Redox reaction and Method for balancing Redox reaction Questions in English

(B) The balanced chemical equation for the reaction of $KMnO_{4}$ with $I^{-}$ in a basic medium is:
$2 KMnO_{4} + H_{2}O + KI \rightarrow 2 MnO_{2} + 2 KOH + KIO_{3}$
From the stoichiometry of the balanced equation,$2 \ mol$ of $KMnO_{4}$ reacts with $1 \ mol$ of $KI$ (which provides $1 \ mol$ of $I^{-}$).
Therefore,the number of moles of $KMnO_{4}$ reduced by $1 \ mol$ of $I^{-}$ is $2$.

(C) In $[Fe(CN)_6]^{-4}$,the oxidation states are $Fe = +2$,$C = +2$,and $N = -3$.
In the products,the oxidation states are $Fe = +3$,$C$ (in $CO_2$) $= +4$,and $N$ (in $NO_3^-$) $= +5$.
Change in oxidation state for $Fe = |3 - 2| \times 1 = 1$.
Change in oxidation state for $C = |4 - 2| \times 6 = 12$.
Change in oxidation state for $N = |5 - (-3)| \times 6 = 48$.
Total $n$-factor $= 1 + 12 + 48 = 61$.

(B) The reaction involves the solid lead sulfide $(PbS)$ reacting with dilute nitric acid $(HNO_3)$ to form soluble lead nitrate $(Pb(NO_3)_2)$,sulfur $(S)$,and nitrogen monoxide $(NO)$ gas.
Since the solid precipitate $(PbS)$ is consumed and converted into soluble products,this is a precipitate dissolution reaction.

(A) The reaction $CuSO_4 + 2KCN \longrightarrow CuCN \downarrow + (CN)_2 \uparrow + K_2SO_4$ involves the formation of copper$(I)$ cyanide $(CuCN)$ as a precipitate.
Therefore,this reaction is classified as a precipitate formation reaction.
The correct option is $A$.

(A) The reaction $Ca(OH)_2 + SO_2 \longrightarrow CaSO_3 \downarrow + H_2O$ involves the formation of calcium sulfite $(CaSO_3)$,which is an insoluble solid that settles out of the solution as a precipitate.
Therefore,this is a precipitate formation reaction.
Thus,the correct option is $A$.

(B) The reaction $PbO_2 + 4HNO_3 (Conc.) \longrightarrow Pb(NO_3)_2 + 2H_2O + O_2$ involves the dissolution of lead$(IV)$ oxide $(PbO_2)$,which is an insoluble solid (precipitate),in concentrated nitric acid to form soluble lead$(II)$ nitrate $(Pb(NO_3)_2)$.
Therefore,this represents a precipitate dissolution reaction.

(D) The given reaction is $C_{(s)} + O_{2(g)} \xrightarrow{\Delta} CO_2 \uparrow$.
In this reaction,carbon $(C)$ is oxidized from $0$ to $+4$ and oxygen $(O_2)$ is reduced from $0$ to $-2$.
Since two elements combine to form a single product upon heating,it is a thermal combination redox reaction.
Therefore,the correct classification is $D$.

(D) In the reaction $3Mg_{(s)} + N_{2(g)} \longrightarrow Mg_3N_{2(s)}$,the oxidation state of $Mg$ changes from $0$ to $+2$ (oxidation) and the oxidation state of $N$ changes from $0$ to $-3$ (reduction).
Since two different elements are undergoing oxidation and reduction,it is an intermolecular redox reaction.
Furthermore,this is a combination reaction where two elements combine to form a single compound.
Therefore,it fits the category $D$ for thermal combination redox reaction.

(C) In the given reaction: $CuSO_4(aq.) + Zn(s) \longrightarrow ZnSO_4(aq.) + Cu(s)$.
$1$. Oxidation state of $Cu$ changes from $+2$ to $0$ (reduction).
$2$. Oxidation state of $Zn$ changes from $0$ to $+2$ (oxidation).
$3$. Since one element is oxidized and another is reduced,it is an intermolecular redox reaction.
$4$. Furthermore,$Zn$ displaces $Cu$ from its salt solution,making it a displacement reaction.
Therefore,the reaction is an intermolecular redox reaction or a displacement reaction,which corresponds to option $C$.

