The eq. wt. of ${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3}$ as reductant, in the reaction, ${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3} + 5{{\rm{H}}_2}{\rm{O}} + 4{\rm{C}}{{\rm{l}}_2} \to 2{\rm{NaHS}}{{\rm{O}}_4} + 8{\rm{HCl}}:$

$\left( {{\rm{Mol}}{\rm{.wt}}{\rm{.}}} \right)/1$

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

31 Easy

Understanding +4 | -1

In nitric oxide (NO), the oxidation state of nitrogen is :

View answer and explanation

Correct option: D

+ 2

Option D (

+2

) is correct because in nitric oxide (NO), oxygen has an oxidation state of -2, and the overall molecule is neutral, so nitrogen must have an oxidation state of +2.

Solution approach

Correct Answer + Core Reason: $+ 2$ . The oxidation state of nitrogen in NO is +2.

Step-by-Step Solution:

Step 1: Assign oxidation states to known elements.

Step 2: Oxygen typically has an oxidation state of -2.

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

Step 4: The sum of the oxidation states in a neutral molecule is zero: $x + (-2) = 0$

Step 5: Solve for x: $x = +2$

Why other options are wrong:
- Option A: -2 would mean oxygen has to be +2, which is not the case.
- Option B: +1 would not balance the -2 charge of oxygen.
- Option C: -1 would not balance the -2 charge of oxygen.

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

Memory Aids:
- Remember that oxygen usually has an oxidation state of -2.

Quick Tips:
- Apply the rules for assigning oxidation states to determine the oxidation state of an unknown element.

32 Easy

32 of 65Question permalink

Understanding +4 | -1

The halogen that most easily reduced is

View answer and explanation

Correct option: A

{F_2}

Option A (

{F_2}

) is correct because fluorine has the highest reduction potential among the halogens, making it the easiest to reduce.

Solution approach

Correct Answer + Core Reason: ${F_2}$ . Fluorine has the highest reduction potential.

Step-by-Step Solution:

Step 1: Recall the trend in reduction potentials for halogens.

Step 2: Reduction potential generally decreases down the group (F > Cl > Br > I).

Step 3: Higher reduction potential means the species is more easily reduced.

Step 4: Therefore, F₂ is the most easily reduced halogen.

Why other options are wrong:
- Option B: Chlorine has a lower reduction potential than fluorine.
- Option C: Bromine has a lower reduction potential than chlorine and fluorine.
- Option D: Iodine has the lowest reduction potential among the given halogens.

Key Concepts:
- Reduction potential
- Halogens
- Periodic trends

Memory Aids:
- Remember that fluorine is the most electronegative element and has the highest tendency to gain electrons (reduction).

Quick Tips:
- Know the trends in reduction potentials for common elements.

33 Easy

33 of 65Question permalink

Application +4 | -1

Carbon is in the lowest oxidation state in

View answer and explanation

Correct option: A

C{H_4}

Option A (

CH_4

) is correct because carbon is bonded to hydrogen, which is less electronegative, giving carbon a -4 oxidation state, the lowest among the options.

Solution approach

Correct Answer + Core Reason: $C{H_4}$ . Carbon has the lowest oxidation state in methane.

Step-by-Step Solution:

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

Step 2: In $CH_4$ , let the oxidation state of C be x.

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

Step 4: $x = -4$

Step 5: In $CCl_4$ , let the oxidation state of C be x.

Step 6: $x + 4(-1) = 0$

Step 7: $x = +4$

Step 8: In $CF_4$ , let the oxidation state of C be x.

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

Step 10: $x = +4$

Step 11: In $CO_2$ , let the oxidation state of C be x.

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

Step 13: $x = +4$

Why other options are wrong:
- Option B: $CC{l_4}$ : Carbon has an oxidation state of +4.
- Option C: $C{F_4}$ : Carbon has an oxidation state of +4.
- Option D: $C{O_2}$ : Carbon has an oxidation state of +4.

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

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

Quick Tips:
- Remember the electronegativity differences between elements.

34 Easy

34 of 65Question permalink

Analyzing +4 | -1

Which reaction involves neither oxidation nor reduction?

View answer and explanation

Correct option: A

\mathrm{CrO}_4^{2 - } \to \mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }

Option A (CrO4^2- → Cr2O7^2-) is correct because the oxidation state of Cr remains +6 in both ions.

Solution approach

Correct Answer + Core Reason: CrO4^2- → Cr2O7^2-. The oxidation number of Cr remains unchanged in this conversion.

Step-by-Step Solution:

Step 1: Determine the oxidation number of Cr in each species:
- CrO4^2-: x + 4(-2) = -2 => x - 8 = -2 => x = +6
- Cr2O7^2-: 2x + 7(-2) = -2 => 2x - 14 = -2 => 2x = +12 => x = +6

Step 2: Compare the oxidation numbers of Cr. In both CrO4^2- and Cr2O7^2-, the oxidation number of Cr is +6.

Step 3: Check the other options:
- Cr → CrCl3: 0 → +3 (oxidation)
- Na → Na+: 0 → +1 (oxidation)
- 2S2O3^2- → S4O6^2-: +2 → +2.5 (oxidation)

Why other options are wrong:
- Option B (Cr → CrCl3): Chromium is oxidized from 0 to +3.
- Option C (Na → Na+): Sodium is oxidized from 0 to +1.
- Option D (2S2O3^2- → S4O6^2-): Sulfur is oxidized from +2 to +2.5.

Key Concepts:
- Oxidation number rules
- Oxidation and reduction

Memory Aids:
- Oxidation is loss of electrons, reduction is gain of electrons (OIL RIG).

Quick Tips:
- Remember to calculate the oxidation number in each species to determine if there is a change.

35 Easy

35 of 65Question permalink

Understanding +4 | -1

Reduction is a process which involves :

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because reduction involves gaining electrons, adding hydrogen/removing oxygen, and adding metal/removing non-metal.

