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

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

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

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

REDOX REACTIONS MCQs for NEET UG CHEMISTRY — Easy

1 Easy

Application +4 | -1

In which

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

acts as oxidant, while reacting with :

View answer and explanation

Correct option: C

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

Option C (

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

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

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

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

2 Easy

2 of 65Question permalink

Understanding +4 | -1

When Fe metal is rusted then Fe is :

View answer and explanation

Correct option: A

Oxidised

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

Solution approach

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

Step-by-Step Solution:

Step 1: Define oxidation and reduction.

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

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

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

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

Key Concepts:
- Oxidation
- Reduction
- Rusting

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

Quick Tips:
- Recognize common oxidation and reduction processes.

3 Easy

3 of 65Question permalink

Application +4 | -1

\frac{{ - 1}}{3}

oxidation state of nitrogen will be obtained in case of

View answer and explanation

Correct option: B

{N_3}H

Option B (

{N_3}H

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

Solution approach

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

Step-by-Step Solution:

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

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

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

Step 4: $3x + 1 = 0$

Step 5: $x = -1/3$

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

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

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

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

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

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

4 Easy

4 of 65Question permalink

Remembering +4 | -1

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

View answer and explanation

Correct option: B

F

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

Key Concepts:
- Oxidation state
- Electronegativity
- Halogens

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

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

5 Easy

5 of 65Question permalink

Analyzing +4 | -1

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

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

View answer and explanation

Correct option: B

Both oxidised and reduced

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

Solution approach

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

Step-by-Step Solution:

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

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

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

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

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

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

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

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

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

6 Easy

6 of 65Question permalink

Understanding +4 | -1

In the aluminothermic process, aluminium acts as :

View answer and explanation

Correct option: C

A reducing agent

Option C (A reducing agent) is correct because aluminum loses electrons to reduce the metal oxide in the aluminothermic process.

Solution approach

Correct Answer + Core Reason: A reducing agent. Aluminum reduces the metal oxide to its elemental form.

Step-by-Step Solution:

Step 1: Understand the aluminothermic process: It involves using aluminum to reduce metal oxides, such as iron oxide ( $\mathrm{Fe}_2\mathrm{O}_3$ ).

Step 2: General reaction: $2\mathrm{Al} + \mathrm{Fe}_2\mathrm{O}_3 \to \mathrm{Al}_2\mathrm{O}_3 + 2\mathrm{Fe}$

Step 3: Determine oxidation states: Al goes from 0 to +3 (oxidation), and Fe goes from +3 to 0 (reduction).

Step 4: Since Al causes reduction, it acts as a reducing agent.

Why other options are wrong:
- Option A: An oxidizing agent gains electrons, which is the opposite of what aluminum does.
- Option B: A flux is used to lower the melting point, which is not the role of aluminum here.
- Option D: A solder is used to join metals, which is not the role of aluminum here.

Key Concepts:
- Reducing agent
- Aluminothermic process
- Redox reactions
- Oxidation states

Memory Aids:
- Reducing agents cause reduction by losing electrons.

Quick Tips:
- Identify the substance that is oxidized to determine the reducing agent.

7 Easy

7 of 65Question permalink

Understanding +4 | -1

The highest oxidation state of Mn is shown by

View answer and explanation

Correct option: B

KMn{o_4}

Option B (

KMn{o_4}

) is correct because in

KMn{o_4}

, Mn has an oxidation state of +7, which is its highest possible oxidation state.

Solution approach

Correct Answer + Core Reason: $KMn{o_4}$ . In $KMn{o_4}$ , Mn has an oxidation state of +7, which is its highest possible oxidation state.

Step-by-Step Solution:

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

Step 2: In K2MnO4: 2(+1) + Mn + 4(-2) = 0 => Mn = +6

Step 3: In KMnO4: +1 + Mn + 4(-2) = 0 => Mn = +7

Step 4: In MnO2: Mn + 2(-2) = 0 => Mn = +4

Step 5: In Mn2O2 (which simplifies to MnO): Mn + (-2) = 0 => Mn = +2

Step 6: Compare the oxidation states: +7 is the highest.

Step 7: Final Answer: $KMn{o_4}$

Why other options are wrong:
- **Option A ( ${K_2}Mn{o_4}$ ): The oxidation state of Mn is +6, which is not the highest.
- Option C ( $Mn{o_2}$ ): The oxidation state of Mn is +4, which is not the highest.
- Option D ( $M{n_2}{O_2}$ ): The oxidation state of Mn is +2, which is not the highest.

Key Concepts:
- Oxidation number determination
- Maximum oxidation state

Memory Aids:
- Remember the common oxidation states of oxygen and potassium.

Quick Tips:**
- Practice calculating oxidation states to quickly identify the highest one.

8 Easy

8 of 65Question permalink

Understanding +4 | -1

Which acts as reducing agent as well as oxidising agent?

View answer and explanation

Correct option: A

{\mathrm{O}_3}

Option A (

{\mathrm{O}_3}

) is correct because ozone can act as both an oxidizing agent (accepting electrons) and a reducing agent (donating electrons).

Solution approach

Correct Answer + Core Reason: ${\mathrm{O}_3}$ . Ozone can act as both a reducing and oxidizing agent.

