
What can be $'X'$ in the following reaction?
$MnO_4^ - + {H^ + } + X \to No\operatorname{Re} action$
1) ${S_2}O_3^{2 - }$
2) $S{O_2}$
3) $SO_4^{2 - }$
4) $SO_3^{2 - }$
Answer
509.7k+ views
Hint: Potassium permanganate is an inorganic compound which has chemical formula $KMn{O_4}$ . It is also called permanganate of potash or Condy’s crystals. It acts as a very good oxidizing agent and can be used as oxidant in many chemical reactions. It shows its oxidizing agent property in acidic conditions.
Complete answer:
Potassium permanganate has chemical formula $KMn{O_4}$ . It is also known as permanganate of potash or Condy’s crystals. It acts as a very good oxidizing agent in the acidic medium.
Now, if we look at the above given reaction:
$MnO_4^ - + {H^ + } + X \to No\operatorname{Re} action$
So, in the above reaction, it is given that $MnO_4^ - $ in presence of ${H^ + }$ . The $MnO_4^ - $ will act as a strong oxidizing agent.
Now, we know that, the compound in lowest or moderate oxidation state (not the highest) is oxidised to a compound having highest oxidation state.
So, in the given equation, if no reaction is taking place, then it can be concluded that compound $'X'$ must be present in the highest oxidation state.
Now, we will calculate oxidation state of each given compound:-
1) ${S_2}O_3^{2 - }$
Let oxidation state of Sulphur is x, then
$2x + 3( - 2) = - 2$
$2x = - 2 + 6$
$2x = 4$
$x = + 2$
2) $S{O_2}$
Let oxidation state of Sulphur is x, then
$x + 2( - 2) = 0$
$x = + 4$
3) $SO_4^{2 - }$
Let oxidation state of Sulphur is x, then
$x + 4( - 2) = - 2$
$x = - 2 + 8$
$x = + 6$
4) $SO_3^{2 - }$
Let oxidation state of Sulphur is x, then
$x + 3( - 2) = - 2$
$x = - 2 + 6$
$x = + 4$
We know that the highest oxidation state of Sulphur is $ + 6$ and from the above calculation, we can see that the highest oxidation state of sulphur is in compound (3) $SO_4^{2 - }$ .
Therefore, the correct option is (C) $SO_4^{2 - }$ .
Note:
Potassium permanganate is used for qualitative analysis. It is also used as a disinfectant to cure many skin problems and can be used in treating bacterial infections, dermatitis, etc. It is also used in tanning leathers, printing fabrics. It can also be used as a bleaching agent, pesticides and an antiseptic.
Complete answer:
Potassium permanganate has chemical formula $KMn{O_4}$ . It is also known as permanganate of potash or Condy’s crystals. It acts as a very good oxidizing agent in the acidic medium.
Now, if we look at the above given reaction:
$MnO_4^ - + {H^ + } + X \to No\operatorname{Re} action$
So, in the above reaction, it is given that $MnO_4^ - $ in presence of ${H^ + }$ . The $MnO_4^ - $ will act as a strong oxidizing agent.
Now, we know that, the compound in lowest or moderate oxidation state (not the highest) is oxidised to a compound having highest oxidation state.
So, in the given equation, if no reaction is taking place, then it can be concluded that compound $'X'$ must be present in the highest oxidation state.
Now, we will calculate oxidation state of each given compound:-
1) ${S_2}O_3^{2 - }$
Let oxidation state of Sulphur is x, then
$2x + 3( - 2) = - 2$
$2x = - 2 + 6$
$2x = 4$
$x = + 2$
2) $S{O_2}$
Let oxidation state of Sulphur is x, then
$x + 2( - 2) = 0$
$x = + 4$
3) $SO_4^{2 - }$
Let oxidation state of Sulphur is x, then
$x + 4( - 2) = - 2$
$x = - 2 + 8$
$x = + 6$
4) $SO_3^{2 - }$
Let oxidation state of Sulphur is x, then
$x + 3( - 2) = - 2$
$x = - 2 + 6$
$x = + 4$
We know that the highest oxidation state of Sulphur is $ + 6$ and from the above calculation, we can see that the highest oxidation state of sulphur is in compound (3) $SO_4^{2 - }$ .
Therefore, the correct option is (C) $SO_4^{2 - }$ .
Note:
Potassium permanganate is used for qualitative analysis. It is also used as a disinfectant to cure many skin problems and can be used in treating bacterial infections, dermatitis, etc. It is also used in tanning leathers, printing fabrics. It can also be used as a bleaching agent, pesticides and an antiseptic.
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