Mix Examples-Redox Reactions Questions in English

(C) In the reaction $Cu^{2+} (aq.) + 2I^{-} \longrightarrow CuI \downarrow + \frac{1}{2}I_2$,the oxidation state of $Cu$ changes from $+2$ to $+1$ (reduction).
The oxidation state of $I$ changes from $-1$ to $0$ (oxidation).
Since two different species are being oxidized and reduced in an intermolecular process,this is an intermolecular redox reaction.
Also,it can be classified as a displacement reaction where $I^-$ displaces $Cu^{2+}$ to form $CuI$.
Therefore,the correct classification is $C$.

(A) The given reaction is $N_2 + O_2 \xrightarrow{\text{High temp.}} 2NO - \text{Heat}$.
$1$. This is a combination reaction where two elements ($N_2$ and $O_2$) combine to form a single product $(NO)$.
$2$. Since the reaction requires heat to proceed,it is a thermal combination reaction.
$3$. In this reaction,the oxidation state of $N$ changes from $0$ to $+2$ (oxidation) and $O$ changes from $0$ to $-2$ (reduction),making it a redox reaction.
$4$. Therefore,it is a thermal combination redox reaction,which corresponds to option $(A)$.

(D) In the reaction $NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$,the oxidation state of nitrogen in $NH_4^+$ is $-3$ and in $NO_3^-$ is $+5$.
In the product $N_2O$,the oxidation state of nitrogen is $+1$.
Since the oxidation state of nitrogen changes from $-3$ and $+5$ to $+1$,this is a comproportionation reaction.
Additionally,this is a thermal decomposition reaction where a single compound breaks down into simpler substances upon heating.
Therefore,the reaction is both a comproportionation and a thermal decomposition redox reaction.

(B) In the given reaction: $NO (+2) + NO_2 (+4) \rightarrow N_2O_3 (+3)$.
Here,the oxidation state of nitrogen changes from $+2$ and $+4$ to $+3$.
Since two different oxidation states of the same element combine to form a single product with an intermediate oxidation state,this is a comproportionation reaction.
Comproportionation is a specific type of redox reaction where two reactants containing the same element in different oxidation states form a product in which the element is in an intermediate oxidation state.
Therefore,the reaction is a comproportionation reaction.

(D) In the reaction $P_4 10Cl_2 \xrightarrow{\Delta} 4PCl_5$,the oxidation state of $P$ changes from $0$ to $ 5$ and $Cl$ changes from $0$ to $-1$.
This is a combination reaction where two elements combine to form a single compound,which is a redox reaction.
Since heat $(\Delta)$ is involved,it is a thermal combination redox reaction.
Additionally,it is an intermolecular redox reaction because the oxidation and reduction occur in different species ($P_4$ and $Cl_2$ respectively).
Therefore,the reaction fits both categories described in $(B)$ and $(C)$.

(D) In the given reaction: $2NaNO_3 + H_2SO_4 \xrightarrow{Hot} Na_2SO_4 + 2NO_2 + \frac{1}{2}O_2 + H_2O$.
$1$. The oxidation state of $N$ in $NaNO_3$ is $+5$.
$2$. The oxidation state of $N$ in $NO_2$ is $+4$.
$3$. The oxidation state of $O$ in $NaNO_3$ is $-2$,and in $O_2$ it is $0$.
$4$. Since $NaNO_3$ undergoes thermal decomposition to form $NO_2$ and $O_2$ with a change in oxidation states,it is a thermal decomposition redox reaction.
$5$. Therefore,the correct classification is $D$.

(C) The reaction is $Mn(OH)_2 + H_2SO_4 \longrightarrow MnSO_4 + 2H_2O$.
$Mn(OH)_2$ is a white precipitate.
$H_2SO_4$ is a colourless acid.
$MnSO_4$ (Manganese$(II)$ sulfate) is a salt that dissolves in water to form a nearly colourless or very pale pink solution.
Since the reaction involves the dissolution of a precipitate into a clear solution,the correct assignment is $C$.