Solution approach

Correct Answer + Core Reason: All of the above. Reduction encompasses electronation, hydrogen addition/oxygen removal, and metal addition/non-metal removal.

Step-by-Step Solution:

Step 1: Define reduction in terms of electrons: Gain of electrons (electronation).

Step 2: Define reduction in terms of hydrogen and oxygen: Addition of hydrogen or removal of oxygen.

Step 3: Define reduction in terms of metals and non-metals: Addition of metal or removal of non-metal.

Step 4: Consider all the given options.

Why other options are wrong:
- Option A: Electronation: This is a part of the reduction process, but not the complete definition.
- Option B: Addition of hydrogen or removal of oxygen: This is a part of the reduction process, but not the complete definition.
- Option C: Addition of metal or removal of non-metal: This is a part of the reduction process, but not the complete definition.

Key Concepts:
- Reduction
- Oxidation

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

Quick Tips:
- Remember the different ways to define reduction.

36 Easy

36 of 65Question permalink

Application +4 | -1

Which compound shows highest oxidation number for chlorine?

View answer and explanation

Correct option: D

\mathrm{KCl}{\mathrm{O}_4}

Option D (

\mathrm{KCl}{\mathrm{O}_4}

) is correct because in KClO4, the oxidation state of chlorine is +7, which is the highest among the given options.

Solution approach

Correct Answer + Core Reason: $\mathrm{KCl}{\mathrm{O}_4}$ . Chlorine has the highest oxidation number (+7) in KClO4.

Step-by-Step Solution:

Step 1: Determine the oxidation state of K and O.
- Potassium (K) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

Step 2: Calculate the oxidation state of Cl in each compound:
- HCl: +1 + x = 0 => x = -1
- KClO: +1 + x + (-2) = 0 => x = +1
- KClO3: +1 + x + 3(-2) = 0 => x = +5
- KClO4: +1 + x + 4(-2) = 0 => x = +7

Step 3: Identify the compound with the highest oxidation state for Cl.

Why other options are wrong:
- Option A: The oxidation state of Cl in HCl is -1.
- Option B: The oxidation state of Cl in KClO is +1.
- Option C: The oxidation state of Cl in KClO3 is +5.

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

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

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

37 Easy

37 of 65Question permalink

Understanding +4 | -1

Stannous chloride gives a white precipitate with a solution of mercuric chloride. In this process mercuric chloride is :

View answer and explanation

Correct option: B

Reduced

Option B (Reduced) is correct because stannous chloride (SnCl2) reduces mercuric chloride (HgCl2) to mercurous chloride (Hg2Cl2) or mercury (Hg).

Solution approach

Correct Answer + Core Reason: Reduced. Stannous chloride reduces mercuric chloride.

Step-by-Step Solution:

Step 1: Write the reaction between stannous chloride and mercuric chloride.
$\mathrm{SnCl}_2 + 2\mathrm{HgCl}_2 \longrightarrow \mathrm{SnCl}_4 + \mathrm{Hg}_2\mathrm{Cl}_2$ (white precipitate)
$\mathrm{SnCl}_2 + \mathrm{Hg}_2\mathrm{Cl}_2 \longrightarrow \mathrm{SnCl}_4 + 2\mathrm{Hg}$ (grey precipitate)

Step 2: Determine the oxidation states of Hg in $\mathrm{HgCl}_2$ and the product.

Step 3: In $\mathrm{HgCl}_2$ , the oxidation state of Hg is +2. In $\mathrm{Hg}_2\mathrm{Cl}_2$ , the oxidation state of Hg is +1. In $\mathrm{Hg}$ , the oxidation state is 0.

Step 4: Since the oxidation state of Hg decreases, $\mathrm{HgCl}_2$ is reduced.

Why other options are wrong:
- Option A: Oxidized: $\mathrm{HgCl}_2$ is reduced, not oxidized.
- Option C: Converted into a complex compound containing $\mathrm{Sn}$ and $\mathrm{Hg}$ : While complex formation is possible, the primary process is reduction.
- Option D: Converted into a chloro complex of $\mathrm{Hg}$ : While chloro complexes can form, the main process is reduction of Hg.

Key Concepts:
- Redox Reactions
- Reduction
- Oxidation State

Memory Aids:
- Remember that stannous chloride is a good reducing agent.

Quick Tips:
- Know the common reactions of stannous chloride.

38 Easy

38 of 65Question permalink

Analyzing +4 | -1

The conversion of

\mathbf{Pb}{\mathbf{O}_\mathbf{2}}\,\,to\,\mathbf{Pb}{\left( {\mathbf{N}{\mathbf{O}_\mathbf{3}}} \right)_\mathbf{2}}

is

View answer and explanation

Correct option: B

Reduction

Reduction. Option B is correct because the oxidation number of Pb decreases from +4 in

\mathbf{Pb}{\mathbf{O}_\mathbf{2}}

to +2 in

\mathbf{Pb}{\left( {\mathbf{N}{\mathbf{O}_\mathbf{3}}} \right)_\mathbf{2}}

.

Solution approach

Correct Answer + Core Reason: Reduction. The oxidation number of Pb decreases.

Step-by-Step Solution:

Step 1: Determine the oxidation number of Pb in $\mathbf{Pb}{\mathbf{O}_\mathbf{2}}$ .
Oxygen has an oxidation number of -2. Therefore, Pb + 2(-2) = 0, so Pb = +4.

Step 2: Determine the oxidation number of Pb in $\mathbf{Pb}{\left( {\mathbf{N}{\mathbf{O}_\mathbf{3}}} \right)_\mathbf{2}}$ .
The nitrate ion ( $\mathbf{N}{\mathbf{O}_\mathbf{3}}^-$ ) has a charge of -1. Therefore, Pb + 2(-1) = 0, so Pb = +2.

Step 3: Compare the oxidation numbers of Pb in the two compounds.
The oxidation number of Pb decreases from +4 to +2.

