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Question: What is the action of \(S{O_2}\)on a) Chlorine b) \({H_2}S\) c) Acidified \({K_2}C{r_2}{O_7}...

What is the action of SO2S{O_2}on
a) Chlorine
b) H2S{H_2}S
c) Acidified K2Cr2O7{K_2}C{r_2}{O_7}

Explanation

Solution

SO2S{O_2} is called Sulphur dioxide. It is an oxide of Sulphur which is in gaseous form. Sulphur dioxide can be produced by heating the elements sulphur and oxygen. The reaction is as follows S+O2SO2S + {O_2} \to S{O_2}. The oxidation number of sulphur is +4 + 4 in SO2S{O_2}. The sulphur is electron deficient in this molecule.

Complete step by step answer:
A) Sulphur dioxide with sulphur having +4 + 4 an oxidation state acts as a reducing agent. Thus it reacts with halogens. On reacting with halogens, it forms sulfuryl halides.With chlorine, SO2S{O_2} form SO2Cl2S{O_2}C{l_2}. The reaction is as follows
B) SO2+Cl2SO2Cl2S{O_2} + C{l_2} \to S{O_2}C{l_2}. The oxidation state of Sulphur SO2Cl2S{O_2}C{l_2} is +6 + 6. We can see that there is an increase in the oxidation number. Thus oxidation has taken place. Sulphur dioxide is oxidized by the halogen.
The action of SO2S{O_2} on H2S{H_2}S can be explained well as the Claus process.
It is the process by which sulphur is recovered from Hydrogen Sulphide by catalytic reaction. In this process, the air is passed to the furnace along with H2S{H_2}S. The first step is the reaction of H2S{H_2}S and oxygen(air) to give SO2S{O_2}. The SO2S{O_2} then reacts with H2S{H_2}S to give elemental Sulphur and water.
The reactions are as follows:- H2S+32O2SO2+H2O{H_2}S + \dfrac{3}{2}{O_2} \to S{O_2} + {H_2}O
SO2+2H2S3S+2H2OS{O_2} + 2{H_2}S \to 3S + 2{H_2}O
C) As we know the oxidation number of Sulphur in +4 + 4. It reacts with an acidified potassium dichromate solution(K2Cr2O7+H2SO4{K_2}C{r_2}{O_7} + {H_2}S{O_4}). The reaction is as follows.
K2Cr2O7+H2SO4+3SO2K2SO4+Cr2(SO4)3+H2O{K_2}C{r_2}{O_7} + {H_2}S{O_4} + 3S{O_2} \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + {H_2}O
We know that the colour of potassium dichromate is orange. Here, In K2Cr2O7{K_2}C{r_2}{O_7} , Cr has an oxidation state +6 + 6 whereas Cr2(SO4)3C{r_2}{(S{O_4})_3} has oxidation state +3 + 3. We can see a decrease in the oxidation number. Thus potassium dichromate is acting as a good oxidizing agent. It will reduce into Cr2(SO4)3C{r_2}{(S{O_4})_3} which is green in colour. Thus, a colour change is observed.

Note: Carbon is the principle component in natural mixes. Carbon can form stable bonds with numerous components, including it. There are four significant sorts of natural mixes: sugars, lipids, proteins, and nucleic acids