Question

Which element from following has smallest ionic size in +3 state?

• A. Lu
• B. La
• C. Dy
• D. Nd

Hint:

For lanthanide +3 ions, size strictly follows atomic number. The higher the atomic number, the smaller the ionic radius. Order of size: $La^{3+} > Nd^{3+} > Dy^{3+} > Lu^{3+}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given four specific inner transition elements (lanthanides) and must identify which one possesses the smallest ionic radius when in the +3 oxidation state.

Step 2: Detailed Explanation:
The elements listed (Lanthanum, Neodymium, Dysprosium, Lutetium) belong to the Lanthanide series (4f block) of the periodic table.
In this series, as the atomic number increases from Lanthanum ($Z=57$) across the period to Lutetium ($Z=71$), electrons are progressively added into the deep inner 4f subshell.
The 4f electrons have highly diffused shapes and offer extremely poor shielding (or screening) against the increasing positive nuclear charge.
Consequently, the effective nuclear charge felt by the outermost electrons increases steadily across the series, pulling the electron clouds closer inward.
This phenomenon causes a steady, cumulative decrease in both atomic and ionic radii across the entire series, a defining trait known as "Lanthanide Contraction".
Because Lutetium (Lu) is the very last element in the lanthanide series, it experiences the maximum cumulative contraction effect, rendering $Lu^{3+}$ the smallest ion among them.

Step 3: Final Answer:
Lutetium (Lu) has the smallest ionic size, corresponding to option (a).