Step-by-Step Solution:

Step 1: Define oxidizing agent: A substance that accepts electrons and gets reduced.

Step 2: Define reducing agent: A substance that donates electrons and gets oxidized.

Step 3: Consider the properties of ozone ( ${\mathrm{O}_3}$ ): It can act as an oxidizing agent by accepting electrons to form oxygen ( ${\mathrm{O}_2}$ ) or oxide ions. It can also act as a reducing agent by donating electrons.

Step 4: Consider the other options.

Why other options are wrong:
- Option B: $\mathrm{ClO}_4^ -$ : Perchlorate ion is a strong oxidizing agent but not a good reducing agent.
- Option C: ${\mathrm{F}_2}$ : Fluorine is a strong oxidizing agent but not a reducing agent.
- Option D: $\mathrm{MnO}_4^ -$ : Permanganate ion is a strong oxidizing agent but not a reducing agent.

Key Concepts:
- Oxidizing Agent
- Reducing Agent
- Ozone

Memory Aids:
- Remember that ozone is a versatile compound that can act as both an oxidizing and reducing agent.

Quick Tips:
- Know the properties of common oxidizing and reducing agents.

9 Easy

9 of 65Question permalink

Analyzing +4 | -1

Which of the following has the highest oxidation number

View answer and explanation

Correct option: D

KI{O_4}

Option D (KIO4) is correct because iodine has the highest oxidation number of +7 in this compound.

Solution approach

Correct Answer + Core Reason: KIO4. The oxidation number of I is the highest in this compound.

Step-by-Step Solution:

Step 1: Determine the oxidation number of I in each compound:
- I3: 0 (elemental form)
- KI: +1 + x = 0 => x = -1
- IF5: x + 5(-1) = 0 => x = +5
- KIO4: +1 + x + 4(-2) = 0 => +1 + x - 8 = 0 => x = +7

Step 2: Compare the oxidation numbers of I. The highest oxidation number is +7 in KIO4.

Why other options are wrong:
- Option A (I3): The oxidation number of I is 0.
- Option B (KI): The oxidation number of I is -1.
- Option C (IF5): The oxidation number of I is +5.

Key Concepts:
- Oxidation number rules

Memory Aids:
- Fluorine is always -1 in compounds.
- Potassium is always +1 in compounds.

Quick Tips:
- Remember to consider the oxidation numbers of other elements when calculating the oxidation number of iodine.

10 Easy

10 of 65Question permalink

Application +4 | -1

\mathrm{HCO}_3^ -

contains carbon in the oxidation state:

View answer and explanation

Correct option: C

4

Option C (4) is correct because the oxidation state of carbon in

\mathrm{HCO}_3^ -

is calculated to be +4 based on the oxidation states of other elements.

Solution approach

Correct Answer + Core Reason: 4. The oxidation state of carbon in bicarbonate ion is +4.

Step-by-Step Solution:

Step 1: Determine the known oxidation states.\nHydrogen (H) has an oxidation state of +1.\nOxygen (O) has an oxidation state of -2.\nThe overall charge of the bicarbonate ion ( $\mathrm{HCO}_3^ -$ ) is -1.

Step 2: Set up the equation to solve for the oxidation state of carbon (C).\nLet x be the oxidation state of carbon.
(+1) + x + 3(-2) = -1

Step 3: Solve for x.
1 + x - 6 = -1\nx - 5 = -1\nx = +4

Step 4: Final Answer\nThe oxidation state of carbon in $\mathrm{HCO}_3^ -$ is +4.

Why other options are wrong:
- Option A (5): This is incorrect. The calculation does not yield +5.
- Option B (1): This is incorrect. The calculation does not yield +1.
- Option D (Zero): This is incorrect. The calculation does not yield 0.

Key Concepts:
- Oxidation state rules
- Algebraic calculation of oxidation states in polyatomic ions

Memory Aids:
- Remember the common oxidation states of H (+1) and O (-2).

Quick Tips:
- Always consider the overall charge of the ion when calculating oxidation states.

11 Easy

11 of 65Question permalink

Analyzing +4 | -1

Which one of the following reactions is not an example of redox reaction

View answer and explanation

Correct option: D

HCl + {H_2}O \to {H_3}{O^ + } + C{l^ - }

Option D (

HCl + {H_2}O \to {H_3}{O^ + } + C{l^ - }

) is correct because there is no change in oxidation states of any element, hence it's not a redox reaction.

Solution approach

Correct Answer + Core Reason: $HCl + {H_2}O \to {H_3}{O^ + } + C{l^ - }$ . This is an acid-base reaction, not a redox reaction.

Step-by-Step Solution:

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

Step 2: In option A: $C{l_2} + 2{H_2}O + S{O_2} \to 4{H^ + } + SO_4^{{2^ - }} + 2C{l^ - }$ , Chlorine is reduced from 0 to -1, and Sulfur is oxidized from +4 to +6. This is a redox reaction.

Step 3: In option B: $C{u^{2 + }} + Zn \to Z{n^{2 + }} + Cu$ , Copper is reduced from +2 to 0, and Zinc is oxidized from 0 to +2. This is a redox reaction.

