The oxidation number of carbon in C3O2  and Mg2C3 are respectively:

1. -4/3, +4/3 

2. +4/3, -4/3

3. -2/3, +2/3

4. -2/3, +4/3

Subtopic:  Oxidizing & Reducing Agents |
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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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The correct statement(s) about the given reaction is -

XeO6(aq)4-+2F-1(aq)+6H+(aq)
XeO3(g)+F2(g)+3H2O(l)

1. XeO64- oxidises F-

2. The oxidation number of F increases from -1  to  zero

3. XeO64- is a stronger oxidizing agent that F-

4. All of the above.

Subtopic:  Introduction to Redox and Oxidation Number | Oxidizing & Reducing Agents |
 86%
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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%
From NCERT
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Oxidation number of oxygen in potassium super oxide (KO2) is-

1. – 2

2. – 1

3. – 1/2

4. – 1/4

Subtopic:  Oxidizing & Reducing Agents |
 83%
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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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The oxidizing agent and reducing agent in the given reaction are :

3N2H4l+4ClO3aq-
6NOg+4Claq-+6H2Ol

1. Oxidising agent = N2H4; Reducing agent = ClO3-

2. Oxidising agent = ClO3-; Reducing agent = N2H4

3. Oxidising agent = N2H4 ; Reducing agent = N2H4

4. Oxidising agent = ClO3- ; Reducing agent = ClO3-

Subtopic:  Oxidizing & Reducing Agents |
 82%
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The oxidising agent and reducing agent in the given reaction are :

Cl2O7g+4H2O2aq+2OHaq-
2ClO2aq-+4O2g+5H2Ol

1. Oxidizing agent = H2O2; Reducing agent = Cl2O7

2. Oxidizing agent = Cl2O7; Reducing agent = H2O2

3. Oxidizing agent = H2O2; Reducing agent = H2O2

4. None of the above

Subtopic:  Oxidizing & Reducing Agents |
 80%
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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 |
 76%
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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:  Oxidizing & Reducing Agents |
 73%
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