Question 1: Palette 1/40

DPT MCQ

Concept: In covalent compounds, oxidation numbers are assigned by considering the electronegativity of bonded atoms. Electrons in a covalent bond are assigned to the more electronegative atom.
* Hydrogen ($\text{H}$) generally has an oxidation state of $+1$.
* Nitrogen ($\text{N}$) is more electronegative than carbon ($\text{C}$) and hydrogen ($\text{H}$), having a characteristic valence contribution of $-3$.
* In $\text{HCN}$ ($\text{H}-\text{C}\equiv\text{N}$): Hydrogen donates $1$ electron to carbon ($+1$ for $\text{H}$), and carbon shares $3$ electrons with the more electronegative nitrogen ($-3$ for $\text{N}$). Since the total charge is $0$:
$(+1) + x + (-3) = 0 \implies x = +2$
* In $\text{HNC}$ ($\text{H}-\text{N}\equiv\text{C}$ or $\text{H}-\text{N}^{+}\equiv\text{C}^{-}$): Nitrogen is bonded to hydrogen and carbon. Nitrogen receives $1$ electron from hydrogen ($+1$ for $\text{H}$) and forms a triple/dative bond structure with carbon. Since nitrogen maintains its oxidation state of $-3$ and hydrogen is $+1$, carbon's oxidation state $x$ is determined by:
$(+1) + (-3) + x = 0 \implies x = +2$
Thus, the oxidation number of carbon ($\text{C}$) is $+2$ in both $\text{HCN}$ and $\text{HNC}$.

1. The oxidation numbers of C in HCN and HNC, respectively, are

A

$+2, +2$

B

$+2, +4$

C

$+4, +4$

D

$-2, -2$

Question 2: Palette 2/40

DPT MCQ

Concept: Potassium ($\text{K}$) is an alkali metal (Group 1 element). Group 1 metals always exhibit an oxidation state of $+1$ in their stable compounds.
In $\text{KO}_2$ (potassium superoxide), the compound consists of a potassium cation $\text{K}^+$ and a superoxide anion $\text{O}_2^-$. Since the overall charge of the neutral molecule is zero, the oxidation state of potassium is $+1$, while oxygen has an average oxidation state of $-\frac{1}{2}$ per atom.

2. The oxidation number of $\text{K}$ in $\text{KO}_2$ is:

A

$+4$

B

$+1$

C

$+\frac{1}{2}$

D

$-\frac{1}{2}$

Question 3: Palette 3/40

DPT MCQ

Concept: In neutral inorganic compounds, the sum of the oxidation numbers of all constituent atoms equals zero.
* Barium ($\text{Ba}$) is a Group 2 alkaline earth metal and always exhibits an oxidation state of $+2$.
* Oxygen ($\text{O}$) typically exhibits an oxidation state of $-2$ in oxide and perxenate salts.
* Xenon ($\text{Xe}$) can expand its octet to form hypervalent species such as the perxenate ion ($\text{XeO}_6^{4-}$), achieving its highest oxidation state of $+8$.

3. What is the oxidation state of Xe in $\text{Ba}_2\text{XeO}_6$?

A

0

B

+4

C

+6

D

+8

Question 4: Palette 4/40

DPT MCQ

Concept: The oxidation state of chromium ($\text{Cr}$) in both dichromate ($\text{Cr}_2\text{O}_7^{2-}$) and chromate ($\text{CrO}_4^{2-}$) ions is $+6$. The conversion of potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$) to potassium chromate ($\text{K}_2\text{CrO}_4$) occurs in basic medium without any electron transfer (it is not a redox reaction). Therefore, there is no change in the oxidation number of chromium.

4. When $\text{K}_2\text{Cr}_2\text{O}_7$ is converted into $\text{K}_2\text{CrO}_4$, the change in oxidation number of $\text{Cr}$ is

A

0

B

6

C

4

D

3

Question 5: Palette 5/40

DPT MCQ

Concept: A disproportionation reaction is a special type of redox reaction in which a single species is simultaneously oxidized and reduced.
To determine the behavior of elements in a redox reaction, track their oxidation states before and after the reaction:
* Elemental bromine $\text{Br}_2$ has an oxidation state of $0$.
* In the bromide ion $\text{Br}^-$, the oxidation state of bromine decreases to $-1$ (reduction).
* In the bromate ion $\text{BrO}_3^-$, the oxidation state of bromine increases to $+5$ (oxidation).
Since bromine is both oxidized and reduced, it acts as both the reducing agent and the oxidizing agent.

5. In the reaction $3\text{Br}_2 + 6\text{CO}_3^{2-} + 3\text{H}_2\text{O} \rightarrow 5\text{Br}^- + \text{BrO}_3^- + 6\text{HCO}_3^-$, which statement is correct?

A

Bromine is oxidized and carbonate is reduced.

B

Bromine is oxidized and water is reduced.

C

Bromine is both oxidized and reduced.

D

Bromine is neither oxidized nor reduced.

Question 6: Palette 6/40

DPT MCQ

Concept: A disproportionation reaction is a type of redox reaction in which an element in a single oxidation state is simultaneously oxidized and reduced, forming two products in different oxidation states.
In contrast, a comproportionation (or synproportionation) reaction occurs when two species containing the same element in different oxidation states react to form a single product in an intermediate oxidation state.

6. Which of the following reaction is not a disproportionation reaction?

A

$\text{Br}_2 + \text{CO}_3^{2-} + \text{H}_2\text{O} \rightarrow \text{Br}^- + \text{BrO}_3^- + \text{HCO}_3^-$

B

$\text{P}_4 + \text{OH}^- + \text{H}_2\text{O} \rightarrow \text{PH}_3 + \text{H}_2\text{PO}_2^-$

C

$\text{H}_2\text{S} + \text{SO}_2 \rightarrow \text{S} + \text{H}_2\text{O}$

D

$\text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + \text{O}_2$

Question 7: Palette 7/40

DPT MCQ

Concept: To balance a redox reaction in a basic medium using the ion-electron (half-reaction) method:
1. Separate the chemical equation into oxidation and reduction half-reactions.
2. Balance all atoms except hydrogen and oxygen.
3. Balance oxygen atoms by adding $\text{H}_2\text{O}$ molecules.
4. Balance hydrogen atoms by adding $\text{H}^+$ ions.
5. Convert to basic medium by adding an equal number of $\text{OH}^-$ ions to both sides as there are $\text{H}^+$ ions, combining $\text{H}^+$ and $\text{OH}^-$ to form $\text{H}_2\text{O}$.
6. Balance charges by adding electrons ($e^-$).
7. Equalize the total number of electrons in both half-reactions and add them together.