Step 4: In option C: $2{H_2} + {O_2} \to 2{H_2}O$ , Hydrogen is oxidized from 0 to +1, and Oxygen is reduced from 0 to -2. This is a redox reaction.

Step 5: In option D: $HCl + {H_2}O \to {H_3}{O^ + } + C{l^ - }$ , The oxidation states of H, Cl, and O remain unchanged. This is an acid-base reaction (proton transfer), not a redox reaction.

Why other options are wrong:
- Option A: $C{l_2} + 2{H_2}O + S{O_2} \to 4{H^ + } + SO_4^{{2^ - }} + 2C{l^ - }$ : Chlorine is reduced, and sulfur is oxidized, making it a redox reaction.
- Option B: $C{u^{2 + }} + Zn \to Z{n^{2 + }} + Cu$ : Copper is reduced, and zinc is oxidized, making it a redox reaction.
- Option C: $2{H_2} + {O_2} \to 2{H_2}O$ : Hydrogen is oxidized, and oxygen is reduced, making it a redox reaction.

Key Concepts:
- Redox Reactions
- Oxidation States
- Acid-Base Reactions

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.
- Acid-base reactions involve proton transfer, not electron transfer.

12 Easy

12 of 65Question permalink

Application +4 | -1

In which of the following compounds Nitrogen exhibits the highest oxidation state

View answer and explanation

Correct option: A

{N_3}H

Option A (

{N_3}H

) is correct because the average oxidation state of nitrogen in hydrazoic acid is -1/3, which is the highest among the given options.

Solution approach

Correct Answer + Core Reason: ${N_3}H$ . Nitrogen exhibits the highest oxidation state in hydrazoic acid.

Step-by-Step Solution:

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

Step 2: In $N_3H$ , let the oxidation state of N be x.

Step 3: $3x + 1 = 0$

Step 4: $x = -1/3$

Step 5: In $NH_2OH$ , let the oxidation state of N be x.

Step 6: $x + 2(1) + (-2) + 1 = 0$

Step 7: $x = -1$

Step 8: In $N_2H_4$ , let the oxidation state of N be x.

Step 9: $2x + 4(1) = 0$

Step 10: $x = -2$

Step 11: In $NH_3$ , let the oxidation state of N be x.

Step 12: $x + 3(1) = 0$

Step 13: $x = -3$

Why other options are wrong:
- Option B: $N{H_2}OH$ : Nitrogen has an oxidation state of -1.
- Option C: ${N_2}{H_4}$ : Nitrogen has an oxidation state of -2.
- Option D: $N{H_3}$ : Nitrogen has an oxidation state of -3.

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

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

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

13 Easy

13 of 65Question permalink

Analyzing +4 | -1

Which of the following change represents a disproportionation reaction

View answer and explanation

Correct option: C

Both 1 and 2

Option C (Both 1 and 2) is correct because in both reactions, the same element undergoes both oxidation and reduction.

Solution approach

Correct Answer + Core Reason: Both 1 and 2. Disproportionation involves simultaneous oxidation and reduction of the same element.

Step-by-Step Solution:

Step 1: Analyze reaction 1: $C{l_2} + 2O{H^ - } \to Cl{O^ - } + C{l^ - } + {H_2}O$ \nOxidation state of Cl in $Cl_2$ is 0.\nOxidation state of Cl in $ClO^-$ is +1.\nOxidation state of Cl in $Cl^-$ is -1.\nSince Cl goes from 0 to +1 (oxidation) and 0 to -1 (reduction), this is a disproportionation reaction.

Step 2: Analyze reaction 2: $C{u_2}O + 2{H^ + } \to Cu + C{u^{2 + }} + {H_2}O$ \nOxidation state of Cu in $Cu_2O$ is +1.\nOxidation state of Cu in $Cu$ is 0.\nOxidation state of Cu in $Cu^{2+}$ is +2.\nSince Cu goes from +1 to 0 (reduction) and +1 to +2 (oxidation), this is a disproportionation reaction.

Step 3: Conclusion.\nBoth reactions are disproportionation reactions.

Why other options are wrong:
- Option A: Reaction 2 is also a disproportionation reaction.
- Option B: Reaction 2 is also a disproportionation reaction.
- Option D: Both reactions are disproportionation reactions.

Key Concepts:
- Disproportionation reaction
- Oxidation state determination

Memory Aids:
- Disproportionation: The same element is both oxidized and reduced.

Quick Tips:
- Look for a single element in the reactant side that appears in two different oxidation states on the product side.

14 Easy

14 of 65Question permalink

Analyzing +4 | -1

when

{\mathbf{K}_\mathbf{2}}\mathbf{C}{\mathbf{r}_\mathbf{2}}{\mathbf{O}_\mathbf{7}}

is converted to

{\mathbf{K}_\mathbf{2}}\mathbf{Cr}{\mathbf{O}_\mathbf{4}}

, the change in the oxidation of chromium is

View answer and explanation

Correct option: A

0

Option A (0) is correct because the oxidation state of chromium remains +6 in both K2Cr2O7 and K2CrO4.