Step 4: Since the oxidation number of Pb decreases, the process is reduction.

Why other options are wrong:
- Option A: Oxidation involves an increase in oxidation number.
- Option C: The process involves a change in oxidation number, so it is either oxidation or reduction.
- Option D: Only reduction occurs.

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.

39 Easy

39 of 65Question permalink

Application +4 | -1

Total number of

\mathrm{Al}{\mathrm{F}_3}

molecules in a sample of

\mathrm{Al}{\mathrm{F}_3}

containing

3.01 \times {10^{23}}

ions of

{\mathrm{F}^ - }

is :

View answer and explanation

Correct option: D

{10^{23}}

Option D (

{10^{23}}

) is correct because each AlF3 molecule contains 3 fluoride ions, so the number of AlF3 molecules is one-third the number of fluoride ions.

Solution approach

Correct Answer + Core Reason: ${10^{23}}$ . The number of $\mathrm{Al}{\mathrm{F}_3}$ molecules is one-third the number of $\mathrm{F}^ -$ ions.

Step-by-Step Solution:

Step 1: Determine the number of $\mathrm{F}^ -$ ions given: $3.01 \times {10^{23}}$ ions.

Step 2: Recognize that each $\mathrm{Al}{\mathrm{F}_3}$ molecule contains 3 $\mathrm{F}^ -$ ions.

Step 3: Calculate the number of $\mathrm{Al}{\mathrm{F}_3}$ molecules by dividing the number of $\mathrm{F}^ -$ ions by 3:\nNumber of $\mathrm{Al}{\mathrm{F}_3}$ molecules = $\frac{{3.01 \times {{10}^{23}}}}{3} = 1.0033 \times {10^{23}} \approx {10^{23}}$

Why other options are wrong:
- Option A: $9 \times {10^{24}}$ is incorrect; it's a result of multiplying the number of fluoride ions by 3 instead of dividing.
- Option B: $3 \times {10^{24}}$ is incorrect; it's a result of multiplying the number of fluoride ions by 10.
- Option C: $7 \times {10^{23}}$ is incorrect and doesn't relate to the correct calculation.

Key Concepts:
- Molecular Formula
- Stoichiometry

Memory Aids:
- Remember the chemical formula $\mathrm{Al}{\mathrm{F}_3}$ and the number of fluoride ions in each molecule.

Quick Tips:
- Always pay attention to the stoichiometry of the compound when relating the number of ions to the number of molecules.

40 Easy

40 of 65Question permalink

Understanding +4 | -1

Oxidation number of sodium in sodium amalgam is :

View answer and explanation

Correct option: D

Zero

Option D (Zero) is correct because sodium in sodium amalgam (NaHg) exists in its metallic state, hence its oxidation number is zero.

Solution approach

Correct Answer + Core Reason: Zero. Sodium in sodium amalgam (NaHg) exists in its metallic state, hence its oxidation number is zero.

Step-by-Step Solution:

Step 1: Identify the compound: Sodium amalgam (NaHg).

Step 2: Understand that sodium amalgam is a solution of sodium in mercury.

Step 3: In this amalgam, sodium exists in its metallic state, not as an ion.

Step 4: Recall that the oxidation number of an element in its elemental state is zero.

Step 5: Therefore, the oxidation number of sodium in sodium amalgam is zero.

Step 6: Final Answer: Zero

Why other options are wrong:
- Option A (+2): Sodium does not exhibit a +2 oxidation state in this compound.
- Option B (+1): Sodium typically forms +1 ions in compounds, but not in its metallic state in an amalgam.
- Option C (-2): Sodium does not exhibit a -2 oxidation state.

Key Concepts:
- Oxidation number rules
- Oxidation state of elements in elemental form
- Amalgams

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

Quick Tips:
- Consider the physical state of the element in the compound when assigning oxidation states.

41 Easy

41 of 65Question permalink

Analyzing +4 | -1

In which reaction there is no change in valency and the oxidation state?

View answer and explanation

Correct option: C

N{a_2}{O_2} + {H_2}S{O_4} \to N{a_2}S{O_4} + {H_2}{O_2}

Option C (Na2O2 + H2SO4 → Na2SO4 + H2O2) is correct because the oxidation states of all elements remain unchanged in this reaction.

Solution approach

Correct Answer + Core Reason: Na2O2 + H2SO4 → Na2SO4 + H2O2. There is no change in oxidation state for any element in this reaction.

Step-by-Step Solution:

Step 1: Determine the oxidation numbers of each element in the reactants and products for each reaction:
- A: SO2 + H2S → 2H2O + 3S
- S in SO2: +4, S in H2S: -2, S in S: 0 (change)
- B: 2Na + O → Na2O
- Na: 0 → +1, O: 0 → -2 (change)
- C: Na2O2 + H2SO4 → Na2SO4 + H2O2
- Na: +1, O: -1, H: +1, S: +6 (no change)
- D: 4KClO3 → 3KClO4 + KCl
- Cl in KClO3: +5, Cl in KClO4: +7, Cl in KCl: -1 (change)

Step 2: Identify the reaction where there is no change in oxidation state.

Step 3: In reaction C, the oxidation numbers of Na, O, H, and S remain the same on both sides of the equation.

Why other options are wrong:
- Option A (SO2 + H2S → 2H2O + 3S): Sulfur undergoes a change in oxidation state.
- Option B (2Na + O → Na2O): Sodium and oxygen undergo a change in oxidation state.
- Option D (4KClO3 → 3KClO4 + KCl): Chlorine undergoes a change in oxidation state.

Key Concepts:
- Oxidation number rules
- Oxidation and reduction

Memory Aids:
- Oxidation is loss of electrons, reduction is gain of electrons (OIL RIG).

Quick Tips:
- Calculate the oxidation number of each element in the reactants and products to determine if there is a change.

