REDOX REACTIONS MCQs for NEET UG CHEMISTRY — Medium, Page 3

61 Medium

Application +4 | -1

When

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

is passed in acidified potassium dichromate solution, the oxidation state of S is changed from :

View answer and explanation

Correct option: C

+4 to +6

Option C (+4 to +6) is correct because the oxidation of

\mathrm{SO}_2

to

\mathrm{SO}_4^{2-}

results in a change in oxidation state from +4 to +6 for sulfur.

Solution approach

Correct Answer + Core Reason: +4 to +6. The oxidation state of S changes from +4 in SO2 to +6 in sulfate ions.

Step-by-Step Solution:

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

Step 2: When $\mathrm{SO}_2$ is passed through acidified potassium dichromate, it is oxidized to $\mathrm{SO}_4^{2-}$ . Determine the oxidation state of S in $\mathrm{SO}_4^{2-}$ .\nLet y be the oxidation state of S.\ny + 4(-2) = -2\ny - 8 = -2\ny = +6

Step 3: The oxidation state of S changes from +4 to +6.

Why other options are wrong:
- Option A (+ 4 to 0): This would be a reduction, not an oxidation.
- Option B (+4 to +2): This would also be a reduction, not an oxidation.
- Option D (+6 to +4): This is the reverse of what happens; the oxidation state increases.

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

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

Quick Tips:
- Remember that oxygen usually has an oxidation state of -2.

62 Medium

62 of 69Question permalink

Remembering +4 | -1

The lanthanoid that does not show + 4 oxidation state is

View answer and explanation

Correct option: C

Eu

Option C (Eu) is correct because Eu readily achieves a stable +2 oxidation state due to its electronic configuration.

Solution approach

Correct Answer + Core Reason: Eu. Europium readily achieves a stable +2 oxidation state due to its electronic configuration.

Step-by-Step Solution:

Step 1: Consider the electronic configurations of the lanthanoids.

Step 2: Recall that lanthanoids exhibit +2, +3, and +4 oxidation states, but +3 is the most common.

Step 3: Europium (Eu) has a tendency to lose two electrons to achieve a stable f7 configuration, resulting in a +2 oxidation state.

Step 4: Cerium (Ce) and Terbium (Tb) are known to exhibit +4 oxidation states.\nDysprosium (Dy) does not readily show +4 oxidation state.

Step 5: Final Answer: Eu

Why other options are wrong:
- Option A (Dy): Dysprosium can exhibit +4 oxidation state, but it is not as common as in Ce or Tb.
- Option B (Ce): Cerium is known to exhibit a +4 oxidation state.
- Option D (Tb): Terbium is known to exhibit a +4 oxidation state.

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

Memory Aids:
- Remember the stable electronic configurations that influence oxidation states.

Quick Tips:
- Consider the electronic configuration of the lanthanoid to predict its preferred oxidation states.

63 Medium

63 of 69Question permalink

Application +4 | -1

Oxidation numbers of Fe in

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

are :

View answer and explanation

Correct option: A

+ 2\mathrm{ }and\mathrm{ } + 3

Option A (

+ 2\mathrm{ }and\mathrm{ } + 3

) is correct because

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

is a mixed oxide containing both

\mathrm{FeO}

and

\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}

, where iron has oxidation states of +2 and +3 respectively.

Solution approach

Correct Answer + Core Reason: $+ 2\mathrm{ }and\mathrm{ } + 3$ . $\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}$ is a mixed oxide of FeO and $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}$ .

Step-by-Step Solution:

Step 1: Recognize that $\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}$ can be considered as a combination of FeO and $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}$ .
$\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4} = \mathrm{FeO} + \mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}$

Step 2: Calculate the oxidation state of Fe in FeO.\nLet x be the oxidation state of Fe.\nx + (-2) = 0\nx = +2

Step 3: Calculate the oxidation state of Fe in $\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}$ .\nLet y be the oxidation state of Fe.
2y + 3(-2) = 0
2y - 6 = 0
2y = 6\ny = +3

Step 4: Therefore, the oxidation states of Fe in $\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}$ are +2 and +3.

Why other options are wrong:
- **Option B ( $+ 1\mathrm{ }and\mathrm{ } + 2$ ): Incorrect oxidation states for iron in this compound.
- Option C ( $+ 1\mathrm{ }and\mathrm{ } + 3$ ): Incorrect oxidation states for iron in this compound.
- Option D (None of these): This is incorrect as the oxidation states are +2 and +3.

Key Concepts:
- Oxidation state calculation
- Mixed oxides

Memory Aids:**
- Remember that $\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}$ is a mixed oxide.

Quick Tips:
- Recognize common mixed oxides and their constituent oxides.