Solution approach

Correct Answer + Core Reason: 0. The oxidation state of Cr remains the same.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Cr in K2Cr2O7.\nLet the oxidation state of Cr be x.\nK2Cr2O7: 2(+1) + 2(x) + 7(-2) = 0
2 + 2x - 14 = 0
2x = 12\nx = +6

Step 2: Determine the oxidation state of Cr in K2CrO4.\nLet the oxidation state of Cr be y.\nK2CrO4: 2(+1) + y + 4(-2) = 0
2 + y - 8 = 0\ny = +6

Step 3: Calculate the change in oxidation state.\nChange = +6 - (+6) = 0

Why other options are wrong:
- Option B (6): This would be the oxidation state of Cr itself, not the change in oxidation state.
- Option C (4): This is an incorrect calculation of the oxidation state or the change.
- Option D (3): This is an incorrect calculation of the oxidation state or the change.

Key Concepts:
- Oxidation state calculation

Memory Aids:
- Remember the common oxidation states of elements like K (+1) and O (-2).

Quick Tips:
- Always set the sum of oxidation states in a neutral compound to zero.

15 Easy

15 of 65Question permalink

Remembering +4 | -1

Which of the following shows highest ox, no. in combined state?

View answer and explanation

Correct option: C

Both (a) and (b)

Option C (Both (a) and (b)) is correct because both Osmium (Os) and Ruthenium (Ru) can exhibit a maximum oxidation state of +8 in their combined states.

Solution approach

Correct Answer + Core Reason: Both (a) and (b). Osmium and Ruthenium can both exhibit +8 oxidation state.

Step-by-Step Solution:

Step 1: Consider the electronic configurations of Os and Ru.\nOs (Z=76): [Xe] 4f¹⁴ 5d⁶ 6s²\nRu (Z=44): [Kr] 4d⁷ 5s¹

Step 2: Analyze the maximum oxidation states of Os and Ru.\nOs can exhibit a maximum oxidation state of +8 in compounds like $OsO_4$ .\nRu can exhibit a maximum oxidation state of +8 in compounds like $RuO_4$ .

Why other options are wrong:
- Option A: Ru can also exhibit +8 oxidation state.
- Option B: Os can also exhibit +8 oxidation state.
- Option D: Both Os and Ru can exhibit +8 oxidation state.

Key Concepts:
- Oxidation States of Transition Metals
- Electronic Configuration

Memory Aids:
- Remember that Os and Ru are in the same group and can exhibit similar oxidation states.

Quick Tips:
- Consider the electronic configurations to determine the maximum oxidation states.

16 Easy

16 of 65Question permalink

Analyzing +4 | -1

Sulphur Containing species that cannot be a reducing agent is

View answer and explanation

Correct option: C

{H_2}S{O_4}

Option C (H2SO4) is correct because sulfur is already in its highest oxidation state (+6) and cannot be further oxidized.

Solution approach

Correct Answer + Core Reason: H2SO4. Sulfur cannot be a reducing agent when it is in its highest oxidation state.

Step-by-Step Solution:

Step 1: Determine the oxidation number of S in each species:
- SO2: x + 2(-2) = 0 => x = +4
- SO3^2-: x + 3(-2) = -2 => x = +4
- H2SO4: 2(+1) + x + 4(-2) = 0 => 2 + x - 8 = 0 => x = +6
- S^2-: -2

Step 2: A reducing agent is a substance that loses electrons and is oxidized. Sulfur can act as a reducing agent if it can be oxidized to a higher oxidation state.

Step 3: In H2SO4, the oxidation number of S is +6, which is its maximum oxidation state. Therefore, it cannot be further oxidized and cannot act as a reducing agent.

Why other options are wrong:
- Option A (SO2): Sulfur can be oxidized to +6.
- Option B (SO3^2-): Sulfur can be oxidized to +6.
- Option D (S^2-): Sulfur can be oxidized to 0, +4, or +6.

Key Concepts:
- Oxidation number rules
- Reducing agent

Memory Aids:
- A reducing agent is oxidized, and an oxidizing agent is reduced.

Quick Tips:
- Determine the oxidation state of the element to see if it can be further oxidized.

17 Easy

17 of 65Question permalink

Understanding +4 | -1

In the preparation of chlorine from

\mathrm{HCl},\mathrm{Mn}{\mathrm{O}_2}

acts as :

View answer and explanation

Correct option: B

oxidising agent

Option B (oxidising agent) is correct because MnO2 accepts electrons from HCl, causing HCl to be oxidized and itself to be reduced.

Solution approach

Correct Answer + Core Reason: Oxidising agent. $\mathrm{MnO}_2$ oxidizes $\mathrm{HCl}$ to $\mathrm{Cl}_2$ .

Step-by-Step Solution:

Step 1: Identify the reaction: $4\mathrm{HCl} + \mathrm{MnO}_2 \to \mathrm{MnCl}_2 + 2{\mathrm{H}_2}\mathrm{O} + \mathrm{Cl}_2$

Step 2: Determine oxidation states: Mn goes from +4 to +2 (reduction), and Cl goes from -1 to 0 (oxidation).

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

Why other options are wrong:
- Option A: Reducing agent donates electrons, which is the opposite of what $\mathrm{MnO}_2$ does.
- Option C: Catalytic agent speeds up the reaction without being consumed, which is not the role of $\mathrm{MnO}_2$ here.
- Option D: Dehydrating agent removes water, which is not the primary function of $\mathrm{MnO}_2$ in this reaction.

