Auto oxidation and Disproportionation Questions in English

(N/A) To analyze the reaction,we calculate the oxidation number of $C$ in $(CN)_2$,$CN^-$,and $CNO^-$:
$1$. In $(CN)_2$: $2(x - 3) = 0 \implies x = +3$
$2$. In $CN^-$: $x - 3 = -1 \implies x = +2$
$3$. In $CNO^-$: $x - 3 - 2 = -1 \implies x = +4$
In the given reaction,the oxidation state of carbon changes from $+3$ to $+2$ (reduction) and $+3$ to $+4$ (oxidation).
Since the same element $(C)$ is simultaneously oxidized and reduced,this is a disproportionation reaction. Thus,the dissolution of cyanogen in a basic medium is a disproportionation reaction.

(N/A) reactant participating in a disproportionation reaction must have at least three oxidation states.
$(a)$ $P$,$Cl$,and $S$ are non-metals that exhibit more than three oxidation states,therefore they can undergo disproportionation reactions.
$(b)$ $Mn$,$Cu$,and $Ga$ are metals that can show disproportionation reactions as they also exhibit more than three oxidation states.

(N/A) The disproportionation reaction of $Mn^{3+}$ involves both oxidation and reduction of the same species.
$1. \text{ Oxidation Half-Reaction (O.H.R.): } Mn^{3+}_{(aq)} \rightarrow MnO_{2(s)}$
$2. \text{ Reduction Half-Reaction (R.H.R.): } Mn^{3+}_{(aq)} \rightarrow Mn^{2+}_{(aq)}$
Balancing the $O$.$H$.$R$.:
$Mn^{3+}_{(aq)} + 2H_2O_{(l)} \rightarrow MnO_{2(s)} + 4H^{+}_{(aq)} + e^-$
Balancing the $R$.$H$.$R$.:
$Mn^{3+}_{(aq)} + e^- \rightarrow Mn^{2+}_{(aq)}$
Adding both half-reactions to cancel the electrons:
$2Mn^{3+}_{(aq)} + 2H_2O_{(l)} \rightarrow MnO_{2(s)} + Mn^{2+}_{(aq)} + 4H^{+}_{(aq)}$

(N/A) Assigning oxidation states to the atoms in the reaction:
$H_{2}^{+1}O^{-2} + F_{2}^{0} \to H^{+1}F^{-1} + H^{+1}O^{-2}F^{+1}$
In this reaction,the oxidation state of fluorine $(F)$ changes from $0$ in $F_{2}$ to $-1$ in $HF$ (reduction) and to $+1$ in $HOF$ (oxidation).
Since the same element $(F)$ is simultaneously oxidized and reduced,the reaction is a disproportionation reaction.

(N/A) disproportionation reaction is a special type of redox reaction in which the same element in a single species is simultaneously oxidized and reduced.
Example: The reaction of white phosphorus $(P_4)$ with sodium hydroxide $(NaOH)$ to form phosphine $(PH_3)$ and sodium hypophosphite $(NaH_2PO_2)$.
$P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$
In this reaction,the oxidation state of phosphorus changes from $0$ in $P_4$ to $-3$ in $PH_3$ (reduction) and to $+1$ in $NaH_2PO_2$ (oxidation).

(D) Statement $(1)$: The presence of $OH^{-}$ indicates a basic medium. Hence,$T$.
Statement $(2)$: $Cl_2$ (oxidation state $0$) goes to $ClO^{-}$ (oxidation state $+1$) and $Cl^{-}$ (oxidation state $-1$). Thus,it undergoes both oxidation and reduction. Hence,$T$.
Statement $(3)$: Since $Cl_2$ is simultaneously oxidized and reduced,it is a disproportionation reaction. Hence,$T$.
Statement $(4)$: In $ClO^{-}$,let the oxidation state of $Cl$ be $x$. Then $x + (-2) = -1$,which gives $x = +1$. Hence,$T$.
The correct sequence is $TTTT$.

(N/A) In the given reaction,the oxidation number of $Cl$ changes from $0$ to $+1$ (in $ClO^{-}$) and from $0$ to $-1$ (in $Cl^{-}$).
This is a disproportionation reaction where $Cl_2$ acts as both an oxidising and a reducing agent.
The species $ClO^{-}$ (hypochlorite ion) is responsible for the bleaching action due to its strong oxidising nature.

