Question

Which from following is a lanthanoid element?

• A. Er
• B. Am
• C. Np
• D. Lr

Hint:

Remember the 4f elements: Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu. Erbium (Er) is right near the end.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The f-block elements in the periodic table are divided into two distinct series: the lanthanoids (or lanthanides) and the actinoids (or actinides). We need to identify which of the given chemical symbols correctly belongs to the lanthanoid series.
Step 2: Key Formula or Approach:
Approach: Locate the given elements by their atomic numbers to distinguish between the $4f$ series (lanthanoids) and the $5f$ series (actinoids).
Step 3: Detailed Explanation:
- Lanthanoids: These are the 14 elements immediately following lanthanum ($\text{La}$, atomic number $57$), ranging from Cerium ($\text{Ce}$, $Z=58$) to Lutetium ($\text{Lu}$, $Z=71$). They involve the progressive filling of the $4f$ subshell. - Actinoids: These are the 14 elements immediately following actinium ($\text{Ac}$, atomic number $89$), ranging from Thorium ($\text{Th}$, $Z=90$) to Lawrencium ($\text{Lr}$, $Z=103$). They involve the filling of the $5f$ subshell. Let's classify each option: - (A) Er (Erbium): Atomic number $68$. It falls squarely in the range $58-71$, so it is a lanthanoid. - (B) Am (Americium): Atomic number $95$. It falls in the range $90-103$, so it is an actinoid. - (C) Np (Neptunium): Atomic number $93$. It falls in the range $90-103$, so it is an actinoid. - (D) Lr (Lawrencium): Atomic number $103$. It is the very last element of the actinoid series.
Step 4: Final Answer:
Erbium (Er) is a lanthanoid element.