In which ${\rm{S}}{{\rm{O}}_2}$ acts as oxidant, while reacting with :

Acidified ${\rm{KMn}}{{\rm{O}}_4}$

Acidified ${{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}$

Acidified ${{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}}$

${\rm{HBr}}$ and ${\rm{HI}}$ reduce ${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$,${\rm{HCl}}$ can reduce ${\rm{KMn}}{{\rm{O}}_4}$ and ${\rm{HF}}$ can reduce:

${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

${{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}$

Which reaction indicates the oxidising behavior of ${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

$2{\rm{HI}} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to {{\rm{I}}_2} + {\rm{S}}{{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{O}}$

$2{\rm{PC}}{{\rm{l}}_5} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to 2{\rm{POC}}{{\rm{l}}_3} + 2{\rm{HCl}} + {\rm{S}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}$

$2{\rm{NaOH}} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to {\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4} + 2{{\rm{H}}_2}{\rm{O}}$

${\rm{NaCl}} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to {\rm{NaHS}}{{\rm{O}}_4} + {\rm{HCl}}$

REDOX REACTIONS MCQs for NEET UG CHEMISTRY — Mixed difficulty

1 Hard

1 of 177Question permalink

Application +4 | -1

The oxidation state of iron in sodium nitroprusside is :

View answer and explanation

Correct option: A

+ 2

Option A (+ 2) is correct because the oxidation state of iron in sodium nitroprusside,

\mathrm{Na}_2\mathrm{[Fe(CN}{\mathrm{)}_5}\mathrm{NO]}

is +2.

Solution approach

Correct Answer + Core Reason: $+ 2$ . The oxidation state of Fe in sodium nitroprusside is +2.

Step-by-Step Solution:

Step 1: Determine the formula of sodium nitroprusside.\nThe formula is $\mathrm{Na}_2\mathrm{[Fe(CN}{\mathrm{)}_5}\mathrm{NO]}$

Step 2: Assign oxidation states to the known ions and ligands.\nNa has an oxidation state of +1.\nCN has an oxidation state of -1.\nNO has an oxidation state of +1 (nitrosonium ion). This is because the complex is stable and diamagnetic.

Step 3: Set up the equation to solve for the oxidation state of Fe.
2(+1) + [x + 5(-1) + (+1)]

2 Medium

2 of 177Question permalink

Understanding +4 | -1

It is found that V forms a double salt isomorphous with Mohr’s salt. The oxidation number of V in this compound is :

View answer and explanation

Correct option: B

+ \mathrm{ }2

Option B (

+ 2

) is correct because Mohr's salt has the formula (NH₄)₂Fe(SO₄)₂.6H₂O, where iron (Fe) has a +2 oxidation state, and since V forms an isomorphous double salt, it will also have a +2 oxidation state.

Solution approach

Correct Answer + Core Reason: $+\mathrm{ }2$ . V forms an isomorphous double salt with Mohr's salt, indicating a +2 oxidation state.

Step-by-Step Solution:

Step 1: Recall the formula of Mohr's salt: (NH₄)₂Fe(SO₄)₂.6H₂O.

Step 2: Determine the oxidation state of iron (Fe) in Mohr's salt.

Step 3: In Mohr's salt, the oxidation state of NH₄ is +1, SO₄ is -2, and H₂O is 0.

Step 4: Let the oxidation state of Fe be x. Then, 2(+1) + x + 2(-2) + 6(0) = 0.

Step 5: Solving for x, we get x = +2.

Step 6: Since V forms an isomorphous double salt, it will have the same structure and oxidation state as Fe in Mohr's salt.

Step 7: Therefore, the oxidation number of V is +2.

Why other options are wrong:
- Option A: +3 is not the oxidation state of Fe in Mohr's salt, so it's unlikely for V to have this oxidation state in an isomorphous salt.
- Option C: +4 is not the oxidation state of Fe in Mohr's salt, so it's unlikely for V to have this oxidation state in an isomorphous salt.
- Option D: -4 is not a realistic oxidation state for a metal like vanadium in this type of compound.

Key Concepts:
- Oxidation state
- Double salts
- Isomorphism
- Mohr's salt

Memory Aids:
- Remember the formula of Mohr's salt and the oxidation state of iron in it.

Quick Tips:
- Use the known oxidation states of ions in a compound to determine the oxidation state of an unknown element.

3 Hard

3 of 177Question permalink

Application +4 | -1

For redox reaction,

\mathrm{MnO}_4^ - + {\mathrm{C}_2}\mathrm{O}_4^{2 - } + {\mathrm{H}^ + } \to \mathrm{M}{\mathrm{n}^{2 + }} + \mathrm{C}{\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{O}

coefficient of reactants in balanced states are

\mathrm{MnO}_4^ -

,

{\mathrm{C}_2}\mathrm{O}_4^{2 - }

,

{\mathrm{H}^ + }

View answer and explanation

Correct option: A

2 5 16

Option A (2, 5, 16) is correct because when the redox reaction is balanced, the coefficients for MnO4-, C2O42-, and H+ are 2, 5, and 16 respectively.

Solution approach

Correct Answer + Core Reason: 2 5 16. The balanced equation requires these coefficients for the reactants.

Step-by-Step Solution:

Step 1: Write the unbalanced equation: $\mathrm{MnO}_4^ - + {\mathrm{C}_2}\mathrm{O}_4^{2 - } + {\mathrm{H}^ + } \to \mathrm{Mn}^{2 + } + \mathrm{CO}_2 + {\mathrm{H}_2}\mathrm{O}$

Step 2: Identify the oxidation and reduction half-reactions:
- Oxidation: ${\mathrm{C}_2}\mathrm{O}_4^{2 - } \to \mathrm{CO}_2$
- Reduction: $\mathrm{MnO}_4^ - \to \mathrm{Mn}^{2 + }$

Step 3: Balance the atoms in each half-reaction:
- Oxidation: ${\mathrm{C}_2}\mathrm{O}_4^{2 - } \to 2\mathrm{CO}_2$
- Reduction: $\mathrm{MnO}_4^ - \to \mathrm{Mn}^{2 + }$

Step 4: Balance the oxygen atoms by adding H2O:
- Oxidation: ${\mathrm{C}_2}\mathrm{O}_4^{2 - } \to 2\mathrm{CO}_2$ (already balanced)
- Reduction: $\mathrm{MnO}_4^ - \to \mathrm{Mn}^{2 + } + 4{\mathrm{H}_2}\mathrm{O}$

Step 5: Balance the hydrogen atoms by adding H+:
- Oxidation: ${\mathrm{C}_2}\mathrm{O}_4^{2 - } \to 2\mathrm{CO}_2$ (already balanced)
- Reduction: $8{\mathrm{H}^ + } + \mathrm{MnO}_4^ - \to \mathrm{Mn}^{2 + } + 4{\mathrm{H}_2}\mathrm{O}$

Step 6: Balance the charge by adding electrons:
- Oxidation: ${\mathrm{C}_2}\mathrm{O}_4^{2 - } \to 2\mathrm{CO}_2 + 2{\mathrm{e}^ - }$
- Reduction: $5{\mathrm{e}^ - } + 8{\mathrm{H}^ + } + \mathrm{MnO}_4^ - \to \mathrm{Mn}^{2 + } + 4{\mathrm{H}_2}\mathrm{O}$

