Question

$C{l_2}$ gas is passed into aqueous solution of $KBr$ and $KI$ and $CHC{l_3}$
added. It is observed that there is:
A. Violet layer in $CHC{l_3}$ layer
B. Yellow layer in $CHC{l_3}$ layer
C. Yellow layer in aqueous layer
D. Violet layer in aqueous layer

Answer 1

We have to know a reaction where an element in a compound is replaced by another element is called single replacement reaction or a substitution reaction. It generally involves a dilute solution of an acid and a metal.

Complete step by step answer:
We must know that the relative activities of halogens could be readily established by single-replacement reactions.
Let us consider the reaction of bromine with aqueous solution of sodium bromide as an example. When we add bromine to aqueous solution of sodium bromide, free iodine is obtained. We can write the chemical equation for this reaction as,
$2NaI\left( {aq} \right) + B{r_2}\left( l \right)\xrightarrow{{}}2NaBr\left( {aq} \right) + {I_2}\left( s \right)$
Here, the liberation of iodine happens because bromine is more active than iodine, and hence it displaces iodine in sodium iodide and solid iodine is produced.
Similarly, when we add fluoride to aqueous solution of potassium chloride, the products formed are potassium fluoride and chlorine gas is liberated. We can write the chemical equation for this reaction as,
$2KCl\left( s \right) + {F_2}\left( g \right)\xrightarrow{{}}2KF\left( s \right) + C{l_2}\left( g \right)$
We must remember the liberation of chlorine happens because fluorine is more active than chlorine, and hence it displaces chlorine in potassium chloride and gaseous chlorine is produced.
We can arrange the order of activity in halogen as,
${F_2} > C{l_2} > B{r_2} > {I_2}$
Now let us get back to the question,
When we pass chlorine into aqueous solution of potassium bromide, the products formed will be potassium chloride, and bromine liquid.
We can write the chemical equation as,
$2KBr + C{l_2}\xrightarrow{{}}2KCl + B{r_2}$
We must remember, the chlorine is more reactive than bromine. Therefore, chlorine would displace bromine in potassium bromide and forms potassium chloride with release of bromine.
Similarly, if we pass iodine into aqueous solution of potassium iodine, the products formed will be potassium chloride, and solid iodine.
We can write the chemical equation as,
$2KI + C{l_2}\xrightarrow{{}}2KCl + {I_2}$
Bromine can displace iodine from potassium iodine. We can write the chemical equation as,
$2KI + B{r_2}\xrightarrow{{}}2KBr + {I_2}$
The liberated iodine is present in the chloroform form that gives violet colour.
So, the correct answer is “Option A”.

Additional Information: The reaction of formation of aluminium oxide and manganese from manganese oxide and aluminium is,
$Al + M{n_2}{O_3}\xrightarrow{{}}Mn + A{l_2}{O_3}$
The two reactants aluminium and manganese oxide are reacted to form manganese and aluminium oxide. The aluminium is substituted by manganese to form the product aluminium oxide and manganese is formed as other products. Thus, this reaction is an example of a single replacement reaction because one of the elements is replaced by another element to form a new product.
The reaction is not balanced. In the reactant side, two moles of aluminium is required to make the reaction a balanced one. So we put 2 besides aluminium and the balanced reaction is given as,
$2Al + M{n_2}{O_3}\xrightarrow{{}}Mn + A{l_2}{O_3}$.

Note: We must remember the activities of nonmetals increase as we go up within a group in the periodic table. This situation is in sharp contrast to the activities of metals that generally increase as we go down within a group. Chlorine, bromine, and iodine used as disinfectants. Atomic radius of halogens increases down the group. Ionization energy decreases down the group. Electronegativity decreases down the group, and electron affinity decreases down the group.