Auto oxidation and Disproportionation Questions in English

(A) The reaction of sulphur $(S_8)$ with hot aqueous sodium hydroxide $(NaOH)$ is a disproportionation reaction.
In this reaction,sulphur is both oxidized and reduced to form sodium sulfide $(Na_2S)$ and sodium thiosulphate $(Na_2S_2O_3)$.
The balanced chemical equation is:
$S_8 + 12 NaOH \longrightarrow 4 Na_2S + 2 Na_2S_2O_3 + 6 H_2O$.
Thus,the products $P$ and $Q$ are $Na_2S$ and $Na_2S_2O_3$ respectively.

(D) disproportionation reaction is a type of redox reaction where the same element in a single oxidation state is simultaneously oxidized and reduced.
In option $A$,$P$ in $H_3 PO_3$ $(+3)$ is oxidized to $H_3 PO_4$ $(+5)$ and reduced to $PH_3$ $(-3)$.
In option $B$,$O$ in $H_2 O_2$ $(-1)$ is oxidized to $O_2$ $(0)$ and reduced to $H_2 O$ $(-2)$.
In option $C$,$P$ in $P_4$ $(0)$ is oxidized to $NaH_2 PO_2$ $(+1)$ and reduced to $PH_3$ $(-3)$.
In option $D$,$P$ in $P_4$ $(0)$ is oxidized to $PCl_3$ $(+3)$ and $S$ in $SOCl_2$ $(+4)$ is reduced to $S_2 Cl_2$ $(+1)$. Since different elements undergo oxidation and reduction,it is a redox reaction but not a disproportionation reaction.

(A) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
$1. 2 H_2\stackrel{-1}{O_2} \rightarrow 2 H_2\stackrel{-2}{O} + \stackrel{0}{O_2}$
Here,the oxidation state of oxygen changes from $-1$ to $-2$ (reduction) and from $-1$ to $0$ (oxidation).
$2. \stackrel{0}{Cl_2} + 2 OH^{-} \rightarrow \stackrel{-1}{Cl^{-}} + \stackrel{+1}{ClO^{-}} + H_2O$
Here,the oxidation state of chlorine changes from $0$ to $-1$ (reduction) and from $0$ to $+1$ (oxidation).
Both reactions involve the same element undergoing both oxidation and reduction,hence they are disproportionation reactions.

(C) In the disproportionation reaction of $Se_2Cl_2$,selenium in the $+1$ oxidation state is converted to $+4$ and $0$ oxidation states.
The balanced chemical equation is:
$2Se_2Cl_2 \rightarrow SeCl_4 + 3Se$
Thus,$P$ is $SeCl_4$ and $Q$ is $Se$.

(A) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
$(A)$ $Cl_2 + 2 NaOH \rightarrow NaCl + NaOCl + H_2O$: The oxidation state of $Cl$ changes from $0$ to $-1$ (reduction) and from $0$ to $+1$ (oxidation). This is a disproportionation reaction.
$(B)$ $2 H_2O_2 \rightarrow 2 H_2O + O_2$: The oxidation state of $O$ in $H_2O_2$ is $-1$. It changes to $-2$ in $H_2O$ (reduction) and to $0$ in $O_2$ (oxidation). This is a disproportionation reaction.
$(C)$ $2 KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$: The oxidation state of $Mn$ changes from $+7$ to $+6$ and $+4$ (both reduction). This is not a disproportionation reaction.
$(D)$ $3 MnO_4^{2-} + 4 H^+ \rightarrow 2 MnO_4^- + MnO_2 + 2 H_2O$: The oxidation state of $Mn$ changes from $+6$ to $+7$ (oxidation) and from $+6$ to $+4$ (reduction). This is a disproportionation reaction.
Thus,reactions $(A)$,$(B)$,and $(D)$ are disproportionation reactions.

(B) disproportionation reaction is a redox reaction in which the same element in a single reactant is both oxidized and reduced.
$(i) \ \stackrel{0}{Cl}_{2} + 2KI \longrightarrow 2KCl + \stackrel{0}{I}_{2}$: This is a displacement reaction where $Cl$ is reduced and $I$ is oxidized. Not disproportionation.
$(ii) \ \stackrel{0}{Cl}_{2} + 2OH^{-} \longrightarrow \stackrel{+1}{ClO}^{-} + \stackrel{-1}{Cl}^{-} + H_2O$: Here,$Cl$ is oxidized from $0$ to $+1$ and reduced from $0$ to $-1$. This is a disproportionation reaction.
$(iii) \ \stackrel{0}{Mg} + 2\stackrel{+1}{H}Cl \longrightarrow \stackrel{+2}{Mg}Cl_{2} + \stackrel{0}{H}_{2}$: This is a displacement reaction where $Mg$ is oxidized and $H$ is reduced. Not disproportionation.
$(iv) \ 2H_2\stackrel{-1}{O}_{2} \longrightarrow 2H_2\stackrel{-2}{O} + \stackrel{0}{O}_{2}$: Here,$O$ is reduced from $-1$ to $-2$ and oxidized from $-1$ to $0$. This is a disproportionation reaction.
Therefore,reactions $(ii)$ and $(iv)$ are disproportionation reactions.