In $MnO_4^{2-}$,the oxidation number of $Mn$ is $+6$. Since it is not in its maximum oxidation state,it can be oxidized to $+7$ and reduced to lower oxidation states like $+4$ (in $MnO_2$). This simultaneous oxidation and reduction is called a disproportionation reaction.
The reaction is: $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^{-} + MnO_2 + 2H_2O$.
In $MnO_4^{-}$,the oxidation number of $Mn$ is $+7$,which is the maximum possible oxidation state for $Mn$ (as it has $d^5s^2$ configuration). Since it cannot be further oxidized,$MnO_4^{-}$ does not undergo a disproportionation reaction.

(NONE) disproportionation reaction occurs when a species in an intermediate oxidation state is simultaneously oxidized and reduced.
$MnO_{4}^{-}$ contains $Mn$ in the $+7$ oxidation state,which is its maximum oxidation state; therefore,it cannot be further oxidized.
$Cr_{2}O_{7}^{2-}$ contains $Cr$ in the $+6$ oxidation state,which is its maximum oxidation state; therefore,it cannot be further oxidized.
$ClO_{4}^{-}$ contains $Cl$ in the $+7$ oxidation state,which is its maximum oxidation state; therefore,it cannot be further oxidized.
$F_{2}$ is the strongest oxidizing agent and can only be reduced to $F^{-}$,so it cannot undergo disproportionation.
Since all provided options contain at least one species that cannot undergo disproportionation,none of the sets are correct.

(B) disproportionation reaction is one in which the same element in a given oxidation state is simultaneously oxidized and reduced.
In $BrO_{4}^{-}$,the oxidation state of $Br$ is $+7$,which is its maximum possible oxidation state (group $17$ valence electrons).
Since it cannot be oxidized further,it can only undergo reduction.
Therefore,$BrO_{4}^{-}$ cannot show a disproportionation reaction.

(A) The disproportionation reaction of manganate $(VI)$ ion $(MnO_{4}^{2-})$ in an acidic medium is given by:
$3MnO_{4}^{2-} + 4H^{+} \longrightarrow 2MnO_{4}^{-} + MnO_{2} + 2H_{2}O$
In $MnO_{4}^{-}$,the oxidation state of $Mn$ is $+7$.
In $MnO_{2}$,the oxidation state of $Mn$ is $+4$.
The difference in oxidation states is $|7 - 4| = 3$.

(A) disproportionation reaction is a type of redox reaction in which the same element is simultaneously oxidized and reduced.
In the reaction $3 MnO_{4}^{2-} + 4 H^{+} \rightarrow 2 MnO_{4}^{-} + MnO_{2} + 2 H_{2}O$:
$1$. The oxidation state of $Mn$ in $MnO_{4}^{2-}$ is $+6$.
$2$. In $MnO_{4}^{-}$,the oxidation state of $Mn$ is $+7$ (Oxidation).
$3$. In $MnO_{2}$,the oxidation state of $Mn$ is $+4$ (Reduction).
Since the same element $(Mn)$ is both oxidized and reduced,this is a disproportionation reaction.

(C) disproportionation reaction is one in which the same element in a given oxidation state is simultaneously oxidized and reduced.
$A$: $2 H_2O_2 \rightarrow 2 H_2O + O_2$: Oxygen in $H_2O_2$ is in $-1$ state. It goes to $-2$ in $H_2O$ (reduction) and $0$ in $O_2$ (oxidation). This is a disproportionation reaction.
$B$: $2 NO_2 + H_2O \rightarrow HNO_3 + HNO_2$: Nitrogen in $NO_2$ is in $+4$ state. It goes to $+5$ in $HNO_3$ (oxidation) and $+3$ in $HNO_2$ (reduction). This is a disproportionation reaction.
$C$: $MnO_4^- + 4 H^+ + 3 e^- \rightarrow MnO_2 + 2 H_2O$: Manganese in $MnO_4^-$ is in $+7$ state and goes to $+4$ in $MnO_2$. This is a simple reduction reaction,not disproportionation.
$D$: $3 MnO_4^{2-} + 4 H^+ \rightarrow 2 MnO_4^- + MnO_2 + 2 H_2O$: Manganese in $MnO_4^{2-}$ is in $+6$ state. It goes to $+7$ in $MnO_4^-$ (oxidation) and $+4$ in $MnO_2$ (reduction). This is a disproportionation reaction.

