Home
Class 12
CHEMISTRY
Statement-1 :Cu^(2+) and Cd^(2+) ions f...

Statement-1 :`Cu^(2+)` and `Cd^(2+)` ions form complexes with excess of potassium cyanide solution.
Statement-2 : On passing `H_(2)S` gas, complex `[Cu(CN)_(4)]^(3-)` is not effected but `[Cd(CN)_(4)]^(2-)` gives yellow precipitated.

A

STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for STATEMENT-1

B

STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation for STATEMENT-2

C

STATEMENT-1 is true, STATEMENT-2 is false

D

STATEMENT-1 is false, STATEMENT-2 is true

Text Solution

AI Generated Solution

The correct Answer is:
To analyze the statements regarding the behavior of `Cu^(2+)` and `Cd^(2+)` ions in the presence of potassium cyanide (KCN) and hydrogen sulfide (H₂S), we can break down the solution into clear steps. ### Step-by-Step Solution: **Step 1: Formation of Complexes with KCN** - When `Cu^(2+)` and `Cd^(2+)` ions are treated with excess potassium cyanide (KCN), they form complexes: - For `Cu^(2+)`: \[ Cu^{2+} + 4CN^- \rightarrow [Cu(CN)_4]^{3-} \] - For `Cd^(2+)`: \[ Cd^{2+} + 4CN^- \rightarrow [Cd(CN)_4]^{2-} \] - Both ions indeed form complexes with KCN, confirming that Statement-1 is correct. **Step 2: Reaction with H₂S** - When hydrogen sulfide (H₂S) gas is passed through the complexes: - The complex `[Cu(CN)₄]^{3-}` is stable and does not react with H₂S. Therefore, it remains unchanged. - The complex `[Cd(CN)₄]^{2-}` is less stable and reacts with H₂S to form cadmium sulfide (CdS), which is a yellow precipitate: \[ [Cd(CN)_4]^{2-} + H_2S \rightarrow CdS \downarrow (yellow \, precipitate) \] - This indicates that Statement-2 is also correct, as it describes the behavior of the cadmium complex in the presence of H₂S. **Step 3: Evaluating the Explanation** - While both statements are correct, Statement-2 does not serve as a correct explanation for Statement-1. Statement-1 simply states that both ions form complexes with KCN, while Statement-2 discusses the stability of these complexes in the presence of H₂S, which is a separate observation. ### Conclusion: - **Statement-1**: True (Both `Cu^(2+)` and `Cd^(2+)` form complexes with KCN). - **Statement-2**: True (The behavior of complexes with H₂S is correctly described). - **Explanation Validity**: Statement-2 is not a correct explanation for Statement-1. ### Final Answer: - Both statements are true, but Statement-2 is not a correct explanation of Statement-1. ---

To analyze the statements regarding the behavior of `Cu^(2+)` and `Cd^(2+)` ions in the presence of potassium cyanide (KCN) and hydrogen sulfide (H₂S), we can break down the solution into clear steps. ### Step-by-Step Solution: **Step 1: Formation of Complexes with KCN** - When `Cu^(2+)` and `Cd^(2+)` ions are treated with excess potassium cyanide (KCN), they form complexes: - For `Cu^(2+)`: \[ ...
Promotional Banner

Similar Questions

Explore conceptually related problems

The complex ion [Ni(CN)_4]^(2-) is :

Cu^(2+) ions give white precipitate with :

[Cu(NH_3)_4]^(2+) ion is coloured while [Cu(CN)_4]^(3-) ion is colourless why ?

When excess of aqueous KCN solution is added to an aqueous solution of copper sulphate , the complex [Cu(CN)_(4)]^(2-) is formed . On passing H_(2)S gas through this solution no precipitate of CuS is formed because

Statement -1 :An original solution containing excess of Ni^(2+) ions gives a yellow coloured solution with potassium cyanide solution. Statement-2 :A solution of Ni^(2+) ions gives red precipitate with dimethylglyoxime solution just made alkaline with ammonia.

The IUPAC name of the complex K_(2)[Ni (CN)_(4)] is

Which of the following statements is correct for complex [Cr(NH_(3))(CN)_(4)(NO)]^(2-) (given that n=1)?

Increasing order EAN of the metals in [Ni(CN)_(4)]^(2-) (II) [Fe(CN)_(6)]^(3-) (III) [Cu(CN)_(4)]^(3-) is .

The hybridization involved in complex [Ni(CN)_(4)]^(2-) is (At. No . Ni = 28)

CuSO_(4)+KCN(excess) to K_(3)[Cu(CN)_(4)]+(1)/(2)(CN)_(2)