Auto oxidation and Disproportionation Questions in English

(B) In the given reaction: $3Br_2 + 6CO_3^{2-} + 3H_2O \to 5Br^- + BrO_3^- + 6HCO_3^-$
The oxidation state of $Br$ in $Br_2$ is $0$.
In $Br^-$,the oxidation state of $Br$ is $-1$ (reduction).
In $BrO_3^-$,the oxidation state of $Br$ is $+5$ (oxidation).
Since bromine undergoes both oxidation and reduction,this is a disproportionation reaction.
Therefore,bromine is both oxidized and reduced.

(D) In the reaction $3NaClO \xrightarrow{\Delta} NaClO_3 + 2NaCl$,the oxidation state of $Cl$ in $NaClO$ is $+1$.
In $NaClO_3$,the oxidation state of $Cl$ is $+5$ (increase).
In $NaCl$,the oxidation state of $Cl$ is $-1$ (decrease).
Since the oxidation state of the same element $(Cl)$ both increases and decreases,it is a disproportionation reaction.
This method is commonly used for the preparation of chlorates (a type of halate).

(C) In the given reaction: $3ClO^- \to ClO_3^- + 2Cl^-$
The oxidation state of chlorine in $ClO^-$ is $+1$.
In $ClO_3^-$,the oxidation state of chlorine is $+5$ (Oxidation).
In $Cl^-$,the oxidation state of chlorine is $-1$ (Reduction).
Since the same element $(Cl)$ is simultaneously oxidized and reduced,this is a disproportionation reaction.

(B) In the reaction $H_2O + \mathop{Br_2}\limits^{0} \to \mathop{HOBr}\limits^{+1} + \mathop{HBr}\limits^{-1}$,the oxidation state of bromine changes from $0$ to $+1$ (oxidation) and from $0$ to $-1$ (reduction).
Since bromine undergoes both oxidation and reduction,it is a disproportionation reaction where bromine is both oxidized and reduced.

(C) In the given reaction: $P_4 + 3NaOH + 3H_2O \to 3NaH_2PO_2 + PH_3$
The oxidation state of phosphorus $(P)$ in $P_4$ is $0$.
In $NaH_2PO_2$,the oxidation state of $P$ is $+1$.
In $PH_3$,the oxidation state of $P$ is $-3$.
Since the same element $(P)$ is simultaneously oxidized (from $0$ to $+1$) and reduced (from $0$ to $-3$),this is a disproportionation reaction.

(C) The oxidation state of $P$ in $H_3PO_3$ is $+3$. In $H_3PO_4$,it is $+5$ (oxidation) and in $PH_3$,it is $-3$ (reduction).
Change in oxidation state for oxidation: $5 - 3 = 2$.
Change in oxidation state for reduction: $3 - (-3) = 6$.
For a disproportionation reaction,the equivalent weight is calculated using the formula:
$E_w = \frac{M}{n_f}$,where $n_f = \frac{n_1 \times n_2}{n_1 + n_2}$.
Here,$n_1 = 2$ and $n_2 = 6$.
$n_f = \frac{2 \times 6}{2 + 6} = \frac{12}{8} = \frac{3}{2}$.
Therefore,$E_w = \frac{M}{3/2} = \frac{2M}{3}$.

(A) The reaction of mercury$(I)$ nitrate with ammonia is a classic example of a disproportionation reaction where $Hg_2^{2+}$ is converted into metallic mercury $(Hg^0)$ and mercury$(II)$ species $(Hg^{2+})$.
This reaction results in the formation of a black precipitate consisting of a mixture of finely divided metallic mercury and an insoluble mercury$(II)$ amido-nitrate complex.
Therefore,this is classified as a precipitate formation reaction.

(D) The reaction is: $\underline{N_2}O_4 + H_2O \longrightarrow HNO_3 + HNO_2$.
In this reaction,$N_2O_4$ undergoes disproportionation upon hydrolysis.
The products are $HNO_3$ (Nitric acid) and $HNO_2$ (Nitrous acid).
$HNO_3$ is an oxyacid with an $-ic$ suffix (Nitric acid).
$HNO_2$ is an oxyacid with an $-ous$ suffix (Nitrous acid).
Therefore,the products are two oxyacids,one with an $-ic$ suffix and the other with an $-ous$ suffix.

