Question:
Excess of KI reacts with $CuSO_4$ solution and then $Na_2S_2O_3$ solution is added to it. Which of the statements is incorrect for this reaction?
Excess of KI reacts with $CuSO_4$ solution and then $Na_2S_2O_3$ solution is added to it. Which of the statements is incorrect for this reaction?
Updated On: Jul 28, 2022
- $Cu_2I_2$ is reduced
- Evolved $I_2$ is reduced
- $Na_2S_2O_3$ is oxidized
- $CuI_2$ is formed
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The Correct Option is D
Solution and Explanation
$ {2 \,CuSO4 + 4KI\, (excess) -> 2K2SO4 + Cu2 I2 + I2 ?}$
$ {Na2S2O3 + I2 -> Na2S4O6 + 2NaI}$
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Concepts Used:
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- P block elements are those in which the last electron enters any of the three p-orbitals of their respective shells. Since a p-subshell has three degenerate p-orbitals each of which can accommodate two electrons, therefore in all there are six groups of p-block elements.
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