Mix Examples-Redox Reactions Questions in English

(A) In acidic medium,the $n$-factor of $KMnO_4$ is $5$.
For $FeSO_4$: The $n$-factor is $1$ $(Fe^{2+} \rightarrow Fe^{3+} + e^-)$.
Equivalents of $FeSO_4 = \text{Equivalents of } KMnO_4$.
$2 \times 1 = X \times 5 \Rightarrow X = 2/5$.
For $FeC_2O_4$: The $n$-factor is $3$ ($Fe^{2+} \rightarrow Fe^{3+} + e^-$ and $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$).
Equivalents of $FeC_2O_4 = \text{Equivalents of } KMnO_4$.
$2 \times 3 = Y \times 5 \Rightarrow Y = 6/5$.
Ratio $X : Y = (2/5) : (6/5) = 2 : 6 = 1 : 3$.

(D) The disproportionation reaction is: $3H_3PO_2 \rightarrow 2H_3PO_3 + PH_3$.
In this reaction,the oxidation state of $P$ in $H_3PO_2$ is $+1$.
In $H_3PO_3$,the oxidation state of $P$ is $+3$ (oxidation).
In $PH_3$,the oxidation state of $P$ is $-3$ (reduction).
For $H_3PO_2 \rightarrow H_3PO_3$,the change in oxidation state is $3 - 1 = 2$. Since there are $2$ molecules of $H_3PO_3$,the total change is $2 \times 2 = 4$.
For $H_3PO_2 \rightarrow PH_3$,the change in oxidation state is $1 - (-3) = 4$.
The total number of electrons transferred is $4$.
The $n$-factor for the disproportionation reaction is calculated as the total change in oxidation state per mole of reactant: $n = 4/3$.
Equivalent weight $= M / n = M / (4/3) = 3M/4$.

(D) The Solvay process involves the following main steps:
$1$. Formation of ammonium bicarbonate: $NH_3 + H_2O + CO_2 \rightarrow NH_4HCO_3$ (No redox change).
$2$. Formation of sodium bicarbonate: $NH_4HCO_3 + NaCl \rightarrow NaHCO_3 + NH_4Cl$ (No redox change).
$3$. Calcination of limestone: $CaCO_3 \rightarrow CaO + CO_2$ (No redox change).
$4$. Recovery of ammonia: $2NH_4Cl + Ca(OH)_2 \rightarrow 2NH_3 + CaCl_2 + 2H_2O$ (No redox change).
Since none of the steps in the Solvay process involve a change in oxidation states of the elements,the correct answer is none of these.

(A) The balanced chemical equation for the reaction between acidic $K_2Cr_2O_7$ and $H_2S$ is:
$K_2Cr_2O_7 + 4H_2SO_4 + 3H_2S \longrightarrow K_2SO_4 + Cr_2(SO_4)_3 + 7H_2O + 3S$
In this redox reaction,the dichromate ion is reduced to chromium$(III)$ sulfate $(Cr_2(SO_4)_3)$,and $H_2S$ is oxidized to elemental sulfur $(S)$.
$K_2SO_4$ is also formed as a byproduct.
$CrSO_4$ (chromium$(II)$ sulfate) is not formed in this reaction.

(C) The oxidation of the components in an acidic medium involves the following $n$-factors:
$FeSO_4 \rightarrow Fe^{3+} + SO_4^{2-} + e^-$,so $n$-factor $= 1$.
$FeC_2O_4 \rightarrow Fe^{3+} + 2CO_2 + 3e^-$,so $n$-factor $= 3$.
$Fe_2(C_2O_4)_3 \rightarrow 2Fe^{3+} + 6CO_2 + 6e^-$,so $n$-factor $= 6$.
$Fe_2(SO_4)_3$ is already in the highest oxidation state $(Fe^{3+})$,so it does not react.
For $KMnO_4$ in acidic medium,the $n$-factor is $5$.
Using the principle of equivalence: $n_{eq}(KMnO_4) = n_{eq}(FeC_2O_4) + n_{eq}(Fe_2(C_2O_4)_3) + n_{eq}(FeSO_4)$.
Let $x$ be the moles of $KMnO_4$: $x \times 5 = (1 \times 3) + (1 \times 6) + (1 \times 1)$.
$5x = 10$,which gives $x = 2$ moles.