(B) The disproportionation reaction of manganate ion $(MnO_{4}^{2-})$ in acidic medium is given by:
$3MnO_{4}^{2-} + 4H^{+} \rightarrow 2MnO_{4}^{-} + MnO_{2} + 2H_{2}O$
Here,the manganese compounds formed are $MnO_{4}^{-}$ (where $Mn$ is in $+7$ oxidation state) and $MnO_{2}$ (where $Mn$ is in $+4$ oxidation state).
Given that the oxidation state of $Mn$ in $B$ is smaller than that of $A$,we identify $A$ as $MnO_{4}^{-}$ $(+7)$ and $B$ as $MnO_{2}$ $(+4)$.
In $MnO_{2}$,$Mn$ is in the $+4$ oxidation state,which corresponds to the electronic configuration $[Ar] 3d^{3}$.
The number of unpaired electrons $(n)$ in $Mn^{4+}$ is $3$.
The spin-only magnetic moment $(\mu)$ is calculated as:
$\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
The nearest integer value is $4$.

(C) species can undergo disproportionation if the central atom exists in an intermediate oxidation state.
$1$. $H_2O_2$: Oxygen is in $-1$ state (can go to $0$ and $-2$).
$2$. $ClO_3^{-}$: Chlorine is in $+5$ state (can go to $+7$ and lower).
$3$. $P_4$: Phosphorus is in $0$ state (can go to $-3$ and $+1, +3, +5$).
$4$. $Cl_2$: Chlorine is in $0$ state (can go to $-1$ and $+1$).
$5$. $Cu^{+}$: Copper is in $+1$ state (can go to $0$ and $+2$).
$6$. $NO_2$: Nitrogen is in $+4$ state (can go to $+3$ and $+5$).
$Ag$,$F_2$,and $K^{+}$ do not undergo disproportionation under standard conditions.
Thus,the total number of species is $6$.

(A) Statement $I$: $S_8$ in alkaline medium undergoes disproportionation: $3S_8 + 24OH^{-} \rightarrow 16S^{2-} + 8S_2O_3^{2-} + 12H_2O$. This statement is correct.
Statement $II$: In $ClO_4^{-}$,the oxidation state of $Cl$ is $+7$,which is its maximum possible oxidation state. Therefore,it cannot be further oxidized and cannot undergo disproportionation. This statement is incorrect.

(A) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
In reaction $(A)$: $2 Cu^{+} \rightarrow Cu^{2+} + Cu$. The oxidation state of $Cu$ changes from $+1$ to $+2$ (oxidation) and from $+1$ to $0$ (reduction). Thus,it is a disproportionation reaction.
In reaction $(B)$: $3 MnO_4^{2-} + 4 H^{+} \rightarrow 2 MnO_4^{-} + MnO_2 + 2 H_2 O$. The oxidation state of $Mn$ changes from $+6$ to $+7$ (oxidation) and from $+6$ to $+4$ (reduction). Thus,it is a disproportionation reaction.
In reaction $(C)$: This is a thermal decomposition reaction,not a disproportionation reaction.
In reaction $(D)$: This is a comproportionation reaction (the reverse of disproportionation),where $Mn$ in $+7$ and $+2$ oxidation states reacts to form $Mn$ in $+4$ oxidation state.

(C) The reaction of white phosphorus $(P_4)$ with aqueous $NaOH$ is given by:
$P_4(s) + 3NaOH(aq) + 3H_2O(l) \rightarrow PH_3(g) + 3NaH_2PO_2(aq)$
In this reaction,the oxidation state of phosphorus changes from $0$ in $P_4$ to $-3$ in $PH_3$ and $+1$ in $NaH_2PO_2$. Since the same element is simultaneously oxidized and reduced,it is a disproportionation reaction.
However,if we consider the final product formed upon further reaction or heating as $Na_3PO_4$,the oxidation state of $P$ in $Na_3PO_4$ is $+5$. Given the options provided,the reaction is a disproportionation reaction where phosphorus is reduced to $-3$ in $PH_3$ and oxidized to $+1$ in $NaH_2PO_2$ (or $+5$ in $Na_3PO_4$ depending on the context of the question). Based on standard textbook options for this specific question,the correct answer is disproportionation reaction with oxidation states $-3$ and $+1$ (or $+5$ if considering the final phosphate product). Given the choices,option $C$ is the most accurate representation of the initial products.