42 Easy

42 of 65Question permalink

Remembering +4 | -1

Oxidation number of carbon in graphite

View answer and explanation

Correct option: A

zero

Option A (zero) is correct because carbon in graphite is bonded to three other carbon atoms, forming a neutral network.

Solution approach

Correct Answer + Core Reason: zero. Carbon in graphite is bonded to three other carbon atoms, forming a neutral network.

Step-by-Step Solution:

Step 1: Understand the structure of graphite.\nGraphite consists of layers of carbon atoms arranged in a hexagonal lattice. Each carbon atom is covalently bonded to three other carbon atoms within the same layer.

Step 2: Determine the oxidation number of carbon.\nSince each carbon atom is bonded to three other carbon atoms of the same element, the bonds are considered nonpolar covalent bonds. Therefore, there is no charge separation, and the oxidation number of carbon is 0.

Why other options are wrong:
- **Option B ( $+ 1$ ): This is incorrect because carbon is not donating electrons to a more electronegative element in graphite.
- Option C ( $+ 4$ ): This is incorrect because carbon is not bonded to four more electronegative atoms. Carbon only forms 3 bonds in graphite.
- Option D ( $+ 2$ ): This is incorrect because carbon is not donating electrons to a more electronegative element in graphite.

Key Concepts:
- Oxidation number
- Covalent bonding
- Structure of graphite

Memory Aids:
- Remember that in elemental forms, the oxidation number of an element is always zero.

Quick Tips:**
- Understand the bonding environment of the element in question before assigning oxidation numbers.

43 Easy

43 of 65Question permalink

Application +4 | -1

In alkaline condition

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

reacts as follows :

2\mathrm{KMn}{\mathrm{O}_4} + 2\mathrm{KOH} \to 2{\mathrm{K}_2}\mathrm{Mn}{\mathrm{O}_4} + {\mathrm{H}_2}\mathrm{O} + \mathrm{O}\mathrm{.}

The eq. wt. of

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

is :

View answer and explanation

Correct option: B

158

Option B (158) is correct because in alkaline conditions, KMnO4 is reduced to K2MnO4, involving a change of 1 in oxidation number.

Solution approach

Correct Answer + Core Reason: 158. The equivalent weight of $\mathrm{KMn}{\mathrm{O}_4}$ is equal to its molecular weight when the change in oxidation state is 1.

Step-by-Step Solution:

Step 1: Determine the oxidation state of Mn in $\mathrm{KMn}{\mathrm{O}_4}$ . The oxidation state of Mn is +7.

Step 2: Determine the oxidation state of Mn in ${\mathrm{K}_2}\mathrm{Mn}{\mathrm{O}_4}$ . The oxidation state of Mn is +6.

Step 3: Calculate the change in oxidation state: |+7 - (+6)| = 1.

Step 4: The equivalent weight is given by: Equivalent weight = (Molecular weight) / (Change in oxidation state).

Step 5: Molecular weight of $\mathrm{KMn}{\mathrm{O}_4}$ = 39 + 55 + (4 * 16) = 39 + 55 + 64 = 158.

Step 6: Equivalent weight = 158 / 1 = 158.

Why other options are wrong:
- Option A: 52.7 is incorrect and doesn't relate to the molecular weight or equivalent weight calculation.
- Option C: 31.6 is incorrect; it would be the equivalent weight if the change in oxidation state was 5, which is not the case in this alkaline condition.
- Option D: 79 is incorrect; it would be the equivalent weight if the change in oxidation state was 2, which is not the case in this alkaline condition.

Key Concepts:
- Equivalent Weight
- Oxidation State

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

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

44 Easy

44 of 65Question permalink

Understanding +4 | -1

Which among the following shows maximum oxidation state?

View answer and explanation

Correct option: C

Mn

Option C (Mn) is correct because manganese can exhibit oxidation states ranging from +2 to +7, which is the highest among the given elements.

Solution approach

Correct Answer + Core Reason: Mn. Manganese exhibits the highest oxidation state among the given elements.

Step-by-Step Solution:

Step 1: Recall the range of oxidation states for each element.

Step 2: Vanadium (V) can exhibit oxidation states from +2 to +5.

Step 3: Iron (Fe) can exhibit oxidation states of +2, +3, and +6.

Step 4: Manganese (Mn) can exhibit oxidation states from +2 to +7.

Step 5: Chromium (Cr) can exhibit oxidation states from +2 to +6.

Step 6: Therefore, Mn shows the maximum oxidation state (+7).

Why other options are wrong:
- Option A: Vanadium's maximum oxidation state is +5, which is less than Mn's +7.
- Option B: Iron's maximum oxidation state is +6, which is less than Mn's +7.
- Option D: Chromium's maximum oxidation state is +6, which is less than Mn's +7.

Key Concepts:
- Oxidation states of transition metals

Memory Aids:
- Remember that Mn can exhibit a wide range of oxidation states, including +7.

Quick Tips:
- Know the common oxidation states of transition metals.

45 Easy

45 of 65Question permalink

Analyzing +4 | -1

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R
Assertion A: The reduction of a metal oxide is easier if the metal formed is in liquid state than solid state.
Reason R: The value of

\Delta \mathrm{G}^{\ominus}

becomes more on negative side as entropy is higher in liquid state than solid state.
In the light of the above statements, choose the most appropriate answer from the options given below

View answer and explanation

Correct option: A

Both A and R are correct and R is the correct explanation of A

Option A (Both A and R are correct and R is the correct explanation of A) is correct because the reduction of a metal oxide is easier when the metal is in the liquid state due to the higher entropy, making the Gibbs free energy change more negative.

Solution approach

Correct Answer + Core Reason: Both A and R are correct and R is the correct explanation of A. Higher entropy in the liquid state makes the reduction easier.

Step-by-Step Solution:

Step 1: Understand Assertion A.\nThe reduction of a metal oxide (e.g., $M_xO_y$ ) to a metal (M) is easier if the metal is formed in the liquid state compared to the solid state.

