` Nd^(2+)="_____"`

Options

A. `4f^(3)`
B. `4f^(4)6s^(2)`
C. `4f^(2)6s^(2)`
D. `4f^(4)`

Answer

To find the outer electronic configuration of Neodymium in the +2 oxidation state (Nd²⁺), we can follow these steps:

### Step 1: Determine the Atomic Number and Ground State Configuration
- Neodymium (Nd) has an atomic number of 60.
- The ground state electronic configuration of Neodymium is:
\[ [Xe] \, 6s^2 \, 4f^4 \]
where [Xe] represents the electron configuration of Xenon.

### Step 2: Identify the Electrons to be Removed for Nd²⁺
- In the +2 oxidation state, Neodymium loses two electrons.
- Electrons are typically removed from the outermost shell first. In this case, the outermost electrons are the ones in the 6s subshell.

### Step 3: Remove the Electrons
- Remove the two electrons from the 6s subshell:
\[ [Xe] \, 6s^2 \, 4f^4 \rightarrow [Xe] \, 4f^4 \]
- Therefore, the electronic configuration of Nd²⁺ is:
\[ [Xe] \, 4f^4 \]

### Conclusion
- The outer electronic configuration of Nd²⁺ is:
\[ 4f^4 \]

### Final Answer
\[ \text{Nd}^{2+} = 4f^4 \]

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