If equal volumes of 1$M$ ${\rm{KMn}}{{\rm{O}}_4}$ and $1M{{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}$ solutions are allowed to oxidise ${{\rm{F}}^{2 + }}$ to ${{\rm{F}}^{3 + }}$ in acidic medium volume of oxidant required for one mole of ${{\rm{F}}^{2 + }}$ will be :

${{\rm{V}}_{{\rm{KMn}}{{\rm{O}}_4}}} > {{\rm{V}}_{{{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}}}$

${{\rm{V}}_{{\rm{KMn}}{{\rm{O}}_4}}} < {{\rm{V}}_{{{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}}}$

${{\rm{V}}_{{\rm{KMn}}{{\rm{O}}_4}}} = {{\rm{V}}_{{{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}}}$

$A,B$ and $C$ are three elements forming a part of compound in oxidation states of $ + 2,{\rm{ }} + 5{\rm{ }}and{\rm{ }} - 2$ respectively. What could be the compound?

${A_2}{\left( {B{C_4}} \right)_3}$

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

31 Hard

31 of 43Question permalink

Application +4 | -1

For the reaction,

\mathrm{N}{\mathrm{H}_3} + \mathrm{OC}{\mathrm{l}^ - } \to {\mathrm{N}_2}{\mathrm{H}_4} + \mathrm{C}{\mathrm{l}^ - }

occurring in basic medium, the coefficient of

{\mathrm{N}_2}{\mathrm{H}_4}

in the balanced equation will be

View answer and explanation

Correct option: A

1

Option A (1) is correct because the balanced equation has a coefficient of 1 for

{\mathrm{N}_2}{\mathrm{H}_4}

.

Solution approach

Correct Answer + Core Reason: 1. The balanced equation has a coefficient of 1 for ${\mathrm{N}_2}{\mathrm{H}_4}$ .

Step-by-Step Solution:

Step 1: Write the unbalanced equation: $\mathrm{N}{\mathrm{H}_3} + \mathrm{OC}{\mathrm{l}^ - } \to {\mathrm{N}_2}{\mathrm{H}_4} + \mathrm{C}{\mathrm{l}^ - }$

Step 2: Balance the nitrogen atoms: $2\mathrm{N}{\mathrm{H}_3} + \mathrm{OC}{\mathrm{l}^ - } \to {\mathrm{N}_2}{\mathrm{H}_4} + \mathrm{C}{\mathrm{l}^ - }$

Step 3: Balance the oxygen atoms by adding water molecules: $2\mathrm{N}{\mathrm{H}_3} + \mathrm{OC}{\mathrm{l}^ - } \to {\mathrm{N}_2}{\mathrm{H}_4} + \mathrm{C}{\mathrm{l}^ - } + {\mathrm{H}_2}\mathrm{O}$

Step 4: Balance the hydrogen atoms by adding hydroxide ions: $2\mathrm{N}{\mathrm{H}_3} + \mathrm{OC}{\mathrm{l}^ - } + 2\mathrm{O}{\mathrm{H}^ - } \to {\mathrm{N}_2}{\mathrm{H}_4} + \mathrm{C}{\mathrm{l}^ - } + {\mathrm{H}_2}\mathrm{O}$

Step 5: Check the charge balance. On the left side, the total charge is -1 + (-2) = -3. On the right side, the total charge is -1. To balance the charge, add 2 more hydroxide ions to the right side: $2\mathrm{N}{\mathrm{H}_3} + \mathrm{OC}{\mathrm{l}^ - } + 2\mathrm{O}{\mathrm{H}^ - } \to {\mathrm{N}_2}{\mathrm{H}_4} + \mathrm{C}{\mathrm{l}^ - } + {\mathrm{H}_2}\mathrm{O} + 2\mathrm{O}{\mathrm{H}^ - }$

Step 6: The balanced equation is $2\mathrm{N}{\mathrm{H}_3} + \mathrm{OC}{\mathrm{l}^ - } \to {\mathrm{N}_2}{\mathrm{H}_4} + \mathrm{C}{\mathrm{l}^ - } + {\mathrm{H}_2}\mathrm{O}$ . The coefficient of ${\mathrm{N}_2}{\mathrm{H}_4}$ is 1.