7. For the redox reaction,
$\text{Zn} + \text{NO}_3^- \rightarrow \text{Zn}^{2+} + \text{NH}_4^+$
in basic medium, the coefficients of $\text{Zn}$, $\text{NO}_3^-$, and $\text{OH}^-$ in the balanced equation, respectively, are:

A

4, 1, 7

B

7, 4, 1

C

4, 1, 10

D

1, 4, 10

Question 8: Palette 8/40

DPT MCQ

Concept: The n-factor is defined as the total change in the oxidation state of the atom involved in the reaction. The formula used is:
n-factor = |Change in oxidation state| $\times$ (Number of atoms involved)
In $\text{KMnO}_4$, the oxidation state of Manganese ($\text{Mn}$) is $+7$.
In $\text{Mn}^{2+}$, the oxidation state of Manganese ($\text{Mn}$) is $+2$.
The change in oxidation state ($\Delta$) is:
$|+7 - (+2)| = 5$
Since there is only one $\text{Mn}$ atom involved in the transition, the n-factor is:
$5 \times 1 = 5$

Find the n-factor for the following chemical change: $\text{KMnO}_4 \xrightarrow{\text{H}^+} \text{Mn}^{2+}$

A

2

B

3

C

5

D

7

Question 9: Palette 9/40

DPT MCQ

Concept: The n-factor is defined as the total change in the oxidation state of the atom involved in the reaction. The formula used is:
n-factor = |Change in oxidation state| $\times$ (Number of atoms involved)
In $\text{KMnO}_4$, the oxidation state of Manganese ($\text{Mn}$) is $+7$.
In $\text{Mn}^{4+}$, the oxidation state of Manganese ($\text{Mn}$) is $+4$.
The change in oxidation state ($\Delta$) is:
$|+7 - (+4)| = 3$
Since there is only one $\text{Mn}$ atom involved in the transition, the n-factor is:
$3 \times 1 = 3$

Find the n-factor for the following chemical change: $\text{KMnO}_4 \xrightarrow{\text{H}_2\text{O}} \text{Mn}^{4+}$

A

1

B

3

C

4

D

7

Question 10: Palette 10/40

DPT MCQ

Concept: The n-factor is defined as the total change in the oxidation state of the atom involved in the reaction. The formula used is:
n-factor = |Change in oxidation state| $\times$ (Number of atoms involved)
In $\text{KMnO}_4$, the oxidation state of Manganese ($\text{Mn}$) is $+7$.
In $\text{Mn}^{6+}$, the oxidation state of Manganese ($\text{Mn}$) is $+6$.
The change in oxidation state ($\Delta$) is:
$|+7 - (+6)| = 1$
Since there is only one $\text{Mn}$ atom involved in the transition, the n-factor is:
$1 \times 1 = 1$

Find the n-factor for the following chemical change: $\text{KMnO}_4 \xrightarrow{\text{OH}^- \text{ (Conc. basic medium)}} \text{Mn}^{6+}$

A

1

B

5

C

6

D

7

Question 11: Palette 11/40

DPT MCQ

Concept: The n-factor is defined as the total change in the oxidation state of the atom involved in the reaction. The formula used is:
n-factor = |Change in oxidation state| $\times$ (Number of atoms involved)
In $\text{K}_2\text{Cr}_2\text{O}_7$, the oxidation state of Chromium ($\text{Cr}$) is $+6$.
In $\text{Cr}^{3+}$, the oxidation state of Chromium ($\text{Cr}$) is $+3$.
The change in oxidation state ($\Delta$) is:
$|+6 - (+3)| = 3$
Since there are $2$ atoms of $\text{Cr}$ in the $\text{K}_2\text{Cr}_2\text{O}_7$ molecule, the n-factor is:
$3 \times 2 = 6$

Find the n-factor for the following chemical change: $\text{K}_2\text{Cr}_2\text{O}_7 \xrightarrow{\text{H}^+} \text{Cr}^{3+}$

A

3

B

6

C

12

D

18

Question 12: Palette 12/40

DPT MCQ

Concept: The n-factor is defined as the total change in the oxidation state of the atom involved in the reaction. The formula used is:
n-factor = |Change in oxidation state| $\times$ (Number of atoms involved)
In the oxalate ion ($\text{C}_2\text{O}_4^{2-}$), the average oxidation state of Carbon ($\text{C}$) is $+3$.
In carbon dioxide ($\text{CO}_2$), the oxidation state of Carbon ($\text{C}$) is $+4$.
The change in oxidation state ($\Delta$) is:
$|+4 - (+3)| = 1$
Since there are $2$ atoms of $\text{C}$ involved in the reactant molecule, the n-factor is:
$1 \times 2 = 2$

Find the n-factor for the following chemical change: $\text{C}_2\text{O}_4^{2-} \rightarrow \text{CO}_2$

A

1

B

2

C

3

D

4

Question 13: Palette 13/40

DPT MCQ

Concept: The n-factor is defined as the total change in the oxidation state of the atom involved in the reaction. The formula used is:
n-factor = |Change in oxidation state| $\times$ (Number of atoms involved)
In $\text{FeSO}_4$, the oxidation state of Iron ($\text{Fe}$) is $+2$.
In $\text{Fe}_2\text{O}_3$, the oxidation state of Iron ($\text{Fe}$) is $+3$.
The change in oxidation state ($\Delta$) is:
$|+3 - (+2)| = 1$
To balance the reaction $\text{FeSO}_4 \rightarrow \text{Fe}_2\text{O}_3$, we have:
$2\text{FeSO}_4 \rightarrow \text{Fe}_2\text{O}_3$
Since there are $2$ atoms of $\text{Fe}$ involved in the reaction unit, the n-factor is:
$1 \times 2 = 2$

