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

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

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

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

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

Which combination is odd with respect to oxidation numbers of ${\rm{S}},{\rm{Cr}},{\rm{N}}$ and ${\rm{H}}$ respectively:

${\rm{N}}{{\rm{H}}_3},{\rm{NH}}_4^ + ,{{\rm{N}}_3}{\rm{H}},{\rm{NO}}_2^ - $

${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_5},{{\rm{H}}_2}{{\rm{S}}_2}{{\rm{O}}_8},{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4},{\rm{S}}{{\rm{F}}_6}$

${{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7},{{\rm{K}}_2}{\rm{Cr}}{{\rm{O}}_4},{\rm{Cr}}{{\rm{O}}_5},{\rm{Cr}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}$

${\rm{Ca}}{{\rm{H}}_2},{\rm{NaH}},{\rm{LiH}},{\rm{Mg}}{{\rm{H}}_2}$

REDOX REACTIONS MCQs for NEET UG CHEMISTRY — Medium

1 Medium

1 of 69Question 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.

2 Medium

2 of 69Question 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.

3 Medium

3 of 69Question 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.

4 Medium

4 of 69Question 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.

5 Medium

5 of 69Question 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.

6 Medium

6 of 69Question 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.

7 Medium

7 of 69Question 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.

8 Medium

8 of 69Question 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.

9 Medium

9 of 69Question 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.

10 Medium

10 of 69Question 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.

11 Medium

11 of 69Question 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.

12 Medium

12 of 69Question 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.

13 Medium

13 of 69Question 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.

14 Medium

14 of 69Question permalink

Application +4 | -1

The oxidation state of Ni in

\mathrm{Ni}{\left( {\mathrm{CO}} \right)_4}

is :

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because carbonyl (CO) is a neutral ligand, so the oxidation state of Ni in the complex is zero.

Solution approach

Correct Answer + Core Reason: Zero. CO is a neutral ligand, so the oxidation state of Ni is zero.

Step-by-Step Solution:

Step 1: Recall the properties of carbonyl ligands.

Step 2: Carbonyl (CO) is a neutral ligand, meaning it has a charge of 0.

Step 3: Let the oxidation state of Ni be x.

Step 4: $x + 4(0) = 0$

Step 5: $x = 0$

Why other options are wrong:
- Option B: $+ 4$ : Incorrect; CO is neutral.
- Option C: $+ 8$ : Incorrect; CO is neutral.
- Option D: $+ 2$ : Incorrect; CO is neutral.

Key Concepts:
- Oxidation state
- Ligands
- Coordination complexes

Memory Aids:
- Carbonyl is a neutral ligand.

Quick Tips:
- Remember the charges of common ligands.

15 Medium

15 of 69Question permalink

Application +4 | -1

Which compound has oxidation number of carbon equal to zero?

View answer and explanation

Correct option: D

{\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6}

Option D (C6H12O6) is correct because the oxidation number of carbon in glucose is zero, calculated by balancing the oxidation states of hydrogen and oxygen.

Solution approach

Correct Answer + Core Reason: C6H12O6. The oxidation number of carbon is calculated based on the oxidation states of hydrogen and oxygen and the overall charge neutrality.

Step-by-Step Solution:

Step 1: Determine the known oxidation numbers. Hydrogen (H) has an oxidation number of +1, and oxygen (O) has an oxidation number of -2.

Step 2: Calculate the average oxidation number of carbon (C) in each compound:
- C6H6: 6x + 6(+1) = 0 => 6x = -6 => x = -1
- CH3: x + 3(+1) = 0 => x = -3
- C2H4: 2x + 4(+1) = 0 => 2x = -4 => x = -2
- C6H12O6: 6x + 12(+1) + 6(-2) = 0 => 6x + 12 - 12 = 0 => 6x = 0 => x = 0

Step 3: Compare the oxidation numbers of C in each compound. The oxidation number of C is 0 in C6H12O6.

Why other options are wrong:
- Option A (C6H6): The oxidation number of C is -1.
- Option B (CH3): The oxidation number of C is -3.
- Option C (C2H4): The oxidation number of C is -2.

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

Memory Aids:
- Hydrogen usually has an oxidation number of +1.
- Oxygen usually has an oxidation number of -2.

Quick Tips:
- Remember to consider the number of atoms when calculating the oxidation number.