Why other options are wrong:
- Option B: Incorrect coefficient for ${\mathrm{N}_2}{\mathrm{H}_4}$ .
- Option C: Incorrect coefficient for ${\mathrm{N}_2}{\mathrm{H}_4}$ .
- Option D: Incorrect coefficient for ${\mathrm{N}_2}{\mathrm{H}_4}$ .

Key Concepts:
- Balancing redox reactions in basic medium

Memory Aids:
- Remember to balance atoms, then charge.

Quick Tips:
- Balance the equation step by step.

32 Hard

32 of 43Question permalink

Analyzing +4 | -1

All the following can undergo disproportion except

View answer and explanation

Correct option: A

MnO_4^ -

Option A (

MnO_4^ -

) is correct because Mn in

MnO_4^ -

is in its highest oxidation state (+7), and therefore cannot be further oxidized, preventing it from undergoing disproportionation.

Solution approach

Correct Answer + Core Reason: $MnO_4^ -$ . Mn is in its highest oxidation state (+7) and cannot be further oxidized.

Step-by-Step Solution:

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

Step 2: Determine the oxidation state of Mn in each species:
- $MnO_4^-$ : Mn is +7
- $MnO_4^{2-}$ : Mn is +6
- $Cu^+$ : Cu is +1
- $Cl_2$ : Cl is 0

Step 3: Analyze each option:
- $MnO_4^-$ : Mn is in its highest possible oxidation state (+7), so it cannot be oxidized further. Therefore, it cannot undergo disproportionation.
- $MnO_4^{2-}$ : Mn is in an intermediate oxidation state (+6), so it can be both oxidized to +7 and reduced to a lower oxidation state. Therefore, it can undergo disproportionation.
- $Cu^+$ : Cu is in an intermediate oxidation state (+1), so it can be both oxidized to +2 and reduced to 0. Therefore, it can undergo disproportionation.
- $Cl_2$ : Cl is in the 0 oxidation state, so it can be both oxidized to a positive oxidation state and reduced to -1. Therefore, it can undergo disproportionation.

Why other options are wrong:
- Option B: $MnO_4^{2 - }$ can undergo disproportionation because Mn can be both oxidized and reduced.
- Option C: $C{u^ + }$ can undergo disproportionation because Cu can be both oxidized and reduced.
- Option D: $C{l_2}$ can undergo disproportionation because Cl can be both oxidized and reduced.

Key Concepts:
- Disproportionation reactions
- Oxidation states

Memory Aids:
- Disproportionation involves simultaneous oxidation and reduction of the same element.

Quick Tips:
- Identify the oxidation state of the element in each species to determine if disproportionation is possible.

33 Hard

33 of 43Question permalink

Understanding +4 | -1

Which of the following is correct representation of a given molecular equation in ionic form?

6 \mathrm{KI}+2 \mathrm{KMnO}_4+4 \mathrm{H}_2 \mathrm{O} \longrightarrow 3 \mathrm{I}_2+2 \mathrm{MnO}_2+8 \mathrm{KOH}

View answer and explanation

Correct option: A

6 \mathrm{I}^{-}+2 \mathrm{MnO}_4^{-}+4 \mathrm{H}_2 \mathrm{O} \longrightarrow 3 \mathrm{I}_2+2 \mathrm{MnO}_2+8 \mathrm{O} \overline{\mathrm{H}}

Option A is correct because it accurately represents the given molecular equation in its ionic form, showing the dissociation of ionic compounds and keeping the spectator ions.

Solution approach

Correct Answer + Core Reason: $6 \mathrm{I}^{-}+2 \mathrm{MnO}_4^{-}+4 \mathrm{H}_2 \mathrm{O} \longrightarrow 3 \mathrm{I}_2+2 \mathrm{MnO}_2+8 \mathrm{O} \overline{\mathrm{H}}$ . This representation correctly shows the ionic species involved in the redox reaction.