64 Medium

64 of 69Question permalink

Remembering +4 | -1

Elements which generally exhibit multiple oxidation states and whose ions are coloured are known as :

View answer and explanation

Correct option: D

Transition metals

Option D (Transition metals) is correct because transition metals are known for exhibiting multiple oxidation states and forming colored ions due to d-d transitions.

Solution approach

Correct Answer + Core Reason: Transition metals. They exhibit multiple oxidation states and colored ions.

Step-by-Step Solution:

Step 1: Recall the properties of transition metals.

Step 2: Transition metals have partially filled d-orbitals, allowing them to exhibit variable oxidation states.

Step 3: The presence of unpaired d-electrons also leads to d-d transitions, which result in the absorption of light and the appearance of color in their ions.

Why other options are wrong:
- Option A: Metalloids have properties intermediate between metals and non-metals and do not typically exhibit multiple oxidation states or colored ions to the same extent as transition metals.
- Option B: Non-metals generally do not exhibit multiple oxidation states or form colored ions in the same way as transition metals.
- Option C: While some metals can exhibit multiple oxidation states, transition metals are particularly known for this property and for forming colored ions.

Key Concepts:
- Properties of transition metals
- Variable oxidation states
- d-d transitions and color

Memory Aids:
- Think of transition metals as being "colorful" and having many "transitions" in oxidation states.

Quick Tips:
- Remember the characteristic properties of transition metals.

65 Medium

65 of 69Question permalink

Understanding +4 | -1

The common positive oxidation states for an element with atomic number 24, are

View answer and explanation

Correct option: A

+2 to +6

Option A (+2 to +6) is correct because chromium (atomic number 24) has a electronic configuration that allows it to exhibit multiple oxidation states ranging from +2 to +6.

Solution approach

Correct Answer + Core Reason: +2 to +6. Chromium (Cr) with atomic number 24 can exhibit oxidation states from +2 to +6.

Step-by-Step Solution:

Step 1: Determine the electronic configuration of chromium.

Step 2: The electronic configuration of Cr (Z=24) is [Ar] 3d5 4s1.

Step 3: Chromium can lose 1 electron from 4s and up to 5 electrons from 3d, resulting in oxidation states from +1 to +6. However, +1 and +3 are less common.

Step 4: The common positive oxidation states are +2, +3, +4, +5, and +6. +2 to +6 is the most appropriate answer.

Why other options are wrong:
- Option B: +1 and +3 to +6: +2 is also a common oxidation state.
- Option C: +1 and +3: This is incomplete; it doesn't include all common oxidation states.
- Option D: +1 to +6: +1 is less common than +2.

Key Concepts:
- Electronic configuration
- Oxidation states of transition metals
- Properties of chromium

Memory Aids:
- Remember the electronic configuration of chromium.

Quick Tips:
- Know the electronic configurations of transition metals to predict their oxidation states.

66 Medium

66 of 69Question permalink

Understanding +4 | -1

Among the properties given below, the set of properties shown by

\mathrm{C}{\mathrm{N}^ - }

ion towards metal species is :1. Reducing; 2. Oxidising ; 3. Complexation

View answer and explanation

Correct option: A

1,3

Option A (1,3) is correct because CN- can act as a reducing agent and form complexes with metals.

Solution approach

Correct Answer + Core Reason: 1,3. $\mathrm{CN}^ -$ can act as a reducing agent and form complexes with metals.

Step-by-Step Solution:

Step 1: Consider the properties of $\mathrm{CN}^ -$ ion.

Step 2: $\mathrm{CN}^ -$ can be oxidized, so it can act as a reducing agent (property 1).

Step 3: $\mathrm{CN}^ -$ is a strong ligand and can form complexes with metal ions (property 3).

Step 4: $\mathrm{CN}^ -$ is not typically an oxidizing agent (property 2).

Why other options are wrong:
- Option B: Includes all properties, but $\mathrm{CN}^ -$ is not an oxidizing agent.
- Option C: Includes both reducing and oxidizing properties, but $\mathrm{CN}^ -$ is not typically an oxidizing agent.
- Option D: Includes only complexation and oxidizing properties, but $\mathrm{CN}^ -$ is primarily a reducing agent and complexing agent.

Key Concepts:
- Reducing agent
- Complexation
- Ligand

Memory Aids:
- $\mathrm{CN}^ -$ is a strong-field ligand and forms stable complexes.

Quick Tips:
- Remember the common properties of $\mathrm{CN}^ -$ as a ligand and its ability to be oxidized.

67 Medium

67 of 69Question permalink

Analyzing +4 | -1

Statement-I: In the reaction,

\frac{1}{2} \mathrm{O}_2+\mathrm{F}_2 \rightarrow \mathrm{OF}_2, \mathrm{~F}_2

is oxidant.