Find the n-factor for the following chemical change: $\text{FeSO}_4 \rightarrow \text{Fe}_2\text{O}_3$

A

1

B

2

C

3

D

4

Question 14: Palette 14/40

DPT MCQ

Concept: The n-factor is defined as the total change in the oxidation state of the atom involved in the reaction. The formula used is:
n-factor = |Change in oxidation state| $\times$ (Number of atoms involved)
In $\text{FeSO}_4$, the oxidation state of Iron ($\text{Fe}$) is $+2$.
In $\text{Fe}_2\text{O}_3$, the oxidation state of Iron ($\text{Fe}$) is $+3$.
The change in oxidation state ($\Delta$) is:
$|+3 - (+2)| = 1$
To balance the reaction, we see that $2$ atoms of $\text{Fe}$ are required to form $\text{Fe}_2\text{O}_3$:
$2\text{FeSO}_4 \rightarrow \text{Fe}_2\text{O}_3$
Since there are $2$ atoms of $\text{Fe}$ involved in the transition, the n-factor is:
$1 \times 2 = 2$

Find the n-factor for the following chemical change: $\text{FeSO}_4 \rightarrow \text{Fe}_2\text{O}_3$

A

1

B

2

C

3

D

4

Question 15: Palette 15/40

DPT MCQ

Concept: In a neutral molecule, the sum of the oxidation states of all constituent atoms must equal zero.
For a neutral compound with formula $\text{A}_x\text{B}_y\text{C}_z$:
$x \times (\text{Oxidation number of A}) + y \times (\text{Oxidation number of B}) + z \times (\text{Oxidation number of C}) = 0$

10. Suppose that there are three atoms $\text{A}$, $\text{B}$, and $\text{C}$ whose oxidation numbers are $+6$, $-1$, and $-2$, respectively. What will be the molecular formula of the compound?

A

$\text{AB}_2\text{C}_2$

B

$\text{A}_2\text{BC}_3$

C

$\text{AB}_3\text{C}_2$

D

$\text{ABC}_3$

Question 16: Palette 16/40

DPT MCQ

Concept: Normality ($N$) is defined as the number of gram equivalents of solute per liter of solution:
$\text{Normality } (N) = \frac{W \times 1000}{E \times V(\text{mL})}$
where:
* $W$ is the mass of the solute in grams.
* $V(\text{mL})$ is the volume of the solution in milliliters.
* $E$ is the equivalent weight of the solute, calculated as:
$E = \frac{\text{Molar mass}}{\text{Valence factor}}$
In an acidic medium, $\text{KMnO}_4$ is reduced from $\text{Mn}^{+7}$ to $\text{Mn}^{2+}$, giving it a valence factor (n-factor) of $5$.

Calculate the normality of a solution containing $15.8\text{ g}$ of $\text{KMnO}_4$ in $50\text{ mL}$ of acidic solution.

A

$5\text{ N}$

B

$10\text{ N}$

C

$2\text{ N}$

D

$1\text{ N}$

Question 17: Palette 17/40

DPT MCQ

Concept: Normality ($N$) is related to molarity ($M$) by the formula:
$\text{Normality} = \text{Molarity} \times \text{Valence factor (n-factor)}$
In an acidic medium, potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$) acts as a strong oxidizing agent where chromium changes its oxidation state from $+6$ in $\text{Cr}_2\text{O}_7^{2-}$ to $+3$ in $\text{Cr}^{3+}$.
Since there are $2$ chromium atoms per molecule of $\text{K}_2\text{Cr}_2\text{O}_7$:
$\text{Valence factor} = 2 \times \vert{}+6 - (+3)\vert{} = 6$

Calculate the normality of a solution containing $50\text{ mL}$ of $5\text{ M}$ solution $\text{K}_2\text{Cr}_2\text{O}_7$ in acidic medium.

A

$15\text{ N}$

B

$30\text{ N}$

C

$5\text{ N}$

D

$10\text{ N}$

Question 18: Palette 18/40

DPT MCQ

Concept: According to the law of equivalence, in a redox reaction, the total number of equivalents of the oxidizing agent equals the total number of equivalents of the reducing agent:
$\text{Equivalents of oxidizing agent} = \text{Equivalents of reducing agent}$
$\text{Moles of } \text{KMnO}_4 \times \text{v.f. of } \text{KMnO}_4 = \text{Moles of } \text{Cu}_2\text{S} \times \text{v.f. of } \text{Cu}_2\text{S}$
where $\text{v.f.}$ is the valence factor (n-factor):
1. In acidic medium, $\text{KMnO}_4$ is reduced from $\text{Mn}^{+7}$ to $\text{Mn}^{2+}$, so its valence factor is $5$.
2. In $\text{Cu}_2\text{S}$, both copper and sulfur undergo oxidation:
* $\text{Cu}^+$ ($2$ atoms) gets oxidized to $\text{Cu}^{2+}$: Change in oxidation state = $2 \times \vert{}+2 - (+1)\vert{} = 2$
* $\text{S}^{2-}$ ($1$ atom) gets oxidized to $\text{SO}_2$ ($\text{S}^{+4}$): Change in oxidation state = $1 \times \vert{}+4 - (-2)\vert{} = 6$
* Total valence factor for $\text{Cu}_2\text{S} = 2 + 6 = 8$

Find the number of moles of $\text{KMnO}_4$ needed to oxidize one mole of $\text{Cu}_2\text{S}$ in acidic medium according to the reaction $\text{KMnO}_4 + \text{Cu}_2\text{S} \rightarrow \text{Mn}^{2+} + \text{Cu}^{2+} + \text{SO}_2$.

A

$0.8$

B

$1.6$

C

$2.4$

D

$3.2$

Question 19: Palette 19/40

DPT MCQ

Concept: Molarity ($M$) is defined as the number of moles of solute dissolved in $1\text{ L}$ ($1000\text{ mL}$) of solution:
$\text{Molarity } (M) = \frac{\text{Moles of solute}}{\text{Volume of solution in L}}$
The mass of solute in grams ($W$) present in $1000\text{ mL}$ of solution can be found using:
$W = M \times \text{Molar mass}$
To find the required volume ($V$) containing a specific mass of solute ($W_{\text{given}}$):
$V = \frac{W_{\text{given}}}{\text{Mass per mL of solution}}$

In a reaction vessel, $1.184\text{ g}$ of $\text{NaOH}$ is required to be added for completing the reaction. How many millilitres of $0.15\text{ M}$ $\text{NaOH}$ should be added for this requirement?

