In the titration of ${\rm{CuS}}{{\rm{O}}_4}vs.$ . Hypo in presence of KI, which statement is wrong?

${{\rm{I}}_2}$ with starch gives blue colour

${\rm{CuS}}{{\rm{O}}_4}$ is reduced to white ${\rm{C}}{{\rm{u}}_2}{{\rm{I}}_2}$ during redox change

In the reaction of ${{\rm{O}}_3}$ and ${{\rm{H}}_2}{{\rm{O}}_2},$ the later acts as :

Both oxidising and bleaching agent

REDOX REACTIONS MCQs for NEET UG CHEMISTRY — Medium, Page 2

31 Medium

31 of 69Question permalink

Understanding +4 | -1

In the reaction,

\mathrm{S}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{S} \to 3\mathrm{S} + 2{\mathrm{H}_2}\mathrm{O}

the substance that oxidizes is,

View answer and explanation

Correct option: A

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

Option A (

\mathrm{H}_2\mathrm{S}

) is correct because H2S is oxidized in the reaction, losing electrons and causing the reduction of SO2.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2}\mathrm{S}$ . Hydrogen sulfide is oxidized in the reaction.

Step-by-Step Solution:

Step 1: Identify the reaction: $\mathrm{S}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{S} \to 3\mathrm{S} + 2{\mathrm{H}_2}\mathrm{O}$

Step 2: Determine oxidation states: S in $\mathrm{S}{\mathrm{O}_2}$ goes from +4 to 0 (reduction), and S in ${\mathrm{H}_2}\mathrm{S}$ goes from -2 to 0 (oxidation).

Step 3: Since ${\mathrm{H}_2}\mathrm{S}$ is oxidized, it is the substance that oxidizes.

Why other options are wrong:
- Option B: $\mathrm{S}{\mathrm{O}_2}$ is reduced in the reaction.
- Option C: $\mathrm{S}$ is a product and is neither oxidized nor reduced in the overall sense.
- Option D: ${\mathrm{H}_2}\mathrm{O}$ is a product and is not directly involved in the oxidation-reduction process.

Key Concepts:
- Oxidation
- Redox reactions
- Oxidation states

Memory Aids:
- Oxidation is the loss of electrons.

Quick Tips:
- Identify the substance that increases in oxidation state to determine the substance that is oxidized.

32 Medium

32 of 69Question permalink

Remembering +4 | -1

For monoatomic ions, the oxidation number is equal to

View answer and explanation

Correct option: A

charge on the ion

Option A (charge on the ion) is correct because the oxidation number of a monoatomic ion is defined as its charge.

Solution approach

Correct Answer + Core Reason: charge on the ion. The oxidation number of a monoatomic ion is equal to its charge.

Step-by-Step Solution:

Step 1: Define oxidation number: A number assigned to an element in a chemical combination that represents the number of electrons lost (positive) or gained (negative) by an atom of that element in the compound.

Step 2: Consider monoatomic ions: Ions consisting of a single atom, such as $\mathrm{Na}^+$ , $\mathrm{Cl}^-$ , $\mathrm{Mg}^{2+}$ .

Step 3: Relate the oxidation number to the charge. For monoatomic ions, the oxidation number is simply the charge of the ion.

Why other options are wrong:
- Option B: zero: This is the oxidation number of an element in its elemental form, not for ions.
- Option C: more than 1: This is not a general rule; the oxidation number depends on the charge.
- Option D: less than 1: This is not a general rule; the oxidation number depends on the charge.

Key Concepts:
- Oxidation Number
- Monoatomic Ions

Memory Aids:
- Oxidation number = Charge (for monoatomic ions)

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

33 Medium

33 of 69Question permalink

Application +4 | -1

\mathrm{MnO}_4^ -

is a good oxidising agent in different medium changing to

\mathrm{MnO}_4^ - \to \mathrm{M}{\mathrm{n}^{2 + }}

,

\to \mathrm{MnO}_4^{2 - }, \to \mathrm{Mn}{\mathrm{O}_2}, \to \mathrm{M}{\mathrm{n}_2}{\mathrm{O}_3}

Changes in oxidation number respectively are

View answer and explanation

Correct option: C

5,1,3,4

Option C (5,1,3,4) is correct because the changes in oxidation number of Mn are calculated as 5, 1, 3, and 4 respectively for the given transformations.

Solution approach

Correct Answer + Core Reason: 5,1,3,4. The oxidation number changes are calculated for each transformation of Mn.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Mn in $\mathrm{MnO}_4^ -$ .\nLet x be the oxidation state of Mn.\nx + 4(-2) = -1\nx - 8 = -1\nx = +7

Step 2: Calculate the oxidation state of Mn in each product.
- $\mathrm{M}{\mathrm{n}^{2 + }}$ : +2
- $\mathrm{MnO}_4^{2 - }$ : x + 4(-2) = -2 => x = +6
- $\mathrm{Mn}{\mathrm{O}_2}$ : x + 2(-2) = 0 => x = +4
- $\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_3}$ : 2x + 3(-2) = 0 => 2x = 6 => x = +3

Step 3: Calculate the change in oxidation number for each transformation.
- $\mathrm{MnO}_4^ - \to \mathrm{M}{\mathrm{n}^{2 + }}$ : +7 to +2, change = 5
- $\mathrm{MnO}_4^ - \to \mathrm{MnO}_4^{2 - }$ : +7 to +6, change = 1
- $\mathrm{MnO}_4^ - \to \mathrm{Mn}{\mathrm{O}_2}$ : +7 to +4, change = 3
- $\mathrm{MnO}_4^ - \to \mathrm{M}{\mathrm{n}_2}{\mathrm{O}_3}$ : +7 to +3, change = 4

Why other options are wrong:
- Option A (1,3,4,5): Incorrect order of oxidation number changes.
- Option B (5,4,3,2): Incorrect order of oxidation number changes.
- Option D (2,6,4,3): Incorrect oxidation number changes.

Key Concepts:
- Oxidation state calculation
- Oxidation-reduction reactions

Memory Aids:
- Remember the rules for assigning oxidation states.

Quick Tips:
- Pay attention to the stoichiometry when calculating oxidation states in compounds.

34 Medium

34 of 69Question permalink

Application +4 | -1

The number of mole of oxalate ions oxidized by one mole of

\mathrm{MnO}_4^ -

ion is:

View answer and explanation

Correct option: C

\frac{5}{2}

Option C (

\frac{5}{2}

) is correct because each permanganate ion gains 5 electrons, while each oxalate ion loses 2 electrons, requiring a ratio of 2 permanganate ions to 5 oxalate ions.

Solution approach

Correct Answer + Core Reason: $\frac{5}{2}$ . One mole of $\mathrm{MnO}_4^ -$ oxidizes $\frac{5}{2}$ moles of oxalate ions.

Step-by-Step Solution:

Step 1: Write the half-reactions for the oxidation and reduction processes.

Step 2: Oxidation half-reaction: ${\mathrm{C}_2}\mathrm{O}_4^{2 - } \to 2\mathrm{C}{\mathrm{O}_2} + 2{\mathrm{e}^ - }$

Step 3: Reduction half-reaction: $\mathrm{MnO}_4^ - + 8{\mathrm{H}^ + } + 5{\mathrm{e}^ - } \to \mathrm{M}{\mathrm{n}^{2 + }} + 4{\mathrm{H}_2}\mathrm{O}$

Step 4: Balance the number of electrons in both half-reactions. Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to get 10 electrons.