(C) The reaction is: $2ClO_2 + H_2O \longrightarrow HClO_2 + HClO_3$.
In this reaction,$ClO_2$ undergoes disproportionation.
The products are $HClO_2$ (chlorous acid) and $HClO_3$ (chloric acid).
$HClO_2$ has the $-ous$ suffix,and $HClO_3$ has the $-ic$ suffix.
Therefore,the products are two oxy acids,one with $-ic$ suffix and the other with $-ous$ suffix.

(A) The reaction is a disproportionation reaction of chlorine dioxide ($ClO_2$,often represented as $Cl_2O_4$ or $ClO_3$ in specific contexts) with water.
$2ClO_2 + H_2O \longrightarrow HClO_2 + HClO_3$
In the given reaction: $\underline{Cl}O_3 + H_2O \longrightarrow HClO_3 + HClO_4$.
The products are $HClO_3$ (chloric acid,$-ic$ suffix) and $HClO_4$ (perchloric acid,$-ic$ suffix).
Since both products are oxy acids with $-ic$ suffixes,the condition matches option $A$.

(C) The reaction is a disproportionation reaction where $N_2O_4$ reacts with water to form two oxyacids of nitrogen.
The reaction is: $N_2O_4 + H_2O \longrightarrow HNO_3 + HNO_2$.
In $HNO_3$ (nitric acid),the oxidation state of nitrogen is $+5$,and it ends with the suffix $-ic$.
In $HNO_2$ (nitrous acid),the oxidation state of nitrogen is $+3$,and it ends with the suffix $-ous$.
Since both products are oxyacids of nitrogen,one with an $-ic$ suffix and the other with an $-ous$ suffix,option $C$ is correct.

(C) The reaction given is the disproportionation of dithionic acid $(H_2S_2O_6)$ upon hydrolysis.
$H_2S_2O_6 + H_2O \longrightarrow H_2SO_3 + H_2SO_4$
In $H_2SO_3$ (sulfurous acid),the oxidation state of $S$ is $+4$,and it ends with the suffix $-ous$.
In $H_2SO_4$ (sulfuric acid),the oxidation state of $S$ is $+6$,and it ends with the suffix $-ic$.
Since the products are two oxy acids,one with an $-ic$ suffix $(H_2SO_4)$ and the other with an $-ous$ suffix $(H_2SO_3)$,option $C$ is correct.

(B) The reaction is: $\underline{Cl}_2 + H_2O \longrightarrow HOCl + HCl$.
In this reaction,$Cl_2$ undergoes disproportionation.
The products are $HOCl$ (hypochlorous acid) and $HCl$ (hydrochloric acid).
$HOCl$ is an oxyacid of chlorine where the oxidation state of $Cl$ is $+1$,which corresponds to the $-ous$ suffix (hypochlorous acid).
$HCl$ is a binary acid (hydrohalic acid),not an oxyacid.
Therefore,the product $HOCl$ is an oxyacid with an $-ous$ suffix.

(D) In the reaction $H_2O_2 \xrightarrow{R.T} H_2O + \frac{1}{2} O_2$,the oxidation state of oxygen in $H_2O_2$ is $-1$.
In the products,the oxidation state of oxygen in $H_2O$ is $-2$ and in $O_2$ is $0$.
Since the oxidation state of oxygen decreases from $-1$ to $-2$ (reduction) and increases from $-1$ to $0$ (oxidation) in the same reaction,it is a disproportionation reaction.
Also,since $H_2O_2$ decomposes into simpler substances upon heating or at room temperature,it is a thermal decomposition redox reaction.
Therefore,both $(A)$ and $(C)$ are correct.

(D) The given reaction is $P_4 + 3NaOH + 3H_2O \longrightarrow PH_3 + 3NaH_2PO_2$.
In this reaction,the oxidation state of phosphorus $(P)$ 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,this is a disproportionation reaction.
None of the categories $A$,$B$,or $C$ describe a redox reaction of this type.
Therefore,the correct assignment is $D$ (no reaction among the given categories).

(A) In the reaction $S_8 + 12NaOH \longrightarrow 4Na_2S + 2Na_2S_2O_3 + 6H_2O$,the oxidation state of sulfur changes from $0$ in $S_8$ to $-2$ in $Na_2S$ and $+2$ in $Na_2S_2O_3$.
Since the same element $(S)$ is simultaneously oxidized and reduced,this is a disproportionation reaction.
Therefore,the correct assignment is $A$.

