What is the observation that will be made when Bromine is added to Potassium Fluoride?
Answer
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Hint: Reactions in which one reactant gets displaced by the other reactant are known as displacement reactions. But there is one condition for such a reaction: only a reactant which is more reactive as compared to another, one can only undergo displacement.
Complete answer:
Here, in this question, we have two halogens, that is, Bromine \[(Br)\]and Fluorine $(F)$. We also know that Fluorine has atomic number $(Z) = 9$and Bromine has atomic number $(Z) = 35$. Also, the reactivity for the halogens, that is, the elements in group $ - VII$, decreases down the group due to the increase in their atomic radii and in their electronic energy levels, increasing their nuclear shielding. So, now we know that Fluorine is more reactive than Bromine since in the halogen group Fluorine comes first and is more reactive than Bromine.
So, we already have Potassium Fluoride, that is, $KF$which is an ionic compound and if we add Bromine to Potassium Fluoride, then Bromine will try to replace Fluorine in order to form Potassium Bromide, that is, $KBr$ but this displacement reaction cannot occur due to the above reason, that is, Bromine is less reactive as compared to Fluorine.
Therefore, if we add Bromine to Potassium Fluoride, then the displacement reaction will not occur but this addition of Bromine will lead to the change in colour of the solution to brown-orange.
Additional information: As we have seen in this question, that the order of reactivity in this group is opposite of the other metal groups, that is, in the groups containing metals, the reactivity increases as we go down the group but in the non-metal containing groups, the reactivity decreases down the group due to high nuclear shielding effect.
The order of reactivity for halogens can be written as:
$F > Cl > Br > I$
Note:
In the reactions like displacement reaction, the displacement can only take place if the reactant is more reactive as compared to the other reactant otherwise the reaction will not take place. Its general reaction can be written as:
$AB + CD \to AC + BD$
Complete answer:
Here, in this question, we have two halogens, that is, Bromine \[(Br)\]and Fluorine $(F)$. We also know that Fluorine has atomic number $(Z) = 9$and Bromine has atomic number $(Z) = 35$. Also, the reactivity for the halogens, that is, the elements in group $ - VII$, decreases down the group due to the increase in their atomic radii and in their electronic energy levels, increasing their nuclear shielding. So, now we know that Fluorine is more reactive than Bromine since in the halogen group Fluorine comes first and is more reactive than Bromine.
So, we already have Potassium Fluoride, that is, $KF$which is an ionic compound and if we add Bromine to Potassium Fluoride, then Bromine will try to replace Fluorine in order to form Potassium Bromide, that is, $KBr$ but this displacement reaction cannot occur due to the above reason, that is, Bromine is less reactive as compared to Fluorine.
Therefore, if we add Bromine to Potassium Fluoride, then the displacement reaction will not occur but this addition of Bromine will lead to the change in colour of the solution to brown-orange.
Additional information: As we have seen in this question, that the order of reactivity in this group is opposite of the other metal groups, that is, in the groups containing metals, the reactivity increases as we go down the group but in the non-metal containing groups, the reactivity decreases down the group due to high nuclear shielding effect.
The order of reactivity for halogens can be written as:
$F > Cl > Br > I$
Note:
In the reactions like displacement reaction, the displacement can only take place if the reactant is more reactive as compared to the other reactant otherwise the reaction will not take place. Its general reaction can be written as:
$AB + CD \to AC + BD$
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