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

Memory Aids:
- Oxidizing agents are reduced, reducing agents are oxidized.

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

18 Easy

18 of 65Question permalink

Remembering +4 | -1

The equivalent weight of a reductant or an oxidant is given by :

View answer and explanation

Correct option: A

\mathrm{Eq}\mathrm{.wt}\mathrm{.} = \frac{{\mathrm{mol}\mathrm{.weight\,\, of\,\, reductant\,\, or\,\, oxidant}}}{{\begin{array}{cccccccccccccccccccc} {\mathrm{no}\mathrm{.of\,\, electrons\,\, lost\,\, or\,\, gained \,\,by}}\\ {1{\rm{molecule\,\, of\,\, reductant\,\, or\,\, oxida

Option A is correct because the equivalent weight is the molecular weight divided by the number of electrons lost or gained per molecule.

Solution approach

Correct Answer + Core Reason: $\mathrm{Eq}\mathrm{.wt}\mathrm{.} = \frac{{\mathrm{mol}\mathrm{.weight\,\, of\,\, reductant\,\, or\,\, oxidant}}}{{\begin{array}{cccccccccccccccccccc} {\mathrm{no}\mathrm{.of\,\, electrons\,\, lost\,\, or\,\, gained \,\,by}}\\ {1\mathrm{molecule\,\, of\,\, reductant\,\, or\,\, oxidant}} \end{array}}}$ . This is the fundamental definition of equivalent weight.

Step-by-Step Solution:

Step 1: Define equivalent weight: The mass of a substance that combines with or is chemically equivalent to one mole of hydrogen atoms.

Step 2: Relate equivalent weight to molecular weight and the number of electrons transferred in a redox reaction.
$\mathrm{Equivalent\ Weight} = \frac\mathrm{Molecular\ Weight}\mathrm{Number\ of\ electrons\ transferred}$

Step 3: Analyze the given options.

Why other options are wrong:
- Option B: $\mathrm{Eq}\mathrm{.wt}\mathrm{.} = \frac{{\mathrm{mol}\mathrm{.wt}\mathrm{.}}}{{\mathrm{valence}}}$ : While valence can relate to the number of electrons, it's not universally applicable to all redox reactions.
- Option C: $\mathrm{Eq}\,\mathrm{.wt}\mathrm{.} = \frac{{\mathrm{mol}\mathrm{.wt}\mathrm{.}}}{{\mathrm{total\,\, charge\,\, on\,\, cation\,\, or\,\, anion}}}$ : This is only applicable in specific cases, not as a general definition.
- Option D: All of the above: This is incorrect because options B and C are not universally correct.

Key Concepts:
- Equivalent Weight
- Molecular Weight
- Redox Reactions
- Number of Electrons Transferred

Memory Aids:
- Equivalent weight is always less than or equal to molecular weight.

Quick Tips:
- Remember the fundamental definition of equivalent weight in terms of electrons transferred.

19 Easy

19 of 65Question permalink

Remembering +4 | -1

Which of the following is the most powerful oxidizing agent?

View answer and explanation

Correct option: A

\mathrm{F}_2

Option A (

\mathrm{F}_2

) is correct because fluorine is the strongest oxidizing agent due to its high electronegativity and tendency to gain electrons.

Solution approach

Correct Answer + Core Reason: $\mathrm{F}_2$ . Fluorine is the strongest oxidizing agent.

Step-by-Step Solution:

Step 1: Define oxidizing agent.\nAn oxidizing agent is a substance that gains electrons in a redox reaction.

Step 2: Consider the electronegativity of halogens.\nThe electronegativity decreases down the group: F > Cl > Br > I.

Step 3: Relate electronegativity to oxidizing power.\nThe higher the electronegativity, the greater the tendency to gain electrons, and the stronger the oxidizing agent.\nTherefore, $F_2$ is the most powerful oxidizing agent.

Why other options are wrong:
- Option B: Chlorine is a weaker oxidizing agent than fluorine.
- Option C: Bromine is a weaker oxidizing agent than fluorine.
- Option D: Iodine is a weaker oxidizing agent than fluorine.

Key Concepts:
- Oxidizing Agent
- Electronegativity
- Redox Reactions

Memory Aids:
- Fluorine is the most electronegative element.

Quick Tips:
- Remember the trend of electronegativity in the halogen group.

20 Easy

20 of 65Question permalink

Remembering +4 | -1

The burning of hydrogen is called :

View answer and explanation

Correct option: C

Oxidation

Option C (Oxidation) is correct because burning involves rapid reaction with oxygen, which is the definition of oxidation.

Solution approach

Correct Answer + Core Reason: Oxidation. Burning is a rapid reaction with oxygen, hence oxidation.

Step-by-Step Solution:

Step 1: Define burning: A combustion reaction involving rapid oxidation.

Step 2: Define oxidation: A process involving the loss of electrons or the gain of oxygen.

Step 3: Define hydrogenation: Addition of hydrogen.

Step 4: Define hydration: Addition of water.

Step 5: Define reduction: A process involving the gain of electrons or the loss of oxygen.