Statement-II: Fluorine cannot show positive oxidation state.

View answer and explanation

Correct option: A

Both the statements are correct

Both statements are correct. In the reaction, fluorine is more electronegative than oxygen and thus acts as an oxidant, and fluorine always has a negative oxidation state.

Solution approach

Correct Answer + Core Reason: Both the statements are correct. Fluorine is the most electronegative element and always has a negative oxidation state.

Step-by-Step Solution:

Step 1: Analyze Statement-I: $\frac{1}{2} \mathrm{O}_2+\mathrm{F}_2 \rightarrow \mathrm{OF}_2$ \nIn this reaction, the oxidation state of O increases from 0 to +2, and the oxidation state of F decreases from 0 to -1. Therefore, F₂ is the oxidant (oxidizing agent) as it is being reduced.

Step 2: Analyze Statement-II: Fluorine cannot show a positive oxidation state.\nFluorine is the most electronegative element and always has a negative oxidation state (-1) in its compounds. It cannot show a positive oxidation state.

Step 3: Determine the correctness of both statements.\nBoth statements are correct.

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

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

Memory Aids:
- OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons)
- Fluorine is the most electronegative element.

Quick Tips:
- Remember that the most electronegative element in a compound will have a negative oxidation state.

68 Medium

68 of 69Question permalink

Remembering +4 | -1

Oxidation state of carbon in graphite is:

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because graphite is an allotrope of carbon in its elemental form, and elements in their standard state have an oxidation state of zero.

Solution approach

Correct Answer + Core Reason: Zero. Elements in their standard state have an oxidation state of zero.

Step-by-Step Solution:

Step 1: Recall the definition of oxidation state.

Step 2: An element in its elemental form has an oxidation state of zero.

Step 3: Graphite is an allotrope of carbon in its elemental form.

Step 4: Therefore, the oxidation state of carbon in graphite is zero.

Why other options are wrong:
- Option B: +1 is not the oxidation state of carbon in graphite.
- Option C: +4 is the oxidation state of carbon in compounds like CO₂, but not in graphite.
- Option D: +2 is the oxidation state of carbon in compounds like CO, but not in graphite.

Key Concepts:
- Oxidation state
- Elemental form
- Allotropes of carbon

Memory Aids:
- Remember that elements in their standard state have an oxidation state of zero.

Quick Tips:
- Recognize elements in their elemental form.

69 Medium

69 of 69Question permalink

Remembering +4 | -1

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

in acid medium is always reduced to :

View answer and explanation

Correct option: B

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

Option B (

\mathrm{Mn}^{2+}

) is correct because in acidic medium, KMnO4 is a strong oxidizing agent and is reduced to Mn2+.

Solution approach

Correct Answer + Core Reason: $\mathrm{M}{\mathrm{n}^{2 + }}$ . In acidic medium, $\mathrm{KMnO}_4$ is reduced to $\mathrm{Mn}^{2+}$ .

Step-by-Step Solution:

Step 1: Recall the reduction half-reaction of $\mathrm{KMnO}_4$ in acidic medium.

Step 2: The balanced half-reaction is: $\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \longrightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$

Step 3: Observe the oxidation state of Mn in $\mathrm{KMnO}_4$ and the product.

Step 4: In $\mathrm{KMnO}_4$ , the oxidation state of Mn is +7. In $\mathrm{Mn}^{2+}$ , the oxidation state is +2.

Step 5: Therefore, $\mathrm{KMnO}_4$ is reduced to $\mathrm{Mn}^{2+}$ in acidic medium.

Why other options are wrong:
- Option A: $\mathrm{M}{\mathrm{n}^{4 + }}$ : $\mathrm{KMnO}_4$ is not reduced to $\mathrm{Mn}^{4+}$ in acidic medium. $\mathrm{MnO}_2$ is formed in neutral or slightly basic medium.
- Option C: $\mathrm{M}{\mathrm{n}^{6 + }}$ : $\mathrm{KMnO}_4$ is not reduced to $\mathrm{Mn}^{6+}$ in acidic medium. $\mathrm{MnO}_4^{2-}$ is formed in strongly alkaline medium.
- Option D: $\mathrm{Mn}$ : $\mathrm{KMnO}_4$ is not directly reduced to Mn in acidic medium.

Key Concepts:
- Redox Reactions
- Oxidation State
- Reduction Half-Reaction
- Potassium Permanganate ( $\mathrm{KMnO}_4$ ) as an Oxidizing Agent

Memory Aids:
- Remember the common reduction products of $\mathrm{KMnO}_4$ in different media (acidic, neutral, basic).

Quick Tips:
- Know the standard reduction potentials of common oxidizing agents.

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