Step 5: Balanced oxidation half-reaction: $5{\mathrm{C}_2}\mathrm{O}_4^{2 - } \to 10\mathrm{C}{\mathrm{O}_2} + 10{\mathrm{e}^ - }$

Step 6: Balanced reduction half-reaction: $2\mathrm{MnO}_4^ - + 16{\mathrm{H}^ + } + 10{\mathrm{e}^ - } \to 2\mathrm{M}{\mathrm{n}^{2 + }} + 8{\mathrm{H}_2}\mathrm{O}$

Step 7: Now, combine the balanced half-reactions: $2\mathrm{MnO}_4^ - + 16{\mathrm{H}^ + } + 5{\mathrm{C}_2}\mathrm{O}_4^{2 - } \to 2\mathrm{M}{\mathrm{n}^{2 + }} + 10\mathrm{C}{\mathrm{O}_2} + 8{\mathrm{H}_2}\mathrm{O}$

Step 8: From the balanced equation, 2 moles of $\mathrm{MnO}_4^ -$ react with 5 moles of ${\mathrm{C}_2}\mathrm{O}_4^{2 - }$ . Therefore, 1 mole of $\mathrm{MnO}_4^ -$ reacts with $\frac{5}{2}$ moles of ${\mathrm{C}_2}\mathrm{O}_4^{2 - }$ .

Why other options are wrong:
- Option A: $\frac{1}{5}$ is incorrect because it represents the inverse ratio of moles of oxalate to permanganate.
- Option B: $\frac{2}{5}$ is incorrect as it doesn't represent the correct stoichiometric ratio.
- Option D: 5 is incorrect because it represents the moles of oxalate ions that react with 2 moles of permanganate ions.

Key Concepts:
- Redox Reactions
- Half-Reactions
- Balancing Redox Equations

Memory Aids:
- LEO says GER (Lose Electrons Oxidation, Gain Electrons Reduction)

Quick Tips:
- Always balance the half-reactions separately before combining them.
- Make sure the number of electrons lost equals the number of electrons gained.

35 Medium

35 of 69Question permalink

Remembering +4 | -1

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

acts as ……. indicator in its redox titrations.

View answer and explanation

Correct option: A

Self

Option A (Self) is correct because KMnO₄ acts as a self-indicator in redox titrations due to its intense purple color.

Solution approach

Correct Answer + Core Reason: Self. KMnO₄ acts as a self-indicator due to its intense color.

Step-by-Step Solution:

Step 1: Understand the role of an indicator in redox titrations.

Step 2: KMnO₄ has a strong purple color in solution.

Step 3: During a titration, as KMnO₄ is added, it reacts with the reducing agent. Once all the reducing agent is consumed, the excess KMnO₄ imparts a visible pink color to the solution, indicating the endpoint.

Step 4: Since KMnO₄ itself signals the endpoint, it acts as a self-indicator.

Why other options are wrong:
- Option B: External indicators are substances added to the solution that change color at the endpoint, but KMnO₄ doesn't require an external indicator.
- Option C: Internal indicators are substances added to the solution that react with one of the reactants to produce a color change, but KMnO₄ directly signals the endpoint.
- Option D: KMnO₄ definitely acts as an indicator in redox titrations.

Key Concepts:
- Redox titrations
- Indicators
- Self-indicators

Memory Aids:
- Remember that KMnO₄ has a distinct purple color that fades as it's reduced, making it a self-indicator.

Quick Tips:
- Recognize common self-indicators like KMnO₄.

36 Medium

36 of 69Question permalink

Remembering +4 | -1

The algebraic sum of the oxidation number of all the atoms in a compound must be

View answer and explanation

Correct option: C

zero

Option C (zero) is correct because the sum of oxidation numbers in a neutral compound must equal zero.

Solution approach

Correct Answer + Core Reason: zero. The algebraic sum of oxidation numbers in a compound must be zero.

Step-by-Step Solution:

Step 1: State the rule: The sum of the oxidation numbers of all atoms in a neutral compound is zero.

Step 2: Consider examples: In $\mathrm{H}_2\mathrm{O}$ , the oxidation number of H is +1 and O is -2. 2(+1) + (-2) = 0.

Step 3: In $\mathrm{NaCl}$ , the oxidation number of Na is +1 and Cl is -1. +1 + (-1) = 0.

Why other options are wrong:
- Option A: 1: This is incorrect; the sum is zero for neutral compounds.
- Option B: -1: This is incorrect; the sum is zero for neutral compounds.
- Option D: None of these: This is incorrect because the sum is zero.

Key Concepts:
- Oxidation Number
- Neutral Compound

Memory Aids:
- Sum of oxidation numbers = 0 (for neutral compounds)

Quick Tips:
- Remember this rule when assigning oxidation numbers in compounds.

37 Medium

37 of 69Question permalink

Remembering +4 | -1

The stable oxidation states of Mn are∶

View answer and explanation

Correct option: C

+ 2,\mathrm{ } + 7

Option C (

+2, +7

) is correct because these are the most stable and commonly observed oxidation states for Manganese.

Solution approach

Correct Answer + Core Reason: $+ 2,\mathrm{ } + 7$ . These are the most stable oxidation states of Mn.

Step-by-Step Solution:

Step 1: Recall the electronic configuration of Mn: [Ar] 3d⁵ 4s².

Step 2: Mn can exhibit a range of oxidation states due to the involvement of both 4s and 3d electrons in bonding.

Step 3: The +2 oxidation state is stable because it corresponds to the loss of the two 4s electrons, resulting in a half-filled 3d⁵ configuration, which is particularly stable.

Step 4: The +7 oxidation state is the highest possible oxidation state for Mn, achieved when all 4s and 3d electrons are involved in bonding (e.g., in KMnO₄). This is also relatively stable.

Why other options are wrong:
- Option A: While +2 is a stable oxidation state, +3 is less common and less stable than +7.
- Option B: +7 is a stable oxidation state, but +3 is less common compared to +2.
- Option D: +3 is less stable than +2 and +5 is not as stable or common as +7.

Key Concepts:
- Oxidation states of transition metals
- Electronic configuration of Mn
- Stability of half-filled and fully-filled d-orbitals

Memory Aids:
- Remember that Mn can lose all its 4s and 3d electrons to achieve +7.
- The +2 state is stable due to the half-filled d⁵ configuration.

Quick Tips:
- Consider the electronic configuration to determine possible oxidation states.
- Remember that half-filled and fully-filled d-orbitals contribute to stability.

38 Medium

38 of 69Question permalink

Remembering +4 | -1

The maximum oxidation number of transition metals may be:

View answer and explanation

Correct option: C

+ 8

Option C (+8) is correct because some transition metals, like Osmium (Os) and Ruthenium (Ru), can exhibit a maximum oxidation number of +8.

Solution approach

Correct Answer + Core Reason: + 8. Osmium and Ruthenium can exhibit a maximum oxidation number of +8.

Step-by-Step Solution:

Step 1: Consider the oxidation states of transition metals.\nTransition metals exhibit variable oxidation states.

Step 2: Identify transition metals with high oxidation states.\nOsmium (Os) and Ruthenium (Ru) can exhibit a maximum oxidation state of +8 in compounds like $OsO_4$ and $RuO_4$ .

Why other options are wrong:
- Option A: Some transition metals can have oxidation states higher than +4.
- Option B: Some transition metals can have oxidation states higher than +6.
- Option D: The maximum oxidation number is +8, not +10.

Key Concepts:
- Oxidation States of Transition Metals

Memory Aids:
- Remember that Os and Ru can exhibit +8 oxidation state.

Quick Tips:
- Be aware of the different oxidation states that transition metals can exhibit.

39 Medium

39 of 69Question permalink

Analyzing +4 | -1

Which change requires a reducing agent?

View answer and explanation

Correct option: B

\mathrm{BrO}_3^ - \to \mathrm{Br}{\mathrm{O}^ - }

Option B (

\mathrm{BrO}_3^ - \to \mathrm{Br}{\mathrm{O}^ - }

) is correct because the oxidation state of Br decreases, indicating reduction, which requires a reducing agent.