Step 7: Multiply the half-reactions to equalize the number of electrons:
- Oxidation: $5({\mathrm{C}_2}\mathrm{O}_4^{2 - } \to 2\mathrm{CO}_2 + 2{\mathrm{e}^ - })$ => $5{\mathrm{C}_2}\mathrm{O}_4^{2 - } \to 10\mathrm{CO}_2 + 10{\mathrm{e}^ - }$
- Reduction: $2(5{\mathrm{e}^ - } + 8{\mathrm{H}^ + } + \mathrm{MnO}_4^ - \to \mathrm{Mn}^{2 + } + 4{\mathrm{H}_2}\mathrm{O})$ => $10{\mathrm{e}^ - } + 16{\mathrm{H}^ + } + 2\mathrm{MnO}_4^ - \to 2\mathrm{Mn}^{2 + } + 8{\mathrm{H}_2}\mathrm{O}$

Step 8: Add the balanced half-reactions:
$2\mathrm{MnO}_4^ - + 5{\mathrm{C}_2}\mathrm{O}_4^{2 - } + 16{\mathrm{H}^ + } \to 2\mathrm{Mn}^{2 + } + 10\mathrm{CO}_2 + 8{\mathrm{H}_2}\mathrm{O}$

Step 9: Identify the coefficients of the reactants: 2 (MnO4-), 5 (C2O42-), and 16 (H+).

Why other options are wrong:
- Option B: The coefficients are incorrect based on the balanced equation.
- Option C: The coefficients are incorrect based on the balanced equation.
- Option D: The coefficients are incorrect based on the balanced equation.

Key Concepts:
- Balancing redox reactions
- Half-reaction method
- Oxidation and reduction

Memory Aids:
- Remember the steps for balancing redox reactions: balance atoms, balance oxygen with water, balance hydrogen with H+, balance charge with electrons, equalize electrons, and add half-reactions.

Quick Tips:
- Always double-check that the final equation is balanced in terms of both atoms and charge.

4 Medium

4 of 177Question permalink

Understanding +4 | -1

Which one of the following reaction is possible at anode?

View answer and explanation

Correct option: D

\mathrm{F}{\mathrm{e}^{2 + }} \to \mathrm{F}{\mathrm{e}^{3 + }} + {\mathrm{e}^ - }

Option D (

\mathrm{F}{\mathrm{e}^{2 + }} \to \mathrm{F}{\mathrm{e}^{3 + }} + {\mathrm{e}^ - }

) is correct because oxidation (loss of electrons) occurs at the anode.

Solution approach

Correct Answer + Core Reason: $\mathrm{F}{\mathrm{e}^{2 + }} \to \mathrm{F}{\mathrm{e}^{3 + }} + {\mathrm{e}^ - }$ . Oxidation occurs at the anode.

Step-by-Step Solution:

Step 1: Recall that oxidation occurs at the anode and reduction occurs at the cathode.

Step 2: Oxidation involves the loss of electrons.

Step 3: Examine each option to see which one represents the loss of electrons.

Step 4: Option D shows $\mathrm{F}{\mathrm{e}^{2 + }}$ losing an electron to become $\mathrm{F}{\mathrm{e}^{3 + }}$ , which is oxidation.

Why other options are wrong:
- Option A: ${\mathrm{F}_2} + 2{\mathrm{e}^ - } \to 2{\mathrm{F}^ - }$ represents the gain of electrons, which is reduction.
- Option B: $2{\mathrm{H}^ + } + \frac{1}{2}{\mathrm{O}_2} + 2{\mathrm{e}^ - } \to {\mathrm{H}_2}\mathrm{O}$ represents the gain of electrons, which is reduction.
- Option C: $2\mathrm{C}{\mathrm{r}^{3 + }} + 7{\mathrm{H}_2}\mathrm{O} \to \mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - } + 14{\mathrm{H}^ + } + 6{\mathrm{e}^ - }$ represents oxidation, but it's less likely to occur at the anode compared to the simpler oxidation of $\mathrm{F}{\mathrm{e}^{2 + }}$ . Also, the question asks for a *possible* reaction.

Key Concepts:
- Anode: Site of oxidation
- Cathode: Site of reduction
- Oxidation: Loss of electrons
- Reduction: Gain of electrons

Memory Aids:
- AN OX (Anode Oxidation)
- RED CAT (Reduction Cathode)

Quick Tips:
- Look for the loss of electrons to identify oxidation reactions.

5 Hard

5 of 177Question permalink

Remembering +4 | -1

The couple having oxidised and reduced forms of a substance taking part in an oxidation or reduction half-reaction is called

View answer and explanation

Correct option: A

redox couple

Option A (redox couple) is correct because a redox couple consists of the oxidized and reduced forms of a substance involved in a half-reaction.

Solution approach

Correct Answer + Core Reason: redox couple. It consists of oxidized and reduced forms of a substance.

Step-by-Step Solution:

Step 1: Define a redox couple.

Step 2: A redox couple is the combination of the oxidized and reduced forms of a substance participating in a half-reaction.

Step 3: For example, $Zn^{2+}/Zn$ is a redox couple.

Why other options are wrong:
- Option B: An oxidized couple only refers to the oxidized form, not the pair.
- Option C: A reduced couple only refers to the reduced form, not the pair.
- Option D: The definition of a redox couple is well-established.

Key Concepts:
- Redox couple
- Oxidation-reduction reactions
- Half-reactions

Memory Aids:
- Think of "redox" as a combination of "reduction" and "oxidation", representing both forms.

Quick Tips:
- Understand the definition of a redox couple in the context of electrochemistry.

6 Hard

6 of 177Question permalink

Analyzing +4 | -1

Assertion: When

\mathrm{Cu}_2 \mathrm{S}

is converted into

\mathrm{Cu}^{+} \,\&\, \mathrm{SO}_2

, then equivalent weight of

\mathrm{Cu}_2 \mathrm{S}

will be

\mathrm{M} / 8\,\left(\mathrm{M}=\mathrm{Mol}\right.

. wt. of

\mathrm{Cu}_2 \mathrm{S}

)
Reason:

\mathrm{Cu}^+

is converted

\mathrm{Cu}^{++}

, during this one electrons is lost.

View answer and explanation

Correct option: B

If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

Option B is correct because the equivalent weight is calculated based on the total change in oxidation state of all copper and sulfur atoms in Cu2S during its conversion to Cu+ and SO2.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion. The equivalent weight of $Cu_2S$ is $M/8$ because of the total change in oxidation state, but the reason provided only considers copper.

Step-by-Step Solution:

Step 1: Determine the oxidation states of Cu and S in $Cu_2S$ . Copper has an oxidation state of +1, and sulfur has an oxidation state of -2.

Step 2: Determine the oxidation states of Cu and S in the products, $Cu^+$ and $SO_2$ . Copper has an oxidation state of +1 (given as $Cu^+$ ), and sulfur has an oxidation state of +4 in $SO_2$ .

Step 3: Calculate the change in oxidation state for each element.\nFor Copper: Since the oxidation state of copper remains +1, the change in oxidation state is 0. However, there are *two* copper atoms in $Cu_2S$ , and the problem states that $Cu_2S$ is converted to $Cu^+$ , so we need to consider the overall reaction. Let's assume the overall reaction is $Cu_2S \to 2Cu^+ + SO_2 + xe^-$ .\nFor Sulfur: The oxidation state changes from -2 to +4, so the change in oxidation state is +6.

