Redox Reactions

In cold KOH, chlorine disproportionates:

$ \mathrm{Cl_2 + 2\,OH^- \rightarrow Cl^- + ClO^- + H_2O} $

Initial moles of $\mathrm{OH^-}$ in $2\text{ L}$ of $2\text{ M}$ KOH:

$ n(\mathrm{OH^-}) = 2 \times 2 = 4\ \text{mol} $

Given $1$ mol $\mathrm{Cl_2}$:

• $\mathrm{OH^-}$ consumed $= 2 \times 1 = 2$ mol

• $\mathrm{Cl^-}$ formed $= 1$ mol

• $\mathrm{ClO^-}$ formed $= 1$ mol

• $\mathrm{OH^-}$ remaining $= 4 - 2 = 2$ mol

Assuming volume stays $2\text{ L}$:

$ [\mathrm{Cl^-}] = \frac{1}{2} = 0.5\text{ M},\quad [\mathrm{ClO^-}] = \frac{1}{2} = 0.5\text{ M},\quad [\mathrm{OH^-}] = \frac{2}{2} = 1\text{ M} $

Answer: Option C — $0.5\text{ M},\ 0.5\text{ M},\ 1\text{ M}$.

In acidified $\mathrm{K_2Cr_2O_7}$, chromium is in the $+6$ oxidation state (in $\mathrm{Cr_2O_7^{2-}}$).

When $\mathrm{KI}$ is added in acidic medium, $\mathrm{I^-}$ is oxidised to $\mathrm{I_2}$ and dichromate is reduced to $\mathrm{Cr^{3+}}$.

Reduction half-reaction (NCERT method):

$ \mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O} $

So, chromium in the final product is $\mathrm{Cr^{3+}}$, i.e. oxidation state is $+3$.

Correct option: B ($+3$)

The reaction of oxygen gas with platinum hexafluoride can be written as:

$\mathrm{O_2 + PtF_6 \rightarrow X^{+}Y^{-}}$

The product formed is:

$\mathrm{O_2^{+}[PtF_6]^{-}}$

Hence, we identify the ions as:

$\mathrm{X^{+} \rightarrow O_2^{+}}$

$\mathrm{Y^{-} \rightarrow [PtF_6]^{-}}$

(A) For $\mathrm{O_2^{+}}$, Bond Order (BO) = 2.5

The molecular orbital electronic configuration of $\mathrm{O_2^{+}}$ is:

$\sigma 1s^2\\ \sigma^{*}1s^2\\ \sigma 2s^2\\ \sigma^{*}2s^2\\ \sigma 2p_z^2\\ \pi 2p_x^2 = \pi 2p_y^2\\ \pi^{*}2p_x^1 = \pi^{*}2p_y^0$

From this configuration, the bond order is calculated as:

$\text{BO} = 2.5$

Therefore, statement A is incorrect as given.

(B) In $\mathrm{O_2^{+}[PtF_6]^{-}}$, oxidation state of Pt = +5

We have $\mathrm{Pt^{+5}}$ with the electronic configuration $5d^5$. The atomic configuration of Pt is $5d^9 6s^1$. Hence, the oxidation state of +5 is correct.

So, statement B is correct.

(C) $\mathrm{PtF_6}$ acts as an oxidising agent

As $\mathrm{O_2}$ is oxidised to $\mathrm{O_2^{+}}$, $\mathrm{PtF_6}$ works as the oxidant that accepts electrons.

Hence, statement C is correct.

(D) No fluorine atom transfer occurs from $\mathrm{PtF_6}$

Since there is no fluorine transfer during the reaction, this statement is incorrect in this context.

In acidic medium, $\mathrm{KMnO_4}$ acts as a strong oxidising agent.

So, we need to find which species in the given mixture can be oxidised further.

The mixture contains $1$ mole each of:

• $\mathrm{FeC_2O_4}$

• $\mathrm{Fe_2(C_2O_4)_3}$

• $\mathrm{FeSO_4}$

• $\mathrm{Fe_2(SO_4)_3}$

Now let us examine each compound.

1. $\mathrm{FeC_2O_4}$

This is ferrous oxalate, so it contains:

• $\mathrm{Fe^{2+}}$ : can be oxidised to $\mathrm{Fe^{3+}}$

• $\mathrm{C_2O_4^{2-}}$ : can be oxidised to $\mathrm{CO_2}$

Oxidation of $\mathrm{Fe^{2+}}$

$ \mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-} $

So, $1$ mole of $\mathrm{Fe^{2+}}$ gives $1$ mole electron.

Oxidation of oxalate

$ \mathrm{C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-} $

So, $1$ mole of $\mathrm{C_2O_4^{2-}}$ gives $2$ moles electrons.

Hence, $1$ mole of $\mathrm{FeC_2O_4}$ gives total electrons:

$ 1 + 2 = 3 $

2. $\mathrm{Fe_2(C_2O_4)_3}$

This contains:

• $\mathrm{Fe^{3+}}$ : already in higher oxidation state, so no oxidation

• $3$ oxalate ions: each can be oxidised

Each oxalate ion gives $2$ electrons, so $3$ oxalates give:

$ 3 \times 2 = 6 $

Thus, $1$ mole of $\mathrm{Fe_2(C_2O_4)_3}$ gives $6$ electrons.