A

$197.33\text{ mL}$

B

$150.00\text{ mL}$

C

$250.50\text{ mL}$

D

$120.25\text{ mL}$

Question 20: Palette 20/40

DPT MCQ

Concept: The mole concept relates the quantity of matter in terms of mass, number of particles, and molar definitions:
* A mole of molecules is defined as the amount of a substance that contains Avogadro's number ($N_A = 6.022 \times 10^{23}$) of molecules.
* Gram molecular weight ($\text{GMW}$) is the mass in grams of one mole of molecules (i.e., the mass of $6.022 \times 10^{23}$ molecules).
* The term "gram-molecule" is an older term that is equivalent to one mole of molecules.

Which of the following statement(s) is/are correct?
(a) $\text{gram molecular weight} = \text{molecular weight in grams} = \text{weight of } 6.022 \times 10^{23} \text{ molecules}$
(b) $1\text{ mole} = N_A \text{ molecules} = 6.022 \times 10^{23} \text{ molecules}$
(c) $1\text{ mole} = 1\text{ gram-molecule}$
(d) All of the above

A

(a) and (b) only

B

(b) and (c) only

C

(a) and (c) only

D

All of the above

Question 21: Palette 21/40

DPT MCQ

Concept: In chemistry, a substance is said to exhibit "oxidising behaviour" if it acts as an oxidising agent. An oxidising agent causes the oxidation of another substance by getting reduced itself.
Oxidation is the loss of electrons, which results in an increase in oxidation state.
Reduction is the gain of electrons, which results in a decrease in oxidation state.
Therefore, for a substance to act as an oxidising agent, its own oxidation state must decrease.
In the given options, we analyze the oxidation state of Sulfur ($\text{S}$) in $\text{H}_2\text{SO}_4$ (where $\text{S}$ is in the $+6$ state):
(a) $\text{2PCl}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{2POCl}_3 + \text{2HCl} + \text{SO}_2\text{Cl}_2$: $\text{S}$ remains $+6$.
(b) $\text{2NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{2H}_2\text{O}$: $\text{S}$ remains $+6$.
(c) $\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}$: $\text{S}$ remains $+6$.
(d) $\text{2HI} + \text{H}_2\text{SO}_4 \rightarrow \text{I}_2 + \text{SO}_2 + \text{2H}_2\text{O}$: $\text{S}$ changes from $+6$ to $+4$.
Since the oxidation state of $\text{S}$ decreases from $+6$ to $+4$ in option (d), $\text{H}_2\text{SO}_4$ is being reduced, meaning it is acting as an oxidising agent.

Which of the following reactions represents the oxidising behaviour of $\text{H}_2\text{SO}_4$?

A

$\text{2PCl}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{2POCl}_3 + \text{2HCl} + \text{SO}_2\text{Cl}_2$

B

$\text{2NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{2H}_2\text{O}$

C

$\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}$

D

$\text{2HI} + \text{H}_2\text{SO}_4 \rightarrow \text{I}_2 + \text{SO}_2 + \text{2H}_2\text{O}$

Question 22: Palette 22/40

DPT MCQ

Concept: A substance acts as a **reducing agent** when it undergoes **oxidation**. Oxidation is characterized by the loss of electrons, which leads to an increase in the oxidation state of the atom involved.
In hydrogen peroxide ($\text{H}_2\text{O}_2$), the oxidation state of Oxygen ($\text{O}$) is $-1$.
For $\text{H}_2\text{O}_2$ to act as a reducing agent, its oxidation state must increase (e.g., from $-1$ to $0$ as it forms $\text{O}_2$).
Let's analyze the options:
(a) $\text{2FeCl}_3 + \text{2HCl} + \text{H}_2\text{O}_2 \rightarrow \text{2FeCl}_2 + \text{2H}_2\text{O}$: Here, $\text{H}_2\text{O}_2$ is reduced to $\text{H}_2\text{O}$ (O state goes from $-1$ to $-2$). It acts as an oxidising agent.
(b) $\text{Cl}_2 + \text{H}_2\text{O}_2 \rightarrow \text{2HCl} + \text{O}_2$: In this reaction, $\text{H}_2\text{O}_2$ is oxidized to $\text{O}_2$. The oxidation state of oxygen increases from $-1$ to $0$. Therefore, $\text{H}_2\text{O}_2$ is the reducing agent.
(c) $\text{2HI} + \text{H}_2\text{O}_2 \rightarrow \text{I}_2 + \text{2H}_2\text{O}$: Here, $\text{H}_2\text{O}_2$ is reduced to $\text{H}_2\text{O}$. It acts as an oxidising agent.
(d) $\text{H}_2\text{SO}_5 + \text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{SO}_4 + \text{H}_2\text{O}$: In this reaction, $\text{H}_2\text{O}_2$ is reduced to $\text{H}_2\text{O}$. It acts as an oxidising agent.

In which of the following reactions does $\text{H}_2\text{O}_2$ act as a reducing agent?

A

$\text{2FeCl}_3 + \text{2HCl} + \text{H}_2\text{O}_2 \rightarrow \text{2FeCl}_2 + \text{2H}_2\text{O}$

B

$\text{Cl}_2 + \text{H}_2\text{O}_2 \rightarrow \text{2HCl} + \text{O}_2$

C

$\text{2HI} + \text{H}_2\text{O}_2 \rightarrow \text{I}_2 + \text{2H}_2\text{O}$

D

$\text{H}_2\text{SO}_5 + \text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{SO}_4 + \text{H}_2\text{O}$

Question 23: Palette 23/40

DPT MCQ

Concept: A chemical reaction involves oxidation or reduction when there is a change in the oxidation numbers of the participating species. If the oxidation state of an element increases, it undergoes oxidation. If it decreases, it undergoes reduction. When the oxidation states of all constituent elements remain constant throughout the transformation, the reaction is neither oxidation nor reduction (non-redox reaction).

