Which of the following Lanthanoid ions is diamagnetic?

[Atomic number of Ce = 58, Am = 62, Eu= 63, Yb = 70]

\(Yb^{2+}\)

The correct answer is option 4. \(Yb^{2+}\).

Lanthanoids are elements with atomic numbers 57 to 71, and they typically have their outermost electrons in the 4f and 5d orbitals. For each lanthanoid ion, we need to determine its electronic configuration after ionization and check for unpaired electrons. Diamagnetic substances have all their electrons paired, resulting in no net magnetic moment.

Here is a step-by-step explanation for each ion:

1. \(Ce^{2+}\) (Cerium Ion)

Atomic Number of Ce: 58

Ground-State Electron Configuration: \([Xe] 4f^1 5d^1 6s^2\)

Here, cerium has 1 electron in the \(4f\) orbital, 1 electron in the \(5d\) orbital, and 2 electrons in the \(6s\) orbital.

For \(Ce^{2+}\): Two electrons are removed: one from the \(6s\) orbital and one from the \(5d\) orbital. The resulting configuration is \([Xe] 4f^1\).

Unpaired Electrons: There is 1 unpaired electron in the \(4f\) orbital.

Magnetic Property: Paramagnetic, due to the presence of an unpaired electron.

2. \(Sm^{2+}\) (Samarium Ion)

Atomic Number of Sm: 62

Ground-State Electron Configuration: \([Xe] 4f^6 6s^2\)

Here, samarium has 6 electrons in the \(4f\) orbitals and 2 electrons in the \(6s\) orbital.

For \(Sm^{2+}\): Two electrons are removed from the \(6s\) orbital. The resulting configuration is \([Xe] 4f^6\).

Unpaired Electrons: In the \(4f\) orbitals, there are 4 unpaired electrons.

Magnetic Property: Paramagnetic, due to the presence of unpaired electrons.

3. \(Eu^{2+}\) (Europium Ion)

Atomic Number of Eu: 63

Ground-State Electron Configuration: \([Xe] 4f^7 6s^2\)

Here, europium has 7 electrons in the \(4f\) orbitals and 2 electrons in the \(6s\) orbital.

For \(Eu^{2+}\): Two electrons are removed from the \(6s\) orbital. The resulting configuration is \([Xe] 4f^7\).

Unpaired Electrons: In the \(4f\) orbitals, there are 7 unpaired electrons.

Magnetic Property: Paramagnetic, due to the presence of unpaired electrons

4. \(Yb^{2+}\) (Ytterbium Ion)

Atomic Number of Yb: 70

Ground-State Electron Configuration: \([Xe] 4f^{14} 6s^2\)

Here, ytterbium has 14 electrons in the \(4f\) orbitals and 2 electrons in the \(6s\) orbital.

For \(Yb^{2+}\): Two electrons are removed from the \(6s\) orbital. The resulting configuration is \([Xe] 4f^{14}\).

Unpaired Electrons: In the \(4f\) orbitals, all electrons are paired.

Magnetic Property: Diamagnetic, as there are no unpaired electrons.

Summary

Diamagnetic ions have all their electrons paired, resulting in no net magnetic moment. \(Ce^{2+}\), \(Sm^{2+}\), and \(Eu^{2+}\) are paramagnetic due to unpaired electrons in their \(4f\) orbitals. \(Yb^{2+}\) is diamagnetic because all electrons in the \(4f\) orbitals are paired.

Therefore, the correct answer is Option 4: \(Yb^{2+}\).