Step-by-Step Solution:

Step 1: Identify the ionic compounds in the given equation: $6 \mathrm{KI}$ and $2 \mathrm{KMnO}_4$ and $8 \mathrm{KOH}$ .

Step 2: Dissociate the ionic compounds into their respective ions: $6 \mathrm{KI} \longrightarrow 6 \mathrm{K}^{+} + 6 \mathrm{I}^{-}$ and $2 \mathrm{KMnO}_4 \longrightarrow 2 \mathrm{K}^{+} + 2 \mathrm{MnO}_4^{-}$ and $8 \mathrm{KOH} \longrightarrow 8 \mathrm{K}^{+} + 8 \mathrm{OH}^{-}$ .

Step 3: Write the ionic equation by including only the ions that participate in the reaction and keeping the spectator ions (ions that do not change their oxidation state) in the equation.
$6 \mathrm{I}^{-}+2 \mathrm{MnO}_4^{-}+4 \mathrm{H}_2 \mathrm{O} \longrightarrow 3 \mathrm{I}_2+2 \mathrm{MnO}_2+8 \mathrm{OH}^{-}$

Why other options are wrong:
- Option B: This option includes all the ions, but it is not the simplest ionic form and is incomplete.
- Option C: This option includes all the ions and the products, but it is not the simplest ionic form and is unnecessarily complex.
- Option D: This option is missing the permanganate ion ( $MnO_4^-$ ) on the reactant side and incorrectly balances the equation.

Key Concepts:
- Ionic equations
- Dissociation of ionic compounds
- Spectator ions
- Redox reactions

Memory Aids:
- Remember to only include ions that participate in the reaction in the net ionic equation.

Quick Tips:
- Always check for the charges of the ions and balance the equation accordingly.

34 Hard

34 of 43Question permalink

Understanding +4 | -1

Assertion: Electrons flow in external circuit of galvanic cell while ions flow in internal circuit.
Reason: Direction of current flow is reverse that of electron flow.

View answer and explanation

Correct option: B

If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

Option B (If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion) is correct because both statements are true, but the direction of current flow does not explain why electrons flow in the external circuit and ions in the internal circuit.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion. Both statements are true, but not related.

Step-by-Step Solution:

Step 1: Analyze the Assertion: Electrons flow in the external circuit of a galvanic cell, while ions flow in the internal circuit. This is true because the external circuit consists of wires where electrons can flow, and the internal circuit (electrolyte) consists of ions that carry the charge.

Step 2: Analyze the Reason: Direction of current flow is reverse that of electron flow. This is also true by convention.

Step 3: Determine if the Reason explains the Assertion: The direction of current flow being opposite to electron flow doesn't explain *why* electrons flow in the external circuit and ions in the internal circuit. The reason is due to the nature of the circuits (wires vs. electrolyte).

Why other options are wrong:
- Option A: The Reason is not the correct explanation of the Assertion.
- Option C: The Reason is correct, not incorrect.
- Option D: Both Assertion and Reason are correct, not incorrect.

Key Concepts:
- Galvanic cell
- Electron flow
- Ion flow
- Current direction

Memory Aids:
- Electrons flow in wires, ions flow in electrolytes.

Quick Tips:
- Carefully analyze the relationship between the Assertion and the Reason to determine if the Reason explains the Assertion.

35 Hard

35 of 43Question permalink

Analyzing +4 | -1

Assertion:

\mathrm{Sn}^{2+}

and

\mathrm{Fe}^{3+}

can't remain together in a solution.
Reason:

\mathrm{Sn}^{2+}

and

\mathrm{Fe}^{3+}

will react mutually to form

\mathrm{Sn}^{4+}

and

\mathrm{Fe}^{2+}

.

View answer and explanation

Correct option: A

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion

Option A is correct because Sn2+ is a good reducing agent and Fe3+ is a good oxidizing agent, leading to their mutual reaction to form Sn4+ and Fe2+.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct and Reason is the correct explanation of Assertion. $Sn^{2+}$ and $Fe^{3+}$ react to form $Sn^{4+}$ and $Fe^{2+}$ due to their redox potentials.

