
Identify the statement which is not correct regarding copper sulphate
(A) It reacts with \[{\mathbf{KI}}\]to give iodine.
(B) It reacts with \[{\mathbf{KCl}}\]to give Cu2Cl2
(C) It reacts with AWN and glucose to give \[{\mathbf{C}}{{\mathbf{u}}_{\mathbf{2}}}{\mathbf{O}}\]
(D) It gives \[{\mathbf{CuO}}\]on strong heating in air.
Answer
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Hint: Copper sulphate is a compound of copper which is a d block element. The d subshell is responsible for its properties.
Complete Step by Step Solution:
Option A: Copper sulphate reacts with potassium iodide to undergo a redox reaction. Potassium iodide is oxidised to iodine and copper sulphate is reduced to copper iodide as by the following reaction:
\[2CuS{O_4} + KI \to 2{K_2}S{O_4} + C{u_2}{I_2} + {I_2}\]
Oxidation state of copper is $ + 2$ in copper sulphate and $ + 1$ in copper iodide. Thus, A is correct.
Option B: Copper Sulphate shows distinct kinds of reaction with potassium chloride and potassium iodide. Reaction with potassium iodide gives copper iodide in which copper is in $ + 1$ oxidation state. But in reaction with potassium chloride, copper chloride is formed in which copper chloride is $ + 2$ oxidation state.
The reaction can be represented as:
\[CuS{O_4}\; + \;2KCl\; \to \;CuC{l_2}\; + \;{K_2}S{O_4}\]
Hence, option B is incorrect.
Option C: Fehling's solution is a mixture of alkaline solution of copper sulphate containing Rochelle salt which is sodium potassium tartrate.
Fehling solution oxidises glucose to carboxylate ion and copper sulphate reduces to cuprous oxide $C{u_2}0$which is the red precipitate used to confirm presence of aldehyde group.
${C_6}{H_{12}}{O_6} + C{u^{ + 2}} + O{H^ - } \to {C_6}{H_{11}}{O_7}^ - + C{u_2}O + 3{H_2}O$.
Option D: On heating copper sulphate, it turns to cupric oxide as the following reaction
$CuS{O_4}\xrightarrow{\Delta }CuO + S{O_2} + {O_2}$.
Thus, the correct answer is B.
Note: Cuprous Halide has formula $C{u_2}{X_2}$ and not $CuX$. Thus, cuprous halide exists in the form $C{u_2}^{2 + }$. It is formed by the combination of $Cu$and $C{u^{ + 2}}$, showing metal-metal ion interaction.
Complete Step by Step Solution:
Option A: Copper sulphate reacts with potassium iodide to undergo a redox reaction. Potassium iodide is oxidised to iodine and copper sulphate is reduced to copper iodide as by the following reaction:
\[2CuS{O_4} + KI \to 2{K_2}S{O_4} + C{u_2}{I_2} + {I_2}\]
Oxidation state of copper is $ + 2$ in copper sulphate and $ + 1$ in copper iodide. Thus, A is correct.
Option B: Copper Sulphate shows distinct kinds of reaction with potassium chloride and potassium iodide. Reaction with potassium iodide gives copper iodide in which copper is in $ + 1$ oxidation state. But in reaction with potassium chloride, copper chloride is formed in which copper chloride is $ + 2$ oxidation state.
The reaction can be represented as:
\[CuS{O_4}\; + \;2KCl\; \to \;CuC{l_2}\; + \;{K_2}S{O_4}\]
Hence, option B is incorrect.
Option C: Fehling's solution is a mixture of alkaline solution of copper sulphate containing Rochelle salt which is sodium potassium tartrate.
Fehling solution oxidises glucose to carboxylate ion and copper sulphate reduces to cuprous oxide $C{u_2}0$which is the red precipitate used to confirm presence of aldehyde group.
${C_6}{H_{12}}{O_6} + C{u^{ + 2}} + O{H^ - } \to {C_6}{H_{11}}{O_7}^ - + C{u_2}O + 3{H_2}O$.
Option D: On heating copper sulphate, it turns to cupric oxide as the following reaction
$CuS{O_4}\xrightarrow{\Delta }CuO + S{O_2} + {O_2}$.
Thus, the correct answer is B.
Note: Cuprous Halide has formula $C{u_2}{X_2}$ and not $CuX$. Thus, cuprous halide exists in the form $C{u_2}^{2 + }$. It is formed by the combination of $Cu$and $C{u^{ + 2}}$, showing metal-metal ion interaction.
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