16 Medium

16 of 69Question permalink

Analyzing +4 | -1

Assuming complete ionization, same moles of which of the following compounds will require the least
Amount of acidified

\mathrm{KMn}{\mathrm{O}_\mathrm{4}}

for complete oxidation?

View answer and explanation

Correct option: C

\mathrm{FeS}{\mathrm{O}_\mathrm{4}}

Option C is correct because FeSO4 requires the least amount of KMnO4 for complete oxidation as it involves the oxidation of only Fe2+ to Fe3+.

Solution approach

Correct Answer + Core Reason: $\mathrm{FeS}{\mathrm{O}_\mathrm{4}}$ . It requires the least amount of acidified $KMnO_4$ for complete oxidation because only $Fe^{2+}$ needs to be oxidized to $Fe^{3+}$ .

Step-by-Step Solution:

Step 1: Analyze each compound and identify the elements that can be oxidized.

Step 2: For $FeC_2O_4$ : $Fe^{2+}$ can be oxidized to $Fe^{3+}$ , and $C_2O_4^{2-}$ can be oxidized to $CO_2$ . The oxidation half-reactions are: $Fe^{2+} \to Fe^{3+} + e^-$ and $C_2O_4^{2-} \to 2CO_2 + 2e^-$ . Total electrons released = 1 + 2 = 3.

Step 3: For $Fe(NO_2)_2$ : $Fe^{2+}$ can be oxidized to $Fe^{3+}$ , and $NO_2^-$ can be oxidized to $NO_3^-$ . The oxidation half-reactions are: $Fe^{2+} \to Fe^{3+} + e^-$ and $NO_2^- + H_2O \to NO_3^- + 2H^+ + 2e^-$ . Total electrons released = 1 + 2(2) = 5.

Step 4: For $FeSO_4$ : $Fe^{2+}$ can be oxidized to $Fe^{3+}$ . The oxidation half-reaction is: $Fe^{2+} \to Fe^{3+} + e^-$ . Total electrons released = 1.

Step 5: For $FeSO_3$ : $Fe^{2+}$ can be oxidized to $Fe^{3+}$ , and $SO_3^{2-}$ can be oxidized to $SO_4^{2-}$ . The oxidation half-reactions are: $Fe^{2+} \to Fe^{3+} + e^-$ and $SO_3^{2-} + H_2O \to SO_4^{2-} + 2H^+ + 2e^-$ . Total electrons released = 1 + 2 = 3.

Step 6: Determine the compound requiring the least amount of $KMnO_4$ . Since $KMnO_4$ acts as an oxidizing agent, the compound that releases the fewest electrons upon oxidation will require the least amount of $KMnO_4$ . $FeSO_4$ releases only 1 electron.

Why other options are wrong:
- Option A: $FeC_2O_4$ requires more $KMnO_4$ because both $Fe^{2+}$ and $C_2O_4^{2-}$ are oxidized, releasing a total of 3 electrons.
- Option B: $Fe(NO_2)_2$ requires even more $KMnO_4$ because both $Fe^{2+}$ and $NO_2^-$ are oxidized, releasing a total of 5 electrons.
- Option D: $FeSO_3$ requires more $KMnO_4$ because both $Fe^{2+}$ and $SO_3^{2-}$ are oxidized, releasing a total of 3 electrons.

Key Concepts:
- Oxidation states
- Redox reactions
- Balancing redox reactions
- Oxidizing agents

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

Quick Tips:
- Identify the species that can be oxidized in each compound.
- Determine the number of electrons released during the oxidation process.
- The compound that releases the fewest electrons will require the least amount of oxidizing agent.

17 Medium

17 of 69Question permalink

Application +4 | -1

The oxidation number of

Fe

in

{\mathrm{K}_4}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}

is :

View answer and explanation

Correct option: A

+ 2

Option A (+2) is correct because K has +1 and CN has -1, balancing the overall charge of the complex.

Solution approach

Correct Answer + Core Reason: +2. The oxidation number is calculated based on the known oxidation states of K and CN and the overall charge neutrality.

Step-by-Step Solution:

Step 1: Determine the known oxidation numbers. Potassium (K) has an oxidation number of +1, and cyanide (CN) has an oxidation number of -1.

Step 2: Set up an equation to solve for the oxidation number of iron (Fe). Let x be the oxidation number of Fe. Since there are four potassium ions and six cyanide ions, the equation is: 4(+1) + x + 6(-1) = 0

Step 3: Solve for x: 4 + x - 6 = 0 => x - 2 = 0 => x = +2

Step 4: Therefore, the oxidation number of Fe in K4Fe(CN)6 is +2.