Step 4: Determine the number of electrons transferred in the overall reaction. The sulfur atom loses 6 electrons. Since the oxidation state of copper does not change in the given products, the number of electrons transferred is determined solely by the sulfur atom.\nHowever, the problem statement is ambiguous. Let's consider the case where $Cu_2S$ is oxidized to $Cu^{2+}$ and $SO_2$ . In this case, the reaction would be: $Cu_2S \to 2Cu^{2+} + SO_2 + 8e^-$ . The oxidation state of copper changes from +1 to +2 (change of +1 per copper atom, so +2 total), and the oxidation state of sulfur changes from -2 to +4 (change of +6). The total change is 2 + 6 = 8.

Step 5: Calculate the equivalent weight of $Cu_2S$ . The equivalent weight is given by $Equivalent \, weight = \frac{Molecular \, weight}{n-factor}$ , where n-factor is the number of electrons transferred. In this case, the n-factor is 8. Therefore, the equivalent weight is $M/8$ .

Step 6: Analyze the reason. The reason states that $Cu^+$ is converted to $Cu^{2+}$ , during this one electron is lost. This is not the complete picture, as it only considers the change in oxidation state of copper and not sulfur. Also, the problem states that $Cu_2S$ is converted into $Cu^+$ , not $Cu^{2+}$ .

Why other options are wrong:
- Option A: If both Assertion and Reason are correct and Reason is the correct explanation of Assertion is incorrect because the reason is not the complete explanation.
- Option C: If Assertion is correct but Reason is incorrect is incorrect because both are correct but not related as explanation.
- Option D: If both Assertion and Reason are incorrect is incorrect because both are correct.

Key Concepts:
- Oxidation states
- Equivalent weight
- Redox reactions

Memory Aids:
- Equivalent weight formula: $Equivalent \, weight = \frac{Molecular \, weight}{n-factor}$

Quick Tips:
- Determine the oxidation states of all elements in the reactants and products.
- Calculate the change in oxidation state for each element.
- Determine the number of electrons transferred in the overall reaction.
- Use the equivalent weight formula to calculate the equivalent weight.

7 Hard

7 of 177Question permalink

Understanding +4 | -1

Titration of

KI

with

{\mathrm{H}_2}{\mathrm{O}_2}

in presence of acid is a :

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because the reaction involves electron transfer (redox), can be used to visually indicate the endpoint (clock reaction), and occurs between different molecules (intermolecular redox).

Solution approach

Correct Answer + Core Reason: All of these. The reaction is a redox reaction, a clock reaction, and an intermolecular redox.

Step-by-Step Solution:

Step 1: Identify the reaction: $2KI + {\mathrm{H}_2}\mathrm{O}_2 + 2{\mathrm{H}^ + } \to {\mathrm{I}_2} + 2{\mathrm{H}_2}\mathrm{O} + 2{\mathrm{K}^ + }$

Step 2: Determine oxidation states: I goes from -1 to 0 (oxidation), and O in ${\mathrm{H}_2}\mathrm{O}_2$ goes from -1 to -2 in ${\mathrm{H}_2}\mathrm{O}$ (reduction).

Step 3: Since there is a change in oxidation states, it is a redox reaction.

Step 4: The reaction can be used as a clock reaction, where the appearance of iodine can be visually detected using starch as an indicator.

Step 5: The reaction involves electron transfer between different molecules, making it an intermolecular redox reaction.

Why other options are wrong:
- Option A: Only identifies one aspect of the reaction.
- Option B: Only identifies one aspect of the reaction.
- Option C: Only identifies one aspect of the reaction.

Key Concepts:
- Redox reaction
- Clock reaction
- Intermolecular redox

Memory Aids:
- Clock reactions involve a sudden change in color or appearance.

Quick Tips:
- Recognize the different types of redox reactions and their characteristics.

8 Hard

8 of 177Question permalink

Application +4 | -1

What volume of 3 molar

\mathrm{HN}{\mathrm{O}_3}

is needed to oxidise 8 g of

\mathrm{F}{\mathrm{e}^{2 + }}

to

\mathrm{F}{\mathrm{e}^{3 + }}

\mathrm{HN}{\mathrm{O}_3}

, gets converted to NO :

View answer and explanation

Correct option: B

16 m L

Option B (16 m L) is correct because it is calculated using stoichiometry based on the balanced redox reaction and the molarity of the HNO3 solution.

Solution approach

Correct Answer + Core Reason: 16 m L. The volume of 3 molar $\mathrm{HN}{\mathrm{O}_3}$ needed to oxidize 8 g of $\mathrm{F}{{\rm{e$

9 Easy

9 of 177Question permalink

Application +4 | -1

In which

\mathrm{S}{\mathrm{O}_2}

acts as oxidant, while reacting with :

View answer and explanation

Correct option: C

{\mathrm{H}_2}\mathrm{S}

Option C (

{\mathrm{H}_2}\mathrm{S}

) is correct because sulfur dioxide acts as an oxidizing agent by accepting electrons from hydrogen sulfide.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2}\mathrm{S}$ . In the reaction with ${\mathrm{H}_2}\mathrm{S}$ , $\mathrm{S}{\mathrm{O}_2}$ acts as an oxidant.

Step-by-Step Solution:

Step 1: Identify the oxidation states of sulfur in $\mathrm{S}{\mathrm{O}_2}$ and ${\mathrm{H}_2}\mathrm{S}$ .

Step 2: In $\mathrm{S}{\mathrm{O}_2}$ , the oxidation state of S is +4.

Step 3: In ${\mathrm{H}_2}\mathrm{S}$ , the oxidation state of S is -2.

Step 4: When $\mathrm{S}{\mathrm{O}_2}$ reacts with ${\mathrm{H}_2}\mathrm{S}$ , the oxidation state of S in $\mathrm{S}{\mathrm{O}_2}$ decreases, and the oxidation state of S in ${\mathrm{H}_2}\mathrm{S}$ increases. This means $\mathrm{S}{\mathrm{O}_2}$ acts as an oxidizing agent.

Step 5: The reaction is: $2{\mathrm{H}_2}\mathrm{S} + \mathrm{S}{\mathrm{O}_2} \to 3\mathrm{S} + 2{\mathrm{H}_2}\mathrm{O}$

Why other options are wrong:
- Option A: Acidified $\mathrm{KMn}{\mathrm{O}_4}$ is a strong oxidizing agent itself and would oxidize $\mathrm{S}{\mathrm{O}_2}$ .
- Option B: Acidified ${\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}$ is also a strong oxidizing agent and would oxidize $\mathrm{S}{\mathrm{O}_2}$ .
- Option D: Acidified ${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}$ can be oxidized by $\mathrm{S}{\mathrm{O}_2}$ , so $\mathrm{S}{\mathrm{O}_2}$ would act as a reducing agent.

Key Concepts:
- Oxidation and Reduction
- Oxidizing and Reducing Agents
- Oxidation States

Memory Aids:
- OIL RIG (Oxidation Is Loss, Reduction Is Gain)

Quick Tips:
- Identify the oxidation states of the elements involved in the reaction to determine which species is oxidized and which is reduced.

10 Hard

10 of 177Question permalink

Understanding +4 | -1

\mathrm{M}{\mathrm{n}^{2 + }}

can be converted into

\mathrm{M}{\mathrm{n}^{7 + }}

by reacting with

View answer and explanation

Correct option: C

\mathrm{Pb}{\mathrm{O}_2}

Option C (

\mathrm{Pb}{\mathrm{O}_2}

) is correct because PbO2 is a strong oxidizing agent that can oxidize Mn2+ to Mn7+.