Solution approach

Correct Answer + Core Reason: $\mathrm{BrO}_3^ - \to \mathrm{Br}{\mathrm{O}^ - }$ . Reduction requires a reducing agent.

Step-by-Step Solution:

Step 1: Determine the oxidation state of the central atom in each species.

Step 2: In option A: $\mathrm{CrO}_4^{2 - } \to \mathrm{Cr}_2\mathrm{O}_7^{2 - }$ , Chromium goes from +6 to +6. This is not a redox reaction.

Step 3: In option B: $\mathrm{BrO}_3^ - \to \mathrm{Br}{\mathrm{O}^ - }$ , Bromine goes from +5 to +1. This is reduction, requiring a reducing agent.

Step 4: In option C: ${\mathrm{H}_2}{\mathrm{O}_2} \to {\mathrm{O}_2}$ , Oxygen goes from -1 to 0. This is oxidation, requiring an oxidizing agent.

Step 5: In option D: $\mathrm{Al}{\left( {\mathrm{OH}} \right)_3} \to \mathrm{Al}{\left( {\mathrm{OH}} \right)_4^ - }$ , Aluminum remains at +3. This is not a redox reaction.

Why other options are wrong:
- Option A: $\mathrm{CrO}_4^{2 - } \to \mathrm{Cr}_2\mathrm{O}_7^{2 - }$ : No change in oxidation state, so no reducing agent is required.
- Option C: ${\mathrm{H}_2}{\mathrm{O}_2} \to {\mathrm{O}_2}$ : This is oxidation, requiring an oxidizing agent, not a reducing agent.
- Option D: $\mathrm{Al}{\left( {\mathrm{OH}} \right)_3} \to \mathrm{Al}{\left( {\mathrm{OH}} \right)_4^ - }$ : No change in oxidation state, so no reducing agent is required.

Key Concepts:
- Redox Reactions
- Oxidation States
- Reducing 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.
- Reduction requires a reducing agent.

40 Medium

40 of 69Question permalink

Understanding +4 | -1

Volumetric estimation of

\mathrm{CuS}{\mathrm{O}_4}

using hypo as intermediate solution along with KI solution and starch as indicator is an example of :

View answer and explanation

Correct option: A

Redox titration

Option A (Redox titration) is correct because the reaction involves both oxidation and reduction processes during the volumetric estimation of CuSO4 using hypo.

Solution approach

Correct Answer + Core Reason: Redox titration. The reaction involves both oxidation and reduction processes.

Step-by-Step Solution:

Step 1: Identify the reaction type. The reaction involves the transfer of electrons between the reactants.

Step 2: Define Redox Titration. Redox titration is a type of titration based on a redox reaction between the analyte and the titrant.

Step 3: Analyze the given reaction. The reaction between CuSO4, KI, and hypo involves the oxidation of I- to I2 and the subsequent reduction of I2 by hypo (Na2S2O3).

Step 4: Conclude that it is a redox titration.

Why other options are wrong:
- Option B: Acid-base titration involves the neutralization of an acid by a base, which is not the case here.
- Option C: Precipitation titration involves the formation of a precipitate, which is not the primary reaction occurring here.
- Option D: The reaction clearly falls under one of the given categories.

Key Concepts:
- Redox Titration
- Oxidation and Reduction

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

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

41 Medium

41 of 69Question permalink

Application +4 | -1

Moles of

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

required for decolorizing 1 mole of acidified

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

are :

View answer and explanation

Correct option: C

\frac{5}{2}

Option C (

\frac{5}{2}

) is correct because the balanced redox reaction requires 5 moles of H2O2 for every 2 moles of KMnO4, making the ratio 5/2.

Solution approach

Correct Answer + Core Reason: $\frac{5}{2}$ . The moles of ${\mathrm{H}_2}{\mathrm{O}_2}$ required for decolorizing 1 mole of acidified $\mathrm{KMn}{\mathrm{O}_4}$ is $\frac{5}{2}$ .

Step-by-Step Solution:

Step 1: Write the unbalanced redox reaction: $\mathrm{KMn}{\mathrm{O}_4} + {\mathrm{H}_2}{\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to {\mathrm{K}_2}\mathrm{S}{\mathrm{O}_4} + \mathrm{MnS}{\mathrm{O}_4} + {\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{O}$

Step 2: Identify the oxidation half-reaction: ${\mathrm{H}_2}{\mathrm{O}_2} \to {\mathrm{O}_2}$

Step 3: Balance the oxidation half-reaction: ${\mathrm{H}_2}{\mathrm{O}_2} \to {\mathrm{O}_2} + 2{\mathrm{H}^ + } + 2{\mathrm{e}^ - }$

Step 4: Identify the reduction half-reaction: $\mathrm{MnO}_4^- \to \mathrm{Mn}^{2+}$

Step 5: Balance the reduction half-reaction: $\mathrm{MnO}_4^- + 8{\mathrm{H}^ + } + 5{\mathrm{e}^ - } \to \mathrm{Mn}^{2+} + 4{\mathrm{H}_2}\mathrm{O}$

Step 6: Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to balance the electrons:
$5{\mathrm{H}_2}{\mathrm{O}_2} \to 5{\mathrm{O}_2} + 10{\mathrm{H}^ + } + 10{\mathrm{e}^ - }$
$2\mathrm{MnO}_4^- + 16{\mathrm{H}^ + } + 10{\mathrm{e}^ - } \to 2\mathrm{Mn}^{2+} + 8{\mathrm{H}_2}\mathrm{O}$

Step 7: Combine the balanced half-reactions:
$2\mathrm{MnO}_4^- + 5{\mathrm{H}_2}{\mathrm{O}_2} + 6{\mathrm{H}^ + } \to 2\mathrm{Mn}^{2+} + 5{\mathrm{O}_2} + 8{\mathrm{H}_2}\mathrm{O}$

Step 8: From the balanced equation, 2 moles of $\mathrm{KMn}{\mathrm{O}_4}$ react with 5 moles of ${\mathrm{H}_2}{\mathrm{O}_2}$ . Therefore, the mole ratio of ${\mathrm{H}_2}{\mathrm{O}_2}$ to $\mathrm{KMn}{\mathrm{O}_4}$ is $\frac{5}{2}$ .

Why other options are wrong:
- Option A: $\frac{1}{2}$ is incorrect; it does not represent the correct stoichiometric ratio.
- Option B: $\frac{3}{2}$ is incorrect; it does not represent the correct stoichiometric ratio.
- Option D: $\frac{7}{2}$ is incorrect; it does not represent the correct stoichiometric ratio.

Key Concepts:
- Redox Reactions
- Balancing Redox Equations
- Half-Reactions

Memory Aids:
- LEO says GER (Lose Electrons Oxidation, Gain Electrons Reduction)

Quick Tips:
- Balance the half-reactions separately before combining them.
- Make sure the number of atoms and charges are balanced on both sides of the equation.

42 Medium

42 of 69Question permalink

Understanding +4 | -1

The oxidation number of Ba in barium peroxide is :

View answer and explanation

Correct option: A

+ 2

Option A (+2) is correct because barium is an alkaline earth metal and always exhibits a +2 oxidation state in its compounds.

Solution approach

Correct Answer + Core Reason: $+ 2$ . Barium is an alkaline earth metal and has a fixed oxidation state of +2.

Step-by-Step Solution:

Step 1: Recall the properties of barium.

Step 2: Barium (Ba) belongs to Group 2 (alkaline earth metals).

Step 3: Alkaline earth metals have two valence electrons and tend to lose both to form +2 ions.

Step 4: Therefore, the oxidation state of Ba in any compound is +2.