Step-by-Step Solution:

Step 1: Understand the redox properties of $Sn^{2+}$ and $Fe^{3+}$ . $Sn^{2+}$ is a reducing agent, and $Fe^{3+}$ is an oxidizing agent.

Step 2: Write the half-reactions for the oxidation of $Sn^{2+}$ and the reduction of $Fe^{3+}$ .\nOxidation: $Sn^{2+} \to Sn^{4+} + 2e^-$ \nReduction: $Fe^{3+} + e^- \to Fe^{2+}$

Step 3: Combine the half-reactions to form the overall redox reaction. To balance the electrons, multiply the reduction half-reaction by 2: $2Fe^{3+} + 2e^- \to 2Fe^{2+}$ . The overall reaction is: $Sn^{2+} + 2Fe^{3+} \to Sn^{4+} + 2Fe^{2+}$ .

Step 4: Analyze the spontaneity of the reaction. The reaction is spontaneous because $Fe^{3+}$ has a higher reduction potential than $Sn^{2+}$ , making $Fe^{3+}$ a stronger oxidizing agent.

Step 5: Conclude that $Sn^{2+}$ and $Fe^{3+}$ cannot coexist in solution because they will react to form $Sn^{4+}$ and $Fe^{2+}$ .

Why other options are wrong:
- Option B: If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion is incorrect because the reason directly explains the assertion.
- Option C: If Assertion is correct but Reason is incorrect is incorrect because both the assertion and the reason are correct.
- Option D: If both Assertion and Reason are incorrect is incorrect because both the assertion and the reason are correct.

Key Concepts:
- Redox reactions
- Oxidizing and reducing agents
- Redox potentials
- Spontaneity of redox reactions

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

Quick Tips:
- Identify the oxidizing and reducing agents in the solution.
- Write the balanced redox reaction to determine the products.
- Consider the redox potentials to determine the spontaneity of the reaction.

36 Hard

36 of 43Question permalink

Analyzing +4 | -1

Given :

\begin{aligned} {XNa}_2 {HAsO}_3+{YNaBrO}_3+{ZHCl} \rightarrow & {NaBr} \\ & +{H}_3 {AsO}_4+{NaCl} \end{aligned}

The values of

{X}, {Y}

and

{Z}

in the above redox reaction are respectively

View answer and explanation

Correct option: C

3,1,6

Option C (3,1,6) is correct because it correctly balances the redox reaction, ensuring that the number of atoms of each element and the total charge are the same on both sides of the equation.

Solution approach

Correct Answer + Core Reason: 3,1,6. Balancing the redox reaction gives these coefficients.

Step-by-Step Solution:

Step 1: Write the unbalanced equation:
${Na}_2 {HAsO}_3 + {NaBrO}_3 + {HCl} \rightarrow {NaBr} + {H}_3 {AsO}_4 + {NaCl}$

Step 2: Identify the oxidation number changes:\nAs goes from +3 in ${HAsO}_3^{2-}$ to +5 in ${H_3AsO_4}$ (oxidation, change of 2)\nBr goes from +5 in ${BrO}_3^-$ to -1 in ${Br^-}$ (reduction, change of 6)

Step 3: Balance the oxidation number changes:\nTo balance the changes, multiply the As-containing species by 3 and the Br-containing species by 1.
$3{Na}_2 {HAsO}_3 + {NaBrO}_3 + {HCl} \rightarrow {NaBr} + 3{H}_3 {AsO}_4 + {NaCl}$

Step 4: Balance Na atoms:
$3{Na}_2 {HAsO}_3$ has 6 Na atoms, and ${NaBrO}_3$ has 1 Na atom, so there are 7 Na atoms on the reactant side. On the product side, we have 1 Na in NaBr and some in NaCl. So we need 6 NaCl.
$3{Na}_2 {HAsO}_3 + {NaBrO}_3 + {HCl} \rightarrow {NaBr} + 3{H}_3 {AsO}_4 + 6{NaCl}$