(C) When $K_2Cr_2O_7$ reacts with $H_2O_2$ in an acidic medium,it forms $CrO_5$ (chromium pentoxide),which has a deep blue colour.
The reaction is: $Cr_2O_7^{2-} + 2H^+ + 4H_2O_2 \rightarrow 2CrO_5 + 5H_2O$.
In $K_2Cr_2O_7$,the oxidation state $(O.S.)$ of $Cr$ is $+6$.
In $CrO_5$,the oxidation state $(O.S.)$ of $Cr$ is also $+6$.
Therefore,the oxidation state of the $Cr$ atom remains constant.
Since the question asks which change does $NOT$ occur,and the oxidation state does not change,option $C$ is the correct statement regarding what does not happen (i.e.,it does not change,so it remains constant is a true statement,but the question asks for the change that does not occur. Actually,the change in $O.S.$ does not occur,so $C$ is the correct answer).

(C) The balanced chemical equation for the reaction is: $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \to K_2SO_4 + 2MnSO_4 + 5O_2 + 8H_2O$.
In a redox reaction,the number of equivalents of reactants are equal.
Equivalents of $KMnO_4 = N \times V(L) = 0.5 \times 0.1 = 0.05 \ eq$.
Since equivalents of $KMnO_4$ = equivalents of $O_2$ produced,we have $0.05 \ eq$ of $O_2$.
The n-factor for $O_2$ in this reaction (from $H_2O_2$ to $O_2$) is $2$.
Moles of $O_2 = \frac{\text{Equivalents}}{\text{n-factor}} = \frac{0.05}{2} = 0.025 \ mol$.
At $STP$ ($1 \ atm$ and $273 \ K$),$1 \ mol$ of gas occupies $22.4 \ L$.
Volume of $O_2 = 0.025 \times 22.4 = 0.56 \ L$.

(D) The oxidation reaction in acidic medium is: $Cu_2S + MnO_4^- \rightarrow Cu^{2+} + SO_2 + Mn^{2+}$.
For $Cu_2S$,the oxidation states change as follows: $Cu$ from $+1$ to $+2$ (change of $1$ per $Cu$,total $2$ for $2Cu$) and $S$ from $-2$ to $+4$ (change of $6$).
Total $n$-factor for $Cu_2S = 2 + 6 = 8$.
For $KMnO_4$,$Mn$ changes from $+7$ to $+2$,so the $n$-factor is $5$.
According to the law of equivalence: $\text{moles of } KMnO_4 \times (n\text{-factor of } KMnO_4) = \text{moles of } Cu_2S \times (n\text{-factor of } Cu_2S)$.
$n \times 5 = 1.5 \times 8$.
$n = \frac{12}{5} = 2.4 \, mol$.

(D) When $Na_2O_2$ is added to an acidified dichromate solution,it reacts to form chromium pentoxide $(CrO_5)$,which is deep blue in colour.
$Cr_2O_7^{2-} + 2H^+ + 4H_2O_2 \rightarrow 2CrO_5 + 5H_2O$
$CrO_5$ is unstable and decomposes rapidly in acidic medium to form chromic sulphate $(Cr_2(SO_4)_3)$ and releases oxygen gas $(O_2)$.
$4CrO_5 + 6H_2SO_4 \rightarrow 2Cr_2(SO_4)_3 + 6H_2O + 7O_2$
Therefore,initially,a blue colour is observed,followed by the evolution of oxygen gas. Thus,both $(A)$ and $(C)$ are correct.