Why other options are wrong:
- Option B (+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 (+6): 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:
- Potassium is always +1 in compounds.
- Cyanide is always -1 in compounds.

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

18 Medium

18 of 69Question permalink

Application +4 | -1

Nitrogen has fractional oxidation number in :

View answer and explanation

Correct option: C

\mathrm{H}{\mathrm{N}_3}

Option C (

\mathrm{H}{\mathrm{N}_3}

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

Solution approach

Correct Answer + Core Reason: $\mathrm{H}{\mathrm{N}_3}$ . Nitrogen has a fractional oxidation number in hydrazoic acid.

Step-by-Step Solution:

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

Step 2: In $N_2H_4$ , let the oxidation state of N be x.

Step 3: $2x + 4(1) = 0$

Step 4: $x = -2$

Step 5: In $NH_4$ , the question is incomplete, assuming it to be $NH_4^+$ , let the oxidation state of N be x.

Step 6: $x + 4(1) = +1$

Step 7: $x = -3$

Step 8: In $HN_3$ , let the oxidation state of N be x.

Step 9: $1 + 3x = 0$

Step 10: $x = -1/3$

Step 11: In $N_2F_2$ , let the oxidation state of N be x.

Step 12: $2x + 2(-1) = 0$

Step 13: $x = +1$

Why other options are wrong:
- Option A: ${\mathrm{N}_2}{\mathrm{H}_4}$ : Nitrogen has an oxidation state of -2.
- Option B: $\mathrm{N}{\mathrm{H}_4}$ : Nitrogen has an oxidation state of -3.
- Option D: ${\mathrm{N}_2}{\mathrm{F}_2}$ : Nitrogen has an oxidation state of +1.

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

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

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

19 Medium

19 of 69Question permalink

Analyzing +4 | -1

Which combination is odd with respect to oxidation numbers of

\mathrm{S},\mathrm{Cr},\mathrm{N}

and

\mathrm{H}

respectively:

View answer and explanation

Correct option: C

\mathrm{N}{\mathrm{H}_3},\mathrm{NH}_4^ + ,{\mathrm{N}_3}\mathrm{H},\mathrm{NO}_2^ -

Option C (

\mathrm{N}{\mathrm{H}_3},\mathrm{NH}_4^ + ,{\mathrm{N}_3}\mathrm{H},\mathrm{NO}_2^ -

) is the odd combination because the oxidation state of H is +1 in the first three compounds, but it is not present in the last compound.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{H}_3},\mathrm{NH}_4^ + ,{\mathrm{N}_3}\mathrm{H},\mathrm{NO}_2^ -$ . The oxidation state of H is not consistent across all compounds.

Step-by-Step Solution:

Step 1: Analyze Option A: ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_5},{\mathrm{H}_2}{\mathrm{S}_2}{\mathrm{O}_8},{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4},\mathrm{S}{\mathrm{F}_6}$
- In ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_5}$ , S has an oxidation state of +6.
- In ${\mathrm{H}_2}{\mathrm{S}_2}{\mathrm{O}_8}$ , S has an oxidation state of +7 (peroxo linkage). However, the maximum oxidation state of S is +6, so the average is +6.
- In ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ , S has an oxidation state of +6.
- In ${\mathrm{S}{\mathrm{F}_6}}$ , S has an oxidation state of +6.\nThe oxidation state of S varies, but H is present in the first three.

Step 2: Analyze Option B: ${\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7},{\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4},\mathrm{Cr}{\mathrm{O}_5},\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$
- In ${\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}$ , Cr has an oxidation state of +6.
- In ${\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4}$ , Cr has an oxidation state of +6.
- In $\mathrm{Cr}{\mathrm{O}_5}$ , Cr has an oxidation state of +6 (butterfly structure with peroxo linkages).
- In $\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}$ , Cr has an oxidation state of +6.\nThe oxidation state of Cr is consistently +6.

Step 3: Analyze Option C: $\mathrm{N}{\mathrm{H}_3},\mathrm{NH}_4^ + ,{\mathrm{N}_3}\mathrm{H},\mathrm{NO}_2^ -$
- In $\mathrm{N}{\mathrm{H}_3}$ , N has an oxidation state of -3, H is +1.
- In $\mathrm{NH}_4^ +$ , N has an oxidation state of -3, H is +1.
- In ${\mathrm{N}_3}\mathrm{H}$ , N has an oxidation state of -1/3, H is +1.
- In $\mathrm{NO}_2^ -$ , N has an oxidation state of +3, H is not present.\nThe oxidation state of H is +1 in the first three, but H is absent in the last compound.