(C) The given reaction is: $2Na_{(s)} + 2H_2O_{(l)} \rightarrow 2NaOH_{(aq)} + H_2(g) \uparrow $.
In this reaction,the oxidation state of $Na$ changes from $0$ to $+1$ (oxidation) and the oxidation state of $H$ in $H_2O$ changes from $+1$ to $0$ (reduction).
This is a displacement reaction where $Na$ displaces $H$ from $H_2O$.
It is also an intermolecular redox reaction because two different species are undergoing oxidation and reduction.
Therefore,the correct classification is $C$.

(C) In the reaction $Ca_{(s)} + 2H_2O_{(l)} \rightarrow Ca(OH)_{2(aq)} + H_{2(g)}$,the oxidation state of $Ca$ changes from $0$ to $+2$ (oxidation) and the oxidation state of $H$ in $H_2O$ changes from $+1$ to $0$ (reduction).
This is a redox reaction where one element $(Ca)$ displaces hydrogen from water.
Therefore,it is classified as a displacement reaction (specifically,a metal displacement reaction) and it is an intermolecular redox reaction.

(C) In the reaction $Mg_{(s)} + 2H_2O_{(l)} \xrightarrow{\text{Warm}} Mg(OH)_{2(aq)} + H_{2(g)} \uparrow$:
$1$. The oxidation state of $Mg$ changes from $0$ to $+2$ (oxidation).
$2$. The oxidation state of $H$ in $H_2O$ changes from $+1$ to $0$ (reduction).
$3$. Since $Mg$ displaces $H$ from water,this is a displacement reaction.
$4$. It is also an intermolecular redox reaction because the oxidation and reduction occur in different species.
$5$. Therefore,the correct classification is $C$.

(C) The given reaction is $3Fe_{(s)} + 4H_2O_{(g)} \xrightarrow{\Delta} Fe_3O_{4(s)} + 4H_{2(g)}$.
In this reaction,the oxidation state of $Fe$ changes from $0$ to $+8/3$ (in $Fe_3O_4$),and the oxidation state of $H$ in $H_2O$ changes from $+1$ to $0$ (in $H_2$).
This is a redox reaction where $Fe$ is oxidized and $H$ is reduced.
Since $Fe$ displaces $H$ from water,it is a displacement reaction.
Also,it is an intermolecular redox reaction because the oxidation and reduction occur in different species.
Therefore,the correct classification is $C$.

(C) In the given reaction: $Zn_{(s)} + 2HCl_{(aq)} \longrightarrow ZnCl_{2(aq)} + H_{2(g)}$
$1$. The oxidation state of $Zn$ changes from $0$ to $+2$ (oxidation).
$2$. The oxidation state of $H$ in $HCl$ changes from $+1$ to $0$ (reduction).
$3$. Since one element is oxidized and another is reduced,it is a redox reaction.
$4$. Specifically,$Zn$ displaces $H$ from $HCl$,making it a displacement reaction.
$5$. It is also an intermolecular redox reaction because the oxidation and reduction occur in different species.
Therefore,the correct classification is for either intermolecular redox reaction or displacement reaction.

(C) In the given reaction: $Mg_{(s)} + 2HCl_{(aq)} \longrightarrow MgCl_{2(aq)} + H_{2(g)}$
$1$. The oxidation state of $Mg$ changes from $0$ to $+2$ (oxidation).
$2$. The oxidation state of $H$ in $HCl$ changes from $+1$ to $0$ (reduction).
$3$. Since the oxidation and reduction occur in different species,it is an intermolecular redox reaction.
$4$. Furthermore,$Mg$ displaces $H$ from $HCl$,making it a displacement reaction.
$5$. Therefore,the correct classification is $C$.

(C) $1$. In the given reaction,$Fe_{(s)} + 2HCl_{(aq)} \longrightarrow FeCl_{2(aq)} + H_{2(g)}$,the oxidation state of $Fe$ changes from $0$ to $+2$ (oxidation).
$2$. The oxidation state of $H$ in $HCl$ changes from $+1$ to $0$ (reduction).
$3$. This is a displacement reaction where a more reactive metal $(Fe)$ displaces hydrogen from an acid $(HCl)$.
$4$. Since both oxidation and reduction occur in different species,it is an intermolecular redox reaction.
$5$. Therefore,the reaction fits the category $C$.

