Question

$ C{{u}^{2+}} $ ion is unstable in aqueous solution because _______________.
(A) it undergoes disproportionation
(B) it undergoes oxidation
(C) it undergoes reduction
(D) it undergoes complex

Answer 1

Hint: Copper is a chemical element with the atomic number 29 and the symbol Cu. It's a ductile, malleable metal with excellent thermal and electrical conductivity. A freshly exposed pure copper surface is pinkish-orange in hue. Copper is used as a heat and electricity conductor, a building material, and a component of numerous metal alloys, including sterling silver for jewellery, cupronickel for nautical hardware and coins, and constantan for strain gauges and thermocouples for temperature monitoring.

Complete answer:
An aqueous solution is one in which water serves as the solvent. Appending (aq) to the applicable chemical formula is the most common way to show it in chemical equations. The term aqueous (derived from the Greek aqua) refers to something that is linked to, comparable to, or dissolved in water. Water is a common solvent in chemistry because it is an excellent solvent that is also naturally abundant.
In an aqueous medium, $ C{{u}^{2+}} $ is more stable than $ C{{u}^{+}} $ . This is because although energy is required to remove one electron from $ C{{u}^{+}} $ to $ C{{u}^{2+}} $ , high hydration energy of $ C{{u}^{2+}} $ compensates for it. Therefore, $ C{{u}^{+}} $ ion in an aqueous solution is unstable. It is disproportionate to give $ C{{u}^{2+}} $ and Cu.
Disproportionation, also known as dismutation in chemistry, is a redox process in which one chemical with an intermediate oxidation state transforms into two compounds with higher and lower oxidation states. The term can be used to describe any desymmetrization reaction of the following types: Regardless of whether it is a redox or other type of process, $ 2\text{ }A\text{ }\to \text{ }A'\text{ }+\text{ }A'' $ .

Note:
Subshells that are half-filled or fully filled have more stability. As a result, one of the $ 4{{s}^{2}} $ electrons makes a $ 3{{d}^{9}} $ leap. This provides us with the (proper) configuration of.
We remove one electron from 4s1 for the $ C{{u}^{+}} $ ion, leaving us with.
We remove two electrons from the $ C{{u}^{2+}} $ ion (one from the 4s1 and one from the $ 3{{d}^{10}} $ ), leaving us with $ 1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{9}} $ .