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Suppose we have a compound, say copper sulfate. I know that sulfate is a poly-atomic ion, having valency 2. And, in this case, copper also has valency 2. So, to fullfill their 'octet' requirements, they both react. The problem is: Suppose we take the metal zinc and bring it in contact with copper sulfate. Now, a displacement reaction occurs, in which zinc replaces copper because of its higher reactivity.

Copper sulfate is a compound that is neutral; both the cation and anion have completed their octets; why then does zinc replace copper? Or, in other words, after completing the octet, why it is necessary for sulfate to leave copper and react with zinc?

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    $\begingroup$ 1) $\ce{Cu^2+}$ and $\ce{SO4^2-}$ ions do not react. Your description of the ionic bonding in copper(II) sulfate sounds like a description of covalent bonding, which it is not. 2) In aqueous solution, which you are probably talking about, the ions are separated and there are hardly any interactions between ions since they are surrounded by water molecules. The reaction is between Zn and $\ce{Cu^2+}$, not Zn and $\ce{CuSO4}$. 3) The octet rule is irrelevant in transition metal compounds like those of copper. 4) The octet rule does not, and cannot, predict whether a reaction takes place. $\endgroup$
    – orthocresol
    Commented Jan 3, 2016 at 15:28

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It is due to that zinc is more reactive than copper which is less reactive metal so, zinc can easily displaces copper from its solution. We can understand it as zinc ion, $\ce{Zn^2+}$ ion form more strong electrostatic attraction with sulphate ion, $\ce{SO4-}$ ion.

Also, we can also consider the electrochemical series or activity series to give a better reason for the same. In activity series, zinc has greater value of standard reduction potential than copper's standard reduction potential value. It means, zinc can easily reduce Copper and it get oxidised.

$$\ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)}$$

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