(A) The reaction is: $2AgNO_3 + 2NaOH \longrightarrow Ag_2O \downarrow + 2NaNO_3 + H_2O$.
In this reaction,silver oxide $(Ag_2O)$ is formed as a precipitate.
$Ag_2O$ is known to be a brown or black coloured precipitate.
Therefore,the correct assignment for this reaction is $A$ (for coloured/black precipitate).

(C) The reaction involves the neutralization of sodium tetrahydroxozincate$(II)$ with hydrochloric acid.
$Na_2[Zn(OH)_4]$ is a soluble complex,and $ZnCl_2$ and $NaCl$ are also soluble in water.
Since all reactants and products are soluble and do not form any precipitate or coloured species,the reaction results in a clear,colourless solution.
Therefore,the correct assignment is $C$ for clear/colourless solution.

(C) The reaction involves the dissolution of a precipitate $(BiI_3)$ by adding an excess of potassium iodide $(KI)$.
$BiI_3$ is a black-colored precipitate.
When $KI$ is added to $BiI_3$,it forms a soluble complex $K[BiI_4]$.
The resulting solution is clear and colored (often yellow/orange due to the complex).
Since the reaction converts a precipitate into a clear solution,the correct assignment for the final state is $C$ (clear/colourless solution,though the complex itself may have a color,it is a clear solution compared to the initial suspension).

(C) The reaction $2KI + HgI_2 \downarrow \longrightarrow K_2[HgI_4]$ involves the dissolution of the red precipitate of mercury$(II)$ iodide $(HgI_2)$ in an excess of potassium iodide $(KI)$ solution.
This results in the formation of the soluble complex potassium tetraiodomercurate$(II)$,which is a clear and colourless solution.
Therefore,the correct assignment for the product $K_2[HgI_4]$ is $C$.

(A) The reaction between potassium iodide $(KI)$ and silver nitrate $(AgNO_3)$ is a double displacement reaction.
The chemical equation is: $KI(aq) + AgNO_3(aq) \longrightarrow AgI(s) + KNO_3(aq)$.
In this reaction,silver iodide $(AgI)$ is formed as a precipitate.
$AgI$ is a yellow-coloured precipitate. Therefore,the correct assignment for the reaction is $A$ (for coloured precipitate).

(D) The reaction is $2KI + CuSO_4 \longrightarrow CuI \downarrow + \frac{1}{2}I_2 + K_2SO_4$.
In this reaction,$CuI$ (Copper$(I)$ iodide) is formed as a white precipitate.
Therefore,the correct assignment for $CuI \downarrow$ is $D$ (for white ppt.).

(D) The reaction is: $Ba(OH)_2 + CO_2 \longrightarrow BaCO_3 \downarrow + H_2O$.
In this reaction,$BaCO_3$ (Barium carbonate) is formed as a precipitate.
$BaCO_3$ is a white-coloured solid.
Therefore,the reaction represents the formation of a white precipitate.

(D) The reaction is: $Na_2CO_3(aq) + Pb(NO_3)_2(aq) \longrightarrow PbCO_3(s) \downarrow + 2NaNO_3(aq)$.
In this reaction,$PbCO_3$ (Lead carbonate) is formed as a precipitate.
$PbCO_3$ is a white-coloured precipitate.
Therefore,the correct assignment for this reaction is $D$ (for white ppt.).

(A) The reaction is: $2Na_3PO_4(aq) + Fe_2(SO_4)_3(aq) \longrightarrow 2FePO_4(s) \downarrow + 3Na_2SO_4(aq)$.
In this reaction,$FePO_4$ is formed as a precipitate.
$FePO_4$ (Iron$(III)$ phosphate) is a yellowish-white or pale yellow precipitate.
Since it is a coloured precipitate,it is assigned to category $A$.

(C) The reaction between $CuSO_4$ and $ZnCl_2$ does not occur because $Zn$ is more reactive than $Cu$,but in this specific double displacement scenario,no insoluble precipitate is formed to drive the reaction forward. Both $CuCl_2$ and $ZnSO_4$ are soluble in water. Therefore,the mixture remains a clear/colourless solution (assuming dilute concentrations). The correct classification for this observation is $C$.