Step 4: Analyze Option D: $\mathrm{Ca}{\mathrm{H}_2},\mathrm{NaH},\mathrm{LiH},\mathrm{Mg}{\mathrm{H}_2}$
- In $\mathrm{Ca}{\mathrm{H}_2}$ , H has an oxidation state of -1.
- In $\mathrm{NaH}$ , H has an oxidation state of -1.
- In $\mathrm{LiH}$ , H has an oxidation state of -1.
- In $\mathrm{Mg}{\mathrm{H}_2}$ , H has an oxidation state of -1.\nThe oxidation state of H is consistently -1.

Step 5: Conclusion.\nOption C is the odd combination because H is not present in all compounds.

Why other options are wrong:
- Option A: S is present in all compounds, though its oxidation state varies.
- Option B: Cr is present in all compounds, and its oxidation state is consistently +6.
- Option D: H is present in all compounds, and its oxidation state is consistently -1.

Key Concepts:
- Oxidation state calculation

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

Quick Tips:
- Pay attention to the presence or absence of elements in the compounds.

20 Medium

20 of 69Question permalink

Application +4 | -1

The eq. wt. of

\mathrm{Fe}{\mathrm{C}_2}{\mathrm{O}_4}

in ,

\mathrm{Fe}{\mathrm{C}_2}{\mathrm{O}_4} \to \mathrm{F}{\mathrm{e}^{3 + }} + 2\mathrm{C}{\mathrm{O}_2}\mathrm{is}:

View answer and explanation

Correct option: B

mol. wt./3

Option B (mol. wt./3) is correct because the equivalent weight is the molecular weight divided by the change in oxidation number, which is 3 in this case.

Solution approach

Correct Answer + Core Reason: mol. wt./3. The equivalent weight of $\mathrm{Fe}{\mathrm{C}_2}{\mathrm{O}_4}$ is its molecular weight divided by 3.

Step-by-Step Solution:

Step 1: Determine the oxidation state changes in the reaction: $\mathrm{Fe}{\mathrm{C}_2}{\mathrm{O}_4} \to \mathrm{F}{\mathrm{e}^{3 + }} + 2\mathrm{C}{\mathrm{O}_2}$

Step 2: The oxidation state of Fe changes from +2 to +3, which is a change of 1.

Step 3: The oxidation state of C in ${\mathrm{C}_2}\mathrm{O}_4^{2-}$ is +3, and it changes to +4 in $\mathrm{C}{\mathrm{O}_2}$ . Since there are two carbon atoms, the total change is 2 * (4-3) = 2.

Step 4: The total change in oxidation number is 1 (from Fe) + 2 (from C) = 3.

Step 5: The equivalent weight is given by: Equivalent weight = (Molecular weight) / (Change in oxidation number).

Step 6: Equivalent weight = (Molecular weight) / 3.

Why other options are wrong:
- Option A: its mol. wt. is incorrect; the equivalent weight is not equal to the molecular weight in this reaction.
- Option C: mol. wt./4 is incorrect; the change in oxidation number is 3, not 4.
- Option D: None of these is incorrect because the equivalent weight can be calculated as mol. wt./3.

Key Concepts:
- Equivalent Weight
- Oxidation State

Memory Aids:
- Equivalent weight = (Molecular weight) / (n-factor)

Quick Tips:
- Determine the change in oxidation state of all elements that are being oxidized or reduced.
- Remember that the equivalent weight depends on the specific reaction.

21 Medium

21 of 69Question permalink

Analyzing +4 | -1

Statement-I: In

{HClO}_4

, Chlorine has the oxidation number of +4.
Statement-II:

{HClO}_4

(perchloric) acid has two peroxide linkages.

View answer and explanation

Correct option: B

Both the statements are incorrect

Both statements are incorrect. In HClO₄, chlorine has an oxidation number of +7, and perchloric acid does not have peroxide linkages.

Solution approach

Correct Answer + Core Reason: Both the statements are incorrect. Chlorine has an oxidation number of +7 in HClO₄, and there are no peroxide linkages.

