.
Complete step by step answer:
In an alkaline medium, $Mn$ has a stable oxidation state $+6$ .
In the presence of \[{KOH},\;{MnO}\] reacts in the presence of atmospheric oxygen to form $KMn{{O}_{4}}$
\[2MnO~+{ }4KOH{ }+{ }O~\to ~2K_2MnO_4~+{ }2H_2O\]
\[K_2MnO_4\] is an intermediate in the industrial synthesis of Potassium permanganate.
On decomposing Potassium permanganate, Oxygen gas is being produced with \[K_2MnO_4\] In a basic solution, such as a solution of Potassium hydroxide - KOH, green colored \[K_2MnO_4\] is seen.
So, the correct answer is Option A.
Note: Water is not used in this reaction, otherwise, it will convert back to purple colored \[KMnO_4\]
\[3{ }K_2MnO_4{ }+{ }2{ }H_2O{ }\to { }2{ }KMnO_4{ }+{ }MnO_2{ }+{ }4{ }KOH\]
Another reason is charge spectra:
As \[Mn\] has one unpaired electron which is responsible for the colour. The unpaired electron absorbs energy from the white light and the reflected light shows complimentary colour. The electron gets excited to a higher level. This is called a d-d transition.
This compound has an ionic bond between the potassium cation and the permanganate anion.
The solution under examination loses its pink colour once all the permanganate ions are used up in the reaction. It suggests the end of the reaction and therefore a self-indicator is called potassium permanganate, as it serves as an indicator apart from being one of the reactants.
Complete step by step answer:
In an alkaline medium, $Mn$ has a stable oxidation state $+6$ .
In the presence of \[{KOH},\;{MnO}\] reacts in the presence of atmospheric oxygen to form $KMn{{O}_{4}}$
\[2MnO~+{ }4KOH{ }+{ }O~\to ~2K_2MnO_4~+{ }2H_2O\]
\[K_2MnO_4\] is an intermediate in the industrial synthesis of Potassium permanganate.
On decomposing Potassium permanganate, Oxygen gas is being produced with \[K_2MnO_4\] In a basic solution, such as a solution of Potassium hydroxide - KOH, green colored \[K_2MnO_4\] is seen.
So, the correct answer is Option A.
Note: Water is not used in this reaction, otherwise, it will convert back to purple colored \[KMnO_4\]
\[3{ }K_2MnO_4{ }+{ }2{ }H_2O{ }\to { }2{ }KMnO_4{ }+{ }MnO_2{ }+{ }4{ }KOH\]
Another reason is charge spectra:
As \[Mn\] has one unpaired electron which is responsible for the colour. The unpaired electron absorbs energy from the white light and the reflected light shows complimentary colour. The electron gets excited to a higher level. This is called a d-d transition.
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