Which of the following reaction involves neither oxidation nor reduction?

A

$\text{CrO}_4^{2-} \rightarrow \text{Cr}_2\text{O}_7^{2-}$

B

$\text{Cr} \rightarrow \text{CrCl}_3$

C

$\text{Na} \rightarrow \text{Na}^+$

D

$2\text{S}_2\text{O}_3^{2-} \rightarrow \text{S}_4\text{O}_6^{2-}$

Question 24: Palette 24/40

DPT MCQ

Concept: To balance a redox reaction in acidic medium using the ion-electron method:
1. Divide the overall reaction into oxidation and reduction half-reactions.
2. In the reduction half-reaction ($\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$), manganese is reduced from $+7$ to $+2$ by accepting $5$ electrons. Balance oxygen atoms by adding $\text{H}_2\text{O}$ and hydrogen atoms by adding $\text{H}^+$.
3. In the oxidation half-reaction ($\text{H}_2\text{O}_2 \rightarrow \text{O}_2$), oxygen is oxidized from $-1$ to $0$ by releasing $2$ electrons. Balance hydrogen atoms by adding $\text{H}^+$.
4. Multiply each half-reaction by suitable integers to equalize the total electrons gained and lost, then combine them.

5. When the reaction $\text{H}_2\text{O}_2 + \text{MnO}_4^- \rightarrow \text{Mn}^{2+} + \text{O}_2$ is balanced in acidic medium, the coefficients of $\text{H}^+$, $\text{H}_2\text{O}_2$, and $\text{MnO}_4^-$, respectively, are:

A

6, 5, 2

B

2, 5, 6

C

8, 3, 2

D

4, 2, 5

Question 25: Palette 25/40

DPT MCQ

Concept: To balance a complex redox reaction involving oxidation of a metal by dilute acid using the ion-electron (half-reaction) method:
1. Identify oxidation and reduction half-reactions in ionic form:
* Oxidation: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$
* Reduction: $\text{NO}_3^- + 10\text{H}^+ + 8e^- \rightarrow \text{NH}_4^+ + 3\text{H}_2\text{O}$
2. Equalize the electrons lost and gained by multiplying the oxidation reaction by 4 and combining:
$4\text{Zn} + \text{NO}_3^- + 10\text{H}^+ \rightarrow 4\text{Zn}^{2+} + \text{NH}_4^+ + 3\text{H}_2\text{O}$
3. Add spectator ions ($\text{NO}_3^-$) to balance the molecular formula:
Add $9\text{NO}_3^-$ to both sides to combine with $10\text{H}^+$ forming $10\text{HNO}_3$, and to balance $4\text{Zn}^{2+}$ and $\text{NH}_4^+$:
$4\text{Zn} + 10\text{HNO}_3\text{(dil)} \rightarrow 4\text{Zn(NO}_3\text{)}_2 + 3\text{H}_2\text{O} + \text{NH}_4\text{NO}_3$

5 (b). Balance the following redox reaction: $\text{Zn} + \text{HNO}_3\text{(dil)} \rightarrow \text{Zn(NO}_3\text{)}_2 + \text{H}_2\text{O} + \text{NH}_4\text{NO}_3$. What are the stoichiometric coefficients of $\text{Zn}$, $\text{HNO}_3$, and $\text{Zn(NO}_3\text{)}_2$ in the balanced chemical equation, respectively?

A

4, 10, 4

B

1, 4, 1

C

4, 8, 4

D

2, 10, 2

Question 26: Palette 26/40

DPT MCQ

Concept: To balance a redox reaction where multiple elements in a single compound undergo oxidation:
1. Identify the species being oxidized and reduced:
* In $\text{CrI}_3$, chromium ($\text{Cr}^{3+}$) is oxidized to $\text{CrO}_4^{2-}$ ($\text{Cr}^{+6}$) and iodine ($\text{I}^-$) is oxidized to $\text{IO}_4^-$ ($\text{I}^{+7}$).
* Chlorine ($\text{Cl}_2$, oxidation state $0$) is reduced to chloride ions ($\text{Cl}^-$, oxidation state $-1$).
2. Calculate the total change in oxidation state (valence factor) for $\text{CrI}_3$:
* $\text{Cr}^{3+} \rightarrow \text{Cr}^{+6}$: loss of $3$ electrons.
* $3\text{I}^- \rightarrow 3\text{I}^{+7}$: loss of $3 \times 8 = 24$ electrons.
* Total electrons lost per formula unit of $\text{CrI}_3 = 3 + 24 = 27$ electrons.
3. Calculate the valence factor for $\text{Cl}_2$:
* $\text{Cl}_2 \rightarrow 2\text{Cl}^-$: gain of $2$ electrons per $\text{Cl}_2$ molecule.
4. Equalize the total number of electrons lost and gained by multiplying $\text{CrI}_3$ by $2$ and $\text{Cl}_2$ by $27$, then balance the remaining potassium, hydrogen, and oxygen atoms.

5 (c). Balance the following redox reaction: $\text{CrI}_3 + \text{KOH} + \text{Cl}_2 \rightarrow \text{K}_2\text{CrO}_4 + \text{KIO}_4 + \text{KCl} + \text{H}_2\text{O}$. What are the stoichiometric coefficients of $\text{CrI}_3$, $\text{KOH}$, and $\text{Cl}_2$ in the balanced chemical equation, respectively?

A

2, 64, 27

B

1, 32, 14

C

2, 32, 27

D

4, 64, 54

Question 27: Palette 27/40

DPT MCQ

Concept: A disproportionation reaction is a redox reaction in which the same element undergoes both oxidation and reduction simultaneously.
1. Determine the oxidation state of phosphorus in each compound:
* In $\text{P}_2\text{H}_4$, oxidation state of $\text{P} = -1$.
* In $\text{PH}_3$, oxidation state of $\text{P} = -3$.
* In $\text{P}_4$, elemental oxidation state of $\text{P} = 0$.
2. Identify the oxidation and reduction half-reactions:
* Reduction: $\text{P}^{-1} \rightarrow \text{P}^{-3}$ (gain of $2e^-$ per $\text{P}$ atom)
* Oxidation: $4\text{P}^{-1} \rightarrow \text{P}_4^0$ (loss of $4e^-$ total for 4 $\text{P}$ atoms)
3. Equalize the electrons gained and lost to find the balanced coefficients.

