
Which reaction is not feasible?
(A)- $2KI+B{{r}_{2}}\to 2KBr+{{I}_{2}}$
(B)- $2KBr+{{I}_{2}}\to 2KI+B{{r}_{2}}$
(C)- $2KBr+C{{l}_{2}}\to 2KCl+B{{r}_{2}}$
(D)- $2{{H}_{2}}O+2{{F}_{2}}\to 4HF+{{O}_{2}}$
Answer
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Hint: The displacement reaction to occur, more the reduction potential of the atom, more is its reactivity. So, on reaction it will displace the less reactive ion present in the solution.
Complete answer:
In the given reactions, it is seen that the solution is reacted with the halogen atom. So, for the displacement reaction to occur, the halogen displaces the less reactive ion in the solution. Thus, making the reaction feasible.
The tendency of the halogen atom to displace the ions in the solution, is determined from its reduction potential. As in the halogen group, with increase in the atomic number, the electronegativity and the electron affinity of the halogens decreases, as the nuclear attraction decreases down the group.
Due to which the oxidizing ability and the reactivity of the atoms decreases. So, the order of the halogen as an oxidising agent is as follows: F > Cl > Br > I
Then, in the reaction (A), the bromine atom being a better oxidising agent than iodine, is more reactive and displaces it in the potassium iodide salt solution.
Similarly, in reaction (C), chlorine atom will displace the bromine ion in the potassium bromide salt solution.
Also, in reaction (D), the fluorine being more electronegative than the oxygen atom, is more reactive and displaces it in the water molecule.
Whereas, in reaction (B), the iodine being a good reducing agent, will not displace the bromine ion in the salt solution. Thus, the reaction will not occur.
Therefore, the reaction that is not feasible is option (B)- $2KBr+{{I}_{2}}\to 2KI+B{{r}_{2}}$.
Note:
Better the oxidising agent, more the reduction potential. Fluorine has the highest reduction potential, and the iodine has a lower reduction potential, so it is a better reducing agent.
The fluorine has less electron affinity than chlorine, due to its small size. But due to the presence of greater repulsion between the outer electrons, the bond energy decreases, making it more oxidising in nature than the chlorine atom.
Complete answer:
In the given reactions, it is seen that the solution is reacted with the halogen atom. So, for the displacement reaction to occur, the halogen displaces the less reactive ion in the solution. Thus, making the reaction feasible.
The tendency of the halogen atom to displace the ions in the solution, is determined from its reduction potential. As in the halogen group, with increase in the atomic number, the electronegativity and the electron affinity of the halogens decreases, as the nuclear attraction decreases down the group.
Due to which the oxidizing ability and the reactivity of the atoms decreases. So, the order of the halogen as an oxidising agent is as follows: F > Cl > Br > I
Then, in the reaction (A), the bromine atom being a better oxidising agent than iodine, is more reactive and displaces it in the potassium iodide salt solution.
Similarly, in reaction (C), chlorine atom will displace the bromine ion in the potassium bromide salt solution.
Also, in reaction (D), the fluorine being more electronegative than the oxygen atom, is more reactive and displaces it in the water molecule.
Whereas, in reaction (B), the iodine being a good reducing agent, will not displace the bromine ion in the salt solution. Thus, the reaction will not occur.
Therefore, the reaction that is not feasible is option (B)- $2KBr+{{I}_{2}}\to 2KI+B{{r}_{2}}$.
Note:
Better the oxidising agent, more the reduction potential. Fluorine has the highest reduction potential, and the iodine has a lower reduction potential, so it is a better reducing agent.
The fluorine has less electron affinity than chlorine, due to its small size. But due to the presence of greater repulsion between the outer electrons, the bond energy decreases, making it more oxidising in nature than the chlorine atom.
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