Step 2: Understand Reason R.\nThe value of $\Delta G^{\ominus}$ becomes more negative as entropy is higher in the liquid state than in the solid state. Since $\Delta G = \Delta H - T\Delta S$ , a larger (more positive) $\Delta S$ will make $\Delta G$ more negative, favoring the reaction.

Step 3: Connect Assertion and Reason.\nA more negative $\Delta G$ indicates a more spontaneous (and therefore easier) reaction. The higher entropy of the liquid state directly contributes to a more negative $\Delta G$ , making the reduction easier.

Step 4: Conclusion.\nBoth the assertion and the reason are correct, and the reason correctly explains why the assertion is true.

Why other options are wrong:
- Option B: While both A and R are correct, R *is* the correct explanation of A.
- Option C: R is correct, so A cannot be correct while R is incorrect.
- Option D: A is correct, so A cannot be incorrect while R is correct.

Key Concepts:
- Gibbs Free Energy ( $\Delta G = \Delta H - T\Delta S$ )
- Entropy (S) and its relation to the state of matter
- Spontaneity of reactions

Memory Aids:
- Remember that liquids have higher entropy than solids.
- More negative $\Delta G$ = more spontaneous reaction.

Quick Tips:
- When evaluating assertion-reason questions, first determine if each statement is true. Then, see if the reason logically explains the assertion.

46 Easy

46 of 65Question permalink

Analyzing +4 | -1

In

\mathrm{C} + {\mathrm{H}_2}\mathrm{O} \to \mathrm{CO} + {\mathrm{H}_2};{\mathrm{H}_2}\mathrm{O acts as}:

View answer and explanation

Correct option: A

Oxidant

Option A (Oxidant) is correct because in the reaction, H2O accepts electrons, thus acting as an oxidizing agent.

Solution approach

Correct Answer + Core Reason: Oxidant. H2O acts as an oxidant by accepting electrons.

Step-by-Step Solution:

Step 1: Identify the oxidation states of the elements in the reaction: $\mathrm{C} + {\mathrm{H}_2}\mathrm{O} \to \mathrm{CO} + {\mathrm{H}_2}$

Step 2: Carbon goes from 0 to +2 (oxidation).

Step 3: Hydrogen in H2O has an oxidation state of +1. In H2, it has an oxidation state of 0. Thus, hydrogen is reduced.

Step 4: Since H2O causes the oxidation of carbon, it itself gets reduced and acts as an oxidizing agent (oxidant).

Why other options are wrong:
- Option B: Reductant: A reductant donates electrons and gets oxidized. H2O accepts electrons, so it's not a reductant.
- Option C: Both (a) and (b): H2O cannot act as both oxidant and reductant simultaneously in this reaction.
- Option D: None of these: H2O clearly acts as an oxidant in this reaction.

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

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

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

47 Easy

47 of 65Question permalink

Application +4 | -1

When the ion

\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }

acts as an oxidant in acidic aqueous solution the ion

\mathrm{C}{\mathrm{r}^{3 + }}\,\mathrm{is}\,\mathrm{formed}.

. How many mole of

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

) would be oxidised to

\mathrm{S}{\mathrm{n}^{4 + }}

by one of

\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }

ions?

View answer and explanation

Correct option: D

3

Option D (3) is correct because each dichromate ion gains 6 electrons, while each tin(II) ion loses 2 electrons, requiring 3 tin(II) ions per dichromate ion.

Solution approach

Correct Answer + Core Reason: 3. One mole of $\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }$ oxidizes 3 moles of $\mathrm{S}{\mathrm{n}^{2 + }}$ to $\mathrm{S}{\mathrm{n}^{4 + }}$ .

Step-by-Step Solution:

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

Step 2: Oxidation half-reaction: $\mathrm{S}{\mathrm{n}^{2 + }} \to \mathrm{S}{\mathrm{n}^{4 + }} + 2{\mathrm{e}^ - }$

Step 3: Reduction half-reaction: $\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - } + 14{\mathrm{H}^ + } + 6{\mathrm{e}^ - } \to 2\mathrm{C}{\mathrm{r}^{3 + }} + 7{\mathrm{H}_2}\mathrm{O}$

Step 4: Balance the number of electrons in both half-reactions. Multiply the oxidation half-reaction by 3 to get 6 electrons.

Step 5: Balanced oxidation half-reaction: $3\mathrm{S}{\mathrm{n}^{2 + }} \to 3\mathrm{S}{\mathrm{n}^{4 + }} + 6{\mathrm{e}^ - }$

Step 6: Now, combine the balanced half-reactions: $\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - } + 14{\mathrm{H}^ + } + 3\mathrm{S}{\mathrm{n}^{2 + }} \to 2\mathrm{C}{\mathrm{r}^{3 + }} + 3\mathrm{S}{\mathrm{n}^{4 + }} + 7{\mathrm{H}_2}\mathrm{O}$

Step 7: From the balanced equation, 1 mole of $\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }$ reacts with 3 moles of $\mathrm{S}{\mathrm{n}^{2 + }}$ .

Why other options are wrong:
- Option A: 2/3 is incorrect because it represents the inverse ratio of moles of dichromate to tin(II).
- Option B: 3/2 is incorrect as it doesn't represent the correct stoichiometric ratio.
- Option C: 2 is incorrect because the stoichiometry requires 3 moles of tin(II) per mole of dichromate.

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

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

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

48 Easy

48 of 65Question permalink

Application +4 | -1

The eq. wt. of

\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_2}{\mathrm{O}_3}

as reductant, in the reaction,

\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_2}{\mathrm{O}_3} + 5{\mathrm{H}_2}\mathrm{O} + 4\mathrm{C}{\mathrm{l}_2} \to 2\mathrm{NaHS}{\mathrm{O}_4} + 8\mathrm{HCl}:

View answer and explanation

Correct option: D

(Mol. wt.)/8

Option D ((Mol. wt.)/8) is correct because the change in oxidation state of S in Na2S2O3 is from +2 to +6/2 = +3. Therefore, the n-factor is 2 * (3-2) * 2 = 2 * 1 * 2 = 8. Eq. wt = Mol. wt / n-factor = (Mol. wt.)/8

Solution approach

Correct Answer + Core Reason: (Mol. wt.)/8. The equivalent weight is the molecular weight divided by the n-factor, which is 8 in this case.