(C) In the reaction $Cl_{2(g)} + 2KI(aq.) \longrightarrow 2KCl(aq.) + I_2(s)$,the oxidation state of $Cl$ changes from $0$ to $-1$ (reduction) and the oxidation state of $I$ changes from $-1$ to $0$ (oxidation).
This is a displacement reaction where $Cl_2$ displaces $I^-$ from $KI$.
It is also an intermolecular redox reaction because the oxidation and reduction occur in different species.
Therefore,the correct classification is $C$.

(C) The given reaction is $Pb_3O_4 + 8HCl \xrightarrow{\text{Warm}} 3PbCl_2 + Cl_2 + 4H_2O$.
In $Pb_3O_4$,the oxidation state of $Pb$ is $+\frac{8}{3}$.
In $PbCl_2$,the oxidation state of $Pb$ is $+2$.
In $Cl_2$,the oxidation state of $Cl$ is $0$,while in $HCl$ and $PbCl_2$,it is $-1$.
This is an intermolecular redox reaction where $Pb$ is reduced from $+\frac{8}{3}$ to $+2$ and $Cl^-$ is oxidized to $Cl_2$.
Therefore,the correct classification is $C$.

(C) In the given reaction: $Cr_2O_7^{2-} + H^{+} + SO_3^{2-} \longrightarrow Cr^{3+} + SO_4^{2-}$.
The oxidation state of $Cr$ changes from $+6$ in $Cr_2O_7^{2-}$ to $+3$ in $Cr^{3+}$ (reduction).
The oxidation state of $S$ changes from $+4$ in $SO_3^{2-}$ to $+6$ in $SO_4^{2-}$ (oxidation).
Since two different elements are undergoing oxidation and reduction in separate species,this is an intermolecular redox reaction.
Therefore,the correct classification is $C$.

(C) $1$. Analyze the oxidation states of the elements in the reaction: $MnO_4^- + H^+ + Br^- \longrightarrow Mn^{2+}(aq.) + Br_2 \uparrow$.
$2$. In $MnO_4^-$,the oxidation state of $Mn$ is $+7$. In $Mn^{2+}$,it is $+2$. Thus,$Mn$ is reduced.
$3$. In $Br^-$,the oxidation state of $Br$ is $-1$. In $Br_2$,it is $0$. Thus,$Br$ is oxidized.
$4$. Since two different elements ($Mn$ and $Br$) are undergoing oxidation and reduction in separate species,this is an intermolecular redox reaction.
$5$. According to the definitions provided,the correct assignment is $C$.

(C) In the given reaction: $Fe^{2+} \longrightarrow Fe^{3+} + e^-$ (Oxidation of $Fe^{2+}$)
$Cr_2O_7^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$ (Reduction of $Cr_2O_7^{2-}$)
Since two different species ($Fe^{2+}$ and $Cr_2O_7^{2-}$) are undergoing oxidation and reduction respectively,this is an intermolecular redox reaction.
Therefore,the correct classification is $C$.

(C) In the given reaction: $I_2 + S_2O_3^{2-} \longrightarrow I^{-} + S_4O_6^{2-}$
$1$. Oxidation state of $I$ in $I_2$ is $0$,and in $I^-$ it is $-1$. Thus,$I_2$ is reduced.
$2$. Oxidation state of $S$ in $S_2O_3^{2-}$ is $+2$,and in $S_4O_6^{2-}$ it is $+2.5$. Thus,$S$ is oxidized.
$3$. Since both oxidation and reduction occur between different species,this is an intermolecular redox reaction.
$4$. Therefore,the correct classification is an intermolecular redox reaction,which corresponds to option $C$.