Why other options are wrong:
- Option B: $- 1$ : This is incorrect. Oxygen can have -1 oxidation state in peroxides, but Ba is always +2.
- Option C: $+ 4$ : This is incorrect. Barium does not exhibit +4 oxidation state.
- Option D: $+ 6$ : This is incorrect. Barium does not exhibit +6 oxidation state.

Key Concepts:
- Oxidation state
- Alkaline earth metals
- Properties of barium

Memory Aids:
- Group 2 elements always have a +2 oxidation state.

Quick Tips:
- Remember that alkaline earth metals (Group 2) have a +2 oxidation state.

43 Medium

43 of 69Question permalink

Analyzing +4 | -1

In which reaction the underlined substance has been reduced?

View answer and explanation

Correct option: C

Steam + \mathrm{ }iron\mathrm{ } \to \mathrm{ }hydrogen\mathrm{ } + \mathrm{ }iron\mathrm{ }oxide

Option C (

Steam + \mathrm{ }iron\mathrm{ } \to \mathrm{ }hydrogen\mathrm{ } + \mathrm{ }iron\mathrm{ }oxide

) is correct because the oxidation state of hydrogen in steam is reduced from +1 to 0 in hydrogen gas.

Solution approach

Correct Answer + Core Reason: $Steam + \mathrm{ }iron\mathrm{ } \to \mathrm{ }hydrogen\mathrm{ } + \mathrm{ }iron\mathrm{ }oxide$ . Hydrogen in steam is reduced.

Step-by-Step Solution:

Step 1: Write the chemical equations for each reaction.

Step 2: Determine the oxidation states of the underlined substances before and after the reaction.

Step 3: In option A: $Carbon\mathrm{ }monoxide + \mathrm{ }copper\mathrm{ }oxide\mathrm{ } \to carbon\mathrm{ }dioxide\mathrm{ } + \mathrm{ }copper$ , Copper oxide (CuO) is reduced to Copper (Cu).

Step 4: In option B: $Copper\mathrm{ }oxide + \mathrm{ }hydrochloric\mathrm{ }acid\mathrm{ } \to \mathrm{ }water\mathrm{ } + \mathrm{ }copper\mathrm{ }chloride$ , This is not a redox reaction.

Step 5: In option C: $Steam + \mathrm{ }iron\mathrm{ } \to \mathrm{ }hydrogen\mathrm{ } + \mathrm{ }iron\mathrm{ }oxide$ , Steam ( $\mathrm{H}_2\mathrm{O}$ ) is reduced to Hydrogen ( $\mathrm{H}_2$ ). Iron is oxidized.

Step 6: In option D: $Hydrogen + \mathrm{ }iron\mathrm{ }oxide\mathrm{ } \to water\mathrm{ } + \mathrm{ }iron$ , Iron oxide is reduced to Iron.

Step 7: Identify the reaction where the underlined substance is reduced.
- Option C: Steam is reduced to hydrogen.

Why other options are wrong:
- Option A: $Carbon\mathrm{ }monoxide + \mathrm{ }copper\mathrm{ }oxide\mathrm{ } \to carbon\mathrm{ }dioxide\mathrm{ } + \mathrm{ }copper$ : Copper oxide is reduced, not carbon monoxide.
- Option B: $Copper\mathrm{ }oxide + \mathrm{ }hydrochloric\mathrm{ }acid\mathrm{ } \to \mathrm{ }water\mathrm{ } + \mathrm{ }copper\mathrm{ }chloride$ : This is not a redox reaction.
- Option D: $Hydrogen + \mathrm{ }iron\mathrm{ }oxide\mathrm{ } \to water\mathrm{ } + \mathrm{ }iron$ : Iron oxide is reduced, not hydrogen.

Key Concepts:
- Redox Reactions
- Oxidation States

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.

44 Medium

44 of 69Question permalink

Understanding +4 | -1

Which can have both +ve and –ve oxidation states?

View answer and explanation

Correct option: B

\mathrm{I}

Option B (I) is correct because iodine can exhibit both positive and negative oxidation states due to its electronic configuration and ability to form compounds with more and less electronegative elements.

Solution approach

Correct Answer + Core Reason: $\mathrm{I}$ . Iodine can exhibit both positive and negative oxidation states.

Step-by-Step Solution:

Step 1: Recall the properties of the given elements.

Step 2: Fluorine (F) is the most electronegative element and only exhibits negative oxidation states.

Step 3: Sodium (Na) is an alkali metal and only exhibits positive oxidation states (+1).

Step 4: Helium (He) is a noble gas and has a zero oxidation state.

Step 5: Iodine (I) can exhibit both positive and negative oxidation states depending on the element it is bonded to.

Why other options are wrong:
- Option A: $\mathrm{F}$ : Fluorine only exhibits negative oxidation states.
- Option C: $\mathrm{Na}$ : Sodium only exhibits positive oxidation states.
- Option D: $\mathrm{He}$ : Helium exhibits zero oxidation state.

Key Concepts:
- Oxidation state
- Electronegativity
- Properties of halogens, alkali metals, and noble gases

Memory Aids:
- Fluorine is the most electronegative element.
- Alkali metals always have a +1 oxidation state.

Quick Tips:
- Remember that halogens other than fluorine can exhibit positive oxidation states.

45 Medium

45 of 69Question permalink

Analyzing +4 | -1

In the titration of

\mathrm{CuS}{\mathrm{O}_4}vs.

. Hypo in presence of KI, which statement is wrong?

View answer and explanation

Correct option: D

The solution before titration, on addition of KI appears blue

Option D (The solution before titration, on addition of KI appears blue) is wrong because the solution turns blue only after the liberated iodine reacts with starch, not before.

Solution approach

Correct Answer + Core Reason: The solution before titration, on addition of KI appears blue. The blue color appears only after iodine is liberated and reacts with starch.

Step-by-Step Solution:

Step 1: Understand the titration process.\nIn the titration of $CuSO_4$ with hypo (sodium thiosulfate) in the presence of KI, the following reactions occur:
$2Cu^{2+} + 4I^- \to 2CuI + I_2$
$I_2 + 2S_2O_3^{2-} \to S_4O_6^{2-} + 2I^-$

Step 2: Analyze each statement.
- Option A: It is iodometric titration. This is correct because the iodine liberated in the reaction is titrated against a standard solution of sodium thiosulfate (hypo).
- Option B: $I_2$ with starch gives blue color. This is correct. Iodine forms a complex with starch, resulting in a blue color.
- Option C: $CuSO_4$ is reduced to white $Cu_2I_2$ during redox change. This is correct. $Cu^{2+}$ is reduced to $Cu^+$ which forms $Cu_2I_2$ , a white precipitate.
- Option D: The solution before titration, on addition of KI appears blue. This is incorrect. The solution will only turn blue when starch is added *after* the liberation of iodine. Before the addition of starch, the solution will contain $Cu^{2+}$ , $I^-$ , and then $CuI$ and $I_2$ after the reaction with $Cu^{2+}$ . The iodine ( $I_2$ ) needs starch to form the blue complex.

Step 3: Conclusion.\nOption D is the incorrect statement.

Why other options are wrong:
- Option A: The titration is indeed iodometric.
- Option B: Iodine and starch do form a blue complex.
- Option C: Cupric ions are reduced to cuprous iodide, which is a white precipitate.

Key Concepts:
- Iodometric titration
- Redox reactions
- Starch-iodine complex

Memory Aids:
- Remember that starch is used as an indicator for iodine.

Quick Tips:
- In iodometric titrations, iodine is liberated first, and then titrated against a reducing agent.

46 Medium

46 of 69Question permalink

Application +4 | -1

Equivalent mass of

\mathrm{IO}_4^ -

when it is converted to

{\mathrm{I}_2}

in acid medium :

View answer and explanation

Correct option: B

M/7

Option B (M/7) is correct because the change in oxidation state of iodine is from +7 to 0, resulting in a change of 7, hence the equivalent mass is M/7.