5 (d). Balance the disproportionation reaction equation: $\text{P}_2\text{H}_4 \rightarrow \text{PH}_3 + \text{P}_4$. What are the stoichiometric coefficients of $\text{P}_2\text{H}_4$, $\text{PH}_3$, and $\text{P}_4$ in the balanced equation, respectively?

A

6, 8, 1

B

3, 4, 1

C

6, 4, 2

D

2, 3, 1

Question 28: Palette 28/40

DPT MCQ

Concept: In the industrial extraction of white phosphorus, calcium phosphate ($\text{Ca}_3(\text{PO}_4)_2$) is reduced using coke ($\text{C}$) in the presence of silica ($\text{SiO}_2$):
1. Identify oxidation state changes:
* Phosphorus ($\text{P}$) in $\text{PO}_4^{3-}$ is reduced from $+5$ to $0$ in elemental $\text{P}_4$.
* Carbon ($\text{C}$) is oxidized from $0$ to $+2$ in carbon monoxide ($\text{CO}$).
* Calcium ($\text{Ca}$), silicon ($\text{Si}$), and oxygen ($\text{O}$) do not undergo changes in oxidation state.
2. Equalize the total number of electrons lost and gained:
* Each phosphorus atom undergoes a change from $+5$ to $0$ (gains $5e^-$). Since $4$ phosphorus atoms are needed to form one molecule of $\text{P}_4$, $20$ electrons are gained per $\text{P}_4$ formed.
* Each carbon atom undergoes a change from $0$ to $+2$ (loses $2e^-$). Therefore, $10$ carbon atoms are required to donate the $20$ electrons.
3. Balance calcium, silicon, and oxygen atoms accordingly.

5 (e). Balance the redox reaction equation: $\text{Ca}_3(\text{PO}_4)_2 + \text{SiO}_2 + \text{C} \rightarrow \text{CaSiO}_3 + \text{P}_4 + \text{CO}$. What are the stoichiometric coefficients of $\text{Ca}_3(\text{PO}_4)_2$, $\text{SiO}_2$, $\text{C}$, $\text{CaSiO}_3$, $\text{P}_4$, and $\text{CO}$ in the balanced chemical equation, respectively?

A

2, 6, 10, 6, 1, 10

B

1, 3, 5, 3, 1, 5

C

2, 3, 10, 3, 1, 10

D

4, 12, 20, 12, 2, 20

Question 29: Palette 29/40

DPT MCQ

Concept: In a redox half-reaction, both atomic mass and electrical charge must be balanced.
To balance the total charge on both sides of a half-reaction:
$\text{Total charge on left side} + \text{Charge of added electrons} = \text{Total charge on right side}$
Electrons carry a unit negative charge ($-1$) and are added to the side with the higher positive charge to equalize the net electrical charge across the reaction.

6. The number of electrons required to balance the following equation are:
$\text{NO}_3^- + 4\text{H}^+ \rightarrow 2\text{H}_2\text{O} + \text{NO}$

A

2 on right side

B

3 on left side

C

3 on right side

D

5 on left side

Question 30: Palette 30/40

DPT MCQ

Concept: In a redox reaction, the reductant (reducing agent) undergoes oxidation by losing electrons.
The number of moles of electrons involved per mole of reductant is determined by the total change in oxidation state of the element undergoing oxidation within one molecule/formula unit of that reductant:
$\text{Moles of electrons per mole of reductant} = \vert{}\text{Final oxidation state} - \text{Initial oxidation state}\vert{} \times \text{Number of oxidizing atoms}$

7. $2\text{KMnO}_4 + 5\text{H}_2\text{S} + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 2\text{K}^+ + 5\text{S} + 8\text{H}_2\text{O}$. In the above reaction, how many moles of electrons would be involved in the oxidation of $1\text{ mole}$ of reductant?

A

Two

B

Five

C

Ten

D

One

Question 31: Palette 31/40

DPT MCQ

Concept: In a redox reaction, the reductant (reducing agent) undergoes oxidation by losing electrons.
The number of moles of electrons involved per mole of reductant is determined by the total change in oxidation state of the element undergoing oxidation within one molecule/formula unit of that reductant:
$\text{Moles of electrons per mole of reductant} = \vert{}\text{Final oxidation state} - \text{Initial oxidation state}\vert{} \times \text{Number of oxidizing atoms}$

7. $2\text{KMnO}_4 + 5\text{H}_2\text{S} + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 2\text{K}^+ + 5\text{S} + 8\text{H}_2\text{O}$. In the above reaction, how many moles of electrons would be involved in the oxidation of $1\text{ mole}$ of reductant?

A

Two

B

Five

C

Ten

D

One

Question 32: Palette 32/40

DPT MCQ

Concept: To balance the redox reaction using the ion-electron method:
1. Identify the oxidation and reduction half-reactions:
* Oxidation: $\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$
* Reduction: $\text{NO}_3^- + 10\text{H}^+ + 8e^- \rightarrow \text{NH}_4^+ + 3\text{H}_2\text{O}$
2. Equalize the total number of electrons lost and gained:
Multiply the oxidation half-reaction by $4$ so that $8e^-$ are transferred in both half-reactions:
$4\text{Fe} + \text{NO}_3^- + 10\text{H}^+ \rightarrow 4\text{Fe}^{2+} + \text{NH}_4^+ + 3\text{H}_2\text{O}$
3. Add spectator nitrate ions ($\text{NO}_3^-$) to obtain the complete molecular equation:
Add $9\text{NO}_3^-$ to both sides:
$4\text{Fe} + 10\text{HNO}_3 \rightarrow 4\text{Fe(NO}_3\text{)}_2 + \text{NH}_4\text{NO}_3 + 3\text{H}_2\text{O}$

