Which cation is larger and why: cu+ or cu2+

Result:
$C{u}^{+}$ is larger than $C{u}^{2+}$ because $C{u}^{+}$ has only one valence electron, whereas $C{u}^{2+}$ has nine valence electrons. The additional electrons in $C{u}^{2+}$ result in increased electron-electron repulsion and an expanded electron cloud.
Solution:
First, let's examine the electronic configurations of $C{u}^{+}$ and $C{u}^{2+}$:
- $C{u}^{+}$: The electronic configuration of neutral copper ($Cu$) is $[Ar]3d^{10}4s^1$. When one electron is removed to form $C{u}^{+}$, it becomes $[Ar]3d^{10}$.
- $C{u}^{2+}$: Starting with the electronic configuration of neutral copper ($Cu$) ($[Ar]3d^{10}4s^1$), we remove two electrons to form $C{u}^{2+}$, resulting in $[Ar]3d^9$.
Now, let's analyze the effective nuclear charge experienced by these cations. The effective nuclear charge is the net positive charge experienced by the valence electrons due to the attraction from the nucleus. It determines the size of an atom or ion.
Since both $C{u}^{+}$ and $C{u}^{2+}$ have the same number of protons (29) in their nucleus, the effective nuclear charge experienced by the valence electrons will be the same.
However, the key difference lies in the number of valence electrons. $C{u}^{+}$ has only one valence electron in the $3d$ orbital, while $C{u}^{2+}$ has nine valence electrons in the $3d$ orbital.
According to the principles of atomic structure, electrons in the same energy level experience a greater repulsion from each other compared to the attraction from the nucleus. Consequently, the additional electrons in $C{u}^{2+}$ will increase the electron-electron repulsion and cause the electron cloud to expand.
Therefore, based on the above analysis, we can conclude that $C{u}^{+}$ is larger than $C{u}^{2+}$ due to the presence of fewer electrons and less electron-electron repulsion.

To determine which cation is larger, Cu+ or Cu2+, we can compare their electronic configurations.
The Cu+ cation has lost one electron from the neutral copper atom (Cu), resulting in the electronic configuration: $[Ar]3d^{10}$.
The Cu2+ cation has lost two electrons from the neutral copper atom (Cu), resulting in the electronic configuration: $[Ar]3d^9$.
In both cases, the 3d sublevel is involved. The 3d sublevel has a higher energy than the 4s sublevel, and it is further from the nucleus. Therefore, electrons in the 3d sublevel experience less effective nuclear charge and are less strongly attracted to the nucleus.
Since Cu2+ has one less electron in the 3d sublevel compared to Cu+, its effective nuclear charge is slightly higher, resulting in a stronger attraction between the remaining electrons and the nucleus. This increased attraction causes the Cu2+ cation to have a smaller size compared to the Cu+ cation.
Hence, Cu+ is larger than Cu2+.

The ionic radius of a cation depends on the number of electrons it has and its electronic configuration. When comparing the sizes of cations, we need to consider their charges.
The ${\text{Cu}}^{+}$ cation has a charge of $+1$, while the ${\text{Cu}}^{2+}$ cation has a charge of $+2$.
According to the general trend, as the charge on a cation increases, its ionic radius decreases. This is because the increased positive charge attracts the surrounding electrons more strongly, leading to a smaller size.
Therefore, we can conclude that the ${\text{Cu}}^{+}$ cation is larger than the ${\text{Cu}}^{2+}$ cation.