Question

Which of the following is a displacement reaction?
A)$NaOH+HN{{O}_{3}}\to NaN{{O}_{3}}+{{H}_{2}}O$
B)$Cu+2AgN{{O}_{3}}\to Cu{{(N{{O}_{3}})}_{2}}+2Ag$
C)$2Hg+{{O}_{2}}\to 2HgO$
D)$FeC{{l}_{3}}+2NaOH\to 3NaCl+Fe{{(OH)}_{3}}$

Answer 1

We can easily predict the product of the displacement reaction since in such type of reaction a more reactive metal replaces a less reactive metal.
It is a type of oxidation-reduction reaction.

Complete step by step solution:
In the question, four reactions are given and we have to say that which of this satisfies the definition of displacement reaction.
We know that displacement reaction is a reaction in which one of the ions in the compound is replaced by more reactive metal, which is called a displacement reaction or single displacement reaction.
$A-B+C\to A-C+B$
If the ions are replaced by another ion of a compound, then it is called as the double displacement reaction.
$A-B+C-D\to A-C+B-D$
We could predict the products of the displacement reaction with the help of the electrochemical series, i.e. with the aid of the redox potential of each constituent in the reaction.
We know that the displacement reaction is basically a redox reaction, one compound will get oxidized and others will get reduced. And the element which undergoes the oxidation and reduction is decided using the redox potential. The tendency of the element to loss or gain electrons to an electrode.
The electrochemical series in the decreasing order of oxidizing potential,
Li>\,Cs\,>\,Rb\,>\,K\,>\,Ba\,>\,Sr\,>\,Na\,>\,Ca\,>\,Mg\,>\,Be\,>\,Al\,>\,H(water)\,>\,Mn\,>\,Zn\,>\,C{{r}^{+3}}\,>F{{e}^{+2}}\,>\,Cd\,>\,Co\,>\,Ni\,>\,Sn\,>\,Pb\,>\,H\left( in\,acids \right)\,>$$$$Cu\,>\,F{{e}^{+3}}\,>\,Hg\,>\,Ag\,>Pd\,>\,Ir\,>\,P{{t}^{+2}}\,>\,Au
So now let’s see the options.
Option (A), $NaOH+HN{{O}_{3}}\to NaN{{O}_{3}}+{{H}_{2}}O$
Here the reaction is occurring between a strong base NaOH and a strong acid $HN{{O}_{3}}$ and the respective produced formed is a salt and water molecule. Hence this is a neutralization reaction.
Option (B), $Cu+2AgN{{O}_{3}}\to Cu{{(N{{O}_{3}})}_{2}}+2Ag$
This reaction is a displacement reaction. As from the electrochemical series we know that the oxidation potential of Cu is more than the Ag.
So Cu will have a tendency to get oxidized whereas Ag will have the tendency to get reduced.
Reduction reaction-$2A{{g}^{+}}_{(aq)}+2{{e}^{-}}\to 2Ag$
Oxidation reaction-$Cu\to C{{u}^{2+}}+2{{e}^{-}}$
Option (C), $2Hg+{{O}_{2}}\to 2HgO$
This is a combination reaction.
Option (D) $FeC{{l}_{3}}+2NaOH\to 3NaCl+Fe{{(OH)}_{3}}$
This reaction may be considered as the double displacement reaction, as the ions of the two compounds are inter-exchanged. The products formed here are NaCl and$Fe{{(OH)}_{3}}$.
$Fe{{(OH)}_{3}}$ is insoluble in the solution and hence it forms precipitation. This reaction is a precipitation reaction.

Hence the correct answer for the question is an option (B).

Note:
We may get confused between the options as these options are tricky, but we should know to judge the type of reactions by comparing the reactants and the products formed.
We should know the electrochemical series order so that if in the question, we are provided with some displacement reactions and have to predict the possible reactions, then it may help us to know the reduction potential of each atom.