(A) Nitrous acid $(HNO_2)$ is unstable in aqueous solution and undergoes disproportionation reaction at room temperature.
The chemical equation for this reaction is:
$3HNO_{2(aq)} \rightleftharpoons H_3O^{+} + NO_3^- + 2NO$
Thus,the species formed are $H_3O^{+}$,$NO_3^-$,and $NO$.

(B) disproportionation reaction is one in which the same element in a given oxidation state is simultaneously oxidized and reduced.
For an element to undergo disproportionation,it must be in an intermediate oxidation state.
In $ClO_4^{-}$,the oxidation state of $Cl$ is calculated as: $x + (-2 \times 4) = -1$,which gives $x = +7$.
Since $+7$ is the maximum possible oxidation state for Chlorine,it cannot be further oxidized.
Therefore,$ClO_4^{-}$ cannot undergo disproportionation.

(A) Disproportionation is a redox reaction in which the same element in an intermediate oxidation state is simultaneously oxidized and reduced.
Let us analyze the oxidation states of the central atoms in each species:
$(A) \ I^{-}$: Iodine is in its lowest oxidation state $(-1)$,so it can only be oxidized. It cannot disproportionate.
$(B) \ ClO_{4}^{-}$: Chlorine is in its highest oxidation state $(+7)$,so it can only be reduced. It cannot disproportionate.
$(C) \ NO_{2}$: Nitrogen is in the $+4$ oxidation state. It can be oxidized to $+5$ $(NO_{3}^{-})$ and reduced to $+3$ $(HNO_{2})$. It can disproportionate.
$(D) \ HNO_{2}$: Nitrogen is in the $+3$ oxidation state. It can be oxidized to $+5$ $(NO_{3}^{-})$ and reduced to $+2$ $(NO)$. It can disproportionate.
$(E) \ Cu^{+2}$: Copper is in its highest stable oxidation state $(+2)$,so it can only be reduced. It cannot disproportionate.
$(F) \ Cu^{+}$: Copper is in the $+1$ oxidation state. It can be oxidized to $+2$ $(Cu^{+2})$ and reduced to $0$ $(Cu)$. It can disproportionate.
Therefore,the species that can undergo disproportionation are $C, D,$ and $F$.

(B) In aqueous solution,$Cu^{+}$ ions are unstable and undergo disproportionation reaction to form $Cu^{2+}$ and $Cu$ metal.
The reaction is: $2Cu^{+}(aq) \rightarrow Cu^{2+}(aq) + Cu(s)$.
This occurs because the hydration enthalpy of $Cu^{2+}$ is much more negative than that of $Cu^{+}$,which compensates for the energy required to remove the second electron from $Cu^{+}$.

(C) In Statement-$1$,the oxidation state of $I$ in $I_2$ is $0$. In $IO_3^-$,the oxidation state of $I$ is $+5$ (calculated as $x + 3(-2) = -1 \Rightarrow x = +5$). In $I^-$,the oxidation state of $I$ is $-1$. Since the same element $(I)$ is simultaneously oxidized ($0$ to $+5$) and reduced ($0$ to $-1$),this is a disproportionation reaction. Thus,Statement-$1$ is correct.
In Statement-$2$,the oxidation state of Iodine $(I)$ ranges from $-1$ (in $I^-$) to $+7$ (in $IO_4^-$). Thus,Statement-$2$ is also correct.

(B) disproportionation reaction is a type of redox reaction in which the same element in a single oxidation state is simultaneously oxidized and reduced.
In the reaction $Cl_2 + 2 OH^{-} \rightarrow ClO^{-} + Cl^{-} + H_2O$:
- The oxidation state of $Cl$ in $Cl_2$ is $0$.
- In $ClO^{-}$,the oxidation state of $Cl$ is $+1$ (oxidation).
- In $Cl^{-}$,the oxidation state of $Cl$ is $-1$ (reduction).
Since the same element $(Cl)$ is both oxidized and reduced,this is a disproportionation reaction.

