Question

$Nd ^{2+}=$ ___________

• A. $4 f ^3$
• B. $4 f^4 6 s^2$
• C. $4 f ^2 6 s ^2$
• D. $4 f ^4$

Hint:

The electron configuration for transition metal ions is derived by removing electrons from the highest energy orbitals, typically starting from the outermost \( s \)-orbital.

Answer 1

The Correct Option is D

The question asks about the electronic configuration of the neodymium ion with a 2+ charge, represented as \(Nd^{2+}\). Neodymium is a lanthanide element with the atomic number 60. In its neutral state, the electronic configuration of neodymium is:

\([Xe] 4f^4 6s^2\)

When neodymium forms a 2+ ion, it loses two electrons. Typically, electrons are removed first from the outermost orbitals, which are the highest energy level orbitals. For neodymium, the 6s electrons are the highest in energy and thus removed first. Therefore, the loss of two electrons from Nd results in the following electron configuration:

• Remove two electrons from the 6s orbital, leaving the 4f electrons intact.

Hence, the configuration of \(Nd^{2+}\) is:

\([Xe] 4f^4\)

Thus, the correct option is: $4 f ^4$.

To conclude, the electron configuration confirms that the neodymium ion with a 2+ charge has the electronic configuration \(4f^4\).