Step-by-Step Solution:

Step 1: Analyze Statement-I: In ${HClO}_4$ , Chlorine has the oxidation number of +4.\nTo determine the oxidation number of chlorine, we use the following rules:
- H has an oxidation number of +1.
- O has an oxidation number of -2.\nLet the oxidation number of Cl be x.
+1 + x + 4(-2) = 0
1 + x - 8 = 0\nx - 7 = 0\nx = +7\nThe oxidation number of chlorine is +7, not +4. Therefore, Statement-I is incorrect.

Step 2: Analyze Statement-II: ${HClO}_4$ (perchloric) acid has two peroxide linkages.\nThe structure of perchloric acid is $HOClO_3$ . There are no O-O (peroxide) linkages in the structure. 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
- Structure of Oxyacids

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.
- Draw the structure of the molecule to identify peroxide linkages.

22 Medium

22 of 69Question permalink

Analyzing +4 | -1

Statement-I: O.N. of carbon in

\mathrm{H}-\mathrm{C} \equiv \mathrm{N}

is +4.

Statement-II: Carbon always shows an O.N. of +4.

View answer and explanation

Correct option: B

Both the statements are incorrect

Both statements are incorrect. The oxidation number of carbon in HCN is -1, and carbon can exhibit various oxidation states, not always +4.

Solution approach

Correct Answer + Core Reason: Both the statements are incorrect. The oxidation number of carbon in HCN is -1, and carbon does not always show an oxidation number of +4.

Step-by-Step Solution:

Step 1: Analyze Statement-I: O.N. of carbon in $\mathrm{H}-\mathrm{C} \equiv \mathrm{N}$ is +4.\nTo determine the oxidation number of carbon, we use the following rules:
- H has an oxidation number of +1.
- N has an oxidation number of -3.\nLet the oxidation number of C be x.
+1 + x + (-3) = 0\nx - 2 = 0\nx = +2\nThe oxidation number of carbon is +2, not +4. Therefore, Statement-I is incorrect.

Step 2: Analyze Statement-II: Carbon always shows an O.N. of +4.\nCarbon can exhibit various oxidation states, such as -4 (e.g., in $CH_4$ ), -2 (e.g., in $CH_2Cl_2$ ), 0 (e.g., in elemental carbon), +2 (e.g., in HCN), and +4 (e.g., in $CO_2$ ). 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.

23 Medium

23 of 69Question permalink

Remembering +4 | -1

The nature of oxides

{\mathrm{V}_\mathrm{2}}{\mathrm{O}_\mathrm{3}}

and

\mathrm{CrO}

is indexed as

X'

and

'Y'

type respectively. The correct set of

X

and

Y

is

View answer and explanation

Correct option: D

X = basic, y = basic

Option D (X = basic, y = basic) is correct because oxides of metals in lower oxidation states are generally basic.

Solution approach

Correct Answer + Core Reason: X = basic, y = basic. Oxides of metals in lower oxidation states are generally basic.

Step-by-Step Solution:

Step 1: Determine the oxidation state of V in V2O3.

Step 2: 2(Oxidation state of V) + 3(-2) = 0 => Oxidation state of V = +3

Step 3: Determine the oxidation state of Cr in CrO.

Step 4: Oxidation state of Cr + (-2) = 0 => Oxidation state of Cr = +2

Step 5: Recall the trend: Metal oxides are generally basic. Lower oxidation states tend to be more basic.

Step 6: Since both V and Cr are in relatively low oxidation states, their oxides are basic.

Step 7: Final Answer: X = basic, Y = basic

Why other options are wrong:
- Option A (X = basic, Y = amphoteric): CrO is generally considered basic, not amphoteric.
- Option B (X= amphoteric, Y = basic): V2O3 is generally considered basic, not amphoteric.
- Option C (X = acidic, Y = acidic): Metal oxides are generally not acidic. Non-metal oxides are acidic.

Key Concepts:
- Nature of oxides
- Oxidation state and oxide properties

Memory Aids:
- Metal oxides are generally basic, non-metal oxides are generally acidic.

Quick Tips:
- Consider the position of the element in the periodic table and its oxidation state when determining the nature of the oxide.

24 Medium

24 of 69Question permalink

Remembering +4 | -1

The oxidation state of a metal in a compound is represented according to the notation which is known as

View answer and explanation

Correct option: C

Stock notation

Option C (Stock notation) is correct because it is the notation used to represent the oxidation state of a metal in a compound.