Solution approach

Correct Answer + Core Reason: $\mathrm{Pb}{\mathrm{O}_2}$ . Lead dioxide is a strong oxidizing agent.

Step-by-Step Solution:

Step 1: Recall that $\mathrm{M}{\mathrm{n}^{2 + }}$ needs to be oxidized to $\mathrm{M}{\mathrm{n}^{7 + }}$ .

Step 2: Consider the oxidizing strengths of the given options.

Step 3: $\mathrm{S}{\mathrm{O}_2}$ is a reducing agent.

Step 4: $\mathrm{C}{\mathrm{l}_2}$ can oxidize $\mathrm{M}{\mathrm{n}^{2 + }}$ but not to $\mathrm{M}{\mathrm{n}^{7 + }}$ .

Step 5: $\mathrm{Pb}{\mathrm{O}_2}$ is a strong oxidizing agent and can oxidize $\mathrm{M}{\mathrm{n}^{2 + }}$ to $\mathrm{M}{\mathrm{n}^{7 + }}$ .

Step 6: $\mathrm{SnC}{\mathrm{l}_2}$ is a reducing agent.

Why other options are wrong:
- Option A: $\mathrm{S}{\mathrm{O}_2}$ is a reducing agent.
- Option B: $\mathrm{C}{\mathrm{l}_2}$ is not strong enough to oxidize $\mathrm{M}{\mathrm{n}^{2 + }}$ to $\mathrm{M}{\mathrm{n}^{7 + }}$ .
- Option D: $\mathrm{SnC}{\mathrm{l}_2}$ is a reducing agent.

Key Concepts:
- Oxidizing agent
- Oxidation states

Memory Aids:
- Strong oxidizing agents can cause large changes in oxidation states.

Quick Tips:
- Remember the common oxidizing and reducing agents and their strengths.

11 Medium

11 of 177Question permalink

Remembering +4 | -1

The maximum oxidation state of oxygen is

View answer and explanation

Correct option: C

+2

Option C (+2) is correct because oxygen exhibits a +2 oxidation state only when bonded to fluorine, the most electronegative element.

Solution approach

Correct Answer + Core Reason: +2. Oxygen exhibits a +2 oxidation state only when bonded to fluorine, the most electronegative element.

Step-by-Step Solution:

Step 1: Recall the common oxidation states of oxygen.

Step 2: Oxygen usually has an oxidation state of -2. However, it can have positive oxidation states when bonded to fluorine.

Step 3: In OF2, the oxidation state of oxygen is +2 because fluorine is more electronegative.

Step 4: Final Answer: +2

Why other options are wrong:
- Option A (-1): This is the oxidation state of oxygen in peroxides (e.g., H2O2).
- Option B (+1): Oxygen does not commonly exhibit a +1 oxidation state.
- Option D (0): This is the oxidation state of oxygen in its elemental form (O2).

Key Concepts:
- Oxidation states of oxygen
- Electronegativity

Memory Aids:
- Oxygen is usually -2, but +2 with fluorine.

Quick Tips:
- Consider the electronegativity difference between oxygen and the other element.

12 Hard

12 of 177Question permalink

Remembering +4 | -1

Turn bull’s blue is :

View answer and explanation

Correct option: A

\mathrm{F}{\mathrm{e}_3}{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]_2}

Option A (

\mathrm{F}{\mathrm{e}_3}{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]_2}

) is correct because Turnbull's blue is ferric ferrocyanide, formed by the reaction of ferric ions with ferrocyanide ions.

Solution approach

Correct Answer + Core Reason: $\mathrm{F}{\mathrm{e}_3}{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]_2}$ . Turnbull's blue is ferric ferrocyanide.

Step-by-Step Solution:

Step 1: Define Turnbull's blue: A blue pigment with the chemical formula $\mathrm{F}{\mathrm{e}_3}{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]_2}$ .

Step 2: Describe its formation: Formed by the reaction of ferric ions ( $\mathrm{Fe}^{3+}$ ) with ferrocyanide ions ( ${\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{4-}}$ ).

Step 3: Compare the given options with the chemical formula of Turnbull's blue.

Why other options are wrong:
- Option B: ${\mathrm{K}_4}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}$ : This is potassium ferrocyanide, not Turnbull's blue.
- Option C: ${\mathrm{K}_3}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}$ : This is potassium ferricyanide, not Turnbull's blue.
- Option D: $\mathrm{N}{\mathrm{a}_4}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}$ : This is sodium ferrocyanide, not Turnbull's blue.

Key Concepts:
- Turnbull's Blue
- Ferric Ferrocyanide
- Complex Ions

Memory Aids:
- Remember the formula of Turnbull's blue: $\mathrm{F}{\mathrm{e}_3}{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]_2}$

Quick Tips:
- Know the chemical formulas of common complex ions and pigments.

13 Hard

13 of 177Question permalink

Understanding +4 | -1

Strongest oxidising agent among halogen is

View answer and explanation

Correct option: D

{F_2}

Option D (F2) is correct because fluorine is the most electronegative element and has the highest reduction potential among halogens, making it the strongest oxidizing agent.

Solution approach

Correct Answer + Core Reason: ${F_2}$ . Fluorine is the strongest oxidizing agent among halogens.

Step-by-Step Solution:

Step 1: Define Oxidizing Agent. An oxidizing agent is a substance that gains electrons and gets reduced.

Step 2: Consider the electronegativity of halogens. Electronegativity decreases down the group.

Step 3: Relate electronegativity to oxidizing power. Higher electronegativity means a greater tendency to gain electrons.

Step 4: Conclude that fluorine is the strongest oxidizing agent due to its highest electronegativity.

Why other options are wrong:
- Option A: Iodine is the least electronegative halogen and the weakest oxidizing agent.
- Option B: Bromine is less electronegative than fluorine and a weaker oxidizing agent.
- Option C: Chlorine is less electronegative than fluorine and a weaker oxidizing agent.

Key Concepts:
- Oxidizing Agent
- Electronegativity

Memory Aids:
- Electronegativity decreases down the halogen group.

Quick Tips:
- Remember that the most electronegative element is the strongest oxidizing agent.

14 Easy

14 of 177Question permalink

Understanding +4 | -1

When Fe metal is rusted then Fe is :

View answer and explanation

Correct option: A

Oxidised

Option A (Oxidised) is correct because when iron rusts, it loses electrons and its oxidation state increases, which is the definition of oxidation.

Solution approach

Correct Answer + Core Reason: Oxidised. Rusting involves the oxidation of iron.

Step-by-Step Solution:

Step 1: Define oxidation and reduction.

Step 2: Oxidation is the loss of electrons, resulting in an increase in oxidation state.

Step 3: Rusting is the process where iron reacts with oxygen and water to form iron oxide (Fe₂O₃.xH₂O).

Step 4: In this process, iron (Fe) loses electrons and is oxidized to Fe²⁺ or Fe³⁺.

Why other options are wrong:
- Option B: Reduction is the gain of electrons, which is not what happens to iron during rusting.
- Option C: Hydrolysis is the reaction with water, which is involved in rusting but doesn't describe the change in iron's oxidation state.
- Option D: Precipitation is the formation of a solid, which occurs during rusting, but the key process is the oxidation of iron.

Key Concepts:
- Oxidation
- Reduction
- Rusting

Memory Aids:
- Remember that "OIL RIG" (Oxidation Is Loss, Reduction Is Gain) helps to remember the definitions of oxidation and reduction.