(C) In the reaction $CuO + H_2 \longrightarrow Cu + H_2O$:
$1$. The oxidation state of $Cu$ changes from $+2$ in $CuO$ to $0$ in $Cu$ (reduction).
$2$. The oxidation state of $H$ changes from $0$ in $H_2$ to $+1$ in $H_2O$ (oxidation).
$3$. Since one element is oxidized and another is reduced,it is an intermolecular redox reaction.
$4$. Additionally,$H_2$ displaces $Cu$ from $CuO$,making it a displacement reaction.
$5$. Therefore,the reaction is classified as either an intermolecular redox reaction or a displacement reaction.

(C) In the reaction $H_3PO_2 + 4AgNO_3 + 2H_2O \longrightarrow 4Ag + H_3PO_4 + 4HNO_3$:
$1$. The oxidation state of $P$ in $H_3PO_2$ is $+1$ and in $H_3PO_4$ is $+5$. Thus,$P$ is oxidized.
$2$. The oxidation state of $Ag$ in $AgNO_3$ is $+1$ and in $Ag$ is $0$. Thus,$Ag$ is reduced.
$3$. Since two different elements in different compounds are undergoing oxidation and reduction,this is an intermolecular redox reaction.
$4$. Therefore,the correct classification is $C$.

(C) In the given reaction: $H_3PO_2 + CuSO_4 + H_2O \longrightarrow Cu + H_3PO_4 + H_2SO_4$
The oxidation state of $Cu$ changes from $+2$ in $CuSO_4$ to $0$ in $Cu$ (reduction).
The oxidation state of $P$ changes from $+1$ in $H_3PO_2$ to $+5$ in $H_3PO_4$ (oxidation).
Since two different elements (or species) are undergoing oxidation and reduction in an intermolecular manner,this is an intermolecular redox reaction.
Also,$Cu$ is displaced from $CuSO_4$ by the reducing agent $H_3PO_2$,making it a displacement reaction.
Therefore,the correct classification is $C$.

(D) In the reaction $2NaNO_3 \xrightarrow{\Delta} 2NaNO_2 + O_2$:
$1$. The oxidation state of $N$ in $NaNO_3$ is $+5$.
$2$. The oxidation state of $N$ in $NaNO_2$ is $+3$.
$3$. The oxidation state of $O$ in $NaNO_3$ is $-2$,and in $O_2$ it is $0$.
$4$. Since $N$ is reduced ($+5$ to $+3$) and $O$ is oxidized ($-2$ to $0$),this is a redox reaction.
$5$. Because the reaction is triggered by heat and involves the breakdown of a single reactant into multiple products,it is classified as a thermal decomposition redox reaction.

(C) In the given reaction: $5CO + I_2O_5 \longrightarrow 5CO_2 + I_2$.
$1$. Oxidation state of $C$ in $CO$ is $+2$ and in $CO_2$ is $+4$. Since the oxidation state increases,$CO$ is oxidized.
$2$. Oxidation state of $I$ in $I_2O_5$ is $+5$ and in $I_2$ is $0$. Since the oxidation state decreases,$I_2O_5$ is reduced.
$3$. This is an intermolecular redox reaction where one substance is oxidized and another is reduced. It can also be classified as a displacement reaction where $CO$ displaces $I$ from $I_2O_5$.

(C) The given reaction is: $4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \longrightarrow 2Fe_2O_3 + 8Na_2CrO_4 + 8CO_2$.
In this reaction,the oxidation state of $Fe$ changes from $+2$ to $+3$ (oxidation) and the oxidation state of $O$ in $O_2$ changes from $0$ to $-2$ (reduction).
Since two different reactants are undergoing oxidation and reduction,this is an intermolecular redox reaction.
Therefore,the correct classification is an intermolecular redox reaction.

(D) In the reaction $MnO_2 + 2KOH + \frac{1}{2}O_2 \longrightarrow K_2MnO_4 + H_2O$:
The oxidation state of $Mn$ in $MnO_2$ is $+4$.
The oxidation state of $Mn$ in $K_2MnO_4$ is $+6$.
The oxidation state of $O$ in $O_2$ is $0$,and in $K_2MnO_4$ and $H_2O$ it is $-2$.
Since the oxidation state of $Mn$ increases from $+4$ to $+6$ (oxidation) and the oxidation state of $O$ decreases from $0$ to $-2$ (reduction),this is an intermolecular redox reaction.
Specifically,it is a combination reaction where $MnO_2$ reacts with $KOH$ and $O_2$ to form a single product $K_2MnO_4$ along with water.
Therefore,it fits the category of a thermal combination redox reaction. The correct option is $D$.