Solution approach

Correct Answer + Core Reason: Stock notation. This notation is used to indicate the oxidation state of a metal in a compound.

Step-by-Step Solution:

Step 1: Understand the question. The question asks about the notation used to represent the oxidation state of a metal in a compound.

Step 2: Recall the different types of notations.

Step 3: Identify the correct notation. Stock notation is used to indicate the oxidation state of a metal in a compound by writing the oxidation state as a Roman numeral in parentheses after the name of the metal.

Step 4: Conclusion: The correct answer is Stock notation.

Why other options are wrong:
- Option A: Alfred stock is not a recognized term or notation in chemistry.
- Option B: German stock is not a recognized term or notation in chemistry.
- Option D: Haworth stock is not a recognized term or notation in chemistry. Haworth projections are used for representing cyclic structures of sugars.

Key Concepts:
- Oxidation state
- Nomenclature of inorganic compounds
- Stock notation

Memory Aids:
- Stock notation: Think of \"stock\" as a way to \"take stock\" of the oxidation state.

Quick Tips:
- Remember that Stock notation uses Roman numerals to represent the oxidation state.

25 Medium

25 of 69Question permalink

Application +4 | -1

In which of the following

N

has lowest oxidation number?

View answer and explanation

Correct option: C

{\mathrm{N}_2}\mathrm{O}

The correct answer is C (N2O) because the oxidation number of N in N2O is +1, which is the lowest among the given options.

Solution approach

Correct Answer + Core Reason: N2O. The oxidation number of N is calculated based on the oxidation state of oxygen and the overall charge neutrality.

Step-by-Step Solution:

Step 1: Determine the oxidation number of oxygen (O), which is usually -2.

Step 2: Calculate the oxidation number of nitrogen (N) in each compound:
- NO: x + (-2) = 0 => x = +2
- NO2: x + 2(-2) = 0 => x = +4
- N2O: 2x + (-2) = 0 => 2x = +2 => x = +1
- N2O5: 2x + 5(-2) = 0 => 2x = +10 => x = +5

Step 3: Compare the oxidation numbers of N in each compound. The lowest oxidation number is +1 in N2O.

Why other options are wrong:
- Option A (NO): The oxidation number of N is +2, which is higher than in N2O.
- Option B (NO2): The oxidation number of N is +4, which is higher than in N2O.
- Option D (N2O5): The oxidation number of N is +5, which is higher than in N2O.

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

Memory Aids:
- Oxygen usually has an oxidation number of -2.

Quick Tips:
- Remember to consider the number of atoms when calculating the oxidation number.

26 Medium

26 of 69Question permalink

Remembering +4 | -1

Which of the following metals do not react with steam?

View answer and explanation

Correct option: C

Tin and cadmium

Option C (Tin and cadmium) is correct because tin and cadmium are less reactive and do not react with steam under normal conditions.

Solution approach

Correct Answer + Core Reason: Tin and cadmium. These metals are less reactive and do not react with steam under normal conditions.

Step-by-Step Solution:

Step 1: Consider the reactivity series of metals.

Step 2: Recall that metals higher in the reactivity series can react with steam to produce hydrogen gas and metal oxide.

Step 3: Tin (Sn) and Cadmium (Cd) are relatively low in the reactivity series and do not react with steam.

Step 4: Magnesium (Mg) and Iron (Fe) are more reactive and can react with steam.

Step 5: Final Answer: Tin and cadmium

Why other options are wrong:
- Option A (Cadmium and magnesium): Magnesium reacts with steam.
- Option B (Cadmium and iron): Iron reacts with steam.
- Option D (Magnesium and tin): Magnesium reacts with steam.

Key Concepts:
- Reactivity series of metals
- Reaction of metals with steam

Memory Aids:
- Remember the order of metals in the reactivity series.

Quick Tips:
- Consider the position of the metal in the reactivity series to determine its reactivity with steam.

27 Medium

27 of 69Question permalink

Remembering +4 | -1

An element if present in the free or the uncombined state, its each atom bears an oxidation number of

View answer and explanation

Correct option: D

zero

Option D (zero) is correct because the oxidation number of an element in its free or uncombined state is always zero.

Solution approach

Correct Answer + Core Reason: zero. The oxidation number of an element in its free or uncombined state is always zero.

Step-by-Step Solution:

Step 1: Recall the definition of oxidation number.

