Chemistry - f-block Question with Solution | TestHub

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