(D) The given reaction is: $2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2 \uparrow $
In this reaction,$KMnO_4$ undergoes thermal decomposition upon heating to produce $K_2MnO_4$,$MnO_2$,and $O_2$.
The oxidation state of $Mn$ changes from $+7$ in $KMnO_4$ to $+6$ in $K_2MnO_4$ and $+4$ in $MnO_2$.
The oxidation state of $O$ changes from $-2$ in $KMnO_4$ to $0$ in $O_2$.
Since a single reactant breaks down into multiple products upon heating and involves a change in oxidation states,it is classified as a thermal decomposition redox reaction.
Therefore,the correct assignment is $D$.

(D) The given reaction is: $4K_2Cr_2O_7 \xrightarrow{\Delta } 4K_2CrO_4 + 2Cr_2O_3 + 3O_2 \uparrow$.
In this reaction,the oxidation state of $Cr$ changes from $+6$ in $K_2Cr_2O_7$ to $+6$ in $K_2CrO_4$ and $+3$ in $Cr_2O_3$.
The oxidation state of $O$ changes from $-2$ in $K_2Cr_2O_7$ to $0$ in $O_2$.
Since a single reactant $(K_2Cr_2O_7)$ decomposes upon heating into multiple products involving a change in oxidation states,it is a thermal decomposition redox reaction.
Therefore,the correct classification is $D$.

(D) The given reaction is: $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 + Cr_2O_3 + 4H_2O$.
In this reaction,the oxidation state of $N$ changes from $-3$ in $NH_4^+$ to $0$ in $N_2$ (oxidation).
The oxidation state of $Cr$ changes from $+6$ in $Cr_2O_7^{2-}$ to $+3$ in $Cr_2O_3$ (reduction).
Since a single compound undergoes decomposition upon heating to produce redox products,it is classified as a thermal decomposition redox reaction.
Therefore,the correct classification is $D$.

(B) In the given reaction: $NH_4Cl + NaNO_2 \xrightarrow{\Delta} N_2 + NaCl + 2H_2O$.
The oxidation state of $N$ in $NH_4^+$ is $-3$ and in $NO_2^-$ is $+3$.
In the product $N_2$,the oxidation state of $N$ is $0$.
Since the oxidation state of nitrogen changes from $-3$ to $0$ (oxidation) and from $+3$ to $0$ (reduction),this is a comproportionation reaction.
Therefore,the correct classification is $B$.

(D) The given reaction is $Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2 \uparrow$.
In this reaction,the oxidation state of $Ba$ is $+2$ in $Ba(N_3)_2$ and $0$ in $Ba$ (reduction).
The oxidation state of $N$ in $N_3^-$ is $-1/3$ and it becomes $0$ in $N_2$ (oxidation).
Since a single reactant $Ba(N_3)_2$ breaks down into simpler substances upon heating,it is a thermal decomposition reaction.
Since both oxidation and reduction are occurring,it is a thermal decomposition redox reaction.
Therefore,the correct option is $D$.

(D) In the reaction $N_2 + 3H_2 \longrightarrow 2NH_3$,the oxidation state of $N$ changes from $0$ in $N_2$ to $-3$ in $NH_3$ (reduction).
The oxidation state of $H$ changes from $0$ in $H_2$ to $+1$ in $NH_3$ (oxidation).
Since two different elements are undergoing oxidation and reduction to form a single product,this is a combination redox reaction.
Specifically,it is a thermal combination redox reaction where two elements combine to form a compound.
Therefore,the correct classification is $D$.

(C) The reaction is: $2NaNO_2 + 3FeSO_4 + 3H_2SO_4 \longrightarrow Fe_2(SO_4)_3 + Na_2SO_4 + 2H_2O + 2NO$
Then,$FeSO_4 + NO + 5H_2O \longrightarrow [Fe(H_2O)_5NO]SO_4$.
In the first step,the oxidation state of $N$ changes from $+3$ (in $NaNO_2$) to $+2$ (in $NO$),and the oxidation state of $Fe$ changes from $+2$ (in $FeSO_4$) to $+3$ (in $Fe_2(SO_4)_3$).
Since different elements are oxidized and reduced in the same reaction,it is an intermolecular redox reaction.