(B) The reaction between tin $(Sn)$ and concentrated sulfuric acid $(H_2SO_4)$ is a redox reaction.
The chemical equation is: $Sn + 2H_2SO_4 \rightarrow SnSO_4 + SO_2 + 2H_2O$.
As shown in the equation,the gaseous product formed is sulfur dioxide $(SO_2)$.

(C) When $AgCl$ is fused with $Na_2CO_3$,the reaction proceeds as follows:
$4AgCl + 2Na_2CO_3 \rightarrow 4Ag + 4NaCl + 2CO_2 + O_2$
Silver chloride is reduced to metallic silver $(Ag)$ because silver oxide is unstable at high temperatures and decomposes into metallic silver and oxygen.

(C) Let us analyze the reactions of $Cu^{2+}_{(aq)}$ with the given reagents:
$(I)$ Reaction with $NaOH_{(aq)}$: $Cu^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Cu(OH)_{2(s)}$. This is a precipitation reaction where the oxidation state of copper remains $+2$.
$(II)$ Reaction with $Fe_{(s)}$: $Cu^{2+}_{(aq)} + Fe_{(s)} \rightarrow Cu_{(s)} + Fe^{2+}_{(aq)}$. This is a redox reaction where copper is reduced from $+2$ to $0$.
$(III)$ Reaction with $KI_{(aq)}$: $2Cu^{2+}_{(aq)} + 4I^-_{(aq)} \rightarrow Cu_2I_{2(s)} + I_{2(s)}$. This is a redox reaction where copper is reduced from $+2$ to $+1$.
Thus,the oxidation state of copper changes in reactions $(II)$ and $(III)$.

(A) When mercuric chloride $(HgCl_2)$ reacts with stannous chloride $(SnCl_2)$,it is first reduced to mercurous chloride $(Hg_2Cl_2)$:
$2HgCl_2 + SnCl_2 \rightarrow Hg_2Cl_2 + SnCl_4$
When excess stannous chloride is added,the mercurous chloride is further reduced to metallic mercury $(Hg)$:
$Hg_2Cl_2 + SnCl_2 \rightarrow 2Hg(l) + SnCl_4$
Thus,the final products obtained with excess stannous chloride are liquid $Hg$ and $SnCl_4$.

(A) In the reaction $HNO_3 + H_2SO_4 \to NO_2^+ + HSO_4^- + H_2O$,$HNO_3$ acts as a base (donor of $OH^-$) and $H_2SO_4$ acts as an acid (donor of $H^+$).
This is an acid-base reaction,not a redox reaction.
In all other options,$HNO_3$ acts as an oxidizing agent.

(D) The chemical reaction is: $K_2Cr_2O_7 + H_2SO_4 + 3SO_2 \to K_2SO_4 + Cr_2(SO_4)_3 + H_2O$.
In this reaction,the orange-colored dichromate ion $(Cr_2O_7^{2-})$ is reduced to green-colored chromium$(III)$ sulfate $(Cr_2(SO_4)_3)$.
Therefore,the solution turns green.

(A) species is basic if it can accept a proton $(H^+)$ and a reducing agent if it can undergo oxidation (increase in oxidation state).
$SO_3^{2-}$ acts as a base because it can accept a proton to form $HSO_3^-$.
$SO_3^{2-}$ acts as a reducing agent because the sulfur atom is in the $+4$ oxidation state and can be oxidized to the $+6$ oxidation state (e.g.,in $SO_4^{2-}$).
$SO_4^{2-}$ cannot be oxidized further,and $HSO_4^-$ is acidic in nature.

(B) In the reaction,$N_2H_4$ acts as the reductant.
The oxidation state of $N$ in $N_2H_4$ is $-2$,and in $N_2$ it is $0$.
The change in oxidation state per nitrogen atom is $0 - (-2) = 2$.
Since there are $2$ nitrogen atoms in $N_2H_4$,the total $n$-factor is $2 \times 2 = 4$.
The molar mass of $N_2H_4$ is $2 \times 14 + 4 \times 1 = 32 \ g/mol$.
The equivalent weight $(EW)$ is calculated as:
$EW = \frac{\text{Molar Mass}}{n\text{-factor}} = \frac{32}{4} = 8$.