3. $\mathrm{FeSO_4}$

This contains:

• $\mathrm{Fe^{2+}}$ : can be oxidised to $\mathrm{Fe^{3+}}$

• $\mathrm{SO_4^{2-}}$ : sulphur is already at maximum oxidation state $+6$, so no oxidation

Thus, $1$ mole of $\mathrm{FeSO_4}$ gives:

$ 1 \text{ electron} $

4. $\mathrm{Fe_2(SO_4)_3}$

This contains:

• $\mathrm{Fe^{3+}}$ : cannot be oxidised further here

• $\mathrm{SO_4^{2-}}$ : cannot be oxidised further

So, electrons given:

$ 0 $

Total electrons released

$ 3 + 6 + 1 + 0 = 10 $

Electrons accepted by $\mathrm{KMnO_4}$ in acidic medium

In acidic medium:

$ \mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O} $

So, $1$ mole of $\mathrm{KMnO_4}$ accepts $5$ moles of electrons.

Therefore, moles of $\mathrm{KMnO_4}$ required:

$ \frac{10}{5} = 2 $

Final answer

$ \boxed{2} $

So, the correct option is Option B.

The oxidising power of halogens decreases in the order $F_2 > Cl_2 > Br_2 > I_2$. This means that fluorine is the strongest oxidising agent and iodine is the weakest. Therefore, $Cl_2$ can easily oxidise bromide and iodide ions to form $Br_2$ and $I_2$ respectively. This principle is used in the “Layer test”.

In this test, chlorine gas is passed through an aqueous solution containing bromide or iodide ions. A redox (oxidation-reduction) reaction takes place, where chlorine oxidises bromide or iodide ions. The chemical equations are as follows:

$\mathrm{NaBr} + \mathrm{Cl}_2 \rightarrow \mathrm{Br}_2 + \mathrm{NaCl}$

$\mathrm{NaI} + \mathrm{Cl}_2 \rightarrow \mathrm{I}_2 + \mathrm{NaCl}$

So, $Cl_2$ displaces $Br_2$ and $I_2$ from their compounds since chlorine has higher oxidising power. This confirms that both statements (I) and (II) are true, and statement (II) correctly explains statement (I).

Molarity of KI = 0.1 M

$5{I^ - } + IO_3^ - + 6{H^ + } \to 3{I_2} + 3{H_2}O$ (balanced chemical equation)

$5KI + KI{O_3} + 3{H_2}S{O_4} \to 3{I_2} + 3{K_2}S{O_4} + 3{H_2}O$

(A) Volume given (KI) = 200 mL = 0.2 L

Molarity (KI) = 0.1 M or mol L$^{-1}$

The molarity formula is used here to calculate the number of moles of K.I.

Molarity, $M = {{number\,of\,moles} \over {volume\,in\,L}}$

So, number of moles = Molarity $\times$ Volume in L

For the given volume of KI,

number of moles = 0.1 mol L$^{-1}$ $\times$ 0.2 L

= 0.02 mol

The stoichiometric ratio between KI and KIO$_3$ is

$KI:KI{O_3}$

${I^ - }:IO_3^ - $

$5:1$

$1:{1 \over 5}$

For one mol ${I^ - },{1 \over 5}$ mol $IO_3^ - $ is used.

So, for 0.2 mol I$^-$,

Moles of $IO_3^ - = 0.02 \times {1 \over 5} = 0.004$ mol

The statement (A) is correct.

(B) Volume given (KI) = 200 mL = 0.2 L

Molarity (KI) = 0.1 M

number of moles of KI (I$^-$) = Molarity $\times$ Volume (L)

$ = 0.1\,mol\,{L^{ - 1}} \times 0.2\,L = 0.02\,mol$

The stoichiometric ratio between KE and ${H_2}H{O_4}$ is

$KI:{H_2}H{O_4}$

$5:3$

$1:{3 \over 5}$ {3 moles ${H_2}S{O_4}$ can give 6H$^+$}

${I^ - },{3 \over 5}$

For one mol ${I^ - },{3 \over 5}$ mol ${H_2}S{O_4}$ is used.

So, for 0.02 mol I$^-$,

moles of ${H_2}S{O_4} = 0.02 \times {3 \over 5} = 0.012$ mol

This statement is not correct.

(C) Volume gien (KI) = 0.5 L

Molarity (KI) = 0.1 M

Number of moles of KI = Molarity $\times$ Volume

$ = 0.1\,mol\,{L^{ - 1}} \times 0.5\,L = 0.05\,mol$

The stoichiometric ratio between I$^-$ and I$_2$ is

$KI:{I_2}$

${I^ - }:{I_2}$

$5:3$

$1:{3 \over 5}$

For 1 mol ${I^ - },{3 \over 5}$ mol I$_2$ is produced.

So, for 0.05 mol, moles of ${I_2} = 0.05 \times {3 \over 5} = 0.03$ mol

This statement is not correct.

(D) Equivalent weight of $KI{O_3} = {{Molecular\,weight} \over 5}$

This statement is correct.

Equivalent weight $ = {{Molecular\,weight} \over {Valency\,factor\, \to number\,of\,electrons}}$

Here, valency factor = 5

Each KIO$_3$ molecule gains 5 electrons in this vacation.

KIO$_3$ acts as the ordinating agent, gains electrons from the iodide ion in KI.

When KIO$_3$ is redued to I$_2$ each molecule gains 5 electrons. So, the valency factor is 5 and equivalent weight is lower than molecular weight.