Step 6: Relate burning of hydrogen to the definitions. Hydrogen reacts with oxygen to form water, which is an oxidation process.

Why other options are wrong:
- Option A: Hydrogenation: This is the addition of hydrogen, not the burning of hydrogen.
- Option B: Hydration: This is the addition of water, not the burning of hydrogen.
- Option D: Reduction: This is the opposite of oxidation; it involves the gain of electrons or loss of oxygen.

Key Concepts:
- Oxidation
- Reduction
- Combustion

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

Quick Tips:
- Remember that burning is a type of oxidation.

21 Easy

21 of 65Question permalink

Analyzing +4 | -1

Which of the following has been arranged in order of increasing oxidation number of Nitrogen

View answer and explanation

Correct option: C

NH_4^ + < {N_2}{H_4} < N{H_2}OH < {N_2}O

Option C (NH4+ < N2H4 < NH2OH < N2O) is correct because it correctly orders the compounds based on increasing oxidation numbers of nitrogen: -3 < -2 < -1 < +1.

Solution approach

Correct Answer + Core Reason: NH4+ < N2H4 < NH2OH < N2O. The compounds are arranged in increasing order of oxidation number of Nitrogen.

Step-by-Step Solution:

Step 1: Calculate the oxidation number of N in each compound:
- NH3: x + 3(+1) = 0 => x = -3
- N2O5: 2x + 5(-2) = 0 => 2x = +10 => x = +5
- NO: x + (-2) = 0 => x = +2
- N2: 0
- NO2+: x + 2(-2) = +1 => x = +5
- NO3-: x + 3(-2) = -1 => x = +5
- NO2-: x + 2(-2) = -1 => x = +3
- N3-: 3x = -1 => x = -1/3
- NH4+: x + 4(+1) = +1 => x = -3
- N2H4: 2x + 4(+1) = 0 => x = -2
- NH2OH: x + 2(+1) + (-2) = 0 => x = 0
- N2O: 2x + (-2) = 0 => x = +1
- NaN3: +1 + 3x = 0 => x = -1/3

Step 2: Arrange the compounds in increasing order of oxidation number of N:
- NH4+ (-3) < N2H4 (-2) < NH2OH (0) < N2O (+1)

Why other options are wrong:
- Option A (NH3 < N2O5 < NO < N2): Incorrect order of oxidation numbers.
- Option B (NO2+ < NO3- < NO2- < N3-): Incorrect order of oxidation numbers.
- Option D (NO3- < NaN3 < NH4+ < N2O): Incorrect order of oxidation numbers.

Key Concepts:
- Oxidation number rules

Memory Aids:
- Remember the common oxidation numbers of H (+1) and O (-2).

Quick Tips:
- Calculate the oxidation number of N in each compound before comparing.

22 Easy

22 of 65Question permalink

Application +4 | -1

Oxidation numbers of two cl atoms in bleaching powder

\left( {CaOC{l_2}} \right)

are,

View answer and explanation

Correct option: B

+ 1,\mathrm{ } - 1

Option B (+1, -1) is correct because bleaching powder is a mixed salt containing both hypochlorite (OCl-) and chloride (Cl-) ions, resulting in +1 and -1 oxidation states for the two chlorine atoms.

Solution approach

Correct Answer + Core Reason: $+ 1,\mathrm{ } - 1$ . Bleaching powder contains $OCl^-$ and $Cl^-$ ions, leading to +1 and -1 oxidation states for the two Cl atoms.

Step-by-Step Solution:

Step 1: Understand the composition of bleaching powder.

Step 2: Bleaching powder is $Ca(OCl)Cl$ .

Step 3: It contains $OCl^-$ and $Cl^-$ ions.

Step 4: In $OCl^-$ , let the oxidation state of Cl be x.

Step 5: $x + (-2) = -1$

Step 6: $x = +1$

Step 7: In $Cl^-$ , the oxidation state of Cl is -1.

Why other options are wrong:
- Option A: $- 1,\mathrm{ } - 1$ : Incorrect; one Cl is in $OCl^-$ .
- Option C: $+ 1,\mathrm{ } + 1$ : Incorrect; one Cl is $Cl^-$ .
- Option D: $0,\mathrm{ } - 1$ : Incorrect; Cl in $OCl^-$ is +1.

Key Concepts:
- Oxidation state
- Composition of bleaching powder
- Oxidation states in ions

Memory Aids:
- Bleaching powder is a mixed salt.

Quick Tips:
- Remember the ions present in bleaching powder.

23 Easy

23 of 65Question permalink

Remembering +4 | -1

The oxidation number of

\mathrm{H}

in sodium hydride is

View answer and explanation

Correct option: C

-1

Option C (-1) is correct because hydrogen has an oxidation number of -1 when bonded to a more electropositive element like sodium.

Solution approach

Correct Answer + Core Reason: -1. Hydrogen has an oxidation number of -1 when bonded to a more electropositive element.

Step-by-Step Solution:

Step 1: Identify the compound: Sodium hydride (NaH).

Step 2: Recall the rule: Hydrogen usually has an oxidation number of +1, but when bonded to a more electropositive element (like alkali metals), it has an oxidation number of -1.

Step 3: Apply the rule: Since Na is more electropositive than H, the oxidation number of H in NaH is -1.