(D) An intramolecular redox reaction is a process where one atom is oxidized and another atom is reduced within the same molecule.
In $NH_4NO_2$,$N$ in $NH_4^+$ $(-3)$ and $N$ in $NO_2^-$ $(+3)$ undergo comproportionation to $N_2$ $(0)$,which is an intramolecular redox reaction.
In $2Mn_2O_7 \to 4MnO_2 + 3O_2$,$Mn$ is reduced ($+7$ to $+4$) and $O$ is oxidized ($-2$ to $0$),which is intramolecular.
In $2KClO_3 \to 2KCl + 3O_2$,$Cl$ is reduced ($+5$ to $-1$) and $O$ is oxidized ($-2$ to $0$),which is intramolecular.
In $2H_2O_2 \to 2H_2O + O_2$,$O$ in $H_2O_2$ $(-1)$ is both oxidized to $O_2$ $(0)$ and reduced to $H_2O$ $(-2)$,which is an intermolecular disproportionation reaction,not intramolecular.

(D) Assigning oxidation states to each element:
$\mathop {Ba}\limits^{ + 2} \mathop {Cl_2}\limits^{ - 1} + \mathop {Na_2}\limits^{ + 1} \mathop {S}\limits^{ + 6} \mathop {O_4}\limits^{ - 2} \to \mathop {Ba}\limits^{ + 2} \mathop {S}\limits^{ + 6} \mathop {O_4}\limits^{ - 2} + 2\mathop {Na}\limits^{ + 1} \mathop {Cl}\limits^{ - 1}$
In this reaction,there is no change in the oxidation number of any element.
Since no oxidation or reduction occurs,this is a precipitation reaction,not a redox reaction.
Therefore,none of the substances act as an oxidant.

(C) In an acidic medium,$K_2Cr_2O_7$ acts as a strong oxidizing agent and is reduced to $Cr^{3+}$ ions,which are green in colour.
$KBr$ provides $Br^-$ ions,which are oxidized to $Br_2$.
$NaNO_2$ provides $NO_2^-$ ions,which are oxidized to $NO_3^-$.
$FeSO_4$ provides $Fe^{2+}$ ions,which are oxidized to $Fe^{3+}$.
However,$Na_2SO_4$ contains $SO_4^{2-}$ ions,which are already in their highest oxidation state $(+6)$ and cannot be further oxidized by $K_2Cr_2O_7$. Therefore,it will not cause the reduction of $K_2Cr_2O_7$ to $Cr^{3+}$ and will not change the colour.

(B) The balanced redox reaction is: $3Fe^{+2} + NO_3^- + 4H^+ \rightarrow 3Fe^{+3} + NO + 2H_2O$.
From the stoichiometry,$3$ moles of $Fe^{+2}$ react with $1$ mole of $HNO_3$.
Moles of $Fe^{+2} = \frac{16 \ g}{56 \ g/mol} = 0.2857 \ mol$.
Moles of $HNO_3$ required = $\frac{1}{3} \times 0.2857 \ mol = 0.09524 \ mol$.
Volume of $HNO_3$ $(V)$ = $\frac{\text{moles}}{\text{Molarity}} = \frac{0.09524 \ mol}{5 \ M} = 0.01905 \ L$.
Converting to $mL$: $0.01905 \ L \times 1000 \ mL/L = 19.05 \ mL$.