Step 5: Balance Cl atoms:\nWe have 6 Cl atoms in 6 NaCl and 1 Cl atom in NaBr, so we need 7 Cl atoms on the product side. Thus, we need 6 HCl on the reactant side.
$3{Na}_2 {HAsO}_3 + {NaBrO}_3 + 6{HCl} \rightarrow {NaBr} + 3{H}_3 {AsO}_4 + 6{NaCl}$

Step 6: Balance H atoms:\nOn the reactant side, we have 3 H atoms in $3{Na}_2 {HAsO}_3$ and 6 H atoms in 6 HCl, for a total of 9 H atoms. On the product side, we have 9 H atoms in $3{H}_3 {AsO}_4$ . The H atoms are balanced.

Step 7: Balance O atoms:\nOn the reactant side, we have 9 O atoms in $3{Na}_2 {HAsO}_3$ and 3 O atoms in ${NaBrO}_3$ , for a total of 12 O atoms. On the product side, we have 12 O atoms in $3{H}_3 {AsO}_4$ . The O atoms are balanced.

Step 8: The balanced equation is:
$3{Na}_2 {HAsO}_3 + {NaBrO}_3 + 6{HCl} \rightarrow {NaBr} + 3{H}_3 {AsO}_4 + 6{NaCl}$ \nThus, X=3, Y=1, Z=6.

Why other options are wrong:
- Option A: 2,1,2 is incorrect because it does not balance the equation.
- Option B: 2,1,3 is incorrect because it does not balance the equation.
- Option D: 3,1,4 is incorrect because it does not balance the equation.

Key Concepts:
- Balancing redox reactions
- Oxidation numbers

Memory Aids:
- Remember to balance both atoms and charge.

Quick Tips:
- Start by balancing the elements that undergo oxidation and reduction.

37 Hard

37 of 43Question permalink

Analyzing +4 | -1

If equal volumes of 1

M

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

and

1M{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}

solutions are allowed to oxidise

{\mathrm{F}^{2 + }}

to

{\mathrm{F}^{3 + }}

in acidic medium volume of oxidant required for one mole of

{\mathrm{F}^{2 + }}

will be :

View answer and explanation

Correct option: A

{\mathrm{V}_{\mathrm{KMn}{\mathrm{O}_4}}} > {\mathrm{V}_{{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}}}

Option A is correct because KMnO4 requires 5 moles of Fe2+ per mole of KMnO4, while K2Cr2O7 requires 6 moles of Fe2+ per mole of K2Cr2O7, thus KMnO4 will have a larger volume.

Solution approach

Correct Answer + Core Reason: ${\mathrm{V}_{\mathrm{KMn}{\mathrm{O}_4}}} > {\mathrm{V}_{{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}}}$ . $KMnO_4$ requires a larger volume because it oxidizes $Fe^{2+}$ more efficiently per mole.

Step-by-Step Solution:

Step 1: Write the balanced redox reactions for the oxidation of $Fe^{2+}$ by $KMnO_4$ and $K_2Cr_2O_7$ in acidic medium.

Step 2: Reaction with $KMnO_4$ : $MnO_4^- + 5Fe^{2+} + 8H^+ \to Mn^{2+} + 5Fe^{3+} + 4H_2O$ . 1 mole of $KMnO_4$ oxidizes 5 moles of $Fe^{2+}$ .

Step 3: Reaction with $K_2Cr_2O_7$ : $Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \to 2Cr^{3+} + 6Fe^{3+} + 7H_2O$ . 1 mole of $K_2Cr_2O_7$ oxidizes 6 moles of $Fe^{2+}$ .

Step 4: Calculate the volume of each oxidant required to oxidize 1 mole of $Fe^{2+}$ .\nFor $KMnO_4$ : Since 1 mole of $KMnO_4$ oxidizes 5 moles of $Fe^{2+}$ , to oxidize 1 mole of $Fe^{2+}$ , we need $\frac{1}{5}$ moles of $KMnO_4$ . Since the concentration is 1M, the volume required is $\frac{1}{5}$ L.\nFor $K_2Cr_2O_7$ : Since 1 mole of $K_2Cr_2O_7$ oxidizes 6 moles of $Fe^{2+}$ , to oxidize 1 mole of $Fe^{2+}$ , we need $\frac{1}{6}$ moles of $K_2Cr_2O_7$ . Since the concentration is 1M, the volume required is $\frac{1}{6}$ L.