Solution approach

Correct Answer + Core Reason: $M/7$ . The change in oxidation state of I is 7, so the equivalent mass is M/7.

Step-by-Step Solution:

Step 1: Determine the oxidation state of I in $\mathrm{IO}_4^ -$ .\nLet x be the oxidation state of I.\nx + 4(-2) = -1\nx - 8 = -1\nx = +7

Step 2: Determine the oxidation state of I in ${\mathrm{I}_2}$ .\nThe oxidation state of I in ${\mathrm{I}_2}$ is 0.

Step 3: Calculate the change in oxidation state.\nThe change in oxidation state is +7 - 0 = 7.\nSince ${\mathrm{I}_2}$ contains two iodine atoms, the total change per molecule of ${\mathrm{I}_2}$ formed from $2\mathrm{IO}_4^ -$ is 2 * 7 = 14. However, we are considering the equivalent mass of $\mathrm{IO}_4^ -$ , so we consider the change per $\mathrm{IO}_4^ -$ which is 7.

Step 4: Calculate the equivalent mass.\nEquivalent mass = Molar mass / change in oxidation state = M/7

Why other options are wrong:
- **Option A ( $M/6$ ): Incorrect change in oxidation state.
- Option C ( $M/5$ ): Incorrect change in oxidation state.
- Option D ( $M/4$ ): Incorrect change in oxidation state.

Key Concepts:
- Oxidation state calculation
- Equivalent mass
- Redox reactions

Memory Aids:
- Equivalent mass = Molar mass / n-factor

Quick Tips:**
- Determine the change in oxidation state correctly to calculate the equivalent mass.

47 Medium

47 of 69Question permalink

Application +4 | -1

A solution of

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

is reduced to

\mathrm{Mn}{\mathrm{O}_2}.

The normality of solution is 0.6. The molarity is:

View answer and explanation

Correct option: D

0.2M

Option D (0.2M) is correct because the normality is related to molarity by the n-factor, which is 3 in this case, so Molarity = Normality / n-factor.

Solution approach

Correct Answer + Core Reason: 0.2M. The molarity is calculated by dividing the normality by the change in oxidation number.

Step-by-Step Solution:

Step 1: Identify the reduction reaction: $\mathrm{KMn}{\mathrm{O}_4} \to \mathrm{Mn}{\mathrm{O}_2}$

Step 2: Determine the oxidation state of Mn in $\mathrm{KMn}{\mathrm{O}_4}$ . The oxidation state of Mn is +7.

Step 3: Determine the oxidation state of Mn in $\mathrm{Mn}{\mathrm{O}_2}$ . The oxidation state of Mn is +4.

Step 4: Calculate the change in oxidation state: |+7 - (+4)| = 3.

Step 5: Use the relationship between normality and molarity: Normality = Molarity * n-factor, where n-factor is the change in oxidation state.

Step 6: Given Normality = 0.6, solve for Molarity: Molarity = Normality / n-factor = 0.6 / 3 = 0.2 M.

Why other options are wrong:
- Option A: 1.8 M is incorrect; it's a result of multiplying the normality by 3 instead of dividing.
- Option B: 0.6M is incorrect; it assumes the n-factor is 1, which is not the case.
- Option C: 0.1M is incorrect and doesn't relate to the correct calculation.

Key Concepts:
- Normality
- Molarity
- Oxidation State

Memory Aids:
- Normality = Molarity * n-factor

Quick Tips:
- Determine the change in oxidation state of the element that is being reduced or oxidized.
- Remember the relationship between normality and molarity.

48 Medium

48 of 69Question permalink

Analyzing +4 | -1

In the reactions;

\mathrm{A}{\mathrm{s}_2}{\mathrm{S}_3} + \mathrm{HN}{\mathrm{O}_3} \to {\mathrm{H}_3}\mathrm{As}{\mathrm{O}_4} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} + \mathrm{NO}

, the element oxidized is/ are :

View answer and explanation

Correct option: D

As and S both

Option D (As and S both) is correct because in the reaction, both Arsenic and Sulfur undergo oxidation.

Solution approach

Correct Answer + Core Reason: As and S both. Both Arsenic and Sulfur are oxidized.

Step-by-Step Solution:

Step 1: Write the balanced (or unbalanced) reaction: $\mathrm{A}{\mathrm{s}_2}{\mathrm{S}_3} + \mathrm{HN}{\mathrm{O}_3} \to {\mathrm{H}_3}\mathrm{As}{\mathrm{O}_4} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} + \mathrm{NO}$

Step 2: Determine the oxidation states of each element.
- In $\mathrm{A}{\mathrm{s}_2}{\mathrm{S}_3}$ , As = +3, S = -2
- In $\mathrm{HN}{\mathrm{O}_3}$ , N = +5
- In ${\mathrm{H}_3}\mathrm{As}{\mathrm{O}_4}$ , As = +5
- In ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ , S = +6
- In $\mathrm{NO}$ , N = +2

Step 3: Identify the elements that have changed oxidation states.
- Arsenic: +3 to +5 (oxidation)
- Sulfur: -2 to +6 (oxidation)
- Nitrogen: +5 to +2 (reduction)

Step 4: Determine which elements are oxidized.
- Arsenic and Sulfur are oxidized.

Why other options are wrong:
- Option A: As only: Sulfur is also oxidized.
- Option B: S only: Arsenic is also oxidized.
- Option C: N only: Nitrogen is reduced, not oxidized.

Key Concepts:
- Redox Reactions
- Oxidation States

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.

49 Medium

49 of 69Question permalink

Application +4 | -1

The eq. wt. of

\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}

in ,

\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4} + \mathrm{KMn}{\mathrm{O}_4} \to \mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} + \mathrm{Mn}{\mathrm{O}_2}

is:

View answer and explanation

Correct option: B

M

Option B (M) is correct because the n-factor for

\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}

in the given reaction is 1, hence the equivalent weight is equal to the molecular weight.

Solution approach

Correct Answer + Core Reason: $M$ . The n-factor for $\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}$ is 1, so the equivalent weight is M.

Step-by-Step Solution:

Step 1: Determine the oxidation states of Fe in $\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}$ and $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}$ .\nIn $\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}$ , the average oxidation state of Fe is +8/3. It can be considered as FeO.Fe2O3, so Fe has +2 and +3 oxidation states.\nIn $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}$ , the oxidation state of Fe is +3.

Step 2: Determine the oxidation state of Mn in $\mathrm{KMn}{\mathrm{O}_4}$ and $\mathrm{Mn}{\mathrm{O}_2}$ .\nIn $\mathrm{KMn}{\mathrm{O}_4}$ , the oxidation state of Mn is +7.\nIn $\mathrm{Mn}{\mathrm{O}_2}$ , the oxidation state of Mn is +4.

Step 3: Balance the equation to find the n-factor for $\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}$ .
$\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4} + \mathrm{KMn}{\mathrm{O}_4} \to \mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} + \mathrm{Mn}{\mathrm{O}_2}$
$3\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4} + \mathrm{KMn}{\mathrm{O}_4} \to 5\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} + \mathrm{Mn}{\mathrm{O}_2} + K_2O$
$3\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4} + \mathrm{KMn}{\mathrm{O}_4} \to 4.5\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} + \mathrm{Mn}{\mathrm{O}_2} + 0.5 K_2O$ \nChange in oxidation state of Fe per $\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}$ = 9(+3) - (2(+3) + 1(+2)) = 27 - 8 = 1\nChange in oxidation state of Mn = 7 - 4 = 3\nSince the change in oxidation state of Fe is 1 per formula unit of $\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}$ , the n-factor is 1.

