Assertion: When a copper wire is dipped in silver nitrate solution, there is no change in the color of the solution.
Reason: Copper cannot displace silver from its salt solution.
(A) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
(B) Both Assertion and Reason are correct but Reason is not the correct explanation for
(C) Assertion is correct but Reason is incorrect
(D) Both Assertion and Reason are incorrect

Hint: Displacement reactions are those reactions in which less reactive metal displaced by more reaction metal.
Example:
${N_1} + MgC{l_2} \to {N_1}C{l_2} + Mg$
Since, ${N_1}$is more reactive than $Mg$so it displace $Mg$from$MgC{l_2}.$

Complete step by step answer:
When a strip of copper metal is placed in a silver nitrate solution a grayish white deposit of silver can be seen on the copper stripe.
This happens because copper is more reactive than silver and displaces silver easily from solution to form copper nitrate.
$Cu + 2AgN{O_3} \to 2Ag + Cu{(N{O_3})_2}$
Therefore, the statement in question is wrong.
This is a single replacement reaction. In this reaction copper leases elections and from ion, these ion replace silver ions that present in the $AgN{O_3}$solution.
$Cu \to C{u^{ + + }} + 2{e^ - }$
$2A{g^ + } + 2{e^ - } \to Ag$
$Cu + 2A{g^ + } \to C{u^{ + + }} + 2A{g^ + }$
This reaction can be explained on the basis of electrochemical series is defined as arrangement of electrodes (metal or non-metal in contact with their ions) with electrode half reactions in order of decreasing standard potential.
The metal with more standard potential displaces metal with less standard potential.
Since electrode potential of copper is $ + 0.337\nu $and that of silver is $ + 0.222\nu .$
Therefore, copper displaces silver from its solution.
Therefore, from the above explanation Both Assertion and Reason are incorrect.

So, the correct answer is “Option D”.

Additional Information:
The electrode potential of electrodes decreases from top to bottom in electrochemical series.
${F^ - }/{F_2}/Pt \to $Fluorine has the highest potential i.e., $ + 2.8\nu $ it is a strong oxidizing agent.
$L{i^ + }/Li \to $Lithium has lowest partial i.e., $ - 3.045\nu $ it is a strong reducing agent.

Note:
The spontaneity of reaction can be determined by using electrochemical series. If the $E$ cell for a reaction is positive, the reaction is spontaneous.