Step 5: Compare the volumes. $\frac{1}{5} > \frac{1}{6}$ , therefore, the volume of $KMnO_4$ required is greater than the volume of $K_2Cr_2O_7$ required.

Why other options are wrong:
- Option B: ${\mathrm{V}_{\mathrm{KMn}{\mathrm{O}_4}}} < {\mathrm{V}_{{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}}}$ is incorrect because $KMnO_4$ requires a smaller molar amount to oxidize the same amount of $Fe^{2+}$ .
- Option C: ${\mathrm{V}_{\mathrm{KMn}{\mathrm{O}_4}}} = {\mathrm{V}_{{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}}}$ is incorrect as the stoichiometry of the reactions is different.
- Option D: Nothing can be predicted is incorrect because we can predict the relationship based on the balanced redox reactions.

Key Concepts:
- Redox reactions
- Stoichiometry of redox reactions
- Molarity and volume calculations

Memory Aids:
- Remember the balanced redox reactions for common oxidizing agents.

Quick Tips:
- Always balance the redox reactions before comparing the volumes of oxidants required.
- Use the stoichiometry of the balanced reactions to determine the molar ratios.

38 Hard

38 of 43Question permalink

Understanding +4 | -1

The oxidation state of

+ 3

for phosphorus is in:

View answer and explanation

Correct option: D

Phosphorous acid

Option D (Phosphorous acid) is correct because in phosphorous acid (H3PO3), the oxidation state of phosphorus is +3.

Solution approach

Correct Answer + Core Reason: Phosphorous acid. The oxidation state of P is +3 in phosphorous acid (H3PO3).

Step-by-Step Solution:

Step 1: Determine the formula for each acid.

Step 2: Calculate the oxidation state of P in each acid:
- Hypophosphorous acid (H3PO2): 3(+1) + x + 2(-2) = 0 => x = +1
- Meta-phosphoric acid (HPO3): +1 + x + 3(-2) = 0 => x = +5
- Ortho-phosphoric acid (H3PO4): 3(+1) + x + 4(-2) = 0 => x = +5
- Phosphorous acid (H3PO3): 3(+1) + x + 3(-2) = 0 => x = +3

Step 3: Identify the acid with P in the +3 oxidation state.

Why other options are wrong:
- Option A: The oxidation state of P in hypophosphorous acid is +1.
- Option B: The oxidation state of P in meta-phosphoric acid is +5.
- Option C: The oxidation state of P in ortho-phosphoric acid is +5.

Key Concepts:
- Oxidation State
- Acid formulas

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

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

39 Hard

39 of 43Question permalink

Understanding +4 | -1

Oxidants are substances which :

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because oxidants gain electrons, decrease in oxidation number, and oxidize others while reducing themselves.

Solution approach

Correct Answer + Core Reason: All of the above. Oxidants gain electrons and reduce their oxidation number.

Step-by-Step Solution:

Step 1: Recall the definition of an oxidant (oxidizing agent).

Step 2: Oxidants cause oxidation in other substances and are themselves reduced.

Step 3: Reduction involves gaining electrons.

Step 4: Gaining electrons results in a decrease in oxidation number.

Step 5: Therefore, all the given statements are correct.

Why other options are wrong:
- Option A: Only includes one aspect of oxidants.
- Option B: Only includes one aspect of oxidants.
- Option C: Only includes one aspect of oxidants.

Key Concepts:
- Oxidant (Oxidizing agent)
- Oxidation
- Reduction
- Oxidation number

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

Quick Tips:
- Remember the key characteristics of oxidants: they gain electrons and decrease in oxidation number.

