Mineral A- dil-H2SO4 -B-C - D-gas B or C-BaCl2 solution - white ppt E B-K3-FeCN6- blue ppt-F C-aq excessNH3- deep blue solution G c-D gas - black ppt- H soluble in dil-HNO3If compound E is CuS enter 1 else 0-

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Byju's Answer

Standard XII

Chemistry

Colours in Coordination Compounds

Mineral A+ ...

Question

Mineral $(A) + d i l . H_{2} S O_{4} \to (B) + (C)      + (D), (g a s)$
$(B) o r (C) + B a C l_{2} s o l u t i o n \to w h i t e p p t (E)$
$(B) + K_{3} [F e (C N)_{6}] \to b l u e p p t . (F)$
$(C) + a q N H_{3} (e x c e s s) \to d e e p b l u e s o l u t i o n (G)$
$(c) + (D) g a s \to b l a c k p p t . (H) s o l u b l e i n d i l . H N O_{3}$
If compound (E) is $C u S$ enter 1 else 0.

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Solution

$C u S (A) + d i l .4 H_{2} S O_{4} ⟶ C u S O_{4} (B) + 4 S O_{2} (D) + 4 H_{2} O (C)$
$C u S O_{4} + B a {C l}_{2} s o l u t i o n ⟶ C u {C l}_{2} + B a S O_{4} (w h i t e p p t)$
$2 H_{2} O + B a {C l}_{2} ⟶ B a {(O H)}_{2} (w h i t e p p t) + 2 H C l$
$3 C u S O_{4} + K_{3} [F e {(C N)}_{6}] ⟶ {C u}_{3} [F e {(C N)}_{6}] + 3 K S O_{4}$
${C u}_{3} [F e {(C N)}_{6}]$ -Cupric ferricyanide (brownish yellow ppt)
Hence, Ans: 0

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Similar questions

Q. Mineral (A)

d i l . H_{2} S O_{4} ⟶

(B) + (C) + (D) (gas)
(D) +

C H_{3} C O O)_{2} P b △ ⟶

black ppt

(B) + (C) (D) \to

E (black ppt) separated from (C) by filtration
(E) +

H N O_{3} d i l . △ \to

(F) blue coloured solution
(F) +

K_{4} [F e (C N)_{6}] ⟶

K_{3} [F e (C N)_{6}] ⟶

(H) blue

A to H are identified as: (A) :

C u F e S_{2}

, (B) :

C u S O_{4}

F e S O_{4}

(D) :

H_{2} S

(E) :

C u S

(F) :

C u (N O_{3})_{2}

(G) :

C u_{2} [F e (C N)_{6}]

(H) :

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

, Turnbull's blue

Show the above reaction.

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:

(A) :

F e S

, (B) :

S O_{2}

, (C) :

H_{2} S

, (D) :

F e S O_{4}

and (E) :

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

If true enter 1, else enter 0.

Q. Here, (A), (B), (C), (D) and (E) are as follows:
(A) :

N H_{4} N O_{2}

(B) :

N H_{3}

N a N O_{3}

(D) :

N_{2}

(E) :

A l N

If true enter 1, else enter 0.

Q. If true enter 1, else enter 0.
(A) (black) + dil.

H C l △ ⟶ (B) (l)

+ (C)(g). gas (C) turns lead acetate paper black. (B) gives orange ppt (D) soluble in excess of KI forming (E). Above mentioned unknown compounds are (A) : HgS, (B) :

H g C l_{2}

, (C) :

H_{2} S

, (D) :

H g I_{2}

, (E) :

K_{2} H g I_{4}

Q. If the identifications given below are true enter 1, else enter 0.
(A) :

{N H}_{4} {N O}_{2}

(B) :

{N H}_{3}

{N a N O}_{2}

(D) :

N_{2}

(E) :

A l N

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Mineral (A)+dil.H2SO4→(B)+(C)+(D),(gas)(B)or(C)+BaCl2solution→whiteppt(E)(B)+K3[Fe(CN)6]→blueppt.(F)(C)+aqNH3(excess)→deepbluesolution(G)(c)+(D)gas→blackppt.(H)solubleindil.HNO3If compound (E) is CuS enter 1 else 0.

CuS(A)+dil.4H2SO4⟶CuSO4(B)+4SO2(D)+4H2O(C)CuSO4+BaCl2solution⟶CuCl2+BaSO4(whiteppt)2H2O+BaCl2⟶Ba(OH)2(whiteppt)+2HCl3CuSO4+K3[Fe(CN)6]⟶Cu3[Fe(CN)6]+3KSO4Cu3[Fe(CN)6]-Cupric ferricyanide (brownish yellow ppt)Hence, Ans: 0