Step 2: Recall the rule: An element in its free or uncombined state has an oxidation number of zero.

Step 3: Apply the rule: Since the element is in the free or uncombined state, its oxidation number is zero.

Step 4: Final Answer: zero

Why other options are wrong:
- Option A (more than 1): This is incorrect as the oxidation number is zero in the free state.
- Option B (less than 1): This is incorrect as the oxidation number is zero in the free state.
- Option C (more than 2): This is incorrect as the oxidation number is zero in the free state.

Key Concepts:
- Oxidation number rules
- Oxidation state of free elements

Memory Aids:
- Free elements = zero oxidation number.

Quick Tips:
- Always remember the basic rules for assigning oxidation numbers.

28 Medium

28 of 69Question permalink

Understanding +4 | -1

WI ich acts as a reducing agent?

View answer and explanation

Correct option: D

{\left( {\mathrm{COOH}} \right)_2}

Option D (

{\left( {\mathrm{COOH}} \right)_2}

) is correct because oxalic acid can be oxidized, thus acting as a reducing agent.

Solution approach

Correct Answer + Core Reason: ${\left( {\mathrm{COOH}} \right)_2}$ . Oxalic acid can be oxidized, thus acting as a reducing agent.

Step-by-Step Solution:

Step 1: Recall that a reducing agent donates electrons and gets oxidized.

Step 2: Consider the oxidation states of the elements in each compound.

Step 3: $\mathrm{HN}{\mathrm{O}_3}$ is a strong oxidizing agent, not a reducing agent.

Step 4: $\mathrm{KMn}{\mathrm{O}_4}$ is a strong oxidizing agent, not a reducing agent.

Step 5: ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ is an oxidizing agent, especially when concentrated and hot.

Step 6: ${\left( {\mathrm{COOH}} \right)_2}$ (oxalic acid) can be oxidized to $\mathrm{CO}_2$ , thus acting as a reducing agent.

Why other options are wrong:
- Option A: $\mathrm{HN}{\mathrm{O}_3}$ is a strong oxidizing agent.
- Option B: $\mathrm{KMn}{\mathrm{O}_4}$ is a strong oxidizing agent.
- Option C: ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ is an oxidizing agent.

Key Concepts:
- Reducing agent
- Oxidation states

Memory Aids:
- Reducing agents are oxidized.

Quick Tips:
- Look for substances that can be easily oxidized to identify reducing agents.

29 Medium

29 of 69Question permalink

Analyzing +4 | -1

Which of the following is redox reaction?

View answer and explanation

Correct option: D

\mathrm{SnC}{\mathrm{l}_2}\mathrm{ + HgC}{\mathrm{l}_2} \to \mathrm{SnC}{\mathrm{l}_4} + \mathrm{Hg}

Option D (

\mathrm{SnC}{\mathrm{l}_2}\mathrm{ + HgC}{\mathrm{l}_2} \to \mathrm{SnC}{\mathrm{l}_4} + \mathrm{Hg}

) is correct because the oxidation state of Sn increases and Hg decreases, indicating a redox reaction.

Solution approach

Correct Answer + Core Reason: $\mathrm{SnC}{\mathrm{l}_2}\mathrm{ + HgC}{\mathrm{l}_2} \to \mathrm{SnC}{\mathrm{l}_4} + \mathrm{Hg}$ . This reaction involves a change in oxidation states.

Step-by-Step Solution:

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

Step 2: In option A: ${\mathrm{N}_2}{\mathrm{O}_5} + {\mathrm{H}_2}\mathrm{O} \to 2\mathrm{HN}{\mathrm{O}_3}$ , the oxidation states of N, H, and O remain unchanged. This is not a redox reaction.

Step 3: In option B: $\mathrm{AgN}{\mathrm{O}_3} + \mathrm{ }KI \to \mathrm{AgI} + \mathrm{KN}{\mathrm{O}_3}$ , the oxidation states of Ag, N, O, K, and I remain unchanged. This is a precipitation reaction, not a redox reaction.

Step 4: In option C: $\mathrm{Ba}{\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to \mathrm{BaS}{\mathrm{O}_4} + {\mathrm{H}_2}{\mathrm{O}_2}$ , the oxidation states of Ba, H, S, and O remain unchanged. This is not a redox reaction.