Quick Tips:
- Recognize common oxidation and reduction processes.

15 Hard

15 of 177Question permalink

Application +4 | -1

The oxidation state of the most electronegative element in the products of the reaction between

\mathrm{BaO}_2

and

\mathrm{H}_2 \mathrm{SO}_4

are

View answer and explanation

Correct option: B

-1

and

-2

Option B (-1 and -2) is correct because in the reaction of BaO2 with H2SO4, the products are BaSO4 and H2O2. The oxidation state of O in BaSO4 is -2, and in H2O2, it is -1. Oxygen is the most electronegative element in both products.

Solution approach

Correct Answer + Core Reason: $-1$ and $-2$ . The products are $BaSO_4$ and $H_2O_2$ , and oxygen's oxidation states are -2 and -1 respectively.

Step-by-Step Solution:

Step 1: Write the reaction between BaO2 and H2SO4: $\mathrm{BaO}_2 + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{BaSO}_4 + \mathrm{H}_2\mathrm{O}_2$

Step 2: Identify the products: Barium sulfate (BaSO4) and hydrogen peroxide (H2O2).

Step 3: Determine the oxidation state of oxygen in BaSO4:
- Ba has an oxidation state of +2.
- SO4 has a charge of -2.
- Let x be the oxidation state of O. Then, S + 4x = -2. Since S is in group 16, its oxidation state is +6. Therefore, +6 + 4x = -2, so 4x = -8, and x = -2.
- The oxidation state of oxygen in BaSO4 is -2.

Step 4: Determine the oxidation state of oxygen in H2O2:
- H has an oxidation state of +1.
- Let y be the oxidation state of O. Then, 2(+1) + 2y = 0, so 2 + 2y = 0, and 2y = -2, so y = -1.
- The oxidation state of oxygen in H2O2 is -1.

Step 5: Identify the most electronegative element in the products. In both BaSO4 and H2O2, oxygen is the most electronegative element.

Step 6: State the oxidation states of oxygen in the products: -1 and -2.

Why other options are wrong:
- Option A: 0 and -1 are incorrect. Oxygen in BaSO4 has an oxidation state of -2, not 0.
- Option C: -2 and 0 are incorrect. Oxygen in H2O2 has an oxidation state of -1, not 0.
- Option D: -2 and +1 are incorrect. Oxygen in H2O2 has an oxidation state of -1, not +1.

Key Concepts:
- Oxidation state rules
- Electronegativity
- Reaction of peroxides with acids

Memory Aids:
- Remember that oxygen usually has an oxidation state of -2, but in peroxides, it is -1.

Quick Tips:
- When determining oxidation states, remember to consider the overall charge of the compound or ion.

16 Medium

16 of 177Question permalink

Application +4 | -1

Oxidation number of

S

in

{\mathrm{S}_2}\mathrm{C}{\mathrm{l}_2}

is :

View answer and explanation

Correct option: A

+ 1

Option A (+1) is correct because the oxidation number of Cl is -1, and the overall molecule is neutral, leading to +1 for S.

Solution approach

Correct Answer + Core Reason: +1. The oxidation number is calculated based on the electronegativity difference and overall charge neutrality.

Step-by-Step Solution:

Step 1: Determine the known oxidation numbers. Chlorine (Cl) is more electronegative than sulfur (S), so it has an oxidation number of -1.

Step 2: Set up an equation to solve for the oxidation number of sulfur (S). Let x be the oxidation number of S. Since there are two sulfur atoms and two chlorine atoms in S2Cl2, the equation is: 2x + 2(-1) = 0

Step 3: Solve for x: 2x - 2 = 0 => 2x = 2 => x = 1

Step 4: Therefore, the oxidation number of S in S2Cl2 is +1.

Why other options are wrong:
- Option B (+6): This is incorrect because it would require a much larger negative oxidation state for chlorine to balance, which is not possible.
- Option C (Zero): This is incorrect because chlorine is more electronegative and will have a negative oxidation state, forcing sulfur to have a positive one to maintain neutrality.
- Option D (-1): This is incorrect because sulfur is less electronegative than chlorine and cannot have a negative oxidation state in this compound.

Key Concepts:
- Oxidation number rules
- Electronegativity
- Charge neutrality of compounds

Memory Aids:
- Remember that more electronegative elements usually have negative oxidation numbers.

Quick Tips:
- Always start by assigning oxidation numbers to the most electronegative elements first.

17 Easy

17 of 177Question permalink

Application +4 | -1

\frac{{ - 1}}{3}

oxidation state of nitrogen will be obtained in case of

View answer and explanation

Correct option: B

{N_3}H

Option B (

{N_3}H

) is correct because the average oxidation state of nitrogen in hydrazoic acid is -1/3.

Solution approach

Correct Answer + Core Reason: ${N_3}H$ . The oxidation state of nitrogen is -1/3 in hydrazoic acid.

Step-by-Step Solution:

Step 1: Determine the oxidation state of nitrogen in each compound.

Step 2: In $NH_3$ , the oxidation state of N is -3.

Step 3: In $N_3H$ , let the oxidation state of N be x.

Step 4: $3x + 1 = 0$

Step 5: $x = -1/3$

Step 6: In $NO$ , the oxidation state of N is +2.

Step 7: In $N_2O$ , the oxidation state of N is +1.

Why other options are wrong:
- Option A: $N{H_3}$ : Nitrogen has an oxidation state of -3.
- Option C: NO: Nitrogen has an oxidation state of +2.
- Option D: ${N_2}O$ : Nitrogen has an oxidation state of +1.

Key Concepts:
- Oxidation state
- Rules for assigning oxidation states

Memory Aids:
- Hydrogen is usually +1, and oxygen is usually -2.

Quick Tips:
- Remember the rules for assigning oxidation states.

18 Medium

18 of 177Question permalink

Analyzing +4 | -1

Statement-I: Oxidation state of Sulphur in

\mathrm{H}_2 \mathrm{SO}_3

is +6.

Statement-II: Maximum oxidation state of

\mathrm{S}

is +5.

View answer and explanation

Correct option: B

Both the statements are incorrect

Both statements are incorrect. The oxidation state of sulfur in H₂SO₃ is +4, and the maximum oxidation state of sulfur is +6.

Solution approach

Correct Answer + Core Reason: Both the statements are incorrect. The oxidation state of S in H₂SO₃ is +4, and the maximum oxidation state of S is +6.

Step-by-Step Solution:

Step 1: Analyze Statement-I: Oxidation state of Sulphur in $\mathrm{H}_2 \mathrm{SO}_3$ is +6.\nTo determine the oxidation state of sulfur, we use the following rules:
- H has an oxidation number of +1.
- O has an oxidation number of -2.\nLet the oxidation number of S be x.
2(+1) + x + 3(-2) = 0
2 + x - 6 = 0\nx - 4 = 0\nx = +4\nThe oxidation number of sulfur is +4, not +6. Therefore, Statement-I is incorrect.

Step 2: Analyze Statement-II: Maximum oxidation state of $\mathrm{S}$ is +5.\nThe maximum oxidation state of sulfur is +6, as seen in compounds like $H_2SO_4$ and $SF_6$ . Therefore, Statement-II is incorrect.

Why other options are wrong:
- Option A: Both statements are incorrect, so this option is incorrect.
- Option C: Statement-I is incorrect, and Statement-II is also incorrect, so this option is incorrect.
- Option D: Statement-I is incorrect, and Statement-II is also incorrect, so this option is incorrect.