Step 4: Final Answer: -1

Why other options are wrong:
- Option A (+1): This is the common oxidation state of hydrogen, but it's incorrect when hydrogen is bonded to a more electropositive element.
- Option B (0): This is incorrect as hydrogen is not in its elemental form.
- Option D (-1/2): Oxidation numbers are usually integers. This is incorrect.

Key Concepts:
- Oxidation number rules
- Oxidation state of hydrogen

Memory Aids:
- Hydrogen is +1 except when bonded to metals, then it's -1.

Quick Tips:
- Consider the electronegativity difference between the elements when assigning oxidation numbers.

24 Easy

24 of 65Question permalink

Remembering +4 | -1

The correct name for

{NO}_2

using stock notation is

View answer and explanation

Correct option: B

Nitrogen (iv) oxide

Option B (Nitrogen (iv) oxide) is correct because stock notation indicates the oxidation state of the element using Roman numerals.

Solution approach

Correct Answer + Core Reason: Nitrogen (iv) oxide. Stock notation indicates the oxidation state of the element using Roman numerals.

Step-by-Step Solution:

Step 1: Determine the oxidation state of nitrogen in NO2.

Step 2: Oxygen has an oxidation state of -2, so 2 oxygen atoms have a total oxidation state of -4.

Step 3: Since the compound is neutral, the oxidation state of nitrogen must be +4.

Step 4: In stock notation, this is written as Nitrogen (iv) oxide.

Step 5: Final Answer: Nitrogen (iv) oxide

Why other options are wrong:
- Option A (Nitrogen dioxide): This is the common name, but not the stock notation name.
- Option C (Nitrogen per oxide): This is an older, less precise name.
- Option D (All of these): Only option B is correct.

Key Concepts:
- Stock notation
- Oxidation number determination

Memory Aids:
- Stock notation uses Roman numerals to indicate oxidation state.

Quick Tips:
- Calculate the oxidation state of the element first, then write the name using stock notation.

25 Easy

25 of 65Question permalink

Application +4 | -1

The oxidation state of oxygen atom in potassium superoxide is

View answer and explanation

Correct option: B

- \frac{1}{2}

Option B (-1/2) is correct because in potassium superoxide (

KO_2

), potassium has a +1 oxidation state, and the two oxygen atoms share a -1 charge, resulting in each oxygen having an oxidation state of -1/2.

Solution approach

Correct Answer + Core Reason: $- \frac{{1}}{{2}}$ . The oxidation state of oxygen in potassium superoxide is -1/2.

Step-by-Step Solution:

Step 1: Determine the oxidation state of potassium.

Step 2: Potassium (K) has an oxidation state of +1.

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

Step 4: The sum of the oxidation states in a compound is zero.

Step 5: $1 + 2x = 0$

Step 6: $2x = -1$

Step 7: $x = -1/2$

Why other options are wrong:
- Option A: zero: Incorrect calculation.
- Option C: $- 1$ : Incorrect calculation.
- Option D: $- 2$ : Incorrect calculation.

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

Memory Aids:
- Potassium is always +1.

Quick Tips:
- Remember that oxygen can have different oxidation states in different compounds.

26 Easy

26 of 65Question permalink

Analyzing +4 | -1

The most powerful oxidizing agent from the following is

View answer and explanation

Correct option: D

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

Option D (H2SO4) is correct because sulfuric acid is a strong oxidizing agent due to sulfur's ability to be reduced from +6 to lower oxidation states.

Solution approach

Correct Answer + Core Reason: $\mathrm{H}_2 \mathrm{SO}_4$ . Sulfuric acid is a strong oxidizing agent.

Step-by-Step Solution:

Step 1: Understand the concept of oxidizing agents.\nAn oxidizing agent is a substance that gains electrons and gets reduced in a chemical reaction. Stronger oxidizing agents have a greater tendency to accept electrons.

Step 2: Analyze the given options.
- $\mathrm{H}_3 \mathrm{BO}_3$ : Boric acid is a weak acid and not a strong oxidizing agent.
- $\mathrm{HPO}_3$ : Phosphorous acid is a reducing agent.
- $\mathrm{H}_3 \mathrm{PO}_4$ : Phosphoric acid is a relatively stable acid and not a strong oxidizing agent.
- $\mathrm{H}_2 \mathrm{SO}_4$ : Sulfuric acid is a strong acid and a strong oxidizing agent. Sulfur in sulfuric acid is in its highest oxidation state (+6) and can be reduced to lower oxidation states like +4 (in $SO_2$ ) or +2 (in S) or -2 (in $H_2S$ ).

Step 3: Conclusion.
$\mathrm{H}_2 \mathrm{SO}_4$ is the most powerful oxidizing agent among the given options.

Why other options are wrong:
- Option A: Boric acid is a very weak acid and not an oxidizing agent.
- Option B: Phosphorous acid is a reducing agent.
- Option C: Phosphoric acid is a weak acid and not a strong oxidizing agent.

Key Concepts:
- Oxidizing agent
- Oxidation states

Memory Aids:
- Remember that sulfuric acid ( $H_2SO_4$ ) is a strong oxidizing agent.

