Question

$FeC{l_3}$ is yellow in aqueous solution but on passing ${H_2}S$ gas, the solution turns green due to formation of:-
(A) $FeCl$
(B) $FeC{l_2}$
(C) $FeC{l_3}$
(D) None of these.

Answer 1

Hint: The change in the colour of the compound is due to change in the oxidation state. The compounds in such cases either undergo oxidation or reduction. Calculate the change in the oxidation state of iron.

Complete step by step solution:
${H_2}S$ is hydrogen sulphide. The gas is colourless. Hydrogen sulphide is heavier than air. It can lead to an explosion. Oxygen and hydrogen sulphide burns with a blue flame to produce sulphur dioxide and water. In presence of a catalyst and also at a higher temperature, sulphur dioxide reacts with hydrogen sulphide which produces water and elemental sulphur. This method is very useful for disposal of hydrogen sulphide.
In hydrogen sulphide, sulphur is in the oxidation state of -2. The highest oxidation state of sulphur is +6 and the lowest oxidation state is +2. The change in the oxidation number can be from +4 to -2 which indicates reduction. The oxidation state can also change from +4 to +6 which indicates oxidation. In the case of hydrogen, sulphide sulphur can only lose electrons to attain +4 and +6 oxidation state. Thus hydrogen sulphide acts as a reducing agent. Ferric chloride is ferric chloride. The oxidation state of Fe is +3. It is corrosive to metals and tissues. It is non-combustible. Ferric chloride in anhydrous state forms a deliquescent nature.
Ferric chloride is yellow in aqueous solution. Thus yellow coloured ferric chloride when reacted with hydrogen sulphide it forms a green solution. Here, hydrogen sulphide acts as a reducing agent. So it will reduce ferric chloride. After reduction ferric chloride gets reduced to ferrous chloride, ferrous chloride I green in colour.
Ferrous chloride is a paramagnetic solid with a high melting point. The compound is soluble in water. Ferrous sulphate is used as a flocculation agent in water treatment. Here Fe is in +2 oxidation state.
The reaction is as follows:
$2FeC{l_3} + {H_2}S \to 2FeC{l_2} + 2HCl + S$
So the green compound is nothing but ferrous chloride.

Hence option (B) is the correct one.

Note: Since sulphur is present in the lowest oxidation state, therefore it undergoes oxidation. It thus reduces iron. Iron is most stable in +3 oxidation state.