(D) The given reaction is: $2Pb(NO_3)_2(s) \xrightarrow{\Delta} 2PbO(s) + 4NO_2(g) + O_2(g)$.
In this reaction,a single compound,lead$(II)$ nitrate,decomposes upon heating to form multiple products.
This is a thermal decomposition reaction.
Furthermore,the oxidation state of $N$ changes from $+5$ in $Pb(NO_3)_2$ to $+4$ in $NO_2$,and the oxidation state of $O$ changes from $-2$ in $Pb(NO_3)_2$ to $0$ in $O_2$.
Since it involves a change in oxidation states,it is a redox reaction.
Therefore,it is a thermal decomposition redox reaction,which corresponds to option $D$.

(C) In the reaction $Ag + PCl_5 \xrightarrow{\Delta} AgCl + PCl_3$:
$1$. The oxidation state of $Ag$ changes from $0$ to $+1$ (oxidation).
$2$. The oxidation state of $P$ in $PCl_5$ changes from $+5$ to $+3$ in $PCl_3$ (reduction).
$3$. Since two different substances ($Ag$ and $PCl_5$) are involved in the redox process where one is oxidized and the other is reduced,this is an intermolecular redox reaction.
$4$. Therefore,the correct classification is $C$.

(C) In the reaction $Sn + PCl_5 \xrightarrow{\Delta} SnCl_4 + PCl_3$,the oxidation state of $Sn$ changes from $0$ to $+4$ (oxidation).
The oxidation state of $P$ changes from $+5$ to $+3$ (reduction).
Since one reactant $(Sn)$ is oxidized and another reactant $(PCl_5)$ is reduced,this is an intermolecular redox reaction.
Additionally,$Sn$ displaces $P$ from $PCl_5$,making it a displacement reaction.
Therefore,the correct classification is $C$.

(D) In the reaction $PCl_5 \xrightarrow{\Delta} PCl_3 + Cl_2$,the oxidation state of phosphorus $(P)$ in $PCl_5$ is $+5$ and in $PCl_3$ is $+3$.
The oxidation state of chlorine $(Cl)$ in $PCl_5$ is $-1$ and in $Cl_2$ is $0$.
Since $P$ is reduced from $+5$ to $+3$ and $Cl$ is oxidized from $-1$ to $0$,this is a redox reaction.
Because the reaction involves the breakdown of a single compound into two or more substances upon heating,it is classified as a thermal decomposition redox reaction.
Therefore,the correct classification is $D$.

(C) In the given reaction: $MnO_2 + 2NaCl + 2H_2SO_4 \longrightarrow MnSO_4 + Na_2SO_4 + 2H_2O + Cl_2 \uparrow$
$1$. Oxidation state of $Mn$ changes from $+4$ in $MnO_2$ to $+2$ in $MnSO_4$ (Reduction).
$2$. Oxidation state of $Cl$ changes from $-1$ in $NaCl$ to $0$ in $Cl_2$ (Oxidation).
$3$. Since two different elements are undergoing oxidation and reduction in different reactants,this is an intermolecular redox reaction.
$4$. Therefore,the reaction is classified as an intermolecular redox reaction.

(C) In the given reaction: $2NaBr + MnO_2 + 2H_2SO_4 \longrightarrow MnSO_4 + Na_2SO_4 + Br_2 + 2H_2O$
$1$. The oxidation state of $Mn$ changes from $+4$ in $MnO_2$ to $+2$ in $MnSO_4$ (reduction).
$2$. The oxidation state of $Br$ changes from $-1$ in $NaBr$ to $0$ in $Br_2$ (oxidation).
$3$. Since two different elements are undergoing oxidation and reduction in the same reaction,it is an intermolecular redox reaction.
$4$. It can also be classified as a displacement reaction where $Mn$ displaces $Br$ or vice versa in terms of electron transfer context.
Therefore,the correct classification is $C$.

