Question

Which of the following ions has [Xe] 4f$^7$ 6s$^0$ as outer electronic configuration?

• A. Yb$^{2+}$
• B. Eu$^{2+}$
• C. Sm$^{2+}$
• D. Tm$^{2+}$

Hint:

Lanthanide elements typically lose 6s and 4f electrons during ionization; Eu$^{2+}$ has a stable half-filled 4f$^7$ configuration.

Answer 1

The Correct Option is B

Europium (Eu) has the atomic number 63, and its ground state electronic configuration is [Xe] 4f$^7$6s$^2$. When it forms the Eu$^{2+}$ ion, it loses two electrons, both from the 6s orbital.
So, Eu$^{2+}$ = [Xe] 4f$^7$ 6s$^0$.
This matches the required configuration. The other ions do not match this configuration due to different numbers of electrons in the 4f orbitals or retention of the 6s electrons.

Answer 2

Which of the following ions has [Xe] 4f⁷ 6s⁰ as outer electronic configuration?

Step 1: Understand the given electronic configuration:
The configuration [Xe] 4f⁷ 6s⁰ suggests the element is in the lanthanide series (elements with electrons filling the 4f orbitals).
- [Xe] is the noble gas core for atomic number 54.
- 4f⁷ indicates 7 electrons are present in the 4f orbital, which corresponds to a half-filled f-subshell — a particularly stable configuration.

Step 2: Consider europium (Eu):
- The atomic number of Eu is 63.
- Its ground-state electronic configuration is: [Xe] 4f⁷ 6s²

Step 3: Consider the Eu²⁺ ion:
- Eu loses two electrons to form Eu²⁺
- These two electrons are most easily lost from the outermost 6s orbital:
So, Eu²⁺ has the configuration: [Xe] 4f⁷ 6s⁰

Step 4: Final reasoning:
- The configuration perfectly matches the one given in the question.
- It also reflects the stability of the half-filled 4f orbital in Eu²⁺, which is why this ion is relatively stable despite being in a +2 oxidation state (unusual for lanthanides).

Final Answer:
\[ \boxed{\text{Eu}^{2+}} \]