40 Hard

40 of 43Question permalink

Analyzing +4 | -1

A polyvalent metal weighing

0.1 \mathrm{gm}

and having atomic weight of 50 reacted with dilute

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

to give

44.8 \mathrm{~mL}

of hydrogen at STP. The solution containing the metal in this Lower oxidation state, was found to require

60 \mathrm{~mL}

of

0.1 \mathrm{~N}

\mathrm{KMnO}_4

solution for complete oxidation. What are valencies of metal?

View answer and explanation

Correct option: C

+2,+5

Option C (+2, +5) is correct because the metal's valencies are determined by the amount of hydrogen released and the amount of KMnO₄ required for oxidation, leading to +2 and +5 oxidation states.

Solution approach

Correct Answer + Core Reason: +2, +5. The valencies are determined by the hydrogen released and KMnO₄ used.

Step-by-Step Solution:

Step 1: Calculate the number of moles of hydrogen gas released.\nVolume of $H_2$ at STP = 44.8 mL = 0.0448 L\nMoles of $H_2 = \frac{0.0448 \text{ L}}{22.4 \text{ L/mol}} = 0.002 \text{ mol}$

Step 2: Determine the initial valency of the metal (n).\nThe reaction with dilute $H_2SO_4$ is: $M + nH_2SO_4 \rightarrow M^{n+} + nH_2$ \nMoles of metal = $\frac{0.1 \text{ g}}{50 \text{ g/mol}} = 0.002 \text{ mol}$ \nFrom the stoichiometry, moles of $H_2$ released = $\frac{n}{2} \times$ moles of metal
$0.002 = \frac{n}{2} \times 0.002$ \nn = 2\nSo, the initial valency of the metal is +2.

Step 3: Calculate the number of moles of $KMnO_4$ used.\nNormality of $KMnO_4$ = 0.1 N\nVolume of $KMnO_4$ = 60 mL = 0.06 L\nMoles of $KMnO_4 = \text{Normality} \times \text{Volume} = 0.1 \times 0.06 = 0.006 \text{ mol}$

Step 4: Determine the final valency of the metal (m).
$KMnO_4$ oxidizes $M^{2+}$ to $M^{m+}$ . The balanced equation for the reaction is:
$MnO_4^- + M^{2+} \rightarrow Mn^{2+} + M^{m+}$
$MnO_4^-$ gains 5 electrons, and $M^{2+}$ loses (m-2) electrons.
$5 \times \text{moles of } M^{2+} = (m-2) \times \text{moles of } KMnO_4 \times 5$
$5 \times 0.002 = (m-2) \times 0.006 \times 5$
$0.01 = (m-2) \times 0.03$
$\frac{0.01}{0.002} = m - 2$
$5 = m - 2$ \nm = 5\nSo, the final valency of the metal is +5.

Why other options are wrong:
- Option A: Incorrect valencies.
- Option B: Incorrect valencies.
- Option D: Incorrect valencies.

Key Concepts:
- Stoichiometry
- Redox Reactions
- Oxidation Number Calculation

Memory Aids:
- Remember the stoichiometry of redox reactions.

Quick Tips:
- Balance the chemical equations to determine the stoichiometry.

41 Hard

41 of 43Question permalink

Application +4 | -1

A sample of

2.5 \mathrm{~mol}

of hydrazine

\left(\mathrm{N}_2 \mathrm{H}_4\right)

loses

25 \mathrm{~mol}

of electrons on being converted to a new compound

\mathrm{X}.

Assuming that there is no loss of nitrogen in the formation of the new compound, what is the oxidation number of nitrogen in compound

\mathrm{X}?

View answer and explanation

Correct option: C

+3

Option C (+3) is correct because the initial oxidation state of N is -2, and it loses 5 electrons per N atom, resulting in a final oxidation state of +3.

Solution approach

Correct Answer + Core Reason: +3. The oxidation number of nitrogen is calculated based on the number of electrons lost during the conversion.