8. In Redox reaction:
$\text{Fe} + \text{HNO}_3 \rightarrow \text{Fe(NO}_3\text{)}_2 + \text{NH}_4\text{NO}_3 + \text{H}_2\text{O}$
the coefficient of $\text{HNO}_3$, $\text{Fe(NO}_3\text{)}_2$, $\text{NH}_4\text{NO}_3$ is:

A

1 : 10 : 4

B

10 : 4 : 1

C

4 : 10 : 1

D

10 : 1 : 4

Question 33: Palette 33/40

DPT MCQ

Concept: Normality ($N$) is defined as the number of gram equivalents of solute per litre of solution:
$\text{Normality } (N) = \frac{\text{Mass of solute (g)}}{\text{Equivalent weight}} \times \frac{1000}{\text{Volume of solution (mL)}}$
The equivalent weight ($\text{Eq. wt.}$) of a salt/reducing agent is calculated by dividing its molar mass by its n-factor (valence factor):
$\text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}}$
For sodium oxalate ($\text{Na}_2\text{C}_2\text{O}_4$):
* Molar mass of $\text{Na}_2\text{C}_2\text{O}_4 = 2(23) + 2(12) + 4(16) = 134\text{ g/mol}$
* In redox reactions, oxalate ion ($\text{C}_2\text{O}_4^{2-}$) undergoes oxidation to $\text{CO}_2$ where carbon changes its oxidation state from $+3$ to $+4$:
$\text{n-factor} = 2 \times \vert{}+4 - (+3)\vert{} = 2$
* Equivalent weight $= \frac{134}{2} = 67\text{ g/eq}$

9. Calculate the normality of a solution containing $13.4\text{ g}$ of sodium oxalate ($\text{Na}_2\text{C}_2\text{O}_4$) in $100\text{ mL}$ solution.

A

$1\text{ N}$

B

$2\text{ N}$

C

$0.5\text{ N}$

D

$4\text{ N}$

Question 34: Palette 34/40

DPT MCQ

Concept: According to the law of equivalence, in a redox reaction, the total milliequivalents ($\text{Meq}$) of the oxidizing agent equal the total milliequivalents of the reducing agent:
$\text{Meq of oxidizing agent } (\text{HNO}_3) = \text{Meq of reducing agent } (\text{Fe}^{2+})$
The number of milliequivalents can be calculated using:
1. For a solution: $\text{Meq} = \text{Molarity } (M) \times \text{n-factor} \times \text{Volume in mL } (V)$
2. For a substance given by mass: $\text{Meq} = \frac{\text{Mass in grams}}{\text{Equivalent weight}} \times 1000 = \frac{\text{Mass in grams}}{\frac{\text{Molar mass}}{\text{n-factor}}} \times 1000$
The n-factor (valence factor) represents the change in oxidation state per mole of substance:
* For $\text{HNO}_3 \rightarrow \text{NO}$, nitrogen changes from $+5$ to $+2$, so $\text{n-factor} = 3$.
* For $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}$, iron changes from $+2$ to $+3$, so $\text{n-factor} = 1$.

10. What volume of $6\text{ M}$ $\text{HNO}_3$ is needed to oxidize $8\text{ g}$ $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$, if $\text{HNO}_3$ gets converted to $\text{NO}$?

A

$8\text{ mL}$

B

$7.936\text{ mL}$

C

$32\text{ mL}$

D

$64\text{ mL}$

Question 35: Palette 35/40

DPT MCQ

Concept: The relationship between Normality ($N$) and Molarity ($M$) of a solution is given by:
$N = M \times \text{n-factor}$
For potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$) acting as an oxidizing agent in acidic medium:
$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$
The change in oxidation state of chromium is from $+6$ to $+3$ per atom (a change of $3$ per chromium atom, and $2 \times 3 = 6$ per formula unit). Thus, the $\text{n-factor} = 6$.
When two solutions of the same solute are mixed, the final normality ($N_f$) is given by:
$N_f = \frac{N_1 V_1 + N_2 V_2}{V_1 + V_2}$

11. Calculate the normality of a solution obtained by mixing $50\text{ mL}$ of $5\text{ M}$ solution of $\text{K}_2\text{Cr}_2\text{O}_7$ and $50\text{ mL}$ of $2\text{ M}$ solution of $\text{K}_2\text{Cr}_2\text{O}_7$ in acidic medium.

A

$21\text{ N}$

B

$15\text{ N}$

C

$42\text{ N}$

D

$10.5\text{ N}$

Question 36: Palette 36/40

DPT MCQ

Concept: When two solutions containing the same solute at different molarities are mixed, the principle of conservation of moles applies:
$\text{Total moles of solute before mixing} = \text{Total moles of solute after mixing}$
$M_1 V_1 + M_2 V_2 = M_3 V_3$
Where:
* $M_1$ and $V_1$ are the molarity and volume of the first solution.
* $M_2$ and $V_2$ are the molarity and volume of the second solution.
* $M_3$ and $V_3$ are the molarity and total volume of the resulting mixture.
* $V_1 + V_2 = V_3$

12. What volume of $6\text{ M } \text{HCl}$ and $2\text{ M } \text{HCl}$ should be mixed to get $2\text{ litres}$ of $3\text{ M } \text{HCl}$?

A

$0.5\text{ L}$ of $6\text{ M } \text{HCl}$ and $1.5\text{ L}$ of $2\text{ M } \text{HCl}$

B

$1.0\text{ L}$ of $6\text{ M } \text{HCl}$ and $1.0\text{ L}$ of $2\text{ M } \text{HCl}$

C

$1.5\text{ L}$ of $6\text{ M } \text{HCl}$ and $0.5\text{ L}$ of $2\text{ M } \text{HCl}$

D

$0.8\text{ L}$ of $6\text{ M } \text{HCl}$ and $1.2\text{ L}$ of $2\text{ M } \text{HCl}$

Question 37: Palette 37/40

DPT MCQ

Concept: According to the principle of equivalence, during a neutralization reaction, the milliequivalents ($\text{Meq}$) of base react completely with an equal number of milliequivalents of acid:
$\text{Meq of pure } \text{Na}_2\text{CO}_3 = \text{Meq of } \text{H}_2\text{SO}_4$
1. Calculate milliequivalents of acid:
$\text{Meq of acid} = \text{Normality } (N) \times \text{Volume in mL } (V)$
2. Calculate required pure mass ($W_{\text{pure}}$):
$\text{Meq of pure } \text{Na}_2\text{CO}_3 = \frac{W_{\text{pure}}}{\text{Equivalent weight}} \times 1000$
The equivalent weight ($\text{Eq. wt.}$) of $\text{Na}_2\text{CO}_3$ is:
$\text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{106}{2} = 53\text{ g/eq}$
3. Adjust for purity:
$\text{Mass of impure sample} = W_{\text{pure}} \times \frac{100}{\text{Purity } \%}$

13. What weight of $\text{Na}_2\text{CO}_3$ of $85\%$ purity would be required to neutralize $45.6\text{ mL}$ of $0.235\text{ N } \text{H}_2\text{SO}_4$?

