\[\left[ {CuCl_4^{2 - }} \right]\] exists while \[\left[ {CuI_4^{2 - }} \right]\] does not exist, because:
(A) \[{I^ - }\] is stronger reductant than \[C{l^ - }\]
(B) \[{I^ - }\] is weaker reductant than \[C{l^ - }\]
(C) \[{I^ - }\] is a stronger oxidant than \[C{l^ - }\]
(D) None of the above
Answer
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Hint: The above question says that \[\left[ {CuCl_4^{2 - }} \right]\] exists while \[\left[ {CuI_4^{2 - }} \right]\] does not exist. Both chlorine \[C{l^ - }\] and iodine \[{I^ - }\] belong to the same group and most of the properties are similar. But here we will consider one of the property which is responsible for the reason that \[\left[ {CuCl_4^{2 - }} \right]\] exists while \[\left[ {CuI_4^{2 - }} \right]\] does not exist.
Complete answer:First, we will understand some basic terms given in the options reductant and oxidant. The term ‘Oxidant’ refers to the reactants which oxidize or remove the electron from the other reactants during a redox reaction. Simply it is a substance which gains electrons. The term ‘Reductant’ refers to the reducing agent that loses an electron. Now we will consider all the options one by one. In the given options we will compare the reducing nature of chloride \[C{l^ - }\] and iodide \[{I^ - }\] ion.
As we know that iodide \[{I^ - }\] ion is a stronger reducing agent, as it reduces \[C{u^{2 + }}\]into \[CuI\] and gets oxidized into \[{I_2}\]. We can understand it with the help of the reaction \[2Cu{I_2} \to C{u_2}{I_2} + {I_2}\]. Therefore, \[C{u^{2 + }}\] does not exist with Iodine. But in \[\left[ {CuI_4^{2 - }} \right]\] the oxidation state of copper is \[x + 4\left( { - 1} \right) = - 2,x = + 2\]. Therefore, \[\left[ {CuCl_4^{2 - }} \right]\] exists while \[\left[ {CuI_4^{2 - }} \right]\] does not exist.
Hence the correct option is (A).
Note: In the complexes \[\left[ {CuCl_4^{2 - }} \right]\] and \[\left[ {CuI_4^{2 - }} \right]\] the size of chlorine is less than iodine. So, we can conclude that \[\left[ {CuCl_4^{2 - }} \right]\] is more stable complex as compared to \[\left[ {CuI_4^{2 - }} \right]\]. The bigger size of iodine makes the complex unstable. Therefore, \[\left[ {CuCl_4^{2 - }} \right]\] exists while \[\left[ {CuI_4^{2 - }} \right]\] does not exist
Complete answer:First, we will understand some basic terms given in the options reductant and oxidant. The term ‘Oxidant’ refers to the reactants which oxidize or remove the electron from the other reactants during a redox reaction. Simply it is a substance which gains electrons. The term ‘Reductant’ refers to the reducing agent that loses an electron. Now we will consider all the options one by one. In the given options we will compare the reducing nature of chloride \[C{l^ - }\] and iodide \[{I^ - }\] ion.
As we know that iodide \[{I^ - }\] ion is a stronger reducing agent, as it reduces \[C{u^{2 + }}\]into \[CuI\] and gets oxidized into \[{I_2}\]. We can understand it with the help of the reaction \[2Cu{I_2} \to C{u_2}{I_2} + {I_2}\]. Therefore, \[C{u^{2 + }}\] does not exist with Iodine. But in \[\left[ {CuI_4^{2 - }} \right]\] the oxidation state of copper is \[x + 4\left( { - 1} \right) = - 2,x = + 2\]. Therefore, \[\left[ {CuCl_4^{2 - }} \right]\] exists while \[\left[ {CuI_4^{2 - }} \right]\] does not exist.
Hence the correct option is (A).
Note: In the complexes \[\left[ {CuCl_4^{2 - }} \right]\] and \[\left[ {CuI_4^{2 - }} \right]\] the size of chlorine is less than iodine. So, we can conclude that \[\left[ {CuCl_4^{2 - }} \right]\] is more stable complex as compared to \[\left[ {CuI_4^{2 - }} \right]\]. The bigger size of iodine makes the complex unstable. Therefore, \[\left[ {CuCl_4^{2 - }} \right]\] exists while \[\left[ {CuI_4^{2 - }} \right]\] does not exist
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