Key Concepts:
- Oxidation Number Calculation
- Rules for Assigning Oxidation Numbers

Memory Aids:
- Remember the common oxidation numbers of elements.
- The sum of oxidation numbers in a neutral molecule is zero.

Quick Tips:
- Practice calculating oxidation numbers regularly.
- Be aware of the different oxidation states that elements can exhibit.

19 Medium

19 of 177Question permalink

Understanding +4 | -1

Manganese acts as strongest oxidising agent in the oxidation state

View answer and explanation

Correct option: A

+ 7

Option A (+7) is correct because Mn in the +7 oxidation state (e.g., in KMnO4) has a strong tendency to gain electrons and get reduced, thus acting as a strong oxidizing agent.

Solution approach

Correct Answer + Core Reason: + 7. Mn in +7 oxidation state is a strong oxidizing agent.

Step-by-Step Solution:

Step 1: Define Oxidizing Agent. An oxidizing agent is a substance that gains electrons and gets reduced.

Step 2: Consider the stability of different oxidation states of Manganese. Higher oxidation states are generally more oxidizing.

Step 3: Evaluate Mn in +7 oxidation state. Mn+7 has a strong tendency to gain electrons to become more stable.

Step 4: Conclude that Mn in +7 oxidation state is the strongest oxidizing agent.

Why other options are wrong:
- Option B: Mn in +2 oxidation state is relatively stable and does not have a strong tendency to gain electrons.
- Option C: Mn in +4 oxidation state is less oxidizing than Mn in +7 oxidation state.
- Option D: Mn in +5 oxidation state is less oxidizing than Mn in +7 oxidation state.

Key Concepts:
- Oxidizing Agent
- Oxidation State

Memory Aids:
- Higher oxidation state = Stronger oxidizing agent (for transition metals)

Quick Tips:
- Remember that oxidizing agents are substances that gain electrons.

20 Medium

20 of 177Question permalink

Analyzing +4 | -1

\mathrm{HBr}

and

\mathrm{HI}

reduce

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

,

\mathrm{HCl}

can reduce

\mathrm{KMn}{\mathrm{O}_4}

and

\mathrm{HF}

can reduce:

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because HF is a weak reducing agent and cannot reduce any of the given compounds.

Solution approach

Correct Answer + Core Reason: None of these. HF is not a strong enough reducing agent to reduce the given compounds.

Step-by-Step Solution:

Step 1: Consider the reducing strength of $\mathrm{HBr}$ , $\mathrm{HI}$ , $\mathrm{HCl}$ , and $\mathrm{HF}$ .

Step 2: Reducing strength increases down the halogen group: $\mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}$ .

Step 3: $\mathrm{HBr}$ and $\mathrm{HI}$ can reduce ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ . $\mathrm{HCl}$ can reduce $\mathrm{KMn}{\mathrm{O}_4}$ .

Step 4: $\mathrm{HF}$ is the weakest reducing agent among the hydrohalic acids. It cannot reduce ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ , ${\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}$ , or $\mathrm{KMn}{\mathrm{O}_4}$ .

Why other options are wrong:
- Option A: ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ : HF is not a strong enough reducing agent to reduce sulfuric acid.
- Option B: ${\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}$ : HF is not a strong enough reducing agent to reduce potassium dichromate.
- Option C: $\mathrm{KMn}{\mathrm{O}_4}$ : HF is not a strong enough reducing agent to reduce potassium permanganate.

Key Concepts:
- Reducing Strength
- Hydrohalic Acids
- Redox Reactions

Memory Aids:
- Reducing strength increases down the halogen group.

Quick Tips:
- Know the relative reducing strengths of common reducing agents.

21 Easy

21 of 177Question permalink

Remembering +4 | -1

Which of the following halogen shows only one oxidation state in its compounds

View answer and explanation

Correct option: B

F

Option B (F) is correct because fluorine is the most electronegative element and only exhibits a -1 oxidation state in its compounds.

Solution approach

Correct Answer + Core Reason: F. Fluorine is the most electronegative element and only exhibits a -1 oxidation state.

Step-by-Step Solution:

Step 1: Recall the properties of halogens.\nHalogens are highly electronegative elements that tend to gain one electron to achieve a stable noble gas configuration.

Step 2: Consider the electronegativity of fluorine.\nFluorine is the most electronegative element, meaning it has the highest tendency to attract electrons. Therefore, it always has a -1 oxidation state in its compounds.

Step 3: Consider the other halogens.\nChlorine, bromine, and iodine can exhibit multiple oxidation states (e.g., -1, +1, +3, +5, +7) because they are less electronegative than fluorine and can form compounds with oxygen and other more electronegative elements.

Why other options are wrong:
- **Option A ( $Cl$ ): Chlorine can exhibit multiple oxidation states such as -1, +1, +3, +5, and +7.
- Option C ( $Br$ ): Bromine can exhibit multiple oxidation states such as -1, +1, +3, and +5.
- Option D ( $I$ ): Iodine can exhibit multiple oxidation states such as -1, +1, +3, +5, and +7.

Key Concepts:
- Oxidation state
- Electronegativity
- Halogens

Memory Aids:
- Remember that fluorine is the most electronegative element and always has a -1 oxidation state.

Quick Tips:**
- Know the trends in electronegativity within the periodic table.

22 Medium

22 of 177Question permalink

Remembering +4 | -1

The most stable oxidation state of copper is :

View answer and explanation

Correct option: A

+ 2

Option A (

+2

) is correct because the +2 oxidation state is the most stable for copper due to its electronic configuration and hydration enthalpy.

Solution approach

Correct Answer + Core Reason: $+ 2$ . It is the most stable oxidation state of copper.

Step-by-Step Solution:

Step 1: Recall the electronic configuration of Cu: [Ar] 3d¹⁰ 4s¹.

Step 2: While Cu can exhibit +1 and +2 oxidation states, +2 is more stable in aqueous solution.

Step 3: The higher stability of Cu²⁺ is attributed to its higher hydration enthalpy, which compensates for the second ionization enthalpy.

Why other options are wrong:
- Option B: While Cu⁺ exists, it is less stable in aqueous solution and tends to disproportionate into Cu and Cu²⁺.
- Option C: The +3 oxidation state is not a common or stable oxidation state for copper.
- Option D: The +4 oxidation state is not a known oxidation state for copper.

Key Concepts:
- Oxidation states of copper
- Electronic configuration of copper
- Hydration enthalpy
- Disproportionation

Memory Aids:
- Remember that Cu²⁺ is more stable in aqueous solution due to hydration enthalpy.

Quick Tips:
- Consider the electronic configuration and hydration enthalpy when determining the stability of oxidation states.

23 Medium

23 of 177Question permalink

Analyzing +4 | -1

Which of the following arrangements represent increasing oxidation number of the central atom

View answer and explanation

Correct option: A

CrO_2^ - ,\,ClO_3^ - ,CrO_4^{2 - },MnO_4^ -

Option A (

CrO_2^ - ,\,ClO_3^ -,CrO_4^{2 - },MnO_4^ -

) is correct because it represents the increasing order of oxidation numbers of the central atoms: Cr(+3), Cl(+5), Cr(+6), Mn(+7).

Solution approach

Correct Answer + Core Reason: $CrO_2^ - ,\,ClO_3^ -,CrO_4^{2 - },MnO_4^ -$ . The oxidation numbers of the central atoms increase in this order.