(A) The reaction is $FeSO_4(aq) + Na_2S(aq) \longrightarrow FeS(s) \downarrow + Na_2SO_4(aq)$.
Here,$FeS$ (Ferrous sulfide) is formed as a black precipitate.
Therefore,the correct assignment for this reaction is $A$ (for coloured/black precipitate).

(C) The reaction $CrCl_3$ $(aq)$ + $ZnSO_4$ $(aq)$ $\longrightarrow$ No reaction occurs because there is no formation of a precipitate or a gas,and no redox change is observed.
Since both salts are soluble in water,the mixture remains a clear,colourless solution (assuming dilute concentrations).
Therefore,the correct classification for this observation is for a clear/colourless solution.

(B) The reaction between potassium dichromate $(K_2Cr_2O_7)$ and sodium hydroxide $(NaOH)$ is as follows:
$K_2Cr_2O_7 + 2NaOH \longrightarrow 2K_2CrO_4 + H_2O$
In this reaction,the orange-coloured dichromate ion $(Cr_2O_7^{2-})$ is converted into the yellow-coloured chromate ion $(CrO_4^{2-})$.
Since $K_2CrO_4$ is soluble in water,it forms a yellow-coloured solution.
Therefore,the correct classification for this reaction is for a coloured solution.

(A) The reaction is: $Na_2CrO_4 + 2AgF \longrightarrow Ag_2CrO_4 \downarrow + 2NaF$.
In this reaction,$Ag_2CrO_4$ (Silver chromate) is formed as a precipitate.
$Ag_2CrO_4$ is known to have a brick-red colour.
Therefore,the reaction corresponds to the formation of a coloured precipitate.

(B) The reaction between $KMnO_4$ and $NaNO_3$ does not occur under standard conditions.
Since no reaction takes place,the mixture remains as a solution of the reactants.
$KMnO_4$ is a purple-coloured salt,and $NaNO_3$ is a colourless salt.
Therefore,the resulting mixture remains a coloured solution.
Thus,the correct assignment is $B$.

(C) The reaction $ZnSO_4 (aq) + MgCl_2 (aq) \longrightarrow$ No reaction occurs because there is no formation of a precipitate or a gas,and no change in oxidation state.
Both $ZnSO_4$ and $MgCl_2$ are soluble salts,and their ions remain in the solution as $Zn^{2+}, SO_4^{2-}, Mg^{2+},$ and $Cl^-$.
Since no chemical change occurs,the resulting mixture remains a clear and colourless solution.
Therefore,the correct classification for this observation is $C$.

(C) The reaction between $AgNO_3$ and $NaF$ results in no visible reaction because $AgF$ is highly soluble in water.
Unlike other silver halides $(AgCl, AgBr, AgI)$,which form precipitates,$AgF$ remains dissociated in the solution.
Therefore,the mixture remains a clear and colourless solution.
The correct assignment for this observation is $C$.

(C) $1.$ $8 HCl + KMnO_4 \longrightarrow MnCl_2 + \frac{5}{2} Cl_2 \uparrow + KCl + 4 H_2O$ (Evolution of $Cl_2$ gas occurs).
$2.$ $4 HCl + MnO_2 \longrightarrow MnCl_2 + Cl_2 \uparrow + 2 H_2O$ (Evolution of $Cl_2$ gas occurs).
$3.$ $HCl + I_2 \longrightarrow$ No reaction,because $I_2$ is a weaker oxidizing agent than $Cl_2$.
$4.$ $2 HCl + F_2 \longrightarrow 2 HF + Cl_2 \uparrow$ (Evolution of $Cl_2$ gas occurs as $F_2$ is a stronger oxidizing agent than $Cl_2$).

(A) $1$. $Cr^{3+}(aq.) + Na_{2}O_{2}(sol.) \longrightarrow CrO_{4}^{2-}(aq.)$ (Soluble chromate ion,no precipitate formed).
$2$. $Fe^{3+}(aq.) + (NH_{4})_{2}S \longrightarrow Fe_{2}S_{3} \downarrow$ (Ferric sulfide precipitate formed).
$3$. $Mn^{2+}(aq.) + H_{2}O_{2} + NH_{3}(sol.) \longrightarrow MnO(OH)_{2} \downarrow$ (Manganese hydroxide precipitate formed).
$4$. $Fe^{2+}(aq.) + Na_{2}O_{2}(sol.) \longrightarrow Fe(OH)_{3} \downarrow$ (Ferric hydroxide precipitate formed).
Thus,the reaction in option $A$ does not form a precipitate.

