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 drop of KMnO4 is added to the above solution. The pink colour disappears. (iii) Compound A is heated strongly. Gases B and C, with pungent smell, come 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) Residue D from step (iii) is heated on charcoal in a reducing flame. It gives a magnetic substance. Name the compound A, B, C, D and E.

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 drop of KMnO4 is added to the above solution. The pink colour disappears. (iii) Compound A is heated strongly. Gases B and C, with pungent smell, come 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) Residue D from step (iii) is heated on charcoal in a reducing flame. It gives a magnetic substance. Name the compound A, B, C, D and E. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Compound A is a light green crystalline solid, so it may be

F e S O_{4} .

(i)

F e S O_{4}

is a salt of strong acid and weak base, so it hydrolyses in dil.

H_{2} S O_{4}

but no gas is evolved.
(ii)

F e S O_{4}

is a strong reducing agent, thus decolourises

K M n O_{4}

solution :

5 F e^{2 +} + Purple M n O_{4}^{-} + 8 H^{+} ⟶ 5 F e^{3 +} + Colourless M n^{2 +} + 4 H_{2} O

(iii)

F e S O_{4}

on strong heating gives both

S O_{2}

(B) and

S O_{3}

(C) gases along with a residue of

F e_{2} O_{3} (D) .

2 F e S O_{4} Δ D F e_{2} O_{3} + B S O_{2} + C S O_{3}

(iv) The gaseous mixture reduced dichromate solution to green
solution and also gives

H_{2} S O_{4}

in solution :

C r_{2} O_{7}^{2 -} + 3 S O_{2} + 2 H^{+} ⟶ 3 S O_{4}^{2 -} + H_{2} O + Green 2 C r^{3 +}

H_{2} O + S O_{3} ⟶ H_{2} S O_{4}

(v) The sulphuric acid (formed in previous step) gives white ppt.
with

B a (N O_{3})_{2}

due to formation of

B a S O_{4} (E)

:

H_{2} S O_{4} + B a (N O_{3})_{2} ⟶ White (E) B a S O_{4} ↓ + 2 H N O_{3}

(vi) The residue

D

when heated on charcoal in a reducing flame reduces to iron

(F e)

which is a magnetic substance.
Hence,

A = F e S O_{4}, B = S O_{2}, C = S O_{3}, D = F e_{2} O_{3}

and

E = B a S O_{4} .