Step-by-Step Solution:

Step 1: Determine the oxidation number of Cr in $CrO_2^-$ .\nLet the oxidation number of Cr be x.\nx + 2(-2) = -1\nx - 4 = -1\nx = +3

Step 2: Determine the oxidation number of Cl in $ClO_3^-$ .\nLet the oxidation number of Cl be x.\nx + 3(-2) = -1\nx - 6 = -1\nx = +5

Step 3: Determine the oxidation number of Cr in $CrO_4^{2-}$ .\nLet the oxidation number of Cr be x.\nx + 4(-2) = -2\nx - 8 = -2\nx = +6

Step 4: Determine the oxidation number of Mn in $MnO_4^-$ .\nLet the oxidation number of Mn be x.\nx + 4(-2) = -1\nx - 8 = -1\nx = +7

Step 5: Arrange the oxidation numbers in increasing order: +3, +5, +6, +7.\nTherefore, the correct order is $CrO_2^ - ,\,ClO_3^ -,CrO_4^{2 - },MnO_4^ -$ .

Why other options are wrong:
- Option B: This option is incorrect because it does not represent the increasing order of oxidation numbers.
- Option C: This option is incorrect because it does not represent the increasing order of oxidation numbers.
- Option D: This option is incorrect because it does not represent the increasing order of oxidation numbers.

Key Concepts:
- Oxidation number calculation
- Rules for assigning oxidation numbers

Memory Aids:
- Oxygen usually has an oxidation number of -2.
- The sum of oxidation numbers in an ion equals the charge of the ion.

Quick Tips:
- Remember the common oxidation numbers of elements.
- Practice calculating oxidation numbers regularly.

24 Medium

24 of 177Question permalink

Analyzing +4 | -1

Which reaction indicates the oxidising behavior of

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

View answer and explanation

Correct option: D

2\mathrm{HI} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to {\mathrm{I}_2} + \mathrm{S}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}

Option D (

2\mathrm{HI} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to {\mathrm{I}_2} + \mathrm{S}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}

) is correct because H2SO4 oxidizes HI to I2, indicating its oxidizing behavior.

Solution approach

Correct Answer + Core Reason: $2\mathrm{HI} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to {\mathrm{I}_2} + \mathrm{S}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}$ . H2SO4 oxidizes HI to I2.

Step-by-Step Solution:

Step 1: Understand the concept of oxidizing behavior.\nAn oxidizing agent causes oxidation of another substance and itself gets reduced. In other words, it gains electrons.

Step 2: Analyze the given reactions.
- Option A: $2\mathrm{PC}{\mathrm{l}_5} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to 2\mathrm{POC}{\mathrm{l}_3} + 2\mathrm{HCl} + \mathrm{S}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$
In this reaction, H2SO4 is acting as a dehydrating agent, not an oxidizing agent.
- Option B: $2\mathrm{NaOH} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to \mathrm{N}{\mathrm{a}_2}\mathrm{S}{\mathrm{O}_4} + 2{\mathrm{H}_2}\mathrm{O}$
This is a neutralization reaction, not an oxidation-reduction reaction.
- Option C: $\mathrm{NaCl} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to \mathrm{NaHS}{\mathrm{O}_4} + \mathrm{HCl}$
This is an acid-base reaction, not an oxidation-reduction reaction.
- Option D: $2\mathrm{HI} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to {\mathrm{I}_2} + \mathrm{S}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}$
In this reaction, H2SO4 oxidizes HI to I2. The oxidation state of I changes from -1 in HI to 0 in I2. Sulfur in H2SO4 is reduced from +6 to +4 in SO2. This clearly shows the oxidizing behavior of H2SO4.

Step 3: Conclusion.\nOption D correctly indicates the oxidizing behavior of H2SO4.

Why other options are wrong:
- Option A: H2SO4 acts as a dehydrating agent, not an oxidizing agent.
- Option B: This is a neutralization reaction.
- Option C: This is an acid-base reaction.

Key Concepts:
- Oxidizing agent
- Oxidation-reduction reactions

Memory Aids:
- Remember that H2SO4 can act as an oxidizing agent, especially with halides.

Quick Tips:
- Look for changes in oxidation states to identify redox reactions.

25 Medium

25 of 177Question permalink

Understanding +4 | -1

Which one of the following lanthanoids does not form

M{O_2}

? [

M

is lanthanoid metal]

View answer and explanation

Correct option: D

Yb

Option D (

Yb

) is correct because Ytterbium (Yb) typically forms +3 compounds, and its +4 state is less stable, preventing the formation of

MO_2

.

Solution approach

Correct Answer + Core Reason: $Yb$ . Ytterbium typically forms +3 compounds and does not readily form $MO_2$ .

Step-by-Step Solution:

Step 1: Recall the properties of lanthanoids.

Step 2: Pr, Nd, and Dy can form stable $MO_2$ oxides.

Step 3: Yb typically forms +3 compounds, and its +4 state is less stable.

Step 4: Therefore, Yb does not readily form $MO_2$ .

Why other options are wrong:
- Option A: $\Pr$ : Praseodymium can form $PrO_2$ .
- Option B: $Dy$ : Dysprosium can form $DyO_2$ .
- Option C: $Nd$ : Neodymium can form $NdO_2$ .

Key Concepts:
- Lanthanoids
- Oxidation states of lanthanoids
- Properties of lanthanoid oxides

Memory Aids:
- Remember the common oxidation states of lanthanoids.

Quick Tips:
- Know the common oxidation states of lanthanoids to predict oxide formation.

26 Easy

26 of 177Question permalink

Analyzing +4 | -1

Which is the best description of behaviour of bromine in the reaction given below?

{\mathrm{H}_2}\mathrm{O} + \mathrm{B}{\mathrm{r}_2} \to \mathrm{HBr} + \mathrm{HOBr}

View answer and explanation

Correct option: B

Both oxidised and reduced

Option B (Both oxidised and reduced) is correct because bromine undergoes both oxidation and reduction in the given reaction.

Solution approach

Correct Answer + Core Reason: Both oxidised and reduced. Bromine's oxidation state changes from 0 to -1 and +1.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Br in ${\mathrm{H}_2}\mathrm{O} + \mathrm{B}{\mathrm{r}_2} \to \mathrm{HBr} + {\mathrm{HOBr}$ .

Step 2: In ${\mathrm{Br}_2}$ , the oxidation state of Br is 0.

Step 3: In $\mathrm{HBr}$ , the oxidation state of Br is -1.

Step 4: In $\mathrm{HOBr}$ , the oxidation state of Br is +1.

Step 5: Since Br goes from 0 to -1 (reduction) and from 0 to +1 (oxidation), it is both oxidized and reduced.

Why other options are wrong:
- Option A: Proton accepted only: This describes acid-base behavior, not redox.
- Option C: Oxidised only: Bromine is also reduced.
- Option D: Reduced only: Bromine is also oxidized.

Key Concepts:
- Redox reactions
- Oxidation states
- Disproportionation reaction

Memory Aids:
- OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons)

Quick Tips:
- Identify the oxidation states of the element in the reactants and products. If the oxidation state increases, it is oxidation; if it decreases, it is reduction. If an element undergoes both oxidation and reduction, it's a disproportionation reaction.

27 Hard

27 of 177Question permalink

Application +4 | -1

O.N. of

\mathrm{V}

in

\mathrm{Rb}_4 \mathrm{Na}\left[\mathrm{HV}_{10} \mathrm{O}_{28}\right]

is-

View answer and explanation

Correct option: B

+5

Option B (+5) is correct because Rb has +1, Na has +1, H has +1, and O has -2, balancing the charge of the complex ion.