Step-by-Step Solution:

Step 1: Determine the oxidation state of S in Na2S2O3.
- 2(+1) + 2x + 3(-2) = 0 => 2x = +4 => x = +2

Step 2: Determine the oxidation state of S in NaHSO4.
- +1 + x + 4(-2) = 0 => x = +7

Step 3: Determine the change in oxidation state of S.
- The average oxidation state of S in NaHSO4 is +6 (since there are two S atoms in Na2S2O3, and they form 2NaHSO4 molecules, each with one S atom). The change in oxidation state per S atom is 6-2 = 4. Since there are two S atoms, the total change is 2*4 = 8.

Step 4: Calculate the equivalent weight.
- Equivalent weight = (Molecular weight) / (change in oxidation state) = (Mol. wt.)/8

Why other options are wrong:
- Option A: The n-factor is not 1.
- Option B: The n-factor is not 2.
- Option C: The n-factor is not 6.

Key Concepts:
- Equivalent Weight
- Oxidation State

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

Quick Tips:
- Balance the redox reaction to determine the n-factor.

49 Easy

49 of 65Question permalink

Application +4 | -1

Oxidation state of oxygen in

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

is

View answer and explanation

Correct option: A

- 1

Option A (-1) is correct because in hydrogen peroxide (H2O2), the oxidation state of oxygen is -1.

Solution approach

Correct Answer + Core Reason: $- 1$ . The oxidation state of oxygen in H2O2 is -1.

Step-by-Step Solution:

Step 1: Determine the oxidation state of H. Hydrogen (H) has an oxidation state of +1.

Step 2: Calculate the oxidation state of O in H2O2.
- 2(+1) + 2x = 0 => 2x = -2 => x = -1

Step 3: State the oxidation state of oxygen.

Why other options are wrong:
- Option B: The oxidation state of oxygen is not +2.
- Option C: The oxidation state of oxygen is not +1/2.
- Option D: The oxidation state of oxygen is not -2. This is the common oxidation state, but not in peroxides.

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.

50 Easy

50 of 65Question permalink

Application +4 | -1

In alkaline condition

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

reacts as follows,

\begin{array}{llllllllllllllllllll} {2\mathrm{KMn}{\mathrm{O}_4} + 2\mathrm{KOH} \to 2{\mathrm{K}_2}\mathrm{Mn}{\mathrm{O}_4} + {\mathrm{H}_2}\mathrm{O} + \mathrm{O}} \end{array}

Therefore, its equivalent weight will be :

View answer and explanation

Correct option: D

158

Option D (158) is correct because the equivalent weight is the molecular weight divided by the change in oxidation number, which is 1 in this case, but the reaction shows 2KMnO4 reacting.

Solution approach

Correct Answer + Core Reason: 158. The equivalent weight is the molecular weight divided by the change in oxidation number, considering the stoichiometry of the reaction.

Step-by-Step Solution:

Step 1: Determine the molecular weight of $\mathrm{KMn}{\mathrm{O}_4}$ : K (39) + Mn (55) + 4O (4 x 16) = 39 + 55 + 64 = 158 g/mol.

Step 2: Determine the change in oxidation number of Mn: Mn goes from +7 in $\mathrm{KMn}{\mathrm{O}_4}$ to +6 in ${\mathrm{K}_2}\mathrm{Mn}{\mathrm{O}_4}$ . The change is 1.

Step 3: Since 2 moles of $\mathrm{KMn}{\mathrm{O}_4}$ are involved in the reaction, the equivalent weight is the molecular weight divided by the number of moles multiplied by the change in oxidation state. However, since the question asks for the equivalent weight of KMnO4 *in this reaction*, and the change in oxidation state is 1, and we are considering the molar mass of KMnO4, the equivalent weight is simply the molar mass.

Step 4: Equivalent weight = $\frac{{\mathrm{Molecular weight}}}{{\mathrm{n - factor}}} = \frac{{158}}{1} = 158$

Why other options are wrong:
- Option A: 31.6 is obtained by dividing the molecular weight by 5, which is not the correct n-factor for this specific reaction.
- Option B: 52.7 is not a correct calculation based on the molecular weight and n-factor.
- Option C: 79 is obtained by dividing the molecular weight by 2, which is not the correct n-factor for this specific reaction.

Key Concepts:
- Equivalent weight
- Molecular weight
- Oxidation number
- n-factor

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

Quick Tips:
- Determine the change in oxidation number to find the n-factor and calculate the equivalent weight.

51 Easy

51 of 65Question permalink

Understanding +4 | -1

Which of the following elements never show positive oxidation number?

View answer and explanation

Correct option: D

F

Option D (F) is correct because fluorine is the most electronegative element and always has a -1 oxidation state.

Solution approach

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

Step-by-Step Solution:

Step 1: Recall the electronegativity trend in the periodic table.

Step 2: Fluorine is the most electronegative element, meaning it has the highest tendency to attract electrons.

Step 3: Therefore, fluorine always has a -1 oxidation state in its compounds.

Step 4: Final Answer: F

Why other options are wrong:
- Option A (O): Oxygen can have positive oxidation states when bonded to fluorine.
- Option B (Fe): Iron can have multiple positive oxidation states.
- Option C (Ga): Gallium can have a +3 oxidation state.

Key Concepts:
- Electronegativity
- Oxidation states

Memory Aids:
- Fluorine is always -1.

Quick Tips:
- Remember the electronegativity trend to predict oxidation states.