(A) In the reaction $Cl_2 + 2NaOH \longrightarrow NaCl + NaOCl + H_2O$,the oxidation state of chlorine $(Cl)$ changes from $0$ in $Cl_2$ to $-1$ in $NaCl$ and $+1$ in $NaOCl$.
Since the same element $(Cl)$ is simultaneously oxidized and reduced,this is a disproportionation reaction.
Therefore,the correct assignment is $A$.

(A) In the reaction $I_2 + 2NaOH \longrightarrow NaI + NaOI + H_2O$,the oxidation state of iodine $(I)$ changes from $0$ in $I_2$ to $-1$ in $NaI$ and $+1$ in $NaOI$.
Since the same element $(I)$ is simultaneously oxidized and reduced,this is a disproportionation reaction.
Therefore,the correct assignment is $A$.

(A) In the reaction $N_2O_3 \rightarrow NO + NO_2$,the oxidation state of nitrogen in $N_2O_3$ is $+3$.
In the products,the oxidation state of nitrogen in $NO$ is $+2$ and in $NO_2$ is $+4$.
Since the oxidation state of nitrogen increases from $+3$ to $+4$ (oxidation) and decreases from $+3$ to $+2$ (reduction) simultaneously,this is a disproportionation reaction.
Therefore,the correct option is $A$.

(A) The reaction is $2Ca(OH)_2 + 2Cl_2 \longrightarrow Ca(OCl)_2 + CaCl_2 + 2H_2O$.
In this reaction,the oxidation state of $Cl$ in $Cl_2$ is $0$. In the products,the oxidation state of $Cl$ in $Ca(OCl)_2$ is $+1$ and in $CaCl_2$ is $-1$.
Since the same element $(Cl)$ is simultaneously oxidized and reduced,this is a disproportionation reaction.
Therefore,the correct assignment is $A$.

(A) The given reaction is: $3XeF_4 + 6H_2O \longrightarrow Xe + 2XeO_3 + 12HF + 1.5O_2$.
In this reaction,the oxidation state of $Xe$ in $XeF_4$ is $+4$.
In the products,$Xe$ is in $0$ oxidation state (in $Xe$) and $+6$ oxidation state (in $XeO_3$).
Since the same element $(Xe)$ is simultaneously oxidized (from $+4$ to $+6$) and reduced (from $+4$ to $0$),this is a disproportionation reaction.
Therefore,the correct assignment is $A$.

(A) In the reaction $K_2MnO_4 + H^{+} \longrightarrow KMnO_4 + MnO_2 + H_2O$,the oxidation state of $Mn$ in $K_2MnO_4$ is $+6$.
In $KMnO_4$,the oxidation state of $Mn$ is $+7$ (oxidation).
In $MnO_2$,the oxidation state of $Mn$ is $+4$ (reduction).
Since the same element $(Mn)$ in the same oxidation state $(+6)$ is simultaneously oxidized to $+7$ and reduced to $+4$,this is a disproportionation reaction.
Therefore,the correct assignment is $A$.

(A) In the given reaction: $P_4 + 3NaOH + 3H_2O \longrightarrow PH_3 + 3NaH_2PO_2$
The oxidation state of phosphorus $(P)$ in $P_4$ is $0$.
In $PH_3$,the oxidation state of $P$ is $-3$ (reduction).
In $NaH_2PO_2$,the oxidation state of $P$ is $+1$ (oxidation).
Since the same element $(P)$ is simultaneously oxidized and reduced in the same reaction,this is a disproportionation reaction.
Therefore,the correct option is $A$.

(A) In the reaction $4H_3PO_3 \xrightarrow{\Delta} 3H_3PO_4 + PH_3$,the oxidation state of phosphorus $(P)$ 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).
Since the same element $(P)$ is simultaneously oxidized and reduced,this is a disproportionation reaction.
Therefore,the correct classification is $A$.

(A) In the reaction $2Se_2Cl_2 \xrightarrow{\Delta} SeCl_4 + 3Se$,the oxidation state of $Se$ in $Se_2Cl_2$ is $+1$.
In $SeCl_4$,the oxidation state of $Se$ is $+4$.
In $Se$,the oxidation state of $Se$ is $0$.
Since the same element $(Se)$ is simultaneously oxidized (from $+1$ to $+4$) and reduced (from $+1$ to $0$),this is a disproportionation reaction.
Therefore,the correct assignment is $(A)$.

