
To an acidic solution of an anion, a few drops of $KMn{{O}_{4}}$ solution are added. Which of the following if present, will not decolourise the $KMn{{O}_{4}}$ solution?
(A) $CO_{3}^{2-}$
(B) $NO_{2}^{-}$
(C) ${{S}^{2-}}$
(D) $C{{l}^{-}}$
Answer
220.8k+ views
Hint: The colour of potassium permanganate is deep purple. When the oxidation state of manganese in potassium permanganate changes from +7 to +2, we observe decolourisation. This is a reduction reaction. So any ion which will reduce the $KMn{{O}_{4}}$ will be our answer.
Complete step by step answer:
Let’s look at the answer to the question:
We know that decolourization is observed when the $KMn{{O}_{4}}$ gets reduced.
The process can be represented as:
$MnO_{4}^{-}\,+\,5{{e}^{-}}\to \,M{{n}^{+2}}$
Now, we can see that Mn is getting reduced then the other ion should get oxidized.
So, we will look for the ion which will get oxidised, that is its oxidation number will increase.
*Now, in $NO_{2}^{-}$ the oxidation state of N is +3 and it can increase its oxidation state to +5. So, will decolourise the $KMn{{O}_{4}}$solution.
*In ${{S}^{2-}}$the oxidation state of S is -2 and it can increase its oxidation state to maximum +6. So, ${{S}^{2-}}$will decolourise the$KMn{{O}_{4}}$solution.
*In $C{{l}^{-}}$the oxidation state of Cl is -1 and it can increase its oxidation state to maximum +7. So, $C{{l}^{-}}$ will decolourise the $KMn{{O}_{4}}$solution.
*In $CO_{3}^{2-}$the oxidation state of C is +4 which is the maximum oxidation state of C. So, $CO_{3}^{2-}$ will not decolourise the $KMn{{O}_{4}}$ solution. This happens as it will not get oxidized and if oxidation doesn’t take place then reduction of $KMn{{O}_{4}}$will also not take place.
Therefore, addition of $CO_{3}^{2-}$ will not decolourise the $KMn{{O}_{4}}$solution.
Hence, the answer for the given question is option (A).
Note: Permanganate ion is coloured due to d-d electronic transitions whereas $M{{n}^{+2}}$ is colourless as there are no d-d electronic transitions. When electrons gain energy they get excited and when they lose energy, radiations are emitted which provide colour to the metal.
Complete step by step answer:
Let’s look at the answer to the question:
We know that decolourization is observed when the $KMn{{O}_{4}}$ gets reduced.
The process can be represented as:
$MnO_{4}^{-}\,+\,5{{e}^{-}}\to \,M{{n}^{+2}}$
Now, we can see that Mn is getting reduced then the other ion should get oxidized.
So, we will look for the ion which will get oxidised, that is its oxidation number will increase.
*Now, in $NO_{2}^{-}$ the oxidation state of N is +3 and it can increase its oxidation state to +5. So, will decolourise the $KMn{{O}_{4}}$solution.
*In ${{S}^{2-}}$the oxidation state of S is -2 and it can increase its oxidation state to maximum +6. So, ${{S}^{2-}}$will decolourise the$KMn{{O}_{4}}$solution.
*In $C{{l}^{-}}$the oxidation state of Cl is -1 and it can increase its oxidation state to maximum +7. So, $C{{l}^{-}}$ will decolourise the $KMn{{O}_{4}}$solution.
*In $CO_{3}^{2-}$the oxidation state of C is +4 which is the maximum oxidation state of C. So, $CO_{3}^{2-}$ will not decolourise the $KMn{{O}_{4}}$ solution. This happens as it will not get oxidized and if oxidation doesn’t take place then reduction of $KMn{{O}_{4}}$will also not take place.
Therefore, addition of $CO_{3}^{2-}$ will not decolourise the $KMn{{O}_{4}}$solution.
Hence, the answer for the given question is option (A).
Note: Permanganate ion is coloured due to d-d electronic transitions whereas $M{{n}^{+2}}$ is colourless as there are no d-d electronic transitions. When electrons gain energy they get excited and when they lose energy, radiations are emitted which provide colour to the metal.
Recently Updated Pages
Difference Between Alcohol and Phenol: Structure, Tests & Uses

Classification of Drugs in Chemistry: Types, Examples & Exam Guide

Class 12 Chemistry Mock Test Series for JEE Main – Free Online Practice

Is PPh3 a strong ligand class 12 chemistry JEE_Main

Full name of DDT is A 111trichloro22bispchlorophenyl class 12 chemistry JEE_Main

Sodium acetate on heating with soda lime produce A class 12 chemistry JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Other Pages
Solutions Class 12 Chemistry Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions - 2025-26

The D and F Block Elements Class 12 Chemistry Chapter 4 CBSE Notes - 2025-26

NCERT Solutions for Class 12 Chemistry Chapter Chapter 7 Alcohol Phenol and Ether

NCERT Solutions ForClass 12 Chemistry Chapter Chapter 8 Aldehydes Ketones And Carboxylic Acids

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