(C) The Cannizzaro reaction is a redox reaction in which two molecules of an aldehyde (lacking an $\alpha$-hydrogen atom) react in the presence of a strong base to produce a primary alcohol and a carboxylic acid salt. Since the same aldehyde species is simultaneously oxidized and reduced,it is classified as a disproportionation reaction.

(B) The Cannizzaro reaction is a classic example of a disproportionation reaction (also known as an auto-oxidation-reduction reaction).
In this reaction,an aldehyde that lacks an $\alpha$-hydrogen atom undergoes self-oxidation and reduction in the presence of a concentrated base.
One molecule of the aldehyde is oxidized to form a salt of a carboxylic acid,while another molecule is reduced to form a primary alcohol.

(C) In the reactant $NH_4NO_2$,the nitrogen atoms are present in two different oxidation states:
$1$. In the ammonium ion $(NH_4^+)$,the oxidation state of $N$ is $x + 4(+1) = +1$,so $x = -3$.
$2$. In the nitrite ion $(NO_2^-)$,the oxidation state of $N$ is $x + 2(-2) = -1$,so $x = +3$.
In the product $N_2$,the oxidation state of $N$ is $0$.
Since the nitrogen atom in $NH_4^+$ $(-3)$ increases its oxidation state to $0$ in $N_2$,it undergoes oxidation.
Since the nitrogen atom in $NO_2^-$ $(+3)$ decreases its oxidation state to $0$ in $N_2$,it undergoes reduction.
Therefore,nitrogen is both oxidised and reduced in this reaction.

(D) In the given reaction,the oxidation state of $Br$ in $Br_2$ is $0$.
In the products,$Br$ exists as $Br^-$ (oxidation state $-1$) and $BrO_3^-$ (oxidation state $+5$).
Since the oxidation state of $Br$ decreases from $0$ to $-1$ (reduction) and increases from $0$ to $+5$ (oxidation),$Br_2$ is undergoing both oxidation and reduction.
This is a disproportionation reaction.

(D) disproportionation reaction is a redox reaction in which the same species is simultaneously oxidized and reduced.
In the reaction $2 HCHO \xrightarrow{50 \% NaOH_{(aq)}} CH_3OH + HCOO^-Na^+$,formaldehyde $(HCHO)$ undergoes the Cannizzaro reaction.
One molecule of $HCHO$ is reduced to methanol $(CH_3OH)$ and another molecule is oxidized to sodium formate $(HCOONa)$.
Therefore,this is a disproportionation reaction.

(C) The given reaction is $3 ClO_{3}^{-}(aq) \rightarrow ClO^{-}(aq) + 2 ClO_{3}^{-}(aq)$.
Actually,the correct reaction for disproportionation of chlorate is $3 ClO_{3}^{-}(aq) \rightarrow ClO_{4}^{-}(aq) + 2 ClO_{2}(aq)$ or similar,but based on the provided image logic:
In the reaction $ClO_{3}^{-} \rightarrow ClO^{-} + Cl^{-}$,the oxidation state of $Cl$ changes from $+5$ to $+1$ and $-1$.
Since the same element $(Cl)$ is simultaneously reduced (from $+5$ to $+1$ and $-1$) and oxidized (if applicable,though here it is only reduction),this specific type of reaction where an element's oxidation state changes in both directions is called a disproportionation reaction.

(B) Nitrous acid $(HNO_2)$ undergoes disproportionation to form nitric acid $(HNO_3)$,water $(H_2O)$,and nitric oxide $(NO)$. The balanced chemical equation is: $3 HNO_2 \longrightarrow HNO_3 + H_2O + 2 NO$. Thus,$X = NO$.
In the second reaction,sodium nitrite $(NaNO_2)$ reacts with sulfuric acid $(H_2SO_4)$ to form sodium bisulfate $(NaHSO_4)$,nitric acid $(HNO_3)$,water $(H_2O)$,and nitric oxide $(NO)$. The balanced chemical equation is: $2 NaNO_2 + H_2SO_4 \longrightarrow NaHSO_4 + HNO_3 + H_2O + NO$. Thus,$Y = NO$.