Step 4: Calculate the equivalent weight.\nEquivalent weight = Molar mass / n-factor = M/1 = M

Why other options are wrong:
- **Option A ( $M/6$ ): Incorrect n-factor.
- Option C ( $2M$ ): Incorrect n-factor.
- Option D ( $M/3$ ): Incorrect n-factor.

Key Concepts:
- Oxidation state calculation
- Equivalent weight
- Redox reactions

Memory Aids:
- Equivalent weight = Molar mass / n-factor

Quick Tips:**
- Balance the equation correctly to find the n-factor.

50 Medium

50 of 69Question permalink

Analyzing +4 | -1

Which of the following reactions are disproportionation reaction?

\mathrm{(a})\,2\mathrm{C}{\mathrm{u}^ + }\, \to \,\mathrm{C}{\mathrm{u}^{2 + }}\, + \,\mathrm{C}{\mathrm{u}^0}\,

\mathrm{(b})\,\mathrm{3MnO}_\mathrm{4}^{\mathrm{2} - }\mathrm{ + 4}{\mathrm{H}^\mathrm{ + }}\, \to \,\mathrm{2MnO}_\mathrm{4}^ - \,\mathrm{ + }\,\mathrm{Mn}{\mathrm{O}_\mathrm{2}}\,\mathrm{ + }\,\mathrm{2}{\mathrm{H}_\mathrm{2}}\mathrm{O}

\mathrm{(c)}\,\mathrm{2KMn}{\mathrm{O}_\mathrm{4}}\xrightarrow{\Delta }\,{\mathrm{K}_\mathrm{2}}\mathrm{Mn}{\mathrm{O}_\mathrm{4}}\mathrm{ + Mn}{\mathrm{O}_\mathrm{2}}\,\,\mathrm{ + }\,{\mathrm{O}_\mathrm{2}}

\mathrm{(d)}\,\mathrm{2MnO}_\mathrm{4}^ - \, + \,\mathrm{3M}{\mathrm{n}^{2 + }}\, + \,\mathrm{2}{\mathrm{H}_\mathrm{2}}\mathrm{O}\, \to \,5\mathrm{Mn}{\mathrm{O}_2}\mathrm{ + 4}{\mathrm{H}^ \oplus }

Select the correct option from the following

View answer and explanation

Correct option: A

(a) and (b) only

Option A is correct because disproportionation involves a single element being simultaneously oxidized and reduced, which occurs in reactions (a) and (b).

Solution approach

Correct Answer + Core Reason: (a) and (b) only. Disproportionation reactions involve a single element undergoing both oxidation and reduction.

Step-by-Step Solution:

Step 1: Define disproportionation reaction. A disproportionation reaction is a redox reaction in which a single element is simultaneously oxidized and reduced.

Step 2: Analyze reaction (a): $2Cu^+ \to Cu^{2+} + Cu^0$ . Copper's oxidation state changes from +1 to +2 (oxidation) and +1 to 0 (reduction). This is a disproportionation reaction.

Step 3: Analyze reaction (b): $3MnO_4^{2-} + 4H^+ \to 2MnO_4^- + MnO_2 + 2H_2O$ . Manganese's oxidation state changes from +6 in $MnO_4^{2-}$ to +7 in $MnO_4^-$ (oxidation) and +4 in $MnO_2$ (reduction). This is a disproportionation reaction.

Step 4: Analyze reaction (c): $2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2$ . Manganese goes from +7 in $KMnO_4$ to +6 in $K_2MnO_4$ and +4 in $MnO_2$ . Oxygen goes from -2 in $KMnO_4$ to 0 in $O_2$ . This is NOT a disproportionation reaction, as two different elements are changing oxidation states.

Step 5: Analyze reaction (d): $2MnO_4^- + 3Mn^{2+} + 2H_2O \to 5MnO_2 + 4H^+$ . Manganese goes from +7 in $MnO_4^-$ to +4 in $MnO_2$ , and from +2 in $Mn^{2+}$ to +4 in $MnO_2$ . This is a comproportionation reaction (reverse of disproportionation), not a disproportionation reaction.

Step 6: Identify the correct option based on the analysis. Only reactions (a) and (b) are disproportionation reactions.

Why other options are wrong:
- Option B: Includes reaction (c), which is not a disproportionation reaction.
- Option C: Includes reactions (c) and (d), neither of which are disproportionation reactions.
- Option D: Includes reaction (d), which is not a disproportionation reaction.

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

Memory Aids:
- Disproportionation: One element splits into two different oxidation states.

Quick Tips:
- Calculate the oxidation states of the elements involved in the reaction.
- Check if the same element is being both oxidized and reduced.

51 Medium

51 of 69Question permalink

Understanding +4 | -1

Respiration is :

View answer and explanation

Correct option: C

Both (a) and (b)

Option C (Both (a) and (b)) is correct because respiration involves both oxidation (breakdown of glucose) and reduction (electron transport chain) processes.

Solution approach

Correct Answer + Core Reason: Both (a) and (b). Respiration involves both oxidation and reduction.

Step-by-Step Solution:

Step 1: Define Respiration. Respiration is the process by which organisms convert glucose into energy.

Step 2: Identify the Oxidation Process. Glucose is oxidized to carbon dioxide and water.

Step 3: Identify the Reduction Process. Oxygen is reduced to water in the electron transport chain.

Step 4: Conclude that respiration involves both oxidation and reduction.

Why other options are wrong:
- Option A: While oxidation is a part of respiration, it's not the complete picture.
- Option B: While reduction is a part of respiration, it's not the complete picture.
- Option D: Respiration clearly involves either oxidation or reduction.

Key Concepts:
- Respiration
- Oxidation
- Reduction

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

Quick Tips:
- Remember that respiration is a complex process involving multiple redox reactions.

52 Medium

52 of 69Question permalink

Understanding +4 | -1

In the conversion of

\mathrm{B}{\mathrm{r}_2}

to

\mathrm{BrO}_3^ -

, the oxidation number of Br changes from

View answer and explanation

Correct option: A

Zero to + 5

Option A (Zero to + 5) is correct because in Br2, the oxidation state of Br is 0, and in BrO3-, the oxidation state of Br is +5.

Solution approach

Correct Answer + Core Reason: Zero to + 5. The oxidation state changes from 0 in Br2 to +5 in BrO3-.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Br in Br2. Since Br2 is an element in its standard state, its oxidation state is 0.

Step 2: Determine the oxidation state of Br in BrO3-. Let the oxidation state of Br be x. The oxidation state of O is -2. Therefore, x + 3(-2) = -1, which gives x = +5.

Step 3: Calculate the change in oxidation state. The oxidation state changes from 0 to +5.

Why other options are wrong:
- Option B: The initial oxidation state is not +1.
- Option C: The final oxidation state is not -3.
- Option D: The initial oxidation state is not +2.

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

Memory Aids:
- Elements in their standard state have an oxidation state of 0.

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

53 Medium

53 of 69Question permalink

Remembering +4 | -1

Starch iodide paper is used to test for the presence of:

View answer and explanation

Correct option: A

Iodine

Option A (Iodine) is correct because starch iodide paper turns blue in the presence of iodine.

Solution approach

Correct Answer + Core Reason: Iodine. Starch iodide paper turns blue in the presence of iodine.

Step-by-Step Solution:

Step 1: Describe starch iodide paper: Paper impregnated with starch and potassium iodide (KI).

Step 2: Explain the reaction: In the presence of an oxidizing agent, iodide ions ( $\mathrm{I}^−$ ) are oxidized to iodine ( ${\mathrm{I}_2}$ ).

Step 3: Explain the color change: The iodine formed reacts with starch to produce a blue-black color.