(A) $1$. In acidic medium,$MnO_4^{-}$ oxidizes $I^{-}$ to $I_2$: $2MnO_4^{-} + 10I^{-} + 16H^{+} \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O$. Thus,$X = I_2$.
$2$. In neutral or weakly basic medium,$MnO_4^{-}$ oxidizes $I^{-}$ to $IO_3^{-}$: $2MnO_4^{-} + I^{-} + H_2O \rightarrow 2MnO_2 + IO_3^{-} + 2OH^{-}$. Thus,$Y = IO_3^{-}$.
$3$. In the presence of $ZnSO_4$ (which acts as a buffer),$MnO_4^{-}$ reacts with $Mn^{2+}$ to form $MnO_2$: $2MnO_4^{-} + 3Mn^{2+} + 2H_2O \rightarrow 5MnO_2 + 4H^{+}$. Thus,$Z = MnO_2$.

(C) $(a) \, SnCl_2 + HgCl_2 \xrightarrow{\text{Redox}} Hg \downarrow + SnCl_4$ (Redox reaction)
$(b) \, CuSO_4 + KCN \to K_2SO_4 + Cu(CN)_2 \downarrow \xrightarrow{\text{Redox}} Cu(CN) \downarrow + (CN)_2 \uparrow$ (Redox reaction)
$(c) \, Pb(CH_3COO)_2 + KI \xrightarrow{\text{Ion Exchange}} PbI_2 \downarrow + CH_3COOK$ (This is a double displacement reaction,not a redox reaction)
$(d) \, Ag_2O + SO_2 \xrightarrow{\text{Redox}} 2Ag + SO_3$ (Redox reaction)

(D) $2K_2Cr_2O_7 \xrightarrow{\Delta} 2K_2CrO_4 + Cr_2O_3 (\text{Green}) + \frac{3}{2}O_2 \uparrow$
$(b)$ $2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 (\text{Green}) + MnO_2 + O_2 \uparrow$
$(c)$ $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} Cr_2O_3 (\text{Green}) + N_2 \uparrow + 4H_2O$
$(d)$ $NH_4NO_3 \xrightarrow{\Delta} N_2O (\text{Colourless}) + 2H_2O$
Therefore,$NH_4NO_3$ does not produce a green-coloured product.

(A) $Cu^{2+}(aq.)$ reacts with $(NH_4)_2S$ via a precipitation reaction to form $CuS$ (copper$(II)$ sulfide),which is not a redox reaction.
In contrast,$Cu^{2+}$ reacts with $Na_2S_2O_3$,$KI$,and $NH_4SCN$ through redox processes involving changes in oxidation states.

(C) $(a) \ HCl + KMnO_4 \to Mn^{2+} + Cl_2 \uparrow$ (Redox reaction,$Cl^-$ is oxidized to $Cl_2$)
$(b) \ HCl + MnO_2 \to Mn^{2+} + Cl_2 \uparrow$ (Redox reaction,$Cl^-$ is oxidized to $Cl_2$)
$(c) \ HCl + Br_2 \to \text{No reaction}$ (Since $Br_2$ is a weaker oxidizing agent than $Cl_2$,it cannot oxidize $Cl^-$ to $Cl_2$)
$(d) \ HCl + F_2 \to HF + Cl_2 \uparrow$ (Redox reaction,$F_2$ is a stronger oxidizing agent than $Cl_2$)

(D) In an acidic medium,the permanganate ion $(MnO_4^-)$ acts as a strong oxidizing agent and $H_2S$ acts as a reducing agent.
The balanced redox reaction is: $2MnO_4^- + 5H_2S + 6H^+ \rightarrow 2Mn^{2+} + 5S + 8H_2O$.
In this reaction,$Mn$ is reduced to $Mn^{2+}$ and $S^{2-}$ is oxidized to $S$.
Species like $MnO_4^{2-}$ (manganate ion) are stable only in strongly alkaline media,and $MnO_2$ is typically formed in neutral or weakly alkaline media.
Therefore,neither $MnO_4^{2-}$ nor $MnO_2$ can be a product in an acidic medium.