(A) $Ortho-phosphorous$ acid $(H_3PO_3)$ on heating undergoes disproportionation to give $ortho-phosphoric$ acid $(H_3PO_4)$ and phosphine $(PH_3)$.
The chemical equation is:
$4H_3PO_3 \xrightarrow{\Delta} 3H_3PO_4 + PH_3$

(A) In the given reaction,the oxidation state of $Cl$ in $Cl_2$ is $0$.
In $ClO^{-}$,the oxidation state of $Cl$ is $+1$. Since the oxidation state increases from $0$ to $+1$,$Cl_2$ is oxidized to $ClO^{-}$. (Statement $I$ is correct).
In $Cl^{-}$,the oxidation state of $Cl$ is $-1$. Since the oxidation state decreases from $0$ to $-1$,$Cl_2$ is reduced to $Cl^{-}$. (Statement $IV$ is correct).
Therefore,statements $I$ and $IV$ are correct.

(A) Disproportionation reactions are those in which the same element undergoes both oxidation and reduction.
Among the given options,$P_4$ and $S_8$ undergo disproportionation in a strong basic medium according to the following reactions:
$P_4 + 3 NaOH + 3 H_2 O \longrightarrow PH_3 + 3 NaH_2 PO_2$
$S_8 + 12 NaOH \xrightarrow{\Delta} 2 Na_2 S_2 O_3 + 4 Na_2 S + 6 H_2 O$
$N_2$ is very stable and almost inert,therefore it does not undergo this reaction.

(D) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
In option $A$,$Hg_2Cl_2$ undergoes disproportionation where $Hg$ is oxidized from $+1$ to $+2$ and reduced from $+1$ to $0$.
In option $B$,$H_3PO_3$ undergoes disproportionation where $P$ is oxidized from $+3$ to $+5$ and reduced from $+3$ to $-3$.
In option $C$,$Cl_2$ undergoes disproportionation where $Cl$ is oxidized from $0$ to $+1$ and reduced from $0$ to $-1$.
In option $D$,$BaO_2 + H_2SO_4 \longrightarrow BaSO_4 + H_2O_2$ is a double displacement reaction,not a redox reaction,as there is no change in oxidation states of any element.

(A) disproportionation reaction is a redox reaction where the same element is simultaneously oxidized and reduced.
$1$. In $CaCO_3 \longrightarrow CaO + CO_2$,the oxidation states of $Ca$ $(+2)$,$C$ $(+4)$,and $O$ $(-2)$ remain unchanged. This is a thermal decomposition reaction,not a redox reaction.
$2$. In $P_4 + 3 NaOH + 3 H_2O \longrightarrow 3 NaH_2PO_2 + PH_3$,$P$ goes from $0$ to $+1$ (oxidation) and $0$ to $-3$ (reduction).
$3$. In $2 H_2O_2 \longrightarrow 2 H_2O + O_2$,$O$ in $H_2O_2$ $(-1)$ goes to $-2$ in $H_2O$ (reduction) and $0$ in $O_2$ (oxidation).
$4$. In $2 Cu^+ \longrightarrow Cu^{2+} + Cu$,$Cu$ goes from $+1$ to $+2$ (oxidation) and $+1$ to $0$ (reduction).
Therefore,the reaction in option $A$ is not a disproportionation reaction.

(C) disproportionation reaction is a type of redox reaction in which the same element is simultaneously oxidized and reduced.
For an element to undergo disproportionation,it must be in an intermediate oxidation state.
In $ClO_4^{-}$,the oxidation state of $Cl$ is calculated as: $x + 4(-2) = -1$,which gives $x = +7$.
Since $+7$ is the maximum possible oxidation state for chlorine (as it has $7$ valence electrons),it cannot be further oxidized.
Therefore,$ClO_4^{-}$ cannot undergo a disproportionation reaction.

(D) disproportionation reaction is a chemical reaction in which the same species undergoes both oxidation and reduction simultaneously.
In $ClO_4^{-}$,the oxidation state of the $Cl$ atom is $+7$. Since the valence shell configuration of chlorine is $3s^2 3p^5$,the maximum oxidation state it can exhibit is $+7$.
Because $Cl$ is already in its maximum oxidation state,it cannot be further oxidized.
Therefore,$ClO_4^{-}$ can only undergo reduction and cannot participate in a disproportionation reaction.