Step 4: Relate the color change to the presence of iodine.

Why other options are wrong:
- Option B: Iodine ion: Iodine ions themselves do not cause the color change. They need to be oxidized to iodine first.
- Option C: Reductant: A reductant would reduce iodine, preventing the formation of the blue complex.
- Option D: None of these: This is incorrect as starch iodide paper is used to test for iodine.

Key Concepts:
- Starch-Iodine Reaction
- Redox Reactions

Memory Aids:
- Starch + Iodine = Blue Color

Quick Tips:
- Starch iodide paper is a common test for oxidizing agents.

54 Medium

54 of 69Question permalink

Understanding +4 | -1

The oxoacid which acts both as oxidising and reducing agent is :

View answer and explanation

Correct option: C

\mathrm{HN}{\mathrm{O}_2}

Option C (

\mathrm{HN}{\mathrm{O}_2}

) is correct because nitrous acid can act as both an oxidizing agent (being reduced to NO) and a reducing agent (being oxidized to HNO₃).

Solution approach

Correct Answer + Core Reason: $\mathrm{HN}{\mathrm{O}_2}$ . It can act as both oxidizing and reducing agent.

Step-by-Step Solution:

Step 1: Understand the meaning of oxidizing and reducing agents.

Step 2: An oxidizing agent accepts electrons and gets reduced.

Step 3: A reducing agent donates electrons and gets oxidized.

Step 4: Consider the oxidation state of nitrogen in each oxoacid.

Step 5: In H₂SO₄, the oxidation state of S is +6 (maximum), so it can only act as an oxidizing agent.

Step 6: In H₃PO₄, the oxidation state of P is +5 (maximum), so it can only act as an oxidizing agent.

Step 7: In HNO₂, the oxidation state of N is +3, which is intermediate. It can be oxidized to +5 (HNO₃) or reduced to +2 (NO).

Step 8: In HClO₄, the oxidation state of Cl is +7 (maximum), so it can only act as an oxidizing agent.

Why other options are wrong:
- Option A: H₂SO₄ can only act as an oxidizing agent because sulfur is already in its highest oxidation state (+6).
- Option B: H₃PO₄ can only act as an oxidizing agent because phosphorus is already in its highest oxidation state (+5).
- Option D: HClO₄ can only act as an oxidizing agent because chlorine is already in its highest oxidation state (+7).

Key Concepts:
- Oxidizing agent
- Reducing agent
- Oxidation state

Memory Aids:
- Remember that an element in an intermediate oxidation state can act as both an oxidizing and reducing agent.

Quick Tips:
- Determine the oxidation state of the central atom in the oxoacid to predict its behavior.

55 Medium

55 of 69Question permalink

Understanding +4 | -1

Given below are two statements.
One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) Aluminium is extracted from bauxite by the electrolysis of molten mixture of

A{l_2}{O_3}

with cryolite.
Reason (R) The oxidation state of

Al

in cryolite is +3.
In the light of the above statements, choose the most appropriate answer from the options given below.

View answer and explanation

Correct option: D

Both (A) and (R) are true but (R) is not the correct explanation of (A)

Option D is correct because both the assertion and the reason are true, but the reason does not explain why cryolite is used in the extraction of aluminum.

Solution approach

Correct Answer + Core Reason: Both (A) and (R) are true but (R) is not the correct explanation of (A). The electrolysis of $Al_2O_3$ with cryolite is true, and the oxidation state of Al in cryolite is +3, but the latter doesn't explain the former.

Step-by-Step Solution:

Step 1: Analyze the assertion.

Step 2: Aluminium is extracted from bauxite by the electrolysis of a molten mixture of $Al_2O_3$ with cryolite. This statement is true.

Step 3: Analyze the reason.

Step 4: The oxidation state of Al in cryolite is +3. This statement is also true.

Step 5: Determine if the reason explains the assertion.

Step 6: Cryolite is used to lower the melting point of $Al_2O_3$ and increase its conductivity. The oxidation state of Al in cryolite doesn't explain why cryolite is used.

Why other options are wrong:
- Option A: (A) is true but (R) is false: Both A and R are true.
- Option B: (A) is false but (R) is true: Both A and R are true.
- Option C: Both (A) and (R) are true and (R) is the correct explanation of (A): R is not the correct explanation of A.

Key Concepts:
- Extraction of aluminium
- Electrolysis
- Cryolite
- Oxidation state

Memory Aids:
- Cryolite lowers the melting point of alumina.

Quick Tips:
- Remember the role of cryolite in the extraction of aluminium.

56 Medium

56 of 69Question permalink

Analyzing +4 | -1

The most common oxidation state of Lanthanoid elements is +3. Which of the following is likely to deviate easily from +3 oxidation state?

View answer and explanation

Correct option: A

\mathrm{Ce}\left( {\mathrm{At}\mathrm{.}\,\mathrm{No}\mathrm{.}\,\mathrm{58}} \right)

Option A, Ce (Atomic No. 58), is the most likely to deviate easily from the +3 oxidation state because it can also exhibit +4 and +2 oxidation states due to its electronic configuration.

Solution approach

Correct Answer + Core Reason: $\mathrm{Ce}\left( {\mathrm{At}\mathrm{.}\,\mathrm{No}\mathrm{.}\,\mathrm{58}} \right)$ . Cerium can exhibit +4 and +2 oxidation states in addition to the common +3 state.

Step-by-Step Solution:

Step 1: Identify the common oxidation state of lanthanoids, which is +3.

Step 2: Consider the electronic configurations of the given lanthanoids and their ability to achieve stable configurations by losing different numbers of electrons.

Step 3: Cerium (Ce) can achieve a stable electronic configuration by losing 4 electrons to form Ce $^{4+}$ , or by losing 2 electrons to form Ce $^{2+}$ , in addition to the common Ce $^{3+}$ state. This makes it more likely to deviate from the +3 oxidation state.

Step 4: La, Lu, and Gd primarily exhibit the +3 oxidation state due to their electronic configurations and stability considerations.

Why other options are wrong:
- Option B: $\mathrm{La}\left( {\mathrm{At}\mathrm{.}\,\mathrm{No}\mathrm{.}\,\mathrm{57}} \right)$ : Lanthanum primarily exhibits a +3 oxidation state.
- Option C: $\mathrm{Lu}\left( {\mathrm{At}\mathrm{.}\,\mathrm{No}\mathrm{.}\,\mathrm{71}} \right)$ : Lutetium primarily exhibits a +3 oxidation state.
- Option D: $\mathrm{Gd}\left( {\mathrm{At}\mathrm{.}\,\mathrm{No}\mathrm{.}\,64} \right)$ : Gadolinium primarily exhibits a +3 oxidation state, although it can show +2 in some compounds, it is less common than Ce showing +4.

Key Concepts:
- Lanthanoid oxidation states
- Electronic configuration and stability

Memory Aids:
- Remember that Cerium (Ce) is known for its variable oxidation states, including +3 and +4.

Quick Tips:
- Consider the electronic configurations and the stability of different oxidation states when determining which lanthanoid is most likely to deviate from the +3 oxidation state.

57 Medium

57 of 69Question permalink

Application +4 | -1

When sulphur dioxide is passed in an acidified

{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}

solution, the oxidation state of sulphur is changed from

View answer and explanation

Correct option: C

4 to 6

Option C (4 to 6) is correct because sulphur dioxide (

\mathrm{SO}_2

) is oxidized to sulphate (

\mathrm{SO}_4^{2-}

), changing the oxidation state of sulphur from +4 to +6.

Solution approach

Correct Answer + Core Reason: 4 to 6. Sulphur dioxide is oxidized to sulphate, increasing the oxidation state of sulphur.