(C) In the reaction $2NaI + 2H_2SO_4 \,(Conc.) \longrightarrow Na_2SO_4 + I_2 + SO_2 + 2H_2O$:
$1$. The oxidation state of $I$ changes from $-1$ in $NaI$ to $0$ in $I_2$ (oxidation).
$2$. The oxidation state of $S$ changes from $+6$ in $H_2SO_4$ to $+4$ in $SO_2$ (reduction).
$3$. Since two different elements ($I$ and $S$) undergo oxidation and reduction in the same reaction,it is an intermolecular redox reaction.
$4$. Therefore,the correct classification is $C$.

(C) The given reaction is: $2NaI + MnO_2 + 2H_2SO_4 \to MnSO_4 + Na_2SO_4 + I_2 + 2H_2O$.
In this reaction,the oxidation state of $I$ changes from $-1$ in $NaI$ to $0$ in $I_2$ (oxidation).
The oxidation state of $Mn$ changes from $+4$ in $MnO_2$ to $+2$ in $MnSO_4$ (reduction).
Since two different species are undergoing oxidation and reduction,it is an intermolecular redox reaction.
Also,$I^-$ is displaced by the redox process,making it a displacement reaction.
Therefore,the correct classification is for either intermolecular redox reaction or displacement reaction.

(C) In the given reaction: $3PbS + 8HNO_3 \longrightarrow 3Pb(NO_3)_2 + 3S + 2NO + 4H_2O$
$1$. Oxidation state of $S$ in $PbS$ is $-2$ and it changes to $0$ in $S$. Thus,$S$ is oxidized.
$2$. Oxidation state of $N$ in $HNO_3$ is $+5$ and it changes to $+2$ in $NO$. Thus,$N$ is reduced.
$3$. Since the oxidation and reduction occur in different species ($PbS$ and $HNO_3$),this is an intermolecular redox reaction.
$4$. Additionally,$Pb$ displaces $H$ or acts as a displacement-type redox process where $S^{2-}$ is oxidized to $S^0$ and $N^{5+}$ is reduced to $N^{2+}$.
Therefore,the reaction is an intermolecular redox reaction.

(C) The given reaction is: $S + 6HNO_3 \, (Dil.) \longrightarrow H_2SO_4 + 6NO_2 + 2H_2O$ (Note: The balanced equation for dilute $HNO_3$ is $S + 2HNO_3 \longrightarrow H_2SO_4 + 2NO$).
In this reaction,the oxidation state of $S$ changes from $0$ to $+6$ (oxidation) and the oxidation state of $N$ in $HNO_3$ changes from $+5$ to $+2$ in $NO$ (reduction).
Since two different elements are undergoing oxidation and reduction in the same reaction,it is an intermolecular redox reaction.
Therefore,the correct classification is $C$.

(C) In the given reaction: $CuSO_4 + Zn_{(s)} \longrightarrow ZnSO_4 + Cu$,
$Zn$ is oxidized from $0$ to $+2$ state,and $Cu^{2+}$ is reduced from $+2$ to $0$ state.
This is a displacement reaction where a more reactive metal $(Zn)$ displaces a less reactive metal $(Cu)$ from its salt solution.
It is also an intermolecular redox reaction because the oxidation and reduction occur in different species.
Therefore,the correct classification is $C$.

(D) The reaction is $Na_2CO_3 + PbSO_4 \longrightarrow PbCO_3 \downarrow + Na_2SO_4$.
In this reaction,$PbCO_3$ (Lead carbonate) is formed as a precipitate.
$PbCO_3$ is a white-coloured solid precipitate.
Therefore,the correct assignment for this reaction is $D$ (For white ppt.).

(B) The balanced chemical equation is: $2Na_2CrO_4 + 2HCl \longrightarrow H_2Cr_2O_7 + 2NaCl$.
In this reaction,$Na_2CrO_4$ (sodium chromate) is yellow,and $H_2Cr_2O_7$ (dichromic acid) forms an orange-red solution.
Since the product $H_2Cr_2O_7$ is a coloured solution,the correct assignment is $B$.