(C) redox reaction is one in which both oxidation and reduction occur simultaneously,involving a change in the oxidation states of the elements involved.
In option $A$,$S$ in $SO_2$ is $+4$ and in $H_2S$ is $-2$,changing to $0$ in $S$. This is a redox reaction.
In option $B$,$Cl$ in $KClO_3$ is $+5$,changing to $+7$ in $KClO_4$ and $-1$ in $KCl$. This is a disproportionation redox reaction.
In option $C$,$Na_2O + H_2SO_4 \longrightarrow Na_2SO_4 + H_2O$,the oxidation states are: $Na(+1)$,$O(-2)$,$H(+1)$,$S(+6)$. These remain unchanged in the products. This is a neutralization reaction,not a redox reaction.
In option $D$,$Na$ changes from $0$ to $+1$ and $O$ changes from $0$ to $-1$. This is a redox reaction.

(A) The reaction involves the mixing of two aqueous solutions,$Pb(NO_3)_2$ and $NaOH$,resulting in the formation of an insoluble solid,$Pb(OH)_2$,which appears as a precipitate $(downarrow)$.
Therefore,it is a precipitate formation reaction.

(A) The reaction involves the mixing of two aqueous solutions,$Mn(NO_3)_2$ and $NaOH$,which results in the formation of an insoluble solid,$Mn(OH)_2$,known as a precipitate.
Therefore,this is a precipitate formation reaction.

(A) The reaction between $Hg(NO_3)_2$ and $NH_3$ (solution) results in the formation of a white precipitate of $HgO \cdot HgNH_2NO_3$.
Therefore,this is a precipitate formation reaction.

(B) The reaction involves the solid precipitate $MnS$ reacting with hydrochloric acid $(HCl)$ to form soluble manganese chloride $(MnCl_2)$ and hydrogen sulfide gas $(H_2S)$.
Since the solid precipitate $MnS$ is being dissolved by the acid,this is classified as a precipitate dissolution reaction.

(A) In the given reaction,$CuSO_4$ reacts with $KI$ to form $CuI$,which is a white precipitate.
The reaction is: $CuSO_4 + 2KI \longrightarrow CuI \downarrow + \frac{1}{2} I_2 + K_2SO_4$.
Since $CuI$ is formed as a solid precipitate,this reaction is classified as a precipitate formation reaction.
Therefore,the correct option is $A$.

(A) The reaction given is $KNO_2 + AgF \longrightarrow AgNO_2 \downarrow + KF$.
In this reaction,$AgNO_2$ is formed as a precipitate (indicated by the $\downarrow$ symbol).
Since a new precipitate is formed from the reactants,this is classified as a precipitate formation reaction.
Therefore,the correct option is $A$.

(A) The given reaction is $FeCl_3(aq) + Na_3PO_4(aq) \longrightarrow FePO_4(s) \downarrow + 3NaCl(aq)$.
In this reaction,two soluble salts react to form an insoluble salt,$FePO_4$,which appears as a precipitate.
Therefore,this is a precipitate formation reaction.

(B) The reaction involves $2BaCrO_4$ which is a yellow precipitate.
When $HCl$ is added,it reacts with the precipitate to form soluble $BaCl_2$ and $H_2Cr_2O_7$.
Since the solid precipitate $BaCrO_4$ disappears and dissolves into the solution,this is a precipitate dissolution reaction.
Therefore,the correct option is $B$.

(B) The reaction involves the dissolution of the yellow precipitate of lead chromate $(PbCrO_4)$ in an excess of sodium hydroxide $(NaOH)$.
$PbCrO_4$ is insoluble in water but reacts with excess $NaOH$ to form a soluble complex,sodium tetrahydroxoplumbate$(II)$ $(Na_2[Pb(OH)_4])$,and sodium chromate $(Na_2CrO_4)$.
Since a solid precipitate is being converted into a soluble complex,this is a precipitate dissolution reaction.
Therefore,the correct option is $B$.

(A) The given reaction is $Ba(NO_3)_2 + Na_2SO_4 \longrightarrow BaSO_4 \downarrow + 2NaNO_3$.
In this reaction,$BaSO_4$ is formed as an insoluble solid,which is known as a precipitate.
Therefore,this is a precipitate formation reaction.
Thus,the correct option is $A$.