Solution approach

Correct Answer + Core Reason: +5. The oxidation state of V is determined by balancing the charges of the other ions in the complex.

Step-by-Step Solution:

Step 1: Determine the known oxidation numbers. Rb has +1, Na has +1, H has +1, and O has -2.

Step 2: Determine the total positive charge from Rb and Na: 4(+1) + 1(+1) = +5

Step 3: Determine the total negative charge from the complex ion: Rb4Na[HV10O28]. The charge of the complex ion is -5 to balance the +5 charge from Rb and Na.

Step 4: Calculate the oxidation state of V. Let x be the oxidation state of V. The equation is: +1 + 10x + 28(-2) = -5

Step 5: Solve for x: 1 + 10x - 56 = -5 => 10x - 55 = -5 => 10x = 50 => x = +5

Why other options are wrong:
- Option A (+3): This is incorrect because it would not balance the charges of the other ions in the complex.
- Option C (+4): This is incorrect because it would not balance the charges of the other ions in the complex.
- Option D (zero): This is incorrect because it would not balance the charges of the other ions in the complex.

Key Concepts:
- Oxidation number rules
- Charge neutrality of coordination complexes

Memory Aids:
- Alkali metals (Rb, Na) are always +1 in compounds.
- Oxygen is usually -2 in compounds.

Quick Tips:
- Remember to consider the overall charge of the complex ion when calculating oxidation numbers.

28 Medium

28 of 177Question permalink

Understanding +4 | -1

Reaction of acidified

\mathrm{KMn}{\mathrm{O}_4}

with ferrous oxalate gives oxidation products containing :

View answer and explanation

Correct option: C

Both (1) and (2)

Option C (Both (1) and (2)) is correct because ferrous oxalate contains both iron(II) and oxalate ions, which are oxidized to iron(III) and carbon dioxide, respectively.

Solution approach

Correct Answer + Core Reason: Both (1) and (2). Ferrous oxalate contains Fe2+ and C2O42- ions, which are oxidized to Fe3+ and CO2 respectively.

Step-by-Step Solution:

Step 1: Identify the reactants and products.

Step 2: Acidified KMnO4 is a strong oxidizing agent.

Step 3: Ferrous oxalate contains Fe2+ and C2O42- ions.

Step 4: Fe2+ is oxidized to Fe3+.

Step 5: C2O42- is oxidized to CO2.

Step 6: Therefore, the oxidation products contain Fe3+ and CO2.

Why other options are wrong:
- Option A: $\mathrm{F}{\mathrm{e}^{3 + }}$ : This is partially correct, but CO2 is also formed.
- Option B: $\mathrm{C}{\mathrm{O}_2}$ : This is partially correct, but Fe3+ is also formed.
- Option D: None of these: Option C is correct.

Key Concepts:
- Oxidation
- Oxidizing agents
- Redox reactions

Memory Aids:
- KMnO4 is a strong oxidizing agent.

Quick Tips:
- Remember that KMnO4 oxidizes Fe2+ to Fe3+ and oxalate to CO2.

29 Hard

29 of 177Question permalink

Application +4 | -1

The oxidation number of arsenic in arsenate is :

View answer and explanation

Correct option: A

+ 5

Option A (

+ 5

) is correct because the oxidation number of arsenic in arsenate (

\mathrm{AsO}_4^{3 - }

) is +5.

Solution approach

Correct Answer + Core Reason: $+ 5$ . The oxidation number of arsenic in arsenate ( $\mathrm{AsO}_4^{3 - }$ ) is +5.

Step-by-Step Solution:

Step 1: Write the formula for arsenate ion: $\mathrm{AsO}_4^{3 - }$

Step 2: Let x be the oxidation number of arsenic (As).

Step 3: The oxidation number of oxygen (O) is -2.

Step 4: The sum of the oxidation numbers in the ion must equal the charge of the ion, which is -3.

Step 5: Set up the equation: x + 4(-2) = -3

Step 6: Solve for x: x - 8 = -3
x = -3 + 8
x = +5

Step 7: The oxidation number of arsenic in arsenate is +5.

Why other options are wrong:
- Option B: Incorrect oxidation number for arsenic.
- Option C: Incorrect oxidation number for arsenic.
- Option D: Incorrect oxidation number for arsenic.

Key Concepts:
- Oxidation number rules
- Polyatomic ion rule

Memory Aids:
- Oxygen is usually -2

Quick Tips:
- Remember the rules for assigning oxidation numbers.

30 Medium

30 of 177Question permalink

Analyzing +4 | -1

Which of the following chemical reactions depicts the oxidising behaviour of

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

?

View answer and explanation

Correct option: A

2\mathrm{HI} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to {\mathrm{I}_2} + \mathrm{S}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}

Option A (

2\mathrm{HI} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to {\mathrm{I}_2} + \mathrm{S}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}

) is correct because H2SO4 oxidizes HI to I2, and itself gets reduced to SO2, showing its oxidizing behavior.

Solution approach

Correct Answer + Core Reason: $2\mathrm{HI} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to {\mathrm{I}_2} + \mathrm{S}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}$ . H2SO4 oxidizes HI.

Step-by-Step Solution:

Step 1: Determine the oxidation states of each element in each reaction.

Step 2: In option A: $2\mathrm{HI} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to {\mathrm{I}_2} + \mathrm{S}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}$ , Iodine is oxidized from -1 to 0, and Sulfur is reduced from +6 to +4. H2SO4 acts as an oxidizing agent.

Step 3: In option B: $\mathrm{Ca}{\left( {\mathrm{OH}} \right)_2} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to \mathrm{CaS}{\mathrm{O}_4} + 2{\mathrm{H}_2}\mathrm{O}$ , This is an acid-base neutralization reaction. There is no change in oxidation states.

Step 4: In option C: $\mathrm{NaCl} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to \mathrm{NaHS}{\mathrm{O}_4} + \mathrm{HCl}$ , This is an acid-base reaction. There is no change in oxidation states.

Step 5: In option D: $2\mathrm{PC}{\mathrm{l}_5} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to 2\mathrm{POC}{\mathrm{l}_3} + 2\mathrm{HCl} + \mathrm{S}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$ , Sulfur remains at +6. This is not a clear example of oxidizing behavior of H2SO4.

Why other options are wrong:
- Option B: $\mathrm{Ca}{\left( {\mathrm{OH}} \right)_2} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to \mathrm{CaS}{\mathrm{O}_4} + 2{\mathrm{H}_2}\mathrm{O}$ : This is an acid-base reaction, not a redox reaction.
- Option C: $\mathrm{NaCl} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to \mathrm{NaHS}{\mathrm{O}_4} + \mathrm{HCl}$ : This is an acid-base reaction, not a redox reaction.
- Option D: $2\mathrm{PC}{\mathrm{l}_5} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to 2\mathrm{POC}{\mathrm{l}_3} + 2\mathrm{HCl} + \mathrm{S}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$ : Sulfur remains at +6, so it doesn't clearly show oxidizing behavior.

Key Concepts:
- Redox Reactions
- Oxidation States
- Oxidizing Agents

Memory Aids:
- OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons)
- LEO says GER: Lose Electrons Oxidation, Gain Electrons Reduction

Quick Tips:
- Determine oxidation states to identify oxidation and reduction.
- Oxidizing agents get reduced.

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Redox reactions practice · Mixed difficulty

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