This statement is correct.

Correct statements are A and D.

For the titration: Oxalic acid v/s $\mathrm{KMnO}_4$

$\begin{aligned} & 2 \mathrm{MnO}_4^{-}+5(\mathrm{COO})_2^{2-}+16 \mathrm{H}^{+} \rightarrow \\ & 10 \mathrm{CO}_2+2 \mathrm{Mn}^{2+}+8 \mathrm{H}_2 \mathrm{O} \end{aligned}$

This reaction is slow at room temperature, but becomes fast at $60^{\circ} \mathrm{C}$. Manganese(II) ions catalyse the reaction; thus, the reaction is autocatalytic; once manganese(II) ions are formed, it becomes faster and faster.

The titration of FAS v/s $\mathrm{KMnO}_4$ do not require heating because at higher temperature the oxidation of $\mathrm{Fe}^{+2}$ to $\mathrm{Fe}^{+3}$ by atmospheric $\mathrm{O}_2$ will be prominent.

(A) Combustion of hydrocarbon

(B) Decomposition into gaseous product.

(C) Displacement of ' $V$ ' by ' Ca ' atom.

(D) Disproportionation of $\mathrm{H}_2 \mathrm{O}_2^{-1}$ into $\mathrm{O}^{-2}$ and $\mathrm{O}^{\circ}$ oxidation states.

$\mathrm{I}^{-} \xrightarrow{\mathrm{H}^{+}} \mathrm{I}_2\qquad \mathrm{I}^{-} \xrightarrow{\mathrm{OH}^{-}} \mathrm{IO}_3^{-}$

$\mathrm{S}^{-2} \xrightarrow{\mathrm{H}^{+}} \mathrm{S} \quad \mathrm{S}_2 \mathrm{O}_3^{2-} \xrightarrow{\mathrm{OH}^{-}} \mathrm{SO}_4^{2-}$

$\begin{aligned} & \mathrm{Fe}^{+2} \longrightarrow \mathrm{Fe}^{+3} \\ & \mathrm{~S}_2 \mathrm{O}_3^{2-} \xrightarrow{\mathrm{H}^{+}} \mathrm{S} \downarrow+\mathrm{SO}_4^{2-} \end{aligned}$

$ \textbf{Oxidation States:} $

In $\mathrm{ClO_4^-}$ (perchlorate), chlorine is in the +7 oxidation state.

In $\mathrm{ClO_3^-}$ (chlorate), chlorine is in the +5 oxidation state.

In $\mathrm{ClO_2^-}$ (chlorite), chlorine is in the +3 oxidation state.

In $\mathrm{ClO^-}$ (hypochlorite), chlorine is in the +1 oxidation state.

$ \textbf{Disproportionation Reaction:} $

A disproportionation reaction is one in which a species simultaneously undergoes oxidation and reduction. For this to occur, the element must be in an intermediate oxidation state such that it can be oxidized to a higher state and reduced to a lower one.

In $\mathrm{ClO^-}$ and $\mathrm{ClO_2^-}$, chlorine is in lower oxidation states (+1 and +3, respectively), making them susceptible to disproportionation. For example, hypochlorite can disproportionate in basic solution as follows:

$ 3\,\mathrm{ClO^-} \rightarrow 2\,\mathrm{Cl^-} + \mathrm{ClO_3^-} $

In $\mathrm{ClO_3^-}$, chlorine is in an intermediate oxidation state (+5) that, under certain conditions, can undergo disproportionation.

However, in $\mathrm{ClO_4^-}$, chlorine is in its highest possible oxidation state (+7) and cannot be oxidized further. Since disproportionation requires one part of the species to be oxidized and the other reduced, $\mathrm{ClO_4^-}$ is thermodynamically stable and does not undergo disproportionation.

$ \boxed{\mathrm{ClO_4^-} \text{ (perchlorate ion) does not undergo disproportionation.}} $

The complete balance reaction is

$ 2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \longrightarrow 2 \mathrm{CuI}+\mathrm{I}_2 $

CuI further reacts to form $\mathrm{Cu}_2 \mathrm{I}_2$.

Moles of $\mathrm{Cu}^{2+}$

$ \begin{aligned} & =\text { Volume × concentration } \\ & =0.1 \mathrm{~L} \times 0.05=0.005 \mathrm{~mol} \end{aligned} $

Moles of $\mathrm{I}^{-}=\mathrm{IL} \times 0.1=0.1 \mathrm{~mol}$ 2 moles of $\mathrm{Cu}^{2+}$ react with 4 moles of $\mathrm{I}^{-}$.

Ratio of $\mathrm{Cu}^{2+}$ to $\mathrm{I}^{-}$required is $\frac{2}{4}=\frac{1}{2}$

Ratio of available $\mathrm{Cu}^{2+}$ to $\mathrm{I}^{-}=0.05$.

Since $0.05<0.5$. Thus, $\mathrm{Cu}^{2+}$ is limiting reactant.

From the stoichiometry, 2 moles of $\mathrm{Cu}^{2+}$ produces 1 moles of $\mathrm{I}_2$

$ \begin{aligned} \text { Moles of } \mathrm{I}_2 & =\frac{1}{2} \times \text { moles of } \mathrm{Cu}^{2+}=\frac{0.005}{2} \\ & =0.0025 \mathrm{~mol} \\ & =2.5 \times 10^{-3} \end{aligned} $

From the stoichiometry, 2 moles of $\mathrm{Cu}^{2+}$ produces 2 moles of CuI and 2 moles of CuI produces 1 mole of $\mathrm{Cu}_2 \mathrm{I}_2$.