(B) In the reaction $Na_2S + H_2SO_4 \longrightarrow S + SO_2 + Na_2SO_4$,we analyze the oxidation states:
$S$ in $Na_2S$ is $-2$.
$S$ in $H_2SO_4$ is $+6$.
In the products,$S$ in $S\downarrow$ is $0$ and $S$ in $SO_2$ is $+4$.
Here,the sulfur atom with oxidation state $-2$ is oxidized to $0$,and the sulfur atom with oxidation state $+6$ is reduced to $+4$.
Since two different oxidation states of the same element ($S^{-2}$ and $S^{+6}$) combine to form intermediate oxidation states ($S^0$ and $S^{+4}$),this is a comproportionation reaction.
Therefore,the correct assignment is $B$.

(A) In the given reaction: $Na_2C_2O_4 + H_2SO_4 \longrightarrow Na_2SO_4 + CO + CO_2$.
The oxidation state of carbon in $Na_2C_2O_4$ is $+3$.
In the products,the oxidation state of carbon in $CO$ is $+2$ and in $CO_2$ is $+4$.
Since the same element (carbon) is simultaneously oxidized (from $+3$ to $+4$) and reduced (from $+3$ to $+2$),this is a disproportionation reaction.
Therefore,the correct assignment is $A$.

(D) $NO_2$ disproportionates in water: $2NO_2 + H_2O \to HNO_3 + HNO_2$. The $HNO_2$ further disproportionates at room temperature $(R.T.)$ to form $HNO_3$ and $NO$.
$Cu^{+}$ ions are unstable in aqueous solution and undergo disproportionation: $2Cu^{+}(aq) \to Cu(s) + Cu^{2+}(aq)$.
$MnO_4^{2-}$ (manganate ion) is unstable in neutral or acidic medium and disproportionates: $3MnO_4^{2-}(aq) + 2H_2O(l) \to 2MnO_4^-(aq) + MnO_2(s) + 4OH^-(aq)$.
$Ca(OCl)Cl$ (bleaching powder) is a salt that dissociates in water into $Ca^{2+}$,$Cl^-$,and $OCl^-$ ions. It does not undergo disproportionation in water at room temperature.

(C) comproportionation reaction is a type of redox reaction where two reactants,each containing an element in a different oxidation state,form a product in which the element is in an intermediate oxidation state.
$1$. $H_2S (-2) + SO_2 (+4) \to S (0) + H_2O$: This is a comproportionation reaction.
$2$. $I^{-} (-1) + IO_3^{-} (+5) + H^{+} \to I_2 (0) + H_2O$: This is a comproportionation reaction.
$3$. $K_2MnO_4 (+6) + H^{+} \to KMnO_4 (+7) + MnO_2 (+4)$: This is a disproportionation reaction,where a single reactant in an intermediate oxidation state forms products with higher and lower oxidation states.
$4$. $MnO_4^{-} (+7) + Mn^{2+} (+2) \to MnO_2 (+4)$: This is a comproportionation reaction.
Therefore,the reaction in option $C$ is a disproportionation reaction,not a comproportionation reaction.

(A) The green solution of $MnO_4^{2-}$ is stable only in a strongly basic medium.
In a neutral,acidic,or less basic medium,it undergoes disproportionation into $MnO_2$ and $MnO_4^-$.
Since $SO_3$ gas reacts with water to form $H_2SO_4$ (an acid),it creates an acidic medium.
Therefore,$K_2MnO_4$ undergoes disproportionation in the presence of $SO_3$ gas:
$3K_2MnO_4 + 2SO_3 \rightarrow 2KMnO_4 + MnO_2 + 2K_2SO_4$.

(D) comproportionation reaction is a type of redox reaction where two reactants,each containing the same element but in different oxidation states,form a product in which the element is in an intermediate oxidation state.
In the reaction between $MnO_4^{-}$ ($Mn$ is $+7$) and $Mn^{2+}$ ($Mn$ is $+2$),they react to form $MnO_2$ ($Mn$ is $+4$).
Since $+4$ is intermediate between $+7$ and $+2$,this is a comproportionation reaction.
The reaction is: $2MnO_4^{-}(aq.) + 3Mn^{2+}(aq.) + 4OH^{-}(aq.) \rightarrow 5MnO_2(s) + 2H_2O(l)$.

