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Central Atom and Ligands

A certain ino...

Question

A certain inorganic compound $(A)$ on heating loses water of crystallisation. On further heating a blackish brown powder $(B)$ and two oxides of sulphur $(C & D)$ are obtained. The powder $(B)$ on boiling with $H C l$ gives a yellow solution $(E)$ . When $H_{2} S$ is passed in $(E)$ a white turbidity $(F)$ and an apple green solution $(G)$ is obtained. The solution $(E)$ on treatment with thiocyanate ion gives blood red compound $(H)$ . Identify G and H :

(G) - F e C l_{2}

(H) - F e (C N S)_{3}

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(G) - F e C l_{3}

(H) - F e (C N S)_{2}

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(G) - F e C l_{2}

(H) - F e (C N S)_{2}

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(G) - F e C l_{3}

(H) - F e (C N S)_{3}

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Solution

The correct option is A $(G) - F e C l_{2}$ $(H) - F e (C N S)_{3}$
Option (A) is correct
$F e S O_{4} . 7 H_{2} O (A) ⟶ F e S O_{4} + 7 H_{2} O$
$F e S O_{4} ⟶ F e_{2} O_{3} (B) + S O_{2} (C) + S O_{3} (D)$
$F e_{2} O_{3} H C l ⟶ F e C l_{3} (E) + H_{2} O$
$F e C l_{3} + H_{2} S ⟶ S (F) + F e C l_{2} (G) + H C l$
$F e C l_{3} + (C N S)^{2 -} ⟶ F e (C N S)_{3} (H)$

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Similar questions

Q. A black coloured compound

(A)

on reaction with dilute

H_{2} S O_{4}

gives a gas

(B)

which on passing in a solution of an acid

(C)

gives a white turbidity

(D)

. Gas

(B)

when passed in an acidified solution of a compound

(E)

gives a precipitate

(F)

soluble in dil.

H N O_{3}

. After boiling this solution when an excess of

N H_{4} O H

is added, a blue coloured compound

(G)

is formed. To this solution on addition of acetic acid and aqueous

K_{4} [F e (C N)_{6}]

a chocolate precipitate

(H)

is obtained. On addition of an aqueous soluion of

B a C l_{2}

to an aqueous solution of

(E)

, a white precipitate insoluble in

H N O_{3}

is obtained. Identify

(G)

and

(H)

Q. A compound (A) is greenish crystalline salt which gave the following results :
On heating (A), water vapours and two oxides of sulfur, (C) and (D) are liberated leaving a red-brown residue (E).
(E) dissolves in warm concentrated HCl to give a yellow solution (F).

With H

_{2}

S, the solution (F) yields a pale yellow ppt. (G), which when filtered, leaves a greenish filtrate (H).

What is compound H?

Compound (A) is a greenish crystalline salt which gives the following results when tested :

(a) Addition of

B a C l_{2}

solution to a solution of (A) results in the formation of a white precipitate (B) which is insoluble in dilute

H C l

.
(b) On heating, water vapour and two gases (C) and (D) are liberated, leaving a red brown residue (E).
(c) (E) dissolves in

c o n c . H C l

to give a yellow solution (F).
(d) With

H_{2} S

, the solution (F) yields a yellow-white ppt. (G) which when filtrated, leaves a greenish filtrate (H).
(e) The solution (F) gives a green solution with

S n C l_{2}

, and when reacted with

K_{4} [F e (C N)_{6}]

gives a blue ppt.

Identify compound (A) :

Q. Compound

C_{8} H_{9} C l

(A) on treatment with

K C N

followed by hydrolysis gives

C_{9} H_{10} O_{2}

(B). Ammonium salt of (B) on dry distillation gives (C), which reacts with alkaline solution of bromine to give

C_{8} H_{11} N

(D). Another compound (E), is obtained by the action of nitrous acid on (D) or by the action of aqueous potash on (A). (E) on oxidation gives

C_{8} H_{6} O_{4}

(F) which gives an anhydride

C_{8} H_{4} O_{3}

(G) on heating.

If x is the DOE of (G) and y is the DOE of (F), then x-y=

A black colored compound (A) on reaction with dilute

H_{2} S O_{4}

gives a gas (B) which on passing in a solution of acid (C) gives white turbidity (D). Gas (B) when placed in an acidified solution of a compound (E) gives a precipitate (F) soluble in dilute

H N O_{3} .

After boiling this solution, when an excess of

N H_{4} O H

is added, blue colored compound (G) is formed. To this solution on addition of

C H_{3} C O O H

and aqueous

K_{4} [F e (C N)_{6}]

, a chocolate precipitate (H) is obtained. On the addition of an aqueous solution of

B a C l_{2}

to an aqueous solution of E, a white ppt insoluble in

H N O_{3}

is obtained.

Identify B.

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The correct option is A (G)−FeCl2(H)−Fe(CNS)3Option (A) is correctFeSO4.7H2O(A)⟶FeSO4+7H2OFeSO4⟶Fe2O3(B)+SO2(C)+SO3(D)Fe2O3HCl⟶FeCl3(E)+H2OFeCl3+H2S⟶S(F)+FeCl2(G)+HClFeCl3+(CNS)2−⟶Fe(CNS)3(H)