Question

What are the charges of two ions of copper?

Answer 1

Hint: An atom or a molecule turns into an ion when it gains or loses electrons and has a net electrical charge. Also copper (Cu) can have 2 ion formations which is $C{{u}^{+1}}$ and $C{{u}^{+2}}$, and both of them are cations (ions whose net electrical charge is positive).

Complete answer: -Copper (Cu) is a transition metal which belongs to d - block of periodic table and its atomic number is 29. It is soft, ductile and malleable in nature with very high electrical and thermal conductivity.
-It's electronic configuration is as follows:-
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{1}}$
-As we can see that Cu has only one electron in the 4s subshell, therefore its valency is +1 and it will tend to remove this electron to gain extra stability. Hence the formation of $C{{u}^{+1}}$ occurs by oxidation (loss of electron) of Cu. Electronic configuration of $C{{u}^{+1}}$ is as follows:-
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}$
-Since $3{{d}^{10}}$ subshell is fully filled so $C{{u}^{+1}}$is the most stable ion form of copper.
-But we must also know that when sufficient enthalpy is given to $C{{u}^{+1}}$ion, it oxidizes again and turn into $C{{u}^{+2}}$ ion as now the removal of electron is tough due to fully filled subshell. This enthalpy (ligation energy) is compensated by the help of ligands attached to the copper metal ion. Also ligation energy of $C{{u}^{+2}}$ is greater than $C{{u}^{+1}}$, therefore $C{{u}^{+1}}$ can form easily.
- Electronic configuration of $C{{u}^{+2}}$ is as follows:-
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{9}}$
-Hence the 2 copper ions are: $C{{u}^{+1}}$ (Copper (I)) and $C{{u}^{+2}}$(Copper (II)) with charges as +1 and +2 respectively.

Note: -We must not get confused with the electronic configuration of copper.
(A) $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{9}}4{{s}^{2}}$is an incorrect electronic configuration.
(B) $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{1}}$is the correct electronic configuration.