(A) disproportionation reaction is a type of redox reaction in which the same element is simultaneously oxidized and reduced.
In the reaction $Cu_2O + 2H^{+} \to Cu + Cu^{2+} + H_2O$:
$1$. The oxidation state of $Cu$ in $Cu_2O$ is $+1$.
$2$. In the product $Cu$,the oxidation state of $Cu$ is $0$ (reduction).
$3$. In the product $Cu^{2+}$,the oxidation state of $Cu$ is $+2$ (oxidation).
Since the same element $(Cu)$ undergoes both oxidation and reduction,this is a disproportionation reaction.

(D) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
$A) \, P_4 + OH^{-} \to PH_3 + H_2PO_2^-$: Phosphorus $(P)$ is oxidized and reduced. This is a disproportionation reaction.
$B) \, Cl_2 + OH^{-} \to Cl^{-} + ClO^-$: Chlorine $(Cl)$ is oxidized and reduced. This is a disproportionation reaction.
$C) \, Br_2 + OH^{-} \to Br^{-} + BrO_3^-$: Bromine $(Br)$ is oxidized and reduced. This is a disproportionation reaction.
$D) \, 2KClO_3 \to 2KCl + 3O_2$: Here,chlorine is reduced from $+5$ to $-1$ oxidation state,while oxygen is oxidized from $-2$ to $0$ oxidation state. Since two different elements are undergoing oxidation and reduction,it is not a disproportionation reaction.

(B) The reaction of chlorine with dilute and cold sodium hydroxide is a disproportionation reaction:
$Cl_2 + 2NaOH \text{ (dil. and cold)} \longrightarrow NaCl + NaOCl + H_2O$
Here,$A$ is $NaCl$ (anion is $Cl^-$) and $B$ is $NaOCl$ (anion is $OCl^-$).
$Cl^-$ is the chloride ion and $OCl^-$ is the hypochlorite ion.
Therefore,the correct option is $B$.

(C) In the reaction $P_4 + 3KOH + 3H_2O \to PH_3 + 3KH_2PO_2$,the oxidation state of phosphorus changes as follows:
$1$. In $P_4$,the oxidation state of $P$ is $0$.
$2$. In $PH_3$,the oxidation state of $P$ is $-3$ (reduction).
$3$. In $KH_2PO_2$,the oxidation state of $P$ is $+1$ (oxidation).
Since phosphorus is simultaneously oxidized and reduced,this is a disproportionation reaction.

(B) In $HNO_2$ (nitrous acid),the oxidation state of nitrogen is $+3$.
Nitrogen can be oxidized to a higher oxidation state (e.g.,$+5$ in $HNO_3$) or reduced to a lower oxidation state (e.g.,$+2$ in $NO$ or $0$ in $N_2$).
Therefore,$HNO_2$ can act as both an oxidizing and a reducing agent.
$HNO_3$ acts only as an oxidizing agent (nitrogen is in its maximum $+5$ state),$H_2SO_4$ acts primarily as an oxidizing agent,and $He$ is an inert noble gas.

(B) In $SO_2$,the oxidation state of sulfur is $+4$.
Sulfur can be oxidized to $+6$ (e.g.,in $SO_3$ or $H_2SO_4$) and reduced to $0$ (e.g.,in $S$).
Therefore,$SO_2$ can act as both an oxidizing agent and a reducing agent.
$H_2SO_4$ acts only as an oxidizing agent because sulfur is in its maximum oxidation state $(+6)$.
$H_2S$ acts only as a reducing agent because sulfur is in its minimum oxidation state $(-2)$.
$HNO_3$ acts only as an oxidizing agent because nitrogen is in its maximum oxidation state $(+5)$.

(B) The reaction of chlorine dioxide $(ClO_2)$ with sodium hydroxide $(NaOH)$ is a disproportionation reaction.
The balanced chemical equation is:
$2ClO_2 + 2NaOH \rightarrow NaClO_2 + NaClO_3 + H_2O$
In this reaction,the oxidation state of chlorine in $ClO_2$ is $+4$.
It is reduced to $+3$ in $NaClO_2$ and oxidized to $+5$ in $NaClO_3$.
Therefore,the mixture obtained consists of $NaClO_2$ and $NaClO_3$.