$ \begin{aligned} \text { Moles of } \mathrm{Cu}_2 \mathrm{I}_2 & =\frac{0.005}{2}=0.0025 \mathrm{~mol} \\ & =2.5 \times 10^{-3} \end{aligned} $

The balanced chemical equation:

$ \begin{aligned} 2 \mathrm{Cr}(\mathrm{OH})_3 + \mathrm{IO}_3^{-} + 4 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{CrO}_4^{2-} + \mathrm{I}^{-} + 5 \mathrm{H}_2 \mathrm{O} \end{aligned} $

Match the given general form to the balanced equation:

$ \begin{aligned} x \mathrm{Cr}(\mathrm{OH})_3 + y(\mathrm{IO}_3)^{-} + z(\mathrm{OH})^{-} \longrightarrow a(\mathrm{CrO}_4)^{2-} + b(\mathrm{I})^{-} + c(\mathrm{H}_2 \mathrm{O}) \end{aligned} $

Compare the numbers of each substance:

From the balanced equation, $x=2$, $y=1$, $z=4$, $a=2$, $b=1$, $c=5$

Check the answer:

Option (b) is wrong because $a + b = 2 + 1 = 3$, not 7.

When ammonium dichromate is thermally decomposed, it produces nitrogen gas, chromium (III) oxide and water vapour.

$ \begin{aligned} \left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7(s) \longrightarrow \mathrm{Cr}_2 \mathrm{O}_3(s)+ & \mathrm{N}_2(g) +4 \mathrm{H}_2 \mathrm{O}(g) \end{aligned} $

We need to find out where nitrogen has the lowest (most negative) oxidation state among the options. To do this, we will calculate the oxidation state of nitrogen in each compound.

(a) In $\mathrm{NH}_2\mathrm{OH}$ (Hydroxylamine):

Let nitrogen's oxidation state be $x$. Hydrogen (H) is +1 and oxygen (O) is -2.

There are 2 H atoms, so $2 \times (+1) = +2$. There is 1 O atom, so $(-2)$. There is also 1 extra H (in OH), so another $+1$.

$ \begin{array}{r} x+2(+1)+(-2)+(+1)=0 \\ x+1=0 \\ x=-1 \end{array} $

So, in hydroxylamine, nitrogen is at −1.

(b) In $\mathrm{NH}_4\mathrm{Cl}$ (Ammonium chloride):

This compound has an ammonium ion ($\mathrm{NH}_4^+$) and a chloride ion ($\mathrm{Cl}^-$).

Let nitrogen's oxidation state be $x$. Hydrogen is +1, chlorine is -1.

There are 4 H atoms, so $4 \times (+1) = +4$.

$ \begin{aligned} x+4 \times 1-1 & =0 \\ x+4(+1) & =+1 \\ x+4 & =1 \\ x & =-3 \end{aligned} $

So, in ammonium chloride, nitrogen is at −3.

(c) In $\mathrm{N}_2\mathrm{H}_4$ (Hydrazine):

Let nitrogen's oxidation state be $x$. Hydrogen is +1.

There are 2 nitrogen and 4 hydrogen atoms.

$ \begin{aligned} 2x+4(+1) & =0 \\ 2x & =-4 \\ x & =-2 \end{aligned} $

So, in hydrazine, nitrogen is at −2.

(d) In $\mathrm{HNO}_2$ (Nitrous acid):

Let nitrogen's oxidation state be $x$. Hydrogen is +1, oxygen is -2.

There is 1 H and 2 O.

$ \begin{array}{r} (+1)+x+2(-2)=0 \\ 1+x-4=0 \\ x-3=0 \\ x=+3 \end{array} $

So, in nitrous acid, nitrogen is at +3.

Now, let's compare the results: $-1,\ -3,\ -2,\ +3$. The smallest value is −3. This happens in $\mathrm{NH}_4\mathrm{Cl}$ (ammonium chloride).

The number of moles of oxalate ions oxidised by one mole of permanganate ions in acidic medium is 2.5. The complete reaction involved is as follows.

$ \begin{aligned} \frac{5}{2} \mathrm{C}_2 \mathrm{O}_4^{2-}+\mathrm{MnO}_4^{-} & +8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}+5 \mathrm{CO}_2 \end{aligned} $

The reactions that happen are:

$ \begin{aligned} \mathrm{MnO}_4^{-}+5 e^{-} & \longrightarrow \mathrm{Mn}^{2+} \quad \text{(1 mole of } \mathrm{MnO}_4^{-} \text{ takes 5 electrons)} \\ \mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e^{-} & \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} \quad \text{(1 mole of } \mathrm{Cr}_2 \mathrm{O}_7^{2-} \text{ takes 6 electrons)} \\ \mathrm{Fe}^{2+} & \longrightarrow \mathrm{Fe}^{3+} \quad \text{(Each Fe}^{2+} \text{ loses 1 electron.)} \end{aligned} $

Step 1: How many Fe²⁺ ions does 1 mole of MnO₄^- oxidize?