(B) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced,resulting in products with higher and lower oxidation states than the reactant.
In option $A$,$Cl_2$ $(0)$ goes to $ClO^-$ $(+1)$ and $Cl^-$ $(-1)$,which is a disproportionation reaction.
In option $B$,$F_2$ $(0)$ reacts with $OH^-$ to form $OF_2$ and $F^-$. In $OF_2$,the oxidation state of $F$ is $-1$ and in $F^-$,it is also $-1$. The oxidation state of $O$ changes from $-2$ in $OH^-$ to $+2$ in $OF_2$ and remains $-2$ in $H_2O$. Since the oxidation state of fluorine only decreases,this is not a disproportionation reaction.
In option $C$,$ClO_3^-$ $(+5)$ goes to $Cl^-$ $(-1)$ and $ClO_4^-$ $(+7)$,which is a disproportionation reaction.
In option $D$,$ClO^-$ $(+1)$ goes to $Cl^-$ $(-1)$ and $ClO_3^-$ $(+5)$,which is a disproportionation reaction.

(C) disproportionation reaction is a type of redox reaction in which the same element is simultaneously oxidized and reduced.
In the reaction $4 KClO_{3(s)} \stackrel{\Delta}{\longrightarrow} KCl_{(s)} + 3 KClO_{4(s)}$:
The oxidation state of chlorine in $KClO_3$ is $+5$.
In $KCl$,the oxidation state of chlorine is $-1$ (reduction).
In $KClO_4$,the oxidation state of chlorine is $+7$ (oxidation).
Since the same element (chlorine) is both oxidized and reduced,this is a disproportionation reaction.

(A) disproportionation reaction is a special type of redox reaction in which the same element in a given oxidation state is simultaneously oxidized and reduced.
In the reaction $3Cl_{2\text{(g)}} + 6OH^-{_{\text{(aq)}}} \rightarrow ClO_3^-{_{\text{(aq)}}} + 5Cl^-{_{\text{(aq)}}} + 3H_2O_{\text{(l)}}$,the oxidation state of chlorine changes from $0$ in $Cl_2$ to $-1$ in $Cl^-$ (reduction) and to $+5$ in $ClO_3^-$ (oxidation).
Thus,it is a disproportionation reaction.

(D) The disproportionation reaction for $MnO_4^{2-}$ occurs as follows:
$3 \stackrel{+6}{MnO_4^{2-}} + 4 H^{+} \longrightarrow \stackrel{+4}{MnO_2} + 2 \stackrel{+7}{MnO_4^{-}} + 2 H_2 O$
Here,$MnO_4^{2-}$ (oxidation state $+6$) is oxidized to $MnO_4^{-}$ (oxidation state $+7$) and reduced to $MnO_2$ (oxidation state $+4$).
Thus,the oxidation states of $Mn$ in the products are $+7$ and $+4$.
Therefore,option $(D)$ is the correct answer.

(D) When $Cl_2$ gas is passed through a hot and concentrated $NaOH$ solution,the reaction is:
$3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O$
In this reaction,the oxidation state of chlorine changes from $0$ in $Cl_2$ to $-1$ in $NaCl$ (reduction) and $+5$ in $NaClO_3$ (oxidation).
Since the same element is simultaneously oxidized and reduced,this is a disproportionation reaction.

(B) When $Cl_2$ is passed through hot aqueous $NaOH$,a disproportionation reaction occurs,resulting in the formation of sodium chloride $(NaCl)$ and sodium chlorate $(NaClO_3)$:
$6NaOH + 3Cl_2 \longrightarrow 5NaCl + NaClO_3 + 3H_2O$
In $NaCl$,the oxidation state of $Cl$ is $x$:
$(+1) + x = 0 \implies x = -1$
In $NaClO_3$,the oxidation state of $Cl$ is $y$:
$(+1) + y + 3(-2) = 0 \implies y - 5 = 0 \implies y = +5$
Thus,the $Cl$ atoms in the products exist in $-1$ and $+5$ oxidation states.

(A) In an acidic solution,the manganate ion $(MnO_4^{2-})$ is unstable and undergoes a disproportionation reaction.
It disproportionates into the permanganate ion $(MnO_4^-)$ and manganese dioxide $(MnO_2)$.
The balanced chemical equation for this reaction is:
$3MnO_4^{2-} + 4H^+ \to 2MnO_4^- + MnO_2 + 2H_2O$.