Step 5: In option D: $\mathrm{SnC}{\mathrm{l}_2}\mathrm{ + HgC}{\mathrm{l}_2} \to \mathrm{SnC}{\mathrm{l}_4} + \mathrm{Hg}$ , Tin (Sn) is oxidized from +2 to +4, and Mercury (Hg) is reduced from +2 to 0. This is a redox reaction.

Why other options are wrong:
- Option A: ${\mathrm{N}_2}{\mathrm{O}_5} + {\mathrm{H}_2}\mathrm{O} \to 2\mathrm{HN}{\mathrm{O}_3}$ : No change in oxidation states.
- Option B: $\mathrm{AgN}{\mathrm{O}_3} + \mathrm{ }KI \to \mathrm{AgI} + \mathrm{KN}{\mathrm{O}_3}$ : No change in oxidation states.
- Option C: $\mathrm{Ba}{\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to \mathrm{BaS}{\mathrm{O}_4} + {\mathrm{H}_2}{\mathrm{O}_2}$ : No change in oxidation states.

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

30 Medium

30 of 69Question permalink

Application +4 | -1

The number of mole of

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

that will be needed to react completely with one mole of ferrous oxalate in acidic solution is :

View answer and explanation

Correct option: A

\frac{3}{5}

Option A (

\frac{3}{5}

) is correct because the balanced redox reaction requires 3 moles of KMnO4 to react completely with 5 moles of ferrous oxalate.

Solution approach

Correct Answer + Core Reason: $\frac{3}{5}$ . The number of moles of $\mathrm{KMn}{\mathrm{O}_4}$ needed to react completely with one mole of ferrous oxalate is $\frac{3}{5}$ .

Step-by-Step Solution:

Step 1: Write the unbalanced redox reaction: $\mathrm{KMn}{\mathrm{O}_4} + \mathrm{Fe}{\mathrm{C}_2}{\mathrm{O}_4} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to {\mathrm{K}_2}\mathrm{S}{\mathrm{O}_4} + \mathrm{MnS}{\mathrm{O}_4} + \mathrm{F}{\mathrm{e}_2}{(\mathrm{S}{\mathrm{O}_4})_3} + \mathrm{C}{\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{O}$

Step 2: Identify the oxidation half-reaction: $\mathrm{Fe}{\mathrm{C}_2}{\mathrm{O}_4} \to \mathrm{F}{\mathrm{e}^{3 + }} + 2\mathrm{C}{\mathrm{O}_2}$

Step 3: Balance the oxidation half-reaction: $\mathrm{Fe}^{2+} \to \mathrm{Fe}^{3+} + e^-$ and ${\mathrm{C}_2}\mathrm{O}_4^{2-} \to 2\mathrm{CO}_2 + 2e^-$ . Combining these, we get $\mathrm{Fe}{\mathrm{C}_2}{\mathrm{O}_4} \to \mathrm{Fe}^{3+} + 2\mathrm{CO}_2 + 3e^-$

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

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

Step 6: Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 3 to balance the electrons:
$5\mathrm{Fe}{\mathrm{C}_2}{\mathrm{O}_4} \to 5\mathrm{Fe}^{3+} + 10\mathrm{CO}_2 + 15e^-$
$3\mathrm{MnO}_4^- + 24\mathrm{H}^+ + 15e^- \to 3\mathrm{Mn}^{2+} + 12\mathrm{H}_2\mathrm{O}$

Step 7: Combine the balanced half-reactions:
$3\mathrm{MnO}_4^- + 5\mathrm{Fe}{\mathrm{C}_2}{\mathrm{O}_4} + 24\mathrm{H}^+ \to 3\mathrm{Mn}^{2+} + 5\mathrm{Fe}^{3+} + 10\mathrm{CO}_2 + 12\mathrm{H}_2\mathrm{O}$

Step 8: From the balanced equation, 3 moles of $\mathrm{KMn}{\mathrm{O}_4}$ react with 5 moles of $\mathrm{Fe}{\mathrm{C}_2}{\mathrm{O}_4}$ . Therefore, the mole ratio of $\mathrm{KMn}{\mathrm{O}_4}$ to $\mathrm{Fe}{\mathrm{C}_2}{\mathrm{O}_4}$ is $\frac{3}{5}$ .

Why other options are wrong:
- Option B: $\frac{2}{5}$ is incorrect; it does not represent the correct stoichiometric ratio.
- Option C: $\frac{4}{5}$ is incorrect; it does not represent the correct stoichiometric ratio.
- Option D: 1 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.

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