MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Polar Molecules

Compound A is...

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 $K M n O_{4}$ is added to the above solution. the pink colour disappears.
(iii) Compound (A) is heated strongly. gases (B) and (C) with 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): $F e S O_{4}$ , (B): $S O_{2},$ , (C): $S O_{3}$ , (D): $F e_{2} O_{3}$ , (E): $B a S O_{4}$
If true enter 1, else enter 0.

Open in App

Solution

Compounds $^{'} A^{'}$ is a light green crystalline solid, so it may be ferrous sulphate.

(i) $F e S O_{4}$ is a salt of strong acid weak base, so it hydrolysed in dil

$H_{2} S O_{4}$ but no gas is produced.

(ii) Ferrous supphate is a strong reducing agent, thus decolourises the

$K M n O_{4}$ solution

$5 F e^{2 +} + M n O_{4}^{-} (p i n k) + 8 H^{+} \to 5 F e^{3 +} + M n^{2 +} (c o l o u r l e s s) + 4 H_{2} O$

(iii) Ferrous sulphate on strong heating gives both $S O_{2}$ 'B' and $S O_{3}$ 'C' alongwith a residue of $F e_{2} O_{3}$ 'D'

$2 F e S O_{4} \to F e_{2} O_{3} + S O_{2} + S O_{3}$

(iv) This gaseous mixture dichromate to green solution and also give sulphuric acid in the solution.

$C r_{2} O_{7}^{2 -} + 3 S O_{2} + 2 H^{+} \to 3 K_{2} S O_{4} + H_{2} O + 2 C r^{3 +} g r e e n$
$H_{2} O + S O_{3} \to H_{2} S O_{4}$

(v) This sulphuric acid (formed in step(iv)) gives white ppt. with a solution of

$B a (N O_{3})_{2}$ due to formation of $B a S O_{4}^{'} E^{'}$

$H_{2} S O_{4} + B a (N O_{3})_{2} \to 2 H N O_{3} + B a S O_{4} w h i t e ↓$

(vi) The residue 'D' when heated on charcoal in a reducing flame, reduce to iron $(F e)$ which is a magnetic substance.

Hence, $A, B, C . D$ and $E$ are respectively $F e S O_{4}, S O_{2}, S O_{3}, F e_{2} O_{3}$ and $B a S O_{4}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. A hydrated metallic salt (A) , light green in colour gives a white anhydrous residue (B) after being heated gradually. (B) is soluble in water and its aqueous solution reacts with NO to give a dark brown compound (C). (B) on strong heating gives a brown residue (D) and a mixture of two gases (E) and (F). The gaseous mixture , when passed through acidified permanganate, discharged the pink colour and when passed through acidified

B a C l_{2}

solution, gives a white precipitate. Compounds (A), (B), (C) , (D) , (E) and (F) are identified as:

F e S O_{4} . 7 H_{2} O

F e S O_{4}

F e S O_{4} . N O o r [F e (H_{2} O)_{6} N O] S O_{4}]

F e_{2} O_{3}

S O_{2}

S O_{3}

If true enter 1, else enter 0.

Q. A white amorphous powder (A) when heated gives a colourless gas (B), which turns lime water milky and the residue (C) which is yellow when hot but white when cold. The residue (C) dissolves in dilute HCl and the resulting solution gives a white precipitate on addition of potassium ferrocyanide solution. (A) dissolves in dilute HCl with the evolution of a gas which is identical in all respects with (B). The solution of (A) as obtained above gives a white precipitate (D) on addition of excess of

N H_{4} O H

H_{2} S

. Another portion of this solution gives initially a white precipitate (E) on addition of NaOH which dissolves in excess of it. Compounds (A) to (D) are identified as:
(A) :

Z n C O_{3}

C O_{2}

Z n O

Z n S

Z n (O H)_{2}

If true enter 1, else enter 0.

Q. (i) A black mineral (A) on heating in presence of air gives a gas (B).
(ii) The mineral (A) on reaction with dilute

H_{2} S O_{4}

gives a gas (C) and solution of a compound (D).
(iii) On passing gas (C) into an aqueous solution (B), a white turbidity is obtained.
(iv) The aqueous solution of compound (D) on reaction with potassium ferricyanide gives a blue compound (E).

Compounds (A) to (E) are identified as:

F e S

S O_{2}

H_{2} S

F e S O_{4}

K F e^{I} I [F e^{I} I I (C N)_{6}]

If true enter 1, else enter 0.

Q. An aqueous solution of salt (A) gives a white crystalline precipitate (B) with NaCl solution. The filtrate gives a black precipitate (C) when

H_{2} S

is passed through it. Compound (B) dissolves in hot water and the solution gives yellow precipitate (D) on treatment with potassium iodine and on cooling. The compound (A) does not give any gas with dilute HCl but liberates a reddish brown gas on heating. Compounds (A) to (D) are identified as:

P b (N O_{3})_{3}

P b C l_{2}

P b S

P b I_{2}

If true enter 1, else enter 0.

Q. Two solid laboratory reagents (A) and (B) gives the following reactions:
Compound (A):
(i) on strongly heating gives two oxides of sulphur
(ii) on adding aqueous NaOH solution to its aqueous solution, a dirty green precipitate is obtained which starts turning brown on exposure to air
Compound (B) :
(iii) imparts green colour to the flame
(iv) Its solution does not give a precipitate on passing

H_{2} S

(v) When it is heated with solid

K_{2} C r_{2} O_{7}

and concentrated

H_{2} S O_{4}

, a red gas is evolved. The gas when passed in an aq NaOH solution turns it yellow
Compound (A) and (B) are identified as

F e S O_{4}

B a C l_{2}

.
If true enter 1, else enter 0.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Polar Molecules

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Polar Molecules

Standard X Chemistry

Join BYJU'S Learning Program

CrossIcon