52 Easy

52 of 65Question permalink

Remembering +4 | -1

A substance, that by its sharp colour change indicates the completion of reaction is known as :

View answer and explanation

Correct option: C

Indicator

Option C (Indicator) is correct because an indicator is a substance that changes color sharply at the endpoint of a reaction, signaling completion.

Solution approach

Correct Answer + Core Reason: Indicator. An indicator shows a sharp color change at the completion of a reaction.

Step-by-Step Solution:

Step 1: Define the term indicator.\nAn indicator is a substance that undergoes a distinct color change at or near the equivalence point of a titration or at the endpoint of a chemical reaction.

Step 2: Explain the role of an indicator.\nThe sharp color change indicates the completion of the reaction.

Why other options are wrong:
- Option A: An acid is a substance that donates protons or accepts electrons.
- Option B: A base is a substance that accepts protons or donates electrons.
- Option D: This option is incorrect because an indicator is the correct answer.

Key Concepts:
- Indicators
- Titration
- Chemical Reactions

Memory Aids:
- Indicators indicate the end of a reaction.

Quick Tips:
- Remember the definition of an indicator.

53 Easy

53 of 65Question permalink

Application +4 | -1

Eq.wt. of

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

in,

\mathrm{N}{\mathrm{H}_3} + {\mathrm{O}_2} \to \mathrm{NO} + {\mathrm{H}_2}\mathrm{O}

is :

View answer and explanation

Correct option: A

3.4

Option A (3.4) is correct because the equivalent weight is calculated by dividing the molecular weight by the change in oxidation number.

Solution approach

Correct Answer + Core Reason: 3.4. The equivalent weight of $\mathrm{N}{\mathrm{H}_3}$ is calculated based on the change in oxidation number of nitrogen.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $4\mathrm{N}{\mathrm{H}_3} + 5{\mathrm{O}_2} \to 4\mathrm{NO} + 6{\mathrm{H}_2}\mathrm{O}$

Step 2: Determine the change in oxidation number of nitrogen. In $\mathrm{N}{\mathrm{H}_3}$ , the oxidation number of N is -3. In $\mathrm{NO}$ , the oxidation number of N is +2.

Step 3: The change in oxidation number is |+2 - (-3)| = 5.

Step 4: Calculate the equivalent weight using the formula: Equivalent weight = Molecular weight / n-factor, where n-factor is the change in oxidation number.

Step 5: Molecular weight of $\mathrm{N}{\mathrm{H}_3}$ = 14 + 3 = 17.

Step 6: Equivalent weight = 17 / 5 = 3.4.

Why other options are wrong:
- Option B: 17 is the molecular weight of $\mathrm{N}{\mathrm{H}_3}$ , not the equivalent weight.
- Option C: 8.5 is half of the molecular weight, which is not the correct calculation for equivalent weight in this reaction.
- Option D: The equivalent weight can be calculated, so this option is incorrect.

Key Concepts:
- Equivalent Weight
- Oxidation Number
- Balancing Chemical Equations

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

Quick Tips:
- Always balance the chemical equation first to determine the correct stoichiometry.
- Identify the change in oxidation number of the element that is being oxidized or reduced.

54 Easy

54 of 65Question permalink

Application +4 | -1

In which of the following oxygen shows -1 oxidation state?

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because in hydrogen peroxide (H2O2), the oxidation state of oxygen is -1.

Solution approach

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

Step-by-Step Solution:

Step 1: Determine the oxidation state of H. Hydrogen (H) has an oxidation state of +1.

Step 2: Calculate the oxidation state of O in each compound:
- H2O2: 2(+1) + 2x = 0 => x = -1
- CO2: +4 + 2x = 0 => x = -2
- H2O: 2(+1) + x = 0 => x = -2
- OF2: x + 2(-1) = 0 => x = +2

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

Why other options are wrong:
- Option B: The oxidation state of O in CO2 is -2.
- Option C: The oxidation state of O in H2O is -2.
- Option D: The oxidation state of O in OF2 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.

55 Easy

55 of 65Question permalink

Analyzing +4 | -1

Which of the following is not a redox reaction?

View answer and explanation

Correct option: C

\mathrm{AgN}{\mathrm{O}_3} + \mathrm{NaCl} \to \mathrm{AgCl} + \mathrm{NaN}{\mathrm{O}_3}

Option C (

\mathrm{AgN}{\mathrm{O}_3} + \mathrm{NaCl} \to \mathrm{AgCl} + \mathrm{NaN}{\mathrm{O}_3}

) is correct because there is no change in oxidation state of any element in this reaction.

Solution approach

Correct Answer + Core Reason: $\mathrm{AgN}{\mathrm{O}_3} + \mathrm{NaCl} \to \mathrm{AgCl} + \mathrm{NaN}{\mathrm{O}_3}$ . This is a double displacement reaction where no oxidation state changes occur.

Step-by-Step Solution:

Step 1: Identify the oxidation states of each element in the reactants and products.

Step 2: In $\mathrm{AgN}{\mathrm{O}_3}$ , Ag is +1, N is +5, and O is -2. In $\mathrm{NaCl}$ , Na is +1 and Cl is -1.

Step 3: In $\mathrm{AgCl}$ , Ag is +1 and Cl is -1. In $\mathrm{NaN}{\mathrm{O}_3}$ , Na is +1, N is +5, and O is -2.

Step 4: Compare the oxidation states of each element on both sides of the equation. No element has changed its oxidation state.

Step 5: Since there is no change in oxidation state, this is not a redox reaction.

Why other options are wrong:
- Option A: $2\mathrm{Na} + \mathrm{C}{\mathrm{l}_2} \to 2\mathrm{NaCl}$ : Na goes from 0 to +1 (oxidation) and Cl goes from 0 to -1 (reduction).
- Option B: $\mathrm{C} + {\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2}$ : C goes from 0 to +4 (oxidation) and O goes from 0 to -2 (reduction).
- Option D: $\mathrm{Zn} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to \mathrm{ZnS}{\mathrm{O}_4} + {\mathrm{H}_2}$ : Zn goes from 0 to +2 (oxidation) and H goes from +1 to 0 (reduction).

