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Question: \(CHC{l_3}\) and \({H_2}O\) from two separate immiscible layers. \(B{r_2}\)(orange) and \({I_2}\)(vi...

CHCl3CHC{l_3} and H2O{H_2}O from two separate immiscible layers. Br2B{r_2}(orange) and I2{I_2}(violet) are soluble in organic layer to a greater extent. In test tube A (KI+CHCl3+H2O)(KI + CHC{l_3} + {H_2}O)and B (KBr+CHCl3+H2O)(KBr + CHC{l_3} + {H_2}O), Cl2C{l_2} gas is passed then:
Given:
12Cl2+e−→Cl−          E∘=1.40V\dfrac{1}{2}C{l_2} + {e^ - } \to C{l^ - }\;\;\;\;\;{E^\circ } = 1.40V
12Br2+e−→Br−        E∘=1.09V\dfrac{1}{2}B{r_2} + {e^ - } \to B{r^ - }\;\;\;\;{E^\circ } = 1.09V
12I2+e−→I−              E∘=0.62V\dfrac{1}{2}{I_2} + {e^ - } \to {I^ - }\;\;\;\;\;\;\;{E^\circ } = 0.62V
A) Organic layer in A appear violet
B) Organic layer in A appear orange
C) Organic layer in B appear orange
D) Organic layer in B first appear violet then orange

Explanation

Solution

As we all know that the halogen reactivity series have an order where chlorine is the most reactive followed by bromine and iodine as chlorine is more electronegative than the other two and it can gain electrons easily as compared to bromine and iodine.

Complete answer
As we know that a more reactive element can displace a less reactive element so following the same reason we can say that chlorine being the most electronegative can displace the less reactive bromine and iodine. The order of the reactivity is as follows: Chlorine> Bromine> Iodine. Bromine can displace iodine but not the chlorine and iodine being the least reactive among these three will not be able to replace the other two.

Now, in the given question, in test tube A we have (KI+CHCl3+H2O)(KI + CHC{l_3} + {H_2}O) and when chlorine gas is passed through this solution the chlorine being more reactive than iodine can displace the iodide ions from potassium iodide and thus the iodine will which is violet coloured will dissolve in the solution and form an organic layer. Therefore we can say that the organic layer in test tube A appears violet and not orange. Hence, option (A) is correct.

In the second test tube (B), we have (KBr+CHCl3+H2O)(KBr + CHC{l_3} + {H_2}O) and when chlorine gas will be passed through this solution, chlorine will displace the bromine from potassium bromide as it is more reactive than bromine so orange coloured bromine will dissolve in solution and form a layer. Therefore we can say that the organic layer in B appears orange and not violet. Hence the correct answer is (C) in this case.

Therefore, the correct answers are option (A) and (C).

Note: Always remember that reactivity of non-metals is directly proportional to the reduction potential and on moving down the electrochemical series reactivity of non-metals increases. Therefore, Chlorine can displace bromine and iodine but not fluorine and fluorine can displace all the three whereas iodine cannot.