Consider the following reaction:
$MnO_2 + KOH + O_2 \rightarrow A + H_2O$
Product '$A$' in neutral or acidic medium disproportionates to give products '$B$' and '$C$' along with water. The sum of spin-only magnetic moment values of '$B$' and '$C$' is . . . . . $BM$. (nearest integer)
(Given atomic number of $Mn$ is $25$)
$MnO_2 + KOH + O_2 \rightarrow A + H_2O$
Product '$A$' in neutral or acidic medium disproportionates to give products '$B$' and '$C$' along with water. The sum of spin-only magnetic moment values of '$B$' and '$C$' is . . . . . $BM$. (nearest integer)
(Given atomic number of $Mn$ is $25$)
- A$3$
- B$4$
- C$5$
- D$7$
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Similar Questions
Match the List-$I$ with List-$II$
List-$I$ $(\text{Redox Reaction})$ List-$II$ $(\text{Type of Redox Reaction})$ $A$. $CH_{4(g)} + 2O_{2(g)} \xrightarrow{\Delta} CO_{2(g)} + 2H_2O_{(l)}$ $I$. Disproportionation reaction $B$. $2NaH_{(s)} \xrightarrow{\Delta} 2Na_{(s)} + H_{2(g)}$ $II$. Combination reaction $C$. $V_2O_{5(s)} + 5Ca_{(s)} \xrightarrow{\Delta} 2V_{(s)} + 5CaO_{(s)}$ $III$. Decomposition reaction $D$. $2H_2O_{2(aq)} \xrightarrow{\Delta} 2H_2O_{(l)} + O_{2(g)}$ $IV$. Displacement reaction
Choose the correct answer from the options given below:
| List-$I$ $(\text{Redox Reaction})$ | List-$II$ $(\text{Type of Redox Reaction})$ |
|---|---|
| $A$. $CH_{4(g)} + 2O_{2(g)} \xrightarrow{\Delta} CO_{2(g)} + 2H_2O_{(l)}$ | $I$. Disproportionation reaction |
| $B$. $2NaH_{(s)} \xrightarrow{\Delta} 2Na_{(s)} + H_{2(g)}$ | $II$. Combination reaction |
| $C$. $V_2O_{5(s)} + 5Ca_{(s)} \xrightarrow{\Delta} 2V_{(s)} + 5CaO_{(s)}$ | $III$. Decomposition reaction |
| $D$. $2H_2O_{2(aq)} \xrightarrow{\Delta} 2H_2O_{(l)} + O_{2(g)}$ | $IV$. Displacement reaction |
Choose the correct answer from the options given below:
Match each of the reactions given in Column $I$ with the corresponding product$(s)$ given in Column $II$.
Column $I$ Column $II$ $(A)$ $Cu + \text{dil } HNO_3$ $(p)$ $NO$ $(B)$ $Cu + \text{conc } HNO_3$ $(q)$ $NO_2$ $(C)$ $Zn + \text{dil } HNO_3$ $(r)$ $N_2O$ $(D)$ $Zn + \text{conc } HNO_3$ $(s)$ $Cu(NO_3)_2$ $(t)$ $Zn(NO_3)_2$
| Column $I$ | Column $II$ |
| $(A)$ $Cu + \text{dil } HNO_3$ | $(p)$ $NO$ |
| $(B)$ $Cu + \text{conc } HNO_3$ | $(q)$ $NO_2$ |
| $(C)$ $Zn + \text{dil } HNO_3$ | $(r)$ $N_2O$ |
| $(D)$ $Zn + \text{conc } HNO_3$ | $(s)$ $Cu(NO_3)_2$ |
| $(t)$ $Zn(NO_3)_2$ |
$x, y$ and $z$ are respectively:
$MnO_4^- + x e^- \xrightarrow{\text{Alkaline medium}} MnO_4^{2-}$
$MnO_4^- + y e^- \xrightarrow{\text{Acidic medium}} Mn^{2+}$
$MnO_4^- + z e^- \xrightarrow{\text{Neutral medium}} MnO_2$
$MnO_4^- + x e^- \xrightarrow{\text{Alkaline medium}} MnO_4^{2-}$
$MnO_4^- + y e^- \xrightarrow{\text{Acidic medium}} Mn^{2+}$
$MnO_4^- + z e^- \xrightarrow{\text{Neutral medium}} MnO_2$
Difficult
View Solution$A$ solution of $KMnO_4$ on reduction yields either a colourless solution,a brown precipitate,or a green solution depending on the $pH$ of the solution. What different stages of the reduction do these represent and how are they carried out?
Medium
View SolutionThe reaction of $Na_2S_2O_3$ with iodine gives
Easy
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