Step 1: Determine Electronic Configurations

The general electronic configuration of lanthanides is $[Xe] 4f^{n} 5d^{0-1} 6s^{2}$. When forming $3+$ ions, the two $6s$ electrons and one $4f$ or $5d$ electron are removed.

Step 2: Analyze each option

We calculate the $4f$ electrons for each ion:

(A) $Ho^{3+}$ ($Z=67$): Neutral $Ho$ is $[Xe] 4f^{11} 6s^{2}$. $Ho^{3+}$ is $[Xe] 4f^{10}$. Unpaired electrons = $14 - 10 = 4$ (in the second half of the shell).

(B) $Nd^{3+}$ ($Z=60$): Neutral $Nd$ is $[Xe] 4f^{4} 6s^{2}$. $Nd^{3+}$ is $[Xe] 4f^{3}$. Unpaired electrons = 3.

(C) $Ce^{3+}$ ($Z=58$): Neutral $Ce$ is $[Xe] 4f^{1} 5d^{1} 6s^{2}$. $Ce^{3+}$ is $[Xe] 4f^{1}$. Unpaired electrons = 1.

(D) $Tb^{3+}$ ($Z=65$): Neutral $Tb$ is $[Xe] 4f^{9} 6s^{2}$. $Tb^{3+}$ is $[Xe] 4f^{8}$. Unpaired electrons = $8 - 7 = 1$ (or 6 depending on orbital filling, but not 4).

Step 3: Conclusion

The ion $Ho^{3+}$ has a $4f^{10}$ configuration. According to Hund's rule, filling 10 electrons in 7 orbitals results in 4 unpaired electrons.