The magnetic moment of \(Cu^{2+}\) ion (Z = 29) is:

\(\sqrt{3}\)

The correct answer is option 1. \(\sqrt{3}\).

To calculate the magnetic moment of the \(Cu^{2+}\) ion, we can use the following formula:

\(\mu = \sqrt{n(n+2)} \, \text{BM}\)

where:

\(\mu\) is the magnetic moment in Bohr magnetons (BM),

\(n\) is the number of unpaired electrons.

The atomic number of copper (Cu) is 29, and its electronic configuration in the ground state is:

\(\text{Cu}: [\text{Ar}] \, 3d^{10} \, 4s^1\)

For the \(Cu^{2+}\) ion, two electrons are removed. One electron is removed from the 4s orbital and one from the 3d orbital:

\(\text{Cu}^{2+}: [\text{Ar}] \, 3d^9\)

In the \(3d^9\) configuration, there is one unpaired electron.

Using the formula:

\(\mu = \sqrt{n(n+2)} \, \text{BM}\)

where \(n = 1\) (the number of unpaired electrons), we get:

\(\mu = \sqrt{1(1+2)} = \sqrt{3} \, \text{BM}\)

The magnetic moment of \(Cu^{2+}\) is \(\sqrt{3}\) BM.

Thus, the correct answer is (1) \(\sqrt{3}\).