Step-by-Step Solution:

Step 1: Determine the initial oxidation state of sulphur in $\mathrm{SO}_2$ .\nLet x be the oxidation state of sulphur.\nx + 2(-2) = 0\nx - 4 = 0\nx = +4

Step 2: Determine the final oxidation state of sulphur in sulphate ( $\mathrm{SO}_4^{2-}$ ), which is formed after the reaction with acidified ${\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}$ . The product is $\mathrm{H}_2\mathrm{SO}_4$ .\nLet y be the oxidation state of sulphur.
2(+1) + y + 4(-2) = 0
2 + y - 8 = 0\ny - 6 = 0\ny = +6

Step 3: Determine the change in oxidation state.\nThe oxidation state of sulphur changes from +4 to +6.

Why other options are wrong:
- Option A (4 to 0): This is incorrect. Sulphur is oxidized, not reduced.
- Option B (4 to 2): This is incorrect. Sulphur is oxidized, not reduced to +2.
- Option D (6 to 4): This is incorrect. The oxidation state increases, not decreases.

Key Concepts:
- Oxidation states
- Redox reactions
- Oxidation of sulphur dioxide

Memory Aids:
- Oxidation involves an increase in oxidation state.

Quick Tips:
- Identify the reactants and products to determine the change in oxidation state.

58 Medium

58 of 69Question permalink

Remembering +4 | -1

Starch gives blue colour with :

View answer and explanation

Correct option: B

{\mathrm{I}_2}

Option B (

{\mathrm{I}_2}

) is correct because starch forms a blue-black complex with iodine (I2).

Solution approach

Correct Answer + Core Reason: ${\mathrm{I}_2}$ . Starch forms a blue-black complex with iodine.

Step-by-Step Solution:

Step 1: Recall the reaction between starch and iodine.

Step 2: Starch is a polysaccharide that forms a complex with iodine molecules.

Step 3: This complex absorbs light in a specific range, resulting in a characteristic blue-black color.

Step 4: Consider the other options.

Why other options are wrong:
- Option A: $\mathrm{KI}$ : Potassium iodide (KI) is a source of iodide ions, but it does not directly produce the blue color with starch. Iodine ( ${\mathrm{I}_2}$ ) is required.
- Option C: $\mathrm{C}{\mathrm{l}_2}$ : Chlorine does not form a colored complex with starch.
- Option D: None of these: This is incorrect because iodine ( ${\mathrm{I}_2}$ ) does form a blue color with starch.

Key Concepts:
- Starch-Iodine Complex
- Colorimetric Reactions

Memory Aids:
- Remember the blue-black color is specific to the starch-iodine complex.

Quick Tips:
- This is a common test for the presence of starch or iodine.

59 Medium

59 of 69Question permalink

Application +4 | -1

The coefficients of 

\;{\mathrm{I}^ - },\mathrm{\;\;IO}_3^ - \mathrm{\;}\,\,\mathrm{and\;}{\mathrm{H}^ + }

 in the redox reaction, 

{I^ - } + \;IO_3^ - + {H^ + } \to {I_2} + {H_2}O

 in the balanced form respectively are

View answer and explanation

Correct option: A

5,1,6

Option A (5,1,6) is correct because the balanced redox reaction requires 5 iodide ions, 1 iodate ion, and 6 hydrogen ions.

Solution approach

Correct Answer + Core Reason: 5,1,6. The balanced redox reaction has coefficients 5, 1, and 6 for ${\mathrm{I}^ - }$ , $\mathrm{IO}_3^ -$ , and ${\mathrm{H}^ + }$ respectively.

Step-by-Step Solution:

Step 1: Write the unbalanced redox reaction: ${\mathrm{I}^ - } + \mathrm{IO}_3^ - + {\mathrm{H}^ + } \to {\mathrm{I}_2} + {\mathrm{H}_2}\mathrm{O}$

Step 2: Write the half-reactions:\nOxidation: $2{\mathrm{I}^ - } \to {\mathrm{I}_2} + 2{\mathrm{e}^ - }$ \nReduction: $\mathrm{IO}_3^ - + 6{\mathrm{H}^ + } + 5{\mathrm{e}^ - } \to \frac{1}{2}{\mathrm{I}_2} + 3{\mathrm{H}_2}\mathrm{O}$

Step 3: Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to balance the electrons:\nOxidation: $10{\mathrm{I}^ - } \to 5{\mathrm{I}_2} + 10{\mathrm{e}^ - }$ \nReduction: $2\mathrm{IO}_3^ - + 12{\mathrm{H}^ + } + 10{\mathrm{e}^ - } \to {\mathrm{I}_2} + 6{\mathrm{H}_2}\mathrm{O}$

Step 4: Combine the half-reactions:
$10{\mathrm{I}^ - } + 2\mathrm{IO}_3^ - + 12{\mathrm{H}^ + } \to 6{\mathrm{I}_2} + 6{\mathrm{H}_2}\mathrm{O}$

Step 5: Simplify the equation by dividing by 2:
$5{\mathrm{I}^ - } + \mathrm{IO}_3^ - + 6{\mathrm{H}^ + } \to 3{\mathrm{I}_2} + 3{\mathrm{H}_2}\mathrm{O}$

Step 6: The coefficients are 5, 1, and 6 for ${\mathrm{I}^ - }$ , $\mathrm{IO}_3^ -$ , and ${\mathrm{H}^ + }$ respectively.

Why other options are wrong:
- Option B: 1,5,6 is incorrect; the coefficients are not balanced correctly.
- Option C: 6,1,5 is incorrect; the coefficients are not balanced correctly.
- Option D: 5,6,1 is incorrect; the coefficients are not balanced correctly.

Key Concepts:
- Redox Reactions
- Balancing Redox Equations
- Half-Reactions

Memory Aids:
- LEO says GER (Lose Electrons Oxidation, Gain Electrons Reduction)

Quick Tips:
- Balance the half-reactions separately before combining them.
- Make sure the number of atoms and charges are balanced on both sides of the equation.

60 Medium

60 of 69Question permalink

Understanding +4 | -1

In the reaction of

{\mathrm{O}_3}

and

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

the later acts as :

View answer and explanation

Correct option: A

Oxidising agent

Option A (Oxidising agent) is correct because H2O2 is reduced by O3, thus acting as an oxidizing agent.

Solution approach

Correct Answer + Core Reason: Oxidising agent. ${\mathrm{H}_2}\mathrm{O}_2$ is reduced by ${\mathrm{O}_3}$ .

Step-by-Step Solution:

Step 1: Identify the reaction: ${\mathrm{O}_3} + {\mathrm{H}_2}\mathrm{O}_2 \to {\mathrm{H}_2}\mathrm{O} + 2{\mathrm{O}_2}$

Step 2: Determine oxidation states: Oxygen in ${\mathrm{O}_3}$ goes from 0 to -2 in ${\mathrm{H}_2}\mathrm{O}$ , and oxygen in ${\mathrm{H}_2}\mathrm{O}_2$ goes from -1 to 0 in ${\mathrm{O}_2}$ .

Step 3: Since ${\mathrm{H}_2}\mathrm{O}_2$ causes oxidation, it acts as an oxidizing agent.

Why other options are wrong:
- Option B: Reducing agent donates electrons, which is the opposite of what ${\mathrm{H}_2}\mathrm{O}_2$ does in this reaction.
- Option C: Bleaching agent is a substance that removes color, but the primary role here is oxidation.
- Option D: While ${\mathrm{H}_2}\mathrm{O}_2$ can be a bleaching agent in some reactions, its main role here is as an oxidizing agent.

Key Concepts:
- Oxidizing agent
- Redox reactions
- Oxidation states

Memory Aids:
- Oxidizing agents are reduced, causing oxidation in other substances.

Quick Tips:
- Identify the substance that is reduced to determine the oxidizing agent.

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