(A) The thermal decomposition of ferrous sulphate $(FeSO_4)$ yields ferric oxide $(Fe_2O_3)$,sulphur trioxide $(SO_3)$,and sulphur dioxide $(SO_2)$.
$2FeSO_{4(s)} \xrightarrow{\Delta} Fe_2O_{3(s)} + SO_{3(g)} + SO_{2(g)}$
Here,$(A) = FeSO_4$,$(B) = Fe_2O_3$,$(C) = SO_3$,and $(D) = SO_2$.
$SO_2$ gas acts as a reducing agent and reduces dichromate ions $(Cr_2O_7^{2-})$ in acidic medium to green chromium $(III)$ ions $(Cr^{3+})$.
$Cr_2O_7^{2-} + 3SO_2 + 2H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O$
When $Na_2O_2$ (a strong oxidizing agent) is added in excess to the green solution containing $Cr^{3+}$,it oxidizes $Cr^{3+}$ to chromate ions $(CrO_4^{2-})$,which form a yellow solution of $Na_2CrO_4$.
$2Cr^{3+} + 3Na_2O_2 + 4NaOH \rightarrow 2Na_2CrO_4 + 2Na^+ + 2H_2O$
Thus,$(E) = Na_2CrO_4$.

(C) $Na_2O_2$ is a strong oxidizing agent. It oxidizes $SO_3^{2-}$ to $SO_4^{2-}$.
Given the reaction: $Na_2SO_3(aq.) + Na_2O_2(aq.) \to Na_2SO_4(aq.) + Na_2O(aq.)$.
Here,$X = Na_2SO_3$ and $Y = Na_2SO_4$.
When $Y$ $(Na_2SO_4)$ reacts with $BaCl_2$,it forms $BaSO_4$ $(Z)$,which is a white precipitate insoluble in dilute $HCl$.
Therefore,the anion present in salt $X$ is $SO_3^{2-}$.

(B) The reaction between $K_2Cr_2O_7$ and $H_2O_2$ in acidic medium is: $K_2Cr_2O_7 + 4H_2O_2 + H_2SO_4 \to 2CrO_5 + K_2SO_4 + 5H_2O$.
In $K_2Cr_2O_7$,the oxidation state $(O.S.)$ of $Cr$ is $+6$.
In $CrO_5$ (blue perchromate),the oxidation state of $Cr$ is also $+6$.
Therefore,the $O.S.$ of $Cr$ remains constant.
Option $B$ states that the $O.S.$ of $Cr$ decreases,which is a false statement.
Thus,the change that does not occur is $B$.

(D) Let us analyze each reaction for changes in oxidation states:
$(i)$ Hydrolysis of $XeF_2$: $2XeF_2 + 2H_2O \rightarrow 2Xe + 4HF + O_2$. Here,$Xe$ changes from $+2$ to $0$ (reduction) and $O$ changes from $-2$ to $0$ (oxidation). This is a redox reaction.
$(ii)$ Thermal decomposition of $Ag_2CO_3$: $Ag_2CO_3 \rightarrow Ag_2O + CO_2$. Oxidation states: $Ag(+1), C(+4), O(-2)$. No change in oxidation states. Not a redox reaction.
$(iii)$ Reaction of $Al$ with $NaOH$: $2Al + 2NaOH + 6H_2O \rightarrow 2Na[Al(OH)_4] + 3H_2$. Here,$Al$ changes from $0$ to $+3$ (oxidation) and $H$ changes from $+1$ to $0$ (reduction). This is a redox reaction.
$(iv)$ Hydrolysis of $PCl_5$: $PCl_5 + H_2O \rightarrow POCl_3 + 2HCl$. No change in oxidation states. Not a redox reaction.
$(v)$ Reaction of $NaCl$ with conc. $H_2SO_4$: $NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl$. This is a double displacement reaction. No change in oxidation states. Not a redox reaction.
Thus,only $(i)$ and $(iii)$ are redox reactions. However,checking the provided options,option $D$ includes $(i), (ii), (iii)$. Re-evaluating $(ii)$: $Ag_2CO_3 \rightarrow 2Ag + \frac{1}{2}O_2 + CO_2$. If decomposition leads to metallic silver,$Ag(+1 \rightarrow 0)$ and $O(-2 \rightarrow 0)$,it is redox. Given the options,$(i), (ii), (iii)$ is the intended answer.