(C) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
In option $A$: $KO_2 + H_2O + CO_2 \to KHCO_3 + O_2$. The oxidation state of $O$ in $KO_2$ is $-1/2$. In $KHCO_3$,it is $-2$,and in $O_2$,it is $0$. This is a disproportionation reaction.
In option $B$: $KClO_3 \to KClO_4 + KCl$. The oxidation state of $Cl$ changes from $+5$ to $+7$ (oxidation) and $-1$ (reduction). This is a disproportionation reaction.
In option $C$: $PbO_2 + H_2O \to PbO + H_2O_2$. The oxidation state of $Pb$ changes from $+4$ to $+2$ (reduction). The oxidation state of $O$ in $PbO_2$ is $-2$ and in $H_2O_2$ is $-1$. This is $NOT$ a disproportionation reaction because only $Pb$ is reduced and $O$ is oxidized; they are different elements.
In option $D$: This is a Cannizzaro-type reaction where the same carbon atom undergoes oxidation and reduction. This is a disproportionation reaction.

(C) species can act as both an oxidizing and a reducing agent if the central atom is in an intermediate oxidation state,allowing it to either increase (oxidation) or decrease (reduction) its oxidation number.

Species	Oxidation State of Central Atom
$Cl^{-}$	$-1$ (Minimum)
$ClO_4^{-}$	$+7$ (Maximum)
$ClO^{-}$	$+1$ (Intermediate)
$MnO_4^{-}$	$+7$ (Maximum)

In $ClO^{-}$,the chlorine atom is in the $+1$ oxidation state. Since its oxidation state can increase (e.g.,to $+3, +5, +7$) or decrease (e.g.,to $-1$),it can function as both an oxidizing and a reducing agent.

(A) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
$(a)$ $2 Cu^{+} \rightarrow Cu^{2+} + Cu^{0}$
Here,$Cu^{+}$ (oxidation state $+1$) is oxidized to $Cu^{2+}$ $(+2)$ and reduced to $Cu^{0}$ $(0)$. This is a disproportionation reaction.
$(b)$ $3 MnO_{4}^{2-} + 4 H^{+} \rightarrow 2 MnO_{4}^{-} + MnO_{2} + 2 H_{2}O$
Here,$Mn$ in $MnO_{4}^{2-}$ (oxidation state $+6$) is oxidized to $MnO_{4}^{-}$ $(+7)$ and reduced to $MnO_{2}$ $(+4)$. This is a disproportionation reaction.
$(c)$ $2 KMnO_{4} \xrightarrow{\Delta} K_{2}MnO_{4} + MnO_{2} + O_{2}$
This is a decomposition reaction,not a disproportionation reaction.
$(d)$ $2 MnO_{4}^{-} + 3 Mn^{2+} + 2 H_{2}O \rightarrow 5 MnO_{2} + 4 H^{+}$
This is a comproportionation reaction where $Mn(+7)$ and $Mn(+2)$ combine to form $Mn(+4)$.
Therefore,only $(a)$ and $(b)$ are disproportionation reactions.

(D) substance can act as both an oxidising and a reducing agent if the central atom is in an intermediate oxidation state.
$1$. In $H_2O_2$,the oxidation state of $O$ is $-1$,which can increase to $0$ (reducing) or decrease to $-2$ (oxidising).
$2$. In $H_2SO_3$,the oxidation state of $S$ is $+4$,which can increase to $+6$ (reducing) or decrease to lower values (oxidising).
$3$. In $HNO_2$,the oxidation state of $N$ is $+3$,which can increase to $+5$ (reducing) or decrease to lower values (oxidising).
$4$. In $H_3PO_4$,the oxidation state of $P$ is $+5$,which is its maximum oxidation state. Therefore,it can only act as an oxidising agent and cannot be further oxidised to act as a reducing agent.

