Question

Compound A is a light green crystalline solid. It gives the following tests:
i)It dissolves in dilute sulphuric acid. No gas is produced.
ii)A $\text{KMn}{{\text{O}}_{4}}$ drop is added to the above solution. The pink colour disappears.
iii)Compound A is heated strongly. Gases B and C with a pungent smell came out. A brown residue D is left behind.
iv)The gas mixture B and C is passed into a dichromate solution. The solution turns green.
v)The green solution from step (iv) gives a white precipitate E with a solution of barium nitrate.
vi)The residue D from (v) is heated on charcoal in reducing the flame. It gives a magnetic substance.
Compound A to E are identified as :
A)FeS{{O}_{4}}$$$$$$ B)S{{O}{2}}$C)$S{{O}{3}}$D)$F{{e}{2}}{{O}{3}}$E)$BaS{{O}_{4}}$$
If true enter 1, else enter 0.

Answer 1

Ferrous sulphate is the iron salt which is also known as green vitriol. It undergoes displacement reaction while reacting with aluminium and forms aluminium sulphate. It has the tendency to react with potassium permanganate but in presence of sulphuric acid. It also helps in treating the disease called anaemia.

Complete step-by-step answer: As it is given in the question that the colour of the compound A is light green that means it is ferrous sulphate because it is light green in colour. Ferrous sulphate is considered to be a salt of weak base and strong acid. Due to this it has the ability to react with sulphuric acid. But during this reaction no gas is evolved. It has the tendency of being a strong reducing agent so it decolours the$\text{KMn}{{\text{O}}_{4}}$ solution. The reaction for this is the following:
$5F{{e}^{2+}}+Mn{{O}_{4}}^{-}+8{{H}^{+}}\to 5F{{e}^{3+}}+M{{n}^{2+}}+4{{H}_{2}}O$
When the compound ferrous sulphate is heated it gives two gases and they are $S{{O}_{2}}$and $S{{O}_{3}}$and a residue is also formed that is $F{{e}_{2}}{{O}_{3}}$. The reaction for it is the following:${{H}_{2}}S{{O}_{4}}+Ba{{(N{{O}_{3}})}_{2}}\to 2HN{{O}_{3}}+BaS{{O}_{4}}\downarrow$
The gaseous mixture formed in the above reaction has the tendency to give a green colour solution and sulphuric acid in the solution. The reaction is:

& C{{r}_{2}}{{O}_{7}}^{2-}+3S{{O}_{2}}+2{{H}^{+}}\to 3{{K}_{2}}S{{O}_{4}}+{{H}_{2}}O+2C{{r}^{3+}} \\\ & {{H}_{2}}O+S{{O}_{3}}\to {{H}_{2}}S{{O}_{4}} \\\ \end{aligned}$$ Now the acid formed in the above reaction is sulphuric acid which gives white precipitate on reacting with the solution of barium nitrate. The reaction is the following: $${{H}_{2}}S{{O}_{4}}+Ba{{(N{{O}_{3}})}_{2}}\to 2HN{{O}_{3}}+BaS{{O}_{4}}\downarrow $$ The residue formed in the step (iii) is $$F{{e}_{2}}{{O}_{3}}$$has the ability and property that on heating with charcoal in a reducing flame it get reduces to iron which is considered as a magnetic substance. So we will enter 1 as the statements in the questions and the products formed are correct. **Note:** Potassium dichromate is the inorganic compound and an oxidising agent used in laboratories. Potassium dichromate is also used for testing the presence of sulphur dioxide in solution. If we impregnate the filter paper with potassium dichromate solution and the paper turns green that means sulphur dioxide is present and the reduction of chromium occurs.