(A) The given reaction is $Pb(NO_3)_2 + H_2SO_4 \longrightarrow PbSO_4 \downarrow + 2HNO_3$.
In this reaction,lead$(II)$ nitrate reacts with sulfuric acid to form lead$(II)$ sulfate,which is an insoluble salt that settles as a precipitate $(downarrow)$.
Since a solid precipitate is formed in the product side,this is a precipitate formation reaction.
Therefore,the correct option is $A$.

(B) The reaction involves the solid precipitate $SrCrO_4$ reacting with excess acetic acid $(AcOH)$.
Since $SrCrO_4$ is a salt of a weak acid $(H_2CrO_4)$,it dissolves in the presence of a stronger acid like acetic acid to form soluble strontium acetate and dichromic acid.
Therefore,this is a precipitate dissolution reaction.
The correct option is $B$.

(A) The given reaction is $CaSO_4 + Pb(NO_3)_2 \longrightarrow PbSO_4 \downarrow + Ca(NO_3)_2$.
In this reaction,$PbSO_4$ is formed as an insoluble solid,which is represented by the downward arrow $(\downarrow)$.
This indicates that a precipitate is formed during the reaction.
Therefore,the reaction is a precipitate formation reaction.

(B) The reaction involves the solid barium carbonate $(BaCO_3 \downarrow)$ reacting with acetic acid $(CH_3COOH)$ to form a soluble barium acetate salt,carbon dioxide gas,and water.
Since the solid precipitate $(BaCO_3)$ is consumed and converted into soluble products,this is a precipitate dissolution reaction.
Therefore,the correct option is $B$.

(A) The reaction between sodium thiosulfate $(Na_2S_2O_3)$ and barium chloride $(BaCl_2)$ results in the formation of barium thiosulfate $(BaS_2O_3)$,which is insoluble in water and appears as a white precipitate $(downarrow)$.
Since a solid precipitate is formed from the reaction of two aqueous solutions,this is classified as a precipitate formation reaction.

(A) The given reaction is $3AgNO_3(aq) + Na_3PO_4(aq) \longrightarrow Ag_3PO_4(s) \downarrow + 3NaNO_3(aq)$.
In this reaction,silver nitrate reacts with sodium phosphate to form silver phosphate as a solid precipitate $(Ag_3PO_4 \downarrow)$.
Since a solid precipitate is formed from the aqueous reactants,this is classified as a precipitate formation reaction.
Therefore,the correct option is $A$.

(C) In the given reaction,$BaSO_3$ is a white precipitate.
When $H_2SO_4$ is added,$BaSO_3$ reacts to form $BaSO_4$,which is also a white precipitate.
Since one precipitate $(BaSO_3)$ is converted into another precipitate $(BaSO_4)$ while releasing $SO_2$ gas,this is a classic example of a precipitate exchange reaction.
Therefore,the correct classification is precipitate exchange reaction.

(A) The given reaction is $Sr(AcO)_2 + Ag_2SO_4 \longrightarrow 2AcOAg \downarrow + SrSO_4 \downarrow$.
In this reaction,two soluble salts react to form two insoluble precipitates,$AcOAg$ (silver acetate) and $SrSO_4$ (strontium sulfate).
Since both products are precipitates,this is a classic example of a precipitate formation reaction.
Therefore,the correct classification is $A$.

(A) The given reaction is $Ca(OH)_2 + CO_2 \longrightarrow CaCO_3 \downarrow + H_2O$.
In this reaction,calcium hydroxide reacts with carbon dioxide to form calcium carbonate,which is an insoluble solid (precipitate) denoted by the downward arrow $\downarrow$.
Therefore,this is a precipitate formation reaction.

(C) The reaction involves the conversion of calcium sulfite $(CaSO_3)$,which is a solid precipitate,into calcium sulfate $(CaSO_4)$,which is also a solid precipitate.
Since one solid precipitate is being replaced by another solid precipitate in the reaction,this is classified as a precipitate exchange reaction.
Therefore,the correct option is $C$.