(D) Among the oxoanions of chlorine listed,$ClO_{4}^{-}$ does not undergo disproportionation because chlorine is present in its highest possible oxidation state,which is $+7$.
The disproportionation reactions for the other three oxoanions are as follows:
$1. \ 3ClO^{-} \to 2Cl^{-} + ClO_{3}^{-}$
$2. \ 3ClO_{2}^{-} \to 2ClO_{3}^{-} + Cl^{-}$
$3. \ 4ClO_{3}^{-} \to Cl^{-} + 3ClO_{4}^{-}$

(N/A) The oxidation numbers of carbon in $(CN)_{2}$,$CN^{-}$,and $CNO^{-}$ are $+3$,$+2$,and $+4$ respectively.
$1$. In $(CN)_{2}$,let the oxidation number of $C$ be $x$. Since $N$ is $-3$,$2(x - 3) = 0$,so $x = +3$.
$2$. In $CN^{-}$,the oxidation number of $C$ is $+2$ (since $x + (-3) = -1$).
$3$. In $CNO^{-}$,the oxidation number of $C$ is $+4$ (since $x - 3 - 2 = -1$).
The reaction is:
$(C^{+3}N)_{2}(g) + 2OH^{-}(aq) \rightarrow C^{+2}N^{-}(aq) + C^{+4}NO^{-}(aq) + H_{2}O(l)$
It can be observed that the same element (carbon) is being reduced (from $+3$ to $+2$) and oxidized (from $+3$ to $+4$) simultaneously. Reactions in which the same substance is both reduced and oxidized are known as disproportionation reactions. Thus,the alkaline decomposition of cyanogen is an example of a disproportionation reaction.

(N/A) In disproportionation reactions,one of the reacting substances always contains an element that can exist in at least three oxidation states.
$(a)$ $P, Cl$,and $S$ can show disproportionation reactions as these elements can exist in three or more oxidation states.
$(b)$ $Mn, Cu$,and $Ga$ can show disproportionation reactions as these elements can exist in three or more oxidation states.

(N/A) The balanced chemical equation is: $3 Cl_{2} + 6 NaOH \rightarrow 5 NaCl + NaClO_{3} + 3 H_{2}O$
Yes,this is a disproportionation reaction because the oxidation state of chlorine changes from $0$ in $Cl_{2}$ to $-1$ in $NaCl$ and $+5$ in $NaClO_{3}$.

(N/A) reaction in which a species is simultaneously oxidized and reduced is called a disproportionation reaction. In this process,a particular oxidation state becomes less stable relative to other oxidation states,one lower and one higher.
For example,the manganate $(VI)$ ion $(MnO_4^{2-})$ is unstable in acidic solution and undergoes disproportionation to form permanganate $(VII)$ $(MnO_4^-)$ and manganese $(IV)$ oxide $(MnO_2)$.
$3MnO_4^{2-} + 4H^{+} \to 2MnO_4^{-} + MnO_2 + 2H_2O$

(N/A) disproportionation reaction is a special type of redox reaction in which a species in a particular oxidation state is simultaneously oxidized and reduced.
Examples in aqueous solution:
$(i)$ $3CrO_4^{3-} + 8H^{+} \to 2CrO_4^{2-} + Cr^{3+} + 4H_2O$
Here,$Cr(V)$ is oxidized to $Cr(VI)$ and reduced to $Cr(III)$.
$(ii)$ $3MnO_4^{2-} + 4H^{+} \to 2MnO_4^{-} + MnO_2 + 2H_2O$
Here,$Mn(VI)$ is oxidized to $Mn(VII)$ and reduced to $Mn(IV)$.

(C) disproportionation reaction is a special type of redox reaction in which the same element in a given oxidation state is simultaneously oxidized and reduced.
Therefore,the oxidation state of the same element increases as well as decreases in the reaction.

(N/A) To justify that the reaction $H_2O_{(s)} + F_{2(g)} \to HF_{(g)} + HOF_{(g)}$ is a redox reaction,we assign oxidation numbers to each atom:
$1$. In $H_2O_{(s)}$,the oxidation number of $H$ is $+1$ and $O$ is $-2$.
$2$. In $F_{2(g)}$,the oxidation number of $F$ is $0$.
$3$. In $HF_{(g)}$,the oxidation number of $H$ is $+1$ and $F$ is $-1$.
$4$. In $HOF_{(g)}$,the oxidation number of $H$ is $+1$,$O$ is $-2$,and $F$ is $+1$.
Analysis:
- The oxidation number of $F$ changes from $0$ (in $F_2$) to $-1$ (in $HF$),which is a reduction.
- The oxidation number of $F$ also changes from $0$ (in $F_2$) to $+1$ (in $HOF$),which is an oxidation.
Since the same element $(F)$ is simultaneously oxidized and reduced,this is a disproportionation redox reaction.