(D) Disproportionation is a specific type of redox reaction in which a species is simultaneously reduced and oxidized to form two different products.
In a disproportionation reaction,the central atom must be in an intermediate oxidation state so that it can both increase and decrease its oxidation number.
The oxidation state of $Cl$ in the given species is:
$ClO^{-}$: $+1$
$ClO_2^{-}$: $+3$
$ClO_3^{-}$: $+5$
$ClO_4^{-}$: $+7$
Since $Cl$ in $ClO_4^{-}$ is already in its maximum oxidation state of $+7$,it cannot be further oxidized.
Therefore,$ClO_4^{-}$ cannot undergo a disproportionation reaction.
Hence,the correct option is $(D)$.

(C) In the reaction $4 KClO_3 \longrightarrow 3 KClO_4 + KCl$,the oxidation state of chlorine in $KClO_3$ is $+5$.
In $KClO_4$,the oxidation state of chlorine is $+7$ (oxidation).
In $KCl$,the oxidation state of chlorine is $-1$ (reduction).
Since the same element (chlorine) is simultaneously oxidized and reduced,this is a disproportionation reaction.
Hence,the correct option is $(C)$.

(B) The disproportionation reaction of $MnO_4^{2-}$ in acidic medium is given by the following half-reactions:
$2e^{-} + 4H^{+} + MnO_4^{2-} \longrightarrow MnO_2 + 2H_2O$
$(MnO_4^{2-} \longrightarrow MnO_4^{-} + e^{-}) \times 2$
Adding these,the overall balanced equation is:
$3MnO_4^{2-} + 4H^{+} \longrightarrow MnO_2 + 2MnO_4^{-} + 2H_2O$
From the stoichiometry,$3 \ mol$ of $MnO_4^{2-}$ produces $1 \ mol$ of $MnO_2$ and $2 \ mol$ of $MnO_4^{-}$.
Therefore,for $1 \ mol$ of $MnO_4^{2-}$,the products are $\frac{1}{3} \ mol$ of $MnO_2$ and $\frac{2}{3} \ mol$ of $MnO_4^{-}$.
Thus,option $(B)$ is correct.

(D) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
$(A)$ $2 NO_{2(g)} + 2 NaOH_{(aq)} \rightarrow NaNO_{2(aq)} + NaNO_{3(aq)} + H_2O_{(l)}$: Nitrogen in $NO_2$ $(+4)$ is oxidized to $NO_3^-$ $(+5)$ and reduced to $NO_2^-$ $(+3)$. This is a disproportionation reaction.
$(B)$ $Cl_{2(g)} + 2 NaOH_{(aq)} \rightarrow NaClO_{(aq)} + NaCl_{(aq)} + H_2O_{(l)}$: Chlorine in $Cl_2$ $(0)$ is oxidized to $ClO^-$ $(+1)$ and reduced to $Cl^-$ $(-1)$. This is a disproportionation reaction.
$(C)$ $3 ClO^- \rightarrow 2 Cl^- + ClO_3^-$: Chlorine in $ClO^-$ $(+1)$ is reduced to $Cl^-$ $(-1)$ and oxidized to $ClO_3^-$ $(+5)$. This is a disproportionation reaction.
$(D)$ $3 Mg_{(s)} + N_{2(g)} \rightarrow Mg_3N_{2(s)}$: This is a combination reaction,not a disproportionation reaction.

(D) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
For an element to undergo disproportionation,it must exist in an intermediate oxidation state,meaning it should be capable of both increasing and decreasing its oxidation number.
In $ClO_4^-$,the oxidation state of chlorine is $+7$.
Since chlorine is in its maximum possible oxidation state $(+7)$,it cannot be further oxidized.
Therefore,$ClO_4^-$ cannot undergo a disproportionation reaction.

(C) The dibasic oxoacid of phosphorus is phosphorous acid,$H_3PO_3$.
On heating,$H_3PO_3$ undergoes disproportionation reaction as follows:
$4H_3PO_3 \rightarrow 3H_3PO_4 + PH_3$
Here,the oxidation state of $P$ in $H_3PO_3$ is $+3$.
In $H_3PO_4$,the oxidation state of $P$ is $+5$ (oxidation).
In $PH_3$,the oxidation state of $P$ is $-3$ (reduction).
Thus,the products $A$ and $B$ are $H_3PO_4$ and $PH_3$ respectively.