Step-by-Step Solution:

Step 1: Determine the initial oxidation number of nitrogen in hydrazine (N2H4). 2x + 4(+1) = 0 => 2x = -4 => x = -2

Step 2: Calculate the number of moles of nitrogen atoms: 2.5 mol N2H4 * 2 N/N2H4 = 5 mol N

Step 3: Calculate the number of electrons lost per nitrogen atom: 25 mol electrons / 5 mol N = 5 electrons/N

Step 4: Calculate the final oxidation number of nitrogen: -2 + 5 = +3

Why other options are wrong:
- Option A (-1): This is incorrect because it does not account for the loss of 5 electrons per nitrogen atom.
- Option B (-2): This is the initial oxidation number, not the final one.
- Option D (+4): This is incorrect because the calculation shows the final oxidation state is +3.

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 consider the number of atoms when calculating oxidation numbers and electron transfer.

42 Hard

42 of 43Question permalink

Application +4 | -1

A,B

and

C

are three elements forming a part of compound in oxidation states of

+ 2,\mathrm{ } + 5\mathrm{ }and\mathrm{ } - 2

respectively. What could be the compound?

View answer and explanation

Correct option: C

{A_3}{\left( {B{C_4}} \right)_2}

Option C (

{A_3}{\left( {B{C_4}} \right)_2}

) is correct because it satisfies the charge neutrality condition based on the given oxidation states.

Solution approach

Correct Answer + Core Reason: ${A_3}{\left( {B{C_4}} \right)_2}$ . This formula satisfies the charge neutrality condition.

Step-by-Step Solution:

Step 1: Identify the oxidation states of A, B, and C: A = +2, B = +5, C = -2.

Step 2: Check each option to see if the compound is neutral.

Step 3: Option A: ${A_2}{\left( {BC} \right)_2}$ = $2(+2) + 2[+5 + (-2)] = 4 + 2(3) = 4 + 6 = +10$ . Not neutral.

Step 4: Option B: ${A_2}{\left( {B{C_4}} \right)_3}$ = $2(+2) + 3[+5 + 4(-2)] = 4 + 3[5 - 8] = 4 + 3(-3) = 4 - 9 = -5$ . Not neutral.

Step 5: Option C: ${A_3}{\left( {B{C_4}} \right)_2}$ = $3(+2) + 2[+5 + 4(-2)] = 6 + 2[5 - 8] = 6 + 2(-3) = 6 - 6 = 0$ . Neutral.

Step 6: Option D: $ABC$ = $+2 + 5 + (-2) = +5$ . Not neutral.

Step 7: Only option C results in a neutral compound.

Why other options are wrong:
- Option A: The compound is not neutral.
- Option B: The compound is not neutral.
- Option D: The compound is not neutral.

Key Concepts:
- Oxidation states
- Charge neutrality

Memory Aids:
- The sum of oxidation states in a neutral compound is zero.

Quick Tips:
- Check the charge of each compound.

43 Hard

43 of 43Question permalink

Application +4 | -1

A, B, C are three element forming a compound in oxidation state of

+1,+6

, and -2 respectively. the compound is :

View answer and explanation

Correct option: C

\mathrm{A}_2 \mathrm{BC}_4

Option C (A2BC4) is correct because it balances the charges: 2(+1) + (+6) + 4(-2) = 0.

Solution approach

Correct Answer + Core Reason: A2BC4. The compound is formed by balancing the charges of the elements.

Step-by-Step Solution:

Step 1: Determine the charges of the elements: A = +1, B = +6, C = -2.

Step 2: To form a neutral compound, the sum of the charges must be zero. Test each option:
- A2BC: 2(+1) + (+6) + (-2) = 6 ≠ 0
- AB2C4: (+1) + 2(+6) + 4(-2) = 5 ≠ 0
- A2BC4: 2(+1) + (+6) + 4(-2) = 2 + 6 - 8 = 0
- ABC4: (+1) + (+6) + 4(-2) = -1 ≠ 0

Step 3: The only option that results in a neutral compound is A2BC4.

Why other options are wrong:
- Option A (A2BC): The sum of the charges is not zero.
- Option B (AB2C4): The sum of the charges is not zero.
- Option D (ABC4): The sum of the charges is not zero.

Key Concepts:
- Charge neutrality of compounds

Memory Aids:
- The sum of oxidation states in a neutral compound is zero.

Quick Tips:
- Always check if the sum of the charges is zero for a neutral compound.

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