A

$0.668\text{ g}$

B

$0.568\text{ g}$

C

$0.483\text{ g}$

D

$0.785\text{ g}$

Question 38: Palette 38/40

DPT MCQ

Concept: The equivalent weight ($\text{Eq. wt.}$) of a reducing agent in a redox reaction is equal to its molecular weight divided by its valence factor ($\text{v.f.}$ or $\text{n-factor}$):
$\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{v.f.}}$
The valence factor ($\text{v.f.}$) represents the total change in oxidation state per molecule/formula unit:
* In $\text{S}_2\text{O}_3^{2-}$, the average oxidation state of sulfur is $+2$.
* In $\text{S}_4\text{O}_6^{2-}$, the average oxidation state of sulfur is $+2.5$.
* For each $\text{Na}_2\text{S}_2\text{O}_3$ formula unit containing $2$ sulfur atoms, the valence factor is:
$\text{v.f.} = 2 \times \vert{}+2.5 - (+2)\vert{} = 2 \times 0.5 = 1$

14. In the reaction, $2\text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{I}^-$, the equivalent weight of $\text{Na}_2\text{S}_2\text{O}_3$ is equal to:

A

$\text{Mol. wt.}$

B

$\frac{\text{Mol. wt.}}{2}$

C

$2 \times \text{Mol. wt.}$

D

$\frac{\text{Mol. wt.}}{6}$

Question 39: Palette 39/40

DPT MCQ

Concept: This is a back-titration redox problem:
1. Total milliequivalents ($\text{Meq}$) of reducing agent ($\text{As}_2\text{O}_3$) taken equal the sum of $\text{Meq}$ used to reduce $\text{MnO}_2$ and $\text{Meq}$ reacted with $\text{KMnO}_4$:
$\text{Meq of } \text{As}_2\text{O}_3 \text{ total} = \text{Meq of } \text{MnO}_2 + \text{Meq of } \text{KMnO}_4$
2. Calculate n-factors for each reaction:
* For $\text{As}_2\text{O}_3 \rightarrow 2\text{AsO}_4^{3-}$ ($\text{As}^{3+} \rightarrow \text{As}^{5+}$):
Change in oxidation state per arsenic atom $= 2$. Since there are $2$ arsenic atoms in $\text{As}_2\text{O}_3$, $\text{n-factor} = 4$.
* For $\text{MnO}_2 \rightarrow \text{Mn}^{2+}$ ($\text{Mn}^{4+} \rightarrow \text{Mn}^{2+}$):
$\text{n-factor} = 2$.
* For $\text{KMnO}_4 \rightarrow \text{Mn}^{2+}$ in acidic medium ($\text{Mn}^{7+} \rightarrow \text{Mn}^{2+}$):
$\text{n-factor} = 5$.
3. Determine the pure mass of $\text{MnO}_2$ and calculate purity percentage:
$\text{Percentage purity} = \frac{\text{Mass of pure } \text{MnO}_2}{\text{Mass of pyrolusite sample}} \times 100$

15. A sample of Pyrolusite ($\text{MnO}_2$) weighs $0.5\text{ g}$. To this sample, $0.594\text{ g}$ of $\text{As}_2\text{O}_3$ and dilute acid are added. After the reaction has ceased, the excess $\text{As}^{3+}$ is titrated with $45\text{ mL}$ of $\frac{\text{M}}{50}$ $\text{KMnO}_4$ solution. Calculate the percentage purity of $\text{MnO}_2$ in pyrolusite. (Molar mass of $\text{As}_2\text{O}_3 = 198\text{ g/mol}$, $\text{MnO}_2 = 87\text{ g/mol}$)

A

$65.25\%$

B

$87.00\%$

C

$43.50\%$

D

$52.20\%$

Question 40: Palette 40/40

DPT MCQ

Concept: 1. Acid-Base Neutralization with $\text{NaOH}$:
* Oxalic acid ($\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}$) furnishes $2$ acidic protons ($\text{n-factor} = 2$).
* Potassium hydrogen oxalate ($\text{KHC}_2\text{O}_4 \cdot \text{H}_2\text{O}$) furnishes $1$ acidic proton ($\text{n-factor} = 1$).
* Neutral impurities do not react with $\text{NaOH}$.
2. Redox Titration with $\text{KMnO}_4$:
* Both $\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}$ and $\text{KHC}_2\text{O}_4 \cdot \text{H}_2\text{O}$ contain one oxalate ion ($\text{C}_2\text{O}_4^{2-}$).
* In redox titration, $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-$, so the n-factor for both components in redox titration is $2$.
3. Calculate the moles of each component per gram of the mixture and determine their molar ratio.

16. The neutralization of a solution of $1.2\text{ g}$ of a mixture containing oxalic acid ($\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}$), potassium hydrogen oxalate ($\text{KHC}_2\text{O}_4 \cdot \text{H}_2\text{O}$), and neutral impurities required $40.0\text{ mL}$ of $0.25\text{ N } \text{NaOH}$. On the other hand, titration of $0.4\text{ g}$ of the same mixture in acidic medium required $40.0\text{ mL}$ of $0.125\text{ N } \text{KMnO}_4$. Find the molar ratio of $\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}$ to $\text{KHC}_2\text{O}_4 \cdot \text{H}_2\text{O}$ in the mixture.

A

1 : 2

B

2 : 1

C

1 : 1

D

3 : 2