1 mole of MnO₄^- can take in 5 electrons. Each Fe²⁺ gives up 1 electron to get oxidized. So, 1 mole of MnO₄^- can oxidize 5 moles of Fe²⁺. This number is called $x$.

Step 2: How many Fe²⁺ ions does 1 mole of Cr₂O₇^2- oxidize?

1 mole of Cr₂O₇^2- can take in 6 electrons. That means it can oxidize 6 moles of Fe²⁺.

Step 3: What is the answer in terms of $x$?

Since $x = 5$, and Cr₂O₇^2- can oxidize 6 moles, the answer is $\frac{6x}{5}$ moles of Fe²⁺ ions.

In acidic medium,

$ \begin{aligned} 5 \mathrm{Fe}^{2+}+\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+} \longrightarrow & \mathrm{Mn}^{2+} + 4 \mathrm{H}_2 \mathrm{O}+5 \mathrm{Fe}^{3+} \end{aligned} $

1 mole of $\mathrm{MnO}_4^{-}$requires 5 moles of $\mathrm{Fe}^{2+} x=5$

$ \begin{aligned} \mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{Fe}^{2+} \longrightarrow 2 \mathrm{Cr}^{3+} & +7 \mathrm{H}_2 \mathrm{O} +6 \mathrm{Fe}^{3+} \end{aligned} $

1 mole of $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ are needed to give 6 moles of $\mathrm{Fe}^{3+}, y=6$.

$ x+y=5+6=11 $

Moles of $\mathrm{Cu}^{2+}: n_{\mathrm{Cu}^{2+}}=M_{\mathrm{Cu}^{2+}} \times V_{\mathrm{Cu}^{2+}}$

$ =0.05 \times 0.1=0.005 \mathrm{~mol} $

For the reaction,

$ 2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \longrightarrow \mathrm{Cu}_2 \mathrm{I}_2+\mathrm{I}_2 $

2 moles of $\mathrm{Cu}^{2+}$ produces 1 mol of $\mathrm{I}_2$

Moles of $\mathrm{I}_2: n_{1_2}=\frac{1}{2} \times n_{\mathrm{Cu}^{2+}}=0.0025 \mathrm{~mol}$

For the reaction;

$ \mathrm{I}_2+2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3 \longrightarrow 2 \mathrm{NaI}+\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6 $

1 mole $\mathrm{I}_2$ react with 2 moles of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$

$ n_{\mathrm{Na} S_2 S_3}=2 \times 0.0025=0.005 \mathrm{~mol} $

Volume of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3: V_{\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3}=\frac{n_{\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3}}{M_{\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3}}$

$ =\frac{0.005}{0.01}=0.5 \mathrm{~L} \text { or } 500 \mathrm{~mL} $

Statements given in II and III are correct.

The correct balanced equation is,

$ \begin{gathered} \mathrm{P}_4+3 \mathrm{OH}^{-}+3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{PH}_3+3 \mathrm{H}_2 \mathrm{PO}_2^{-} \\ a=1, b=3, c=3, d=1, e=3 \\ a+b+c=7, b+c-e=3 \end{gathered} $

Balanced oxidation half reaction

$ \mathrm{H}_2 \mathrm{~S} \rightarrow \mathrm{~S}+2 \mathrm{H}^{+}+2 e^{-} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Balanced reduction half reaction

$ \begin{aligned} \mathrm{MnO}_4^{-} \rightarrow 8 \mathrm{H}^{+}+5 e^{-} \longrightarrow & \mathrm{Mn}^{2+} + 4 \mathrm{H}_2 \mathrm{O} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Eq. (i) multiply by 5 and Eq. (ii) multiply by 2 then add

$ \begin{aligned} 2 \mathrm{MnO}_4^{-}+5 \mathrm{H}_2 \mathrm{~S}+6 \mathrm{H}^{+} \longrightarrow & 2 \mathrm{Mn}^{2+} + 5 \mathrm{~S}+8 \mathrm{H}_2 \mathrm{O} \end{aligned} $

5 mole $\mathrm{H}_2 \mathrm{~S}$ react with 2 mol of $\mathrm{MnO}_4^{-}$

$1 \mathrm{~mol} \mathrm{H}_2 \mathrm{~S}=\frac{2}{5} \mathrm{~mol} \mathrm{MnO}_4^{-}$or 0.4

In acidic condition, $\mathrm{KMnO}_4$ (Mn $=+7$ oxidation state) reduced by $\mathrm{H}_2 \mathrm{O}_2$ to $\mathrm{Mn}^{2+}$

So, $x=\mathrm{Mn}^{2+}$

In basic condition $\mathrm{KMnO}_4(\mathrm{Mn}=+7$ oxidation state)

is reduced by $\mathrm{H}_2 \mathrm{O}_2$ to $\mathrm{MnO}_2 y=\mathrm{MnO}_2$

Acidic $2 \mathrm{KMnO}_4+5 \mathrm{H}_2 \mathrm{O}_2+3 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow2 \mathrm{MnSO}_4+\mathrm{K}_2 \mathrm{SO}_4+8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{O}_2 $

Basic

$ \begin{array}{r} 2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{MnO}_2+2 \mathrm{KOH} +2 \mathrm{H}_2 \mathrm{O}+3 \mathrm{O}_2 \end{array} $

Among the given reactions, the reaction

$ \mathrm{I}_2(s)+2 \mathrm{KBr}(a q) \longrightarrow 2 \mathrm{KI}(a q)+\mathrm{Br}_2(l) $

is not feasible because iodine is less reactive than bromine. In a displacement reaction, more reactive element can displace a less reactive element in compound.

