Question

Identify the wrong statement regarding copper sulphate:

• A. It reacts with $KI$ to give $I_2$.
• B. It reacts with $KI$ to give ${Cl}_2$.
• C. It's tartarate complex reacts with $NaOH$ and glucose to give ${Cu}_{2}O$.
• D. It gives $CuO$ on strong heating in air.

Answer 1

The correct option is B It reacts with $KI$ to give ${Cl}_2$.

Copper sulfate reacts with potassium iodide to liberate iodine.

$2CuS{O}_4+4KI\to C{u}_2{I}_2\downarrow +I_2+2K-2S{O}_4$

Thus statement A is correct and statement B is incorrect.

Copper(II) ion complexed with tartrate ion in basic solution is known as Fehling solution. It oxidizes glucose and itself gets reduced to cuprous oxide.

$RCHO+2C{u}^{2+}+5O{H}^{-}\to RCO{O}^{-}+\underset{brickredprecipitateofcuprousoxide}{C{u}_{2}O}\downarrow +3H_{2}O$

On strong heating, copper sulfate forms cupric oxide.

$2CuS{O}_4\stackrel{{750}^{o}C}{\to }2CuO+2S{O}_2+O_2$

Thus statement D is correct.

Because only B option is incorrect. Option B is the answer.