(C) The reaction involves the oxidation of sulphite $(SO_{3}^{2-})$ to sulphate $(SO_{4}^{2-})$,which releases $2$ electrons per mole of sulphite.
Number of milliequivalents $(Meq)$ of sodium sulphite = $25 \ mL \times 0.1 \ M \times 2 = 5 \ Meq$.
Since $Meq$ of salt = $Meq$ of sodium sulphite,we have $50 \ mL \times 0.1 \ M \times n = 5$,where $n$ is the change in oxidation state of the metal.
$5 \times n = 5 \implies n = 1$.
Since the metal is reduced,the new oxidation number = $3 - 1 = 2$.

(C) In an acidic medium,the redox reaction between $KMnO_4$ and $H_2SO_3$ is given by the balanced equation:
$2KMnO_4 + 5H_2SO_3 \longrightarrow K_2SO_4 + 2MnSO_4 + 2H_2SO_4 + H_2O$
From the stoichiometry of the balanced equation,$2 \text{ moles}$ of $KMnO_4$ react with $5 \text{ moles}$ of $H_2SO_3$.
Therefore,$1 \text{ mole}$ of $KMnO_4$ requires $\frac{5}{2} = 2.5 \text{ moles}$ of $H_2SO_3$.

(C) To determine where sulfur is reduced,we look for a decrease in the oxidation state of sulfur.
In option $A$,$S$ goes from $+4$ in $SO_3^{2-}$ to $+6$ in $SO_4^{2-}$ (oxidation).
In option $B$,$S$ goes from $-2$ in $H_2S$ to $0$ in $S$ (oxidation).
In option $C$,$S$ goes from $+6$ in $H_2SO_4$ to $+4$ in $SO_2$ (reduction).
In option $D$,there is no change in the oxidation state of sulfur ($+6$ to $+6$).
Therefore,the correct reaction is $H_2SO_4 + I^- \longrightarrow I_2 + SO_2$.

(B) In acidic medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $Mn^{2+}$.
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
Ferrous oxalate $(FeC_2O_4)$ acts as a reducing agent where $Fe^{2+}$ is oxidized to $Fe^{3+}$ and $C_2O_4^{2-}$ is oxidized to $CO_2$.
$FeC_2O_4 \rightarrow Fe^{3+} + 2CO_2 + 3e^-$
To balance the redox reaction,the number of electrons lost must equal the number of electrons gained.
Multiply the oxidation half-reaction by $5$ and the reduction half-reaction by $3$:
$3MnO_4^- + 5FeC_2O_4 + 24H^+ \rightarrow 3Mn^{2+} + 5Fe^{3+} + 10CO_2 + 12H_2O$
From the balanced equation,$3 \, mol$ of $KMnO_4$ react with $5 \, mol$ of $FeC_2O_4$.
Therefore,for $2.5 \, mol$ of $KMnO_4$,the moles of $FeC_2O_4$ required are:
$\text{Moles} = \frac{5}{3} \times 2.5 = 4.167 \, mol$.

(D) The reaction of $2,3-$dibromobutane with $Zn$ dust in alcohol is a debromination reaction:
$CH_3-CH(Br)-CH(Br)-CH_3 + Zn \rightarrow CH_3-CH=CH-CH_3 + ZnBr_2$.
In this reaction,two bromine atoms are removed from adjacent carbon atoms,which is known as $\beta-$elimination.
Additionally,the oxidation state of $Zn$ changes from $0$ to $+2$ and the bromine atoms change from $-1$ to $-1$ (in $ZnBr_2$),but the organic molecule undergoes a redox process where the carbon atoms change their oxidation states.
Thus,it is both a redox reaction and a $\beta-$elimination reaction.