Key Concepts:
- Redox reactions
- Oxidation states
- Double displacement reactions

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

Quick Tips:
- Look for changes in oxidation states to identify redox reactions. If no oxidation state changes, it is not a redox reaction.

56 Easy

56 of 65Question permalink

Remembering +4 | -1

The oxidation number of phosphorus vary from

View answer and explanation

Correct option: A

-3

to

+5

Option A (-3 to +5) is correct because phosphorus can exhibit oxidation states ranging from -3 in phosphides to +5 in phosphates.

Solution approach

Correct Answer + Core Reason: -3 to +5. Phosphorus can exhibit oxidation states from -3 to +5.

Step-by-Step Solution:

Step 1: Consider the possible oxidation states of phosphorus.

Step 2: Recall the range: Phosphorus can have oxidation states ranging from -3 (e.g., in PH3) to +5 (e.g., in P2O5).

Step 3: Final Answer: -3 to +5

Why other options are wrong:
- Option B (-1 to +1): This is a much smaller range than phosphorus can exhibit.
- Option C (-3 to +3): This range does not include the maximum oxidation state of phosphorus (+5).
- Option D (-5 to +1): Phosphorus does not have a -5 oxidation state. Also, the maximum oxidation state is +5, not +1.

Key Concepts:
- Oxidation number range
- Oxidation states of phosphorus

Memory Aids:
- Phosphorus can have a wide range of oxidation states due to its electronic configuration.

Quick Tips:
- Consider the common compounds of the element to determine the range of oxidation states.

57 Easy

57 of 65Question permalink

Remembering +4 | -1

A reducing agent is a substance which can

View answer and explanation

Correct option: B

Donate electrons

Option B (Donate electrons) is correct because a reducing agent is a substance that donates electrons, causing another substance to be reduced.

Solution approach

Correct Answer + Core Reason: Donate electrons. A reducing agent donates electrons.

Step-by-Step Solution:

Step 1: Define a reducing agent.\nA reducing agent is a substance that donates electrons in a redox reaction.

Step 2: Explain the role of a reducing agent.\nBy donating electrons, the reducing agent causes another substance to be reduced.

Why other options are wrong:
- Option A: Accept electrons is the property of an oxidizing agent.
- Option C: Accept protons is the property of a base.
- Option D: Donate protons is the property of an acid.

Key Concepts:
- Reducing Agent
- Redox Reactions

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

Quick Tips:
- Remember the definition of a reducing agent.

58 Easy

58 of 65Question permalink

Understanding +4 | -1

Oxygen can show positive oxidation state in compounds with

View answer and explanation

Correct option: A

Fluorine

Option A (Fluorine) is correct because fluorine is the most electronegative element and forces oxygen to have a positive oxidation state in compounds like

OF_2

.

Solution approach

Correct Answer + Core Reason: Fluorine. Fluorine is the most electronegative element and forces oxygen to have a positive oxidation state in compounds like $OF_2$ .

Step-by-Step Solution:

Step 1: Recall the electronegativity trend.

Step 2: Fluorine is the most electronegative element, meaning it attracts electrons more strongly than any other element.

Step 3: When oxygen is bonded to fluorine, fluorine will have a -1 oxidation state, and oxygen will have a positive oxidation state to balance the charge.

Step 4: Final Answer: Fluorine

Why other options are wrong:
- Option B (Nitrogen): Oxygen is more electronegative than nitrogen, so it will have a negative oxidation state.
- Option C (Hydrogen): Oxygen is more electronegative than hydrogen, so it will have a negative oxidation state.
- Option D (Sulphur): Oxygen is more electronegative than sulphur, so it will have a negative oxidation state.

Key Concepts:
- Electronegativity
- Oxidation states

Memory Aids:
- Fluorine is always -1, and it can force oxygen to be positive.

Quick Tips:
- Consider the electronegativity difference between the elements.

59 Easy

59 of 65Question permalink

Remembering +4 | -1

Oxidation is a process which involves :

View answer and explanation

Correct option: A

de-electronation

Option A (de-electronation) is correct because oxidation is defined as the loss of electrons, which is also known as de-electronation.

Solution approach

Correct Answer + Core Reason: de-electronation. Oxidation is the loss of electrons.

Step-by-Step Solution:

Step 1: Define oxidation.\nOxidation is the process involving the loss of electrons.

Step 2: Relate oxidation to de-electronation.\nDe-electronation means the removal of electrons. Therefore, oxidation is de-electronation.

Why other options are wrong:
- Option B: Electronation is the gain of electrons (reduction).
- Option C: Addition of hydrogen is reduction.
- Option D: Addition of metal is not necessarily oxidation.

Key Concepts:
- Oxidation
- Reduction
- Redox Reactions

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

Quick Tips:
- Remember the definition of oxidation.

60 Easy

60 of 65Question permalink

Remembering +4 | -1

The process in which oxidation number increases is

View answer and explanation

Correct option: C

Oxidation

Option C (Oxidation) is correct because oxidation is the process where the oxidation number increases due to the loss of electrons.

Solution approach

Correct Answer + Core Reason: Oxidation. Oxidation is the process where the oxidation number increases.

Step-by-Step Solution:

Step 1: Define oxidation number.\nThe oxidation number is a measure of the degree of oxidation of an atom in a chemical compound.

Step 2: Relate oxidation to the change in oxidation number.\nIn oxidation, an atom loses electrons, which results in an increase in its oxidation number.

Why other options are wrong:
- Option A: Reduction is the process where the oxidation number decreases.
- Option B: Hydrolysis is the reaction with water.
- Option D: Decomposition is the breakdown of a compound.

Key Concepts:
- Oxidation
- Oxidation Number
- Redox Reactions

Memory Aids:
- Oxidation increases oxidation number.

Quick Tips:
- Remember the definition of oxidation.

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