MnO42- undergoes disproportionation reaction in acidic medium but MnO4- does not. It is:

1.  Due to the highest oxidation state of Mn in  MnO42-
2.  Due to the highest oxidation state of Mn in  MnO4-
3.  Due to the endothermic nature of the disproportionation reaction.
4.  Due to the exothermic nature of the disproportionation reaction.

Subtopic:  Introduction to Redox and Oxidation Number |
 81%
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(a) \(E_{k^+/K}^o = - 2.93\ V\); \(E_{Ag^+/Ag}^o = 0.80\ V\)
(b) \(E_{Hg^{2+}/Hg}^o = 0.79\ V\); \(E_{Mg^{2+}/Mg}^o = - 2.37\ V\)  
(c) \(E_{Cr^{3+}/Cr}^o = -0.74\ V\)

Based on standard electrode potentials given above, the correct arrangement for increasing order of reducing power of
elements is: 

1. \(\mathrm{Ag}<\mathrm{Hg}<\mathrm{Cr}<\mathrm{Mg}<\mathrm{K} \)
2. \(\mathrm{Ag}>\mathrm{Cr}>\mathrm{Mg}>\mathrm{Hg}>\mathrm{K}\)
3. \(\mathrm{K}>\mathrm{Mg}<\mathrm{Cr}<\mathrm{Hg}>\mathrm{Ag} \)
4. \(\mathrm{K}<\mathrm{Mg}<\mathrm{Cr}<\mathrm{Hg}<\mathrm{Ag}\)
Subtopic:  Emf & Electrode Potential |
 74%
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The element that does not show a disproportionation tendency is/are:

1. Cl

2. Br

3. F

4. I

Subtopic:  Oxidizing & Reducing Agents |
 83%
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Given below are two statements: 

Assertion (A): The decomposition of hydrogen peroxide to form water and oxygen is an example of a disproportionation reaction.
Reason (R): The oxygen of peroxide is in –1 oxidation state and it is converted to zero oxidation state in O2
and –2 oxidation state in H2O.

1. Both (A) and (R) are true and (R) is the correct explanation of (A).
2. Both (A) and (R) are true but (R) is not the correct explanation of (A).
3. (A) is true but (R) is false.
4. (A) is false but (R) is true.

Subtopic:  Redox Titration & Type of Redox |
 89%
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The incorrect statement regarding the rule to find the oxidation number among the following is-

1. The oxidation number of hydrogen is always +1.
2. The algebraic sum of all the oxidation numbers carried by elements in a compound is zero.
3. An element in its free or uncombined state has an oxidation number of zero.
4. Generally, in all its compounds, the oxidation number of fluorine is -1.

Subtopic:  Oxidizing & Reducing Agents |
 77%
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The largest oxidation number exhibited by an element depends on its electronic configuration. With which of the following electronic configurations will the element exhibit the largest oxidation number?

1. 3d14s2

2. 3d34s2

3. 3d54s1

4. 3d54s2

Subtopic:  d-Block Elements- Properties & Uses |
 73%
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Given below are two statements: 

Assertion (A): In the reaction between potassium permanganate and potassium iodide, permanganate ions act as an oxidising agent.
Reason (R): The oxidation state of manganese changes from +2 to +7 during the reaction.

1. Both (A) and (R) are true and (R) is the correct explanation of (A).
2. Both (A) and (R) are true but (R) is not the correct explanation of (A).
3. (A) is true but (R) is false.
4. (A) is false but (R) is true.

Subtopic:  Redox Titration & Type of Redox |
 64%
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What is the alteration in the oxidation state of carbon in the given reaction?

\(\mathrm{{CH_4}_{(g)} + 4{Cl_2}_{(g)} \rightarrow {CCl_4}_{(l)} + 4 HCl_{(g)}}\)

1. 0 to +4

2. –4 to +4

3. 0 to –4

4. +4 to +4

Subtopic:  Oxidizing & Reducing Agents |
 82%
From NCERT
NEET - 2020
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Among the following the correct order of acidity is-

1. HClO < HClO2< HClO3< HClO4
2. HClO2< HClO < HClO3< HClO4
3. HClO4< HClO2< HClO < HClO3
4. HClO3< HClO4< HClO2< HClO

Subtopic:  Oxidizing & Reducing Agents |
 85%
From NCERT
NEET - 2016
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Standard reduction potentials of the half-reactions are given below: 

F2(g) + 2e-2F-(aq) ; E° =+2.85 V 

Cl2(g) + 2e-2Cl-(aq) ; E° =+1.36 V 

Br2(g) + 2e-2Br-(aq) ; E° =+1.06 V 

I2(g) + e-2I-(aq) ; E° =+0.53 V 

The strongest oxidizing and reducing agents, respectively, are: 

1. Br2 and Cl-

2. Cl2 and Br-

3. Cl2 and I2

4. F2 and l-

Subtopic:  Emf & Electrode Potential |
 84%
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AIPMT - 2012
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