Question

Cuprous ion is colorless while cupric ion is colored because
(A) Cuprous ion has completed d-orbitals, while cupric ion has incomplete d- orbitals.
(B) Cuprous ion has exactly half-filled d-orbitals
(C) Cupric ion has completely filled d-orbitals, while cuprous ion has d-orbitals which are not completely filled
(D) Cupric ion has half-filled d-orbitals

Answer 1

Hint: Cuprous ion is represented as $C{{u}^{+}}$while cupric ion is represented as $C{{u}^{2+}}$ . Color can be exhibited by any of the elements due to the presence of unpaired electrons or due to d-d transition. It can also be due to charge transfer.

Complete step by step solution:
As you know from your chemistry lesson that copper is found in two states one is cuprous ion $C{{u}^{+}}$ and other is cupric ion $C{{u}^{2+}}$.
- where cuprous ion is present in +1 oxidation state and is also represented as copper (I) while cupric ion is present in +2 oxidation state and also represented as copper(II).
- The colour exhibited by copper is mainly due to d-d transition or presence of unpaired electrons in d orbitals of copper ions.
-Atomic no of copper $(Cu)$ is 29 and its electronic configuration is,
\[Cu=\,\left[ Ar \right]3{{d}^{9}}4{{s}^{2}}\]
But, the more stable electronic configuration is,
\[Cu\,=\,\left[ Ar \right]3{{d}^{10}}4{{s}^{1}}\]
Here, 4s orbital having 2 electrons donate one of them to 3d orbital to complete the d-orbitals having 10 electrons in it. This is done because fully filled d- orbital is more stable than incomplete d-orbital.
- So, the electronic configuration of cuprous ion $C{{u}^{+}}$ will be
\[C{{u}^{+}}\,=\,\left[ Ar \right]3{{d}^{10}}\]
Here, one electron is removed from the outermost orbital which is 4s to form the cuprous ion and due to presence of fully filled d- orbitals or absence of unpaired electron they don’t exhibit any color. So, cuprous ions are colorless.
-Now, the electronic configuration of cupric ion $C{{u}^{2+}}$is,
\[C{{u}^{2+}}\,=\,\left[ Ar \right]3{{d}^{9}}\]
 -In this case you have seen that two electrons are removed from the outermost orbital one from 4s and other from 3d to form cupric ions.
 -Cupric ions show d-d transition due to which it exhibits color or you can say due to presence of unpaired electrons in d-orbital it shows color. The color is mainly shown due to the d-d transition that u have studied in crystal field theory. According to this theory the electrons in the d-orbital get excited when a photon is applied on a metal and the electron in d -orbital jumps to another d-orbital of higher energy level and thus energy is absorbed in the visible region and color is exhibited.

Thus, the correct option is (A).

Note: Remember the electronic configuration of copper so you can find the electronic configuration of cuprous and cupric ions. Do not confuse between + sign and – sign. +sign represents removal of electrons and – sign means addition of electron. If possible try to convert the configuration in stable form rather than leaving it in unstable form.