Quick Tips:
- Consider the oxidation state of the central atom. Elements in high oxidation states tend to be good oxidizing agents.

27 Easy

27 of 65Question permalink

Understanding +4 | -1

The violent reaction between sodium and water is an example of :

View answer and explanation

Correct option: C

Redox reaction

Option C (Redox reaction) is correct because the reaction involves both oxidation (loss of electrons) and reduction (gain of electrons).

Solution approach

Correct Answer + Core Reason: Redox reaction. Sodium is oxidized and water is reduced.

Step-by-Step Solution:

Step 1: Identify the reaction: $2Na(s) + 2H_2O(l) \to 2NaOH(aq) + H_2(g)$

Step 2: Determine oxidation states: Na goes from 0 to +1 (oxidation), and H in water goes from +1 to 0 in H2 (reduction).

Step 3: Since both oxidation and reduction occur, it's a redox reaction.

Why other options are wrong:
- Option A: Reduction alone doesn't describe the entire reaction.
- Option B: Oxidation alone doesn't describe the entire reaction.
- Option D: Neutralization involves acid-base reactions, not electron transfer in this case.

Key Concepts:
- Redox reactions
- Oxidation states
- Oxidation: Loss of electrons
- Reduction: Gain of electrons

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

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

28 Easy

28 of 65Question permalink

Analyzing +4 | -1

Which of the following reactions does not involve either oxidation or reduction

View answer and explanation

Correct option: C

CrO_4^{ - 2} \to C{r_2}O_7^{ - 2}

Option C (

CrO_4^{ - 2} \to C{r_2}O_7^{ - 2}

) is correct because the oxidation state of Cr remains the same (+6) in both species.

Solution approach

Correct Answer + Core Reason: $CrO_4^{ - 2} \to C{r_2}O_7^{ - 2}$ . The oxidation state of Cr remains the same.

Step-by-Step Solution:

Step 1: Determine the oxidation number of Cr in $CrO_4^{ - 2}$ .
Let x be the oxidation number of Cr. Then x + 4(-2) = -2, so x - 8 = -2, and x = +6.

Step 2: Determine the oxidation number of Cr in $C{r_2}O_7^{ - 2}$ .
Let x be the oxidation number of Cr. Then 2x + 7(-2) = -2, so 2x - 14 = -2, and 2x = +12, so x = +6.

Step 3: Compare the oxidation numbers of Cr in the two ions.
The oxidation number of Cr remains +6.

Step 4: Since there is no change in oxidation number, the reaction does not involve either oxidation or reduction.

Why other options are wrong:
- Option A: The oxidation number of V changes from +1 to +3, indicating oxidation.
- Option B: The oxidation number of Na changes from 0 to +1, indicating oxidation.
- Option D: The oxidation number of Zn changes from +2 to 0, indicating reduction.

Key Concepts:
- Oxidation and reduction
- Oxidation number rules

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

Quick Tips:
- Determine the oxidation numbers of the key elements.

29 Easy

29 of 65Question permalink

Application +4 | -1

Oxidation state of oxygen is −1 in the compound :

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because in Na2O2 (sodium peroxide), the oxidation state of oxygen is -1.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}$ . In sodium peroxide, the oxidation state of oxygen is -1.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Na. Sodium (Na) is an alkali metal and has an oxidation state of +1.

Step 2: Calculate the oxidation state of O in each compound:
- NO2: +4 + 2x = 0 => x = -2
- MnO2: +4 + 2x = 0 => x = -2
- PbO2: +4 + 2x = 0 => x = -2
- Na2O2: 2(+1) + 2x = 0 => x = -1

Step 3: Identify the compound with O in the -1 oxidation state.

Why other options are wrong:
- Option A: The oxidation state of O in NO2 is -2.
- Option B: The oxidation state of O in MnO2 is -2.
- Option C: The oxidation state of O in PbO2 is -2.

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

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

Quick Tips:
- Practice calculating oxidation states in different compounds.

30 Easy

30 of 65Question permalink

Understanding +4 | -1

Which quantities are conserved in all oxidation-reduction reactions?

View answer and explanation

Correct option: A

Charge only

Option A (Charge only) is correct because in all chemical reactions, including redox reactions, charge is always conserved, but mass may not be conserved in nuclear reactions.

Solution approach

Correct Answer + Core Reason: Charge only. Charge is always conserved in redox reactions.

Step-by-Step Solution:

Step 1: Recall the fundamental conservation laws in chemical reactions.

Step 2: Charge must be conserved in all chemical reactions, including oxidation-reduction reactions.

Step 3: Mass is conserved in chemical reactions, but not necessarily in nuclear reactions.

Step 4: In redox reactions, electrons are transferred, but the total charge remains constant.

Why other options are wrong:
- Option B: Mass is conserved in chemical reactions, but the question asks about *all* oxidation-reduction reactions, implying only charge is always conserved.
- Option C: Mass is not always conserved in all reactions (e.g., nuclear reactions).
- Option D: Charge is always conserved.

Key Concepts:
- Conservation of charge
- Conservation of mass
- Redox reactions

Memory Aids:
- Remember that charge is a fundamental property that is always conserved.

Quick Tips:
- Consider the fundamental conservation laws when analyzing chemical reactions.

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