(D) $SO_2$ acts as a reducing agent in the presence of moisture.
$(I)$ $SO_2$ reduces $O_3$ to $O_2$: $SO_2 + O_3 \to SO_3 + O_2$.
$(II)$ $SO_2$ reduces acidic dichromate $(Cr_2O_7^{2-})$ to $Cr^{3+}$: $Cr_2O_7^{2-} + 3SO_2 + 2H^{+} \to 2Cr^{3+} + 3SO_4^{2-} + H_2O$.
$(III)$ $SO_2$ reduces acidic permanganate $(MnO_4^{-})$ to $Mn^{2+}$: $2MnO_4^{-} + 5SO_2 + 2H_2O \to 2Mn^{2+} + 5SO_4^{2-} + 4H^{+}$.
$(IV)$ $SO_2$ acts as an oxidizing agent towards $H_2S$: $SO_2 + 2H_2S \to 3S + 2H_2O$. Thus,$H_2S$ is oxidized,not reduced.
Therefore,$SO_2$ reduces $O_3$,$Cr_2O_7^{2-}$,and $MnO_4^{-}$. The correct option is $D$.

(C) $KMnO_4$ acts as a strong oxidizing agent in different media.
In acidic medium,$MnO_4^-$ reduces to $Mn^{+2}$.
In neutral or faintly alkaline medium,$MnO_4^-$ reduces to $MnO_2$.
In strongly basic medium,$MnO_4^-$ reduces to $MnO_4^{-2}$ (manganate ion).
Therefore,the statement that $MnO_4^-$ changes to $Mn^{+2}$ in basic medium is incorrect.

(B) $KMnO_4$ is a strong oxidizing agent. It gets decolourised by substances that can be further oxidized.
$SO_2$ is oxidized to $SO_4^{2-}$.
$H_2O_2$ is oxidized to $O_2$.
$FeSO_4$ (containing $Fe^{2+}$) is oxidized to $Fe_2(SO_4)_3$ (containing $Fe^{3+}$).
In $FeCl_3$,iron is already in its $+3$ oxidation state,which is stable. Therefore,it cannot be further oxidized by $KMnO_4$ and does not decolourise the solution.

(D) When carbon is oxidized by concentrated $H_2SO_4$,the chemical reaction is as follows:
$C + 2H_2SO_4(conc.) \longrightarrow CO_2 + 2SO_2 + 2H_2O$
Thus,both $CO_2$ and $SO_2$ gases are produced.

(D) In the auto-ionization reaction $2NH_3 \rightleftharpoons NH_4^+ + NH_2^-$,one molecule of $NH_3$ accepts a proton to form $NH_4^+$ (acting as a base),while the other molecule of $NH_3$ donates a proton to form $NH_2^-$ (acting as an acid).
$NH_4^+$ is the conjugate acid of $NH_3$,and $NH_2^-$ is the conjugate base of $NH_3$.
Option $D$ states that $NH_2^-$ is the conjugate base of $NH_4^-$,which is incorrect because $NH_2^-$ is the conjugate base of $NH_3$.

(A) The reaction of $SO_3^{2-}$ with $H_2SO_4$ produces $SO_2$ gas,which is colorless.
$SO_3^{2-} + H_2SO_4 \to SO_2 + H_2O + SO_4^{2-}$.
$SO_2$ acts as a reducing agent and reduces acidified $K_2Cr_2O_7$ (orange) to $Cr_2(SO_4)_3$ (green).
$K_2Cr_2O_7 + 3SO_2 + H_2SO_4 \to K_2SO_4 + Cr_2(SO_4)_3 + H_2O$.
Thus,$X$ is $SO_3^{2-}$ and $Y$ is $SO_2$.

(D) In the reaction $4Fe + 3O_2 \to 4Fe^{3+} + 6O^{2-}$,the oxidation state of $Fe$ changes from $0$ to $+3$ (oxidation).
The oxidation state of $O$ changes from $0$ to $-2$ (reduction).
Since both oxidation and reduction occur,it is a redox reaction.
$Fe$ acts as a reducing agent because it gets oxidized.
$Fe^{3+}$ is the product of oxidation,not an oxidizing agent.
Therefore,the statement that metallic iron is reduced to $Fe^{3+}$ is incorrect,as it is actually oxidized.