$\mathrm{MnSO}_4$ acts as the autocatalyst during the titration of $\mathrm{KMnO}_4$ and oxalic acid.

An autocatalyst is the substance that is formed during a chemical reaction and then speeds up the reaction.

Oxidation state of Mn in acidic medium

$\mathrm{KMnO}_4$ reduced to $\mathrm{Mn}^{2+}$

Thus, oxidation state $=+2$

In basic medium,

$\mathrm{KMnO}_4$ reduced to $\mathrm{MnO}_2$

$ 0=x+2(-2) \Rightarrow x=+4 $

Normality of $\mathrm{KMnO}_4$ solution,

$ N_{\mathrm{KMnO}_4}=n_{\text {factor }} \times M=0.4 \times 5=2 \mathrm{~N} $

Normality of $\mathrm{H}_2 \mathrm{O}_2$

$ N_{\mathrm{H}_2 \mathrm{O}_2}=\frac{\text { Volume strength }}{5.6}=\frac{10}{5.6} \mathrm{~N} $

Using normality equation

$ \begin{aligned} & N_1 V_1=N_2 V_2 \\ & N_{\mathrm{KMnO}_4} \times V_{\mathrm{KMnO}_4}=N_{\mathrm{H}_2 \mathrm{O}_2} \times V_{\mathrm{H}_2 \mathrm{O}_2} \\ & 2 \times 200= \frac{10}{5.6} \times V_{\mathrm{H}_2 \mathrm{O}_2} \\ & V_{\mathrm{H}_2 \mathrm{O}_2}= 224 \mathrm{~mL} \end{aligned} $

First, separate into two half-reactions:

Oxidation $\left(\mathrm{Fe}^{2+}\right.$ to $\left.\mathrm{Fe}^{3+}\right)$,

$ \mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+1 e^{-} $

Reduction ( $\mathrm{MnO}_4^{-}$to $\mathrm{Mn}^{2+}$ ),

$ \mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+} $

Balance O by adding $4 \mathrm{H}_2 \mathrm{O}$,

$ \mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} $

Balance H by adding $8 \mathrm{H}^{+}$,

$ 8 \mathrm{H}^{+}+\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} $

Add $5 e^{-}$to balance charge,

$ 8 \mathrm{H}^{+}+\mathrm{MnO}_4^{-}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} $

Now, multiply oxidation half-reaction by 5 :

$ 5 \mathrm{Fe}^{2+} \longrightarrow 5 \mathrm{Fe}^{3+}+5 e^{-} $

Now add both:

$ 8 \mathrm{H}^{+}+\mathrm{MnO}_4^{-}+5 \mathrm{Fe}^{2+} \longrightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_2 \mathrm{O} $

Thus, the stoichiometric coefficient of $\mathrm{H}_2 \mathrm{O}=4$

Balanced reaction,

$ \begin{aligned} & 5 \mathrm{Fe}^{2+}+\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+} \longrightarrow 5 \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \\ & \text { Using, molarity }=\frac{\text { Number of moles of solute }}{\text { Volume of solution }(\mathrm{L})} \\ & \begin{aligned} \text { or } \quad\left(\mathrm{KMnO}_4\right) & =\left(\mathrm{FeSO}_4\right) \\ \frac{M_1 V_1}{n_1} & =\frac{M_2 V_2}{n_2} \\ \mathrm{Fe}^{2+} \text { moles } & =0.02 \times 0.5=0.01 \mathrm{~mol} \end{aligned} \end{aligned} $

5 moles of $\mathrm{Fe}^{2+}$ react with 1 mole of $\mathrm{KMnO}_4$. So,

$\mathrm{KMnO}_4$ required $=0.01 / 5=0.002 \mathrm{~mol}$

Volume $=0.002 / 0.1=0.02 \mathrm{~L}=20 \mathrm{~mL}$

In the given reactions:

$\begin{aligned} & 2 \mathrm{S}_2\mathrm{O}_3^{2-} + \mathrm{I}_2 \rightarrow \mathrm{S}_4\mathrm{O}_6^{2-} + 2 \mathrm{I}^{-} \\\\ & \mathrm{S}_2\mathrm{O}_3^{2-} + 5 \mathrm{Br}_2 + 5 \mathrm{H}_2\mathrm{O} \rightarrow 2 \mathrm{SO}_4^{2-} + 4 \mathrm{Br}^{-} + 10 \mathrm{H}^{+} \end{aligned}$

In the first reaction with iodine ($I_2$), thiosulfate ($S_2O_3^{2-}$) acts as a reducing agent, converting $I_2$ to $I^{-}$, while it is itself oxidized to $S_4O_6^{2-}$. This indicates that iodine is acting as an oxidant (oxidizing agent), getting reduced in the process.

In the second reaction with bromine ($Br_2$), thiosulfate is oxidized more extensively, being converted all the way to sulfate ($SO_4^{2-}$), with hydrogen ions ($H^+$) and bromide ions ($Br^-$) also produced. This shows that bromine acts as a stronger oxidant compared to iodine, as it promotes a complete oxidation of $S_2O_3^{2-}$ to $SO_4^{2-}$.