(A) The reaction between $Na_2SO_3$ (aqueous sodium sulfite) and $BaCl_2$ (aqueous barium chloride) results in the formation of $BaSO_3$ (barium sulfite),which is an insoluble solid that precipitates out of the solution.
Since a solid precipitate $(BaSO_3 \downarrow)$ is formed as a product,this is classified as a precipitate formation reaction.
Therefore,the correct option is $A$.

(D) $PbO_2$ is a lead$(IV)$ oxide,which is an insoluble solid in dilute nitric acid $(HNO_3)$.
Since $PbO_2$ does not react with dilute $HNO_3$,there is no dissolution or chemical change observed.
Therefore,the correct classification for this observation is 'For no reaction'.

(A) The given reaction is $Pb(CH_3COO)_2 + Na_2CrO_4 \longrightarrow PbCrO_4 \downarrow + 2CH_3COONa$.
In this reaction,lead$(II)$ acetate reacts with sodium chromate to form a yellow precipitate of lead$(II)$ chromate $(PbCrO_4)$.
Since a solid precipitate is formed from the aqueous solutions of the reactants,this is a precipitate formation reaction.
Therefore,the correct option is $A$.

(C) The given reaction is the hydrolysis of hypophosphoric acid $(H_4P_2O_6)$:
$H_4P_2O_6 + H_2O \longrightarrow H_3PO_3 + H_3PO_4$
In $H_3PO_3$ (phosphorous acid),the oxidation state of $P$ is $+3$,which corresponds to the $-ous$ suffix.
In $H_3PO_4$ (phosphoric acid),the oxidation state of $P$ is $+5$,which corresponds to the $-ic$ suffix.
Since the products are two oxy acids,one with an $-ic$ suffix and the other with an $-ous$ suffix,the correct option is $C$.

(D) The reaction is $2F_2 + 2H_2O \longrightarrow 4HF + O_2$.
In this reaction,the products are $HF$ (hydrofluoric acid) and $O_2$ (oxygen gas).
$HF$ is a binary acid (hydracid),not an oxy acid.
Since the product is not an oxy acid,it does not have an $-ic$ or $-ous$ suffix.
Therefore,the correct classification is $D$.

(D) In the reaction $NaH + H_2O \longrightarrow NaOH + H_2$,the oxidation states are:
$Na$ changes from $+1$ to $+1$ (no change).
$H$ in $NaH$ is $-1$ and in $H_2O$ is $+1$.
In the product $NaOH$,$H$ is $+1$ and in $H_2$,$H$ is $0$.
Here,$H$ in $NaH$ $(-1)$ is oxidized to $H_2$ $(0)$ and $H$ in $H_2O$ $(+1)$ is reduced to $H_2$ $(0)$.
Since two different species (or atoms in different oxidation states) react to form a product with a common oxidation state,this is a comproportionation reaction.
Additionally,it can be classified as a displacement reaction where $H^-$ displaces $H^+$ from water.
Therefore,both $(B)$ and $(C)$ are correct.

(D) The reaction is: $Pb_3O_4 + 4HNO_3 \rightarrow 2Pb(NO_3)_2 + PbO_2 + 2H_2O$.
In $Pb_3O_4$,the oxidation state of $Pb$ is a mixture of $+2$ and $+4$ (it is $2PbO \cdot PbO_2$).
When treated with dilute $HNO_3$,$PbO$ part reacts to form $Pb(NO_3)_2$ $(Pb^{2+})$,while $PbO_2$ $(Pb^{4+})$ remains as an insoluble precipitate.
This is a non-redox reaction where the mixed oxide is decomposed into its constituent oxides,one of which reacts with the acid.
However,in the context of standard redox classification,this reaction is often categorized as a type of disproportionation or decomposition-like process where the oxidation states do not change for the $Pb$ atoms involved in the product formation relative to their state in the mixed oxide.
Given the options,$D$ is the most appropriate classification as it represents a decomposition-like redox process.