The lanthanide ion having four unpaired electrons is (Given :...

The general electronic configuration of lanthanide ions $\text{Ln}^{3+}$ is $[\text{Xe}] 4f^{Z-57}$, where $Z$ is the atomic number.

For $\text{Ho}^{3+}$ ($Z=67$):
Number of $4f$ electrons $= 67 - 57 = 10$
Configuration is $[\text{Xe}] 4f^{10}$.
Since the $f$-subshell has 7 orbitals, 10 electrons will fill as 3 pairs and 4 unpaired electrons. Number of unpaired electrons $= 14 - 10 = 4$.

For $\text{Nd}^{3+}$ ($Z=60$):
Number of $4f$ electrons $= 60 - 57 = 3$
Configuration is $[\text{Xe}] 4f^{3}$.
Number of unpaired electrons $= 3$.

For $\text{Ce}^{3+}$ ($Z=58$):
Number of $4f$ electrons $= 58 - 57 = 1$
Configuration is $[\text{Xe}] 4f^{1}$.
Number of unpaired electrons $= 1$.

For $\text{Tb}^{3+}$ ($Z=65$):
Number of $4f$ electrons $= 65 - 57 = 8$
Configuration is $[\text{Xe}] 4f^{8}$.
Number of unpaired electrons $= 14 - 8 = 6$.

Thus, $\text{Ho}^{3+}$ has exactly four unpaired electrons.

Answer: $\text{Ho}^{3+}$