Given these observations, the correct option that justifies the dual behavior of thiosulfate reacting with iodine and bromine differently is:

Option C: Bromine is a stronger oxidant than iodine

This statement correctly explains why thiosulfate is oxidized to $S_4O_6^{2-}$ in the presence of iodine, and to $SO_4^{2-}$ in the presence of bromine. Bromine's stronger oxidizing ability leads to a more complete oxidation of thiosulfate compared to the oxidation by iodine.

Due to inert pair effect $\mathrm{Pb}^{2+}$ is more stable than $\mathrm{Pb}^{4+}$ and $\mathrm{Tl}^{+}$ is more stable than $\mathrm{Tl}^{3+}$. Therefore, $\mathrm{Pb}^{4+}$ and $\mathrm{T}^{3+}$ will function as oxidising agents and easily get reduced to $\mathrm{Pb}^{2+}$ and $\mathrm{TI}^{+}$ respectively.

(A) $\stackrel{\circ}{\mathrm{N}}_2+\stackrel{\circ}{\mathrm{O}}_2 \longrightarrow 2 \mathrm{NO}_{(\mathrm{g})}$

Combination reaction

(B) $2 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_{2(\mathrm{~s})} \longrightarrow 2 \mathrm{PbO}+4 \mathrm{NO}_2+\mathrm{O}_2$

Decomposition reaction

(C) $2 \mathrm{Na}_{(\mathrm{s})}+2 \mathrm{H}_2 \mathrm{O}(\mathrm{I}) \longrightarrow 2 \mathrm{NaOH}_{(\mathrm{aq})}+\mathrm{H}_{2(\mathrm{~g})}$

Displacement reaction

(D) $2 \mathrm{NO}_{2(\mathrm{~g})}+2 \mathrm{OH}_{(\mathrm{g})}^{-} \longrightarrow \mathrm{NO}_{2 \text { (aq) }}^{-}+\mathrm{NO}_{3(\text { (qq) }}^{-}+\mathrm{H}_2 \mathrm{O}$

Nitrogen oxidises and reduces both. So it is a disproportionation reaction.

$\mathrm{A} \rightarrow(\mathrm{IV}), \mathrm{B} \rightarrow(\mathrm{I}), \mathrm{C} \rightarrow(\mathrm{II}), \mathrm{D} \rightarrow(\mathrm{III})$

When a salt reacts with $\mathrm{MnO}_2$ and concentrated $\mathrm{H}_2 \mathrm{SO}_4$, the type of gas evolved depends on the anion present in the salt. In this case, a greenish yellow gas is evolved, which indicates the formation of chlorine gas ($\mathrm{Cl}_2$). $\mathrm{Cl}_2$ gas is greenish-yellow in color and is typically produced from halide salts (namely chlorides, bromides, or iodides) when they are oxidized. Among the given options, we need to identify a salt that contains a halide which can produce $\mathrm{Cl}_2$ upon reaction with $\mathrm{MnO}_2$ and concentrated $\mathrm{H}_2 \mathrm{SO}_4$.

Option A ($\mathrm{NH}_4 \mathrm{Cl}$) contains chloride ions, and when heated with $\mathrm{MnO}_2$ and concentrated $\mathrm{H}_2 \mathrm{SO}_4$, it can undergo a redox reaction where the chloride ions are oxidized to $\mathrm{Cl}_2$ gas. The corresponding reaction can be represented as follows:

$\mathrm{MnO}_2 + 4\mathrm{HCl} \longrightarrow \mathrm{MnCl}_2 + \mathrm{Cl}_2 + 2\mathrm{H}_2\mathrm{O}$

Option B ($\mathrm{CaI}_2$) contains iodide ions, which would lead to the liberation of iodine or $\mathrm{I}_2$, not a greenish yellow gas.

Option C ($\mathrm{KNO}_3$) contains nitrate ions and does not produce a halogen gas upon reaction with $\mathrm{MnO}_2$ and $\mathrm{H}_2 \mathrm{SO}_4$.

Option D ($\mathrm{NaBr}$) contains bromide ions which would lead to the formation of bromine ($\mathrm{Br}_2$), a reddish-brown gas, not greenish yellow.

Therefore, the correct answer is Option A ($\mathrm{NH}_4 \mathrm{Cl}$), since it is the salt that can produce a greenish yellow gas ($\mathrm{Cl}_2$) when reacted with $\mathrm{MnO}_2$ and concentrated $\mathrm{H}_2 \mathrm{SO}_4$.

A disproportionation reaction is a chemical reaction in which a single substance is simultaneously reduced and oxidized, forming two different products. To identify disproportionation reactions, one should look for a specified element in the reactant that gains electrons (reduction) and for the same element that lose electrons (oxidation).

Here's an analysis of each reaction :

(A) $\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}$

Here, copper (Cu) is in the +1 oxidation state and forms two products, where it is oxidized to the +2 state in $\mathrm{Cu}^{2+}$ and reduced to the 0 state in Cu. This is a classic example of disproportionation. So, (A) is true for disproportionation.

(B) $3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$

In this reaction, $\mathrm{MnO}_4^{2-}$ (where Mn is in +6 oxidation state) is converted into $\mathrm{MnO}_4^{-}$ (where Mn is in +7 oxidation state, oxidation has occurred) and $\mathrm{MnO}_2$ (where Mn is in +4 oxidation state, reduction has occurred). Here, again, we have the same element, manganese (Mn), undergoing both reduction and oxidation. Hence, (B) is also a disproportionation reaction.

(C) $2 \mathrm{KMnO}_4 \longrightarrow \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2$

For the reaction with potassium permanganate, $\mathrm{KMnO}_4$, manganese starts in the +7 oxidation state. It forms $\mathrm{K}_2 \mathrm{MnO}_4$ where Mn is in +6 state (reduction) and $\mathrm{MnO}_2$ where Mn is in +4 state (further reduction), but no increase in oxidation state of manganese is observed. Instead, oxygen is released, so the manganese is not being oxidized in any of its products; it is only reduced twice. This is not a disproportionation reaction because the same element should be undergoing both oxidation and reduction, which is not the case here. Hence, (C) is not a disproportionation reaction.

(D) $2 \mathrm{MnO}_4^{-}+3 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 5 \mathrm{MnO}_2+4 \mathrm{H}^{+}$

In this reaction, the $\mathrm{MnO}_4^{-}$ ion (where Mn is at +7 oxidation state) and $\mathrm{Mn}^{2+}$ ion (where Mn is at +2 oxidation state) react to form $\mathrm{MnO}_2$ (where Mn is at +4 oxidation state). This reaction is not a case of disproportionation; it is a comproportionation or synproportionation reaction, where two species with the same element in different oxidation states produce a product with the element at an intermediate oxidation state.

Therefore, the disproportionation reactions among the given choices are:

• (A) $\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}$


• (B) $3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$

Option A would be the correct choice: (A), (B).

• Balancing the Half-Reaction

1. Chromium (Cr) : Already balanced with 2 Cr on each side.


2. Oxygen (O): Add 7 H₂O to the right:

$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + \mathrm{XH}^{+} + \mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A} + 7\mathrm{H}_2 \mathrm{O}$

1. Hydrogen (H) : Add 14 H⁺ to the left:

$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + 14\mathrm{H}^{+} + \mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A} + 7\mathrm{H}_2 \mathrm{O}$

1. Charge : Add 6e⁻ to the left:

$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + 14\mathrm{H}^{+} + 6\mathrm{e}^{\ominus} \rightarrow 2 \mathrm{~A} + 7\mathrm{H}_2 \mathrm{O}$

• Identifying A :

The reduction of Cr₂O₇²⁻ in acidic medium forms Cr³⁺ ions.

• Final Balanced Equation :

$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + 14\mathrm{H}^{+} + 6\mathrm{e}^{\ominus} \rightarrow 2 \mathrm{Cr}^{3+} + 7\mathrm{H}_2 \mathrm{O}$

• X = 14


• Y = 6


• Z = 7


• A = Cr³⁺

$$\Rightarrow \mathrm{Cl}_2+2 \overline{\mathrm{O}} \mathrm{H} \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}^{-}+\mathrm{H}_2 \mathrm{O}$$

$$2 \mathrm{MnO}_4^{-}+\mathrm{H}_2 \mathrm{O}+\mathrm{I}^{-} \xrightarrow{\text { alkaline medium }} 2 \mathrm{MnO}_2+2 \mathrm{OH}^{-}+\mathrm{IO}_3^{-}$$

The ability of a species to function as an oxidizing agent depends on its ability to gain electrons (be reduced). For a species to be an oxidizing agent, it must contain an element that has a positive oxidation state and is capable of being reduced (gain electrons), thereby oxidizing another species.

Let's look at each of the options :

Option A: $\mathrm{SO}_4^{2-}$ (sulfate ion) has sulfur in a +6 oxidation state. Sulfate is the fully oxidized form of sulfur in aqueous solution, and while it can undergo certain reductions under specific conditions (to form, for example, $\mathrm{SO}_3^{2-}$ or sulfite), it is generally stable and not commonly considered a strong oxidizing agent.

Option B: $\mathrm{MnO}_4^{-}$ (permanganate ion) has manganese in a +7 oxidation state. This ion is a very strong oxidizing agent and is well known for its ability to oxidize a wide range of organic and inorganic substances, with manganese being reduced to a lower oxidation state (often to +2 as $\mathrm{Mn}^{2+}$ in acidic solution).

Option C: $\mathrm{N}^{3-}$ is the nitride ion, with nitrogen in a -3 oxidation state. Since nitrogen is in its lowest oxidation state, it cannot be further reduced (gain more electrons), and hence cannot act as an oxidizing agent. In fact, it would more likely act as a reducing agent because it has room to be oxidized (lose electrons).

Option D: $\mathrm{BrO}_3^{-}$ (bromate ion) has bromine in a +5 oxidation state. This ion can act as an oxidizing agent because the bromine can be reduced to a lower oxidation state (often to -1 as $\mathrm{Br}^{-}$).

Therefore, the species that cannot function as an oxidizing agent due to its inability to undergo reduction is Option C, the $\mathrm{N}^{3-}$ ion.

The balanced chemical reaction is:

$ \mathrm{C} + 2 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{CO}_{2} + 2 \mathrm{SO}_{2} + 2 \mathrm{H}_{2} \mathrm{O} $

In this reaction:

Product $ X $ is Carbon dioxide $(\mathrm{CO}_{2})$

Product $ Y $ is Sulphur dioxide $(\mathrm{SO}_{2})$