Identify the lanthanoid that exhibits zero effective magnetic moment in +3 state.

A species exhibits zero effective magnetic moment only if it contains no unpaired electrons. For lanthanoid ions in the +3 oxidation state, we analyze their electronic configurations to determine which one has no unpaired electrons.

Holmium (Ho, atomic number 67):
Ground state configuration is $[\mathrm{Xe}] 4f^{11} 6s^2$, so $\mathrm{Ho}^{3+}$ has $[\mathrm{Xe}] 4f^{10}$.
With 10 electrons in 7 orbitals, there are 4 unpaired electrons according to Hund’s rule.

Praseodymium (Pr, atomic number 59):
Ground state is $[\mathrm{Xe}] 4f^{3} 6s^2$, so $\mathrm{Pr}^{3+}$ is $[\mathrm{Xe}] 4f^{2}$.
This yields 2 unpaired electrons.

Erbium (Er, atomic number 68):
Ground state is $[\mathrm{Xe}] 4f^{12} 6s^2$, so $\mathrm{Er}^{3+}$ is $[\mathrm{Xe}] 4f^{11}$.
This gives 3 unpaired electrons.

Lutetium (Lu, atomic number 71):
Ground state is $[\mathrm{Xe}] 4f^{14} 5d^{1} 6s^2$, so $\mathrm{Lu}^{3+}$ has $[\mathrm{Xe}] 4f^{14}$.
This is a completely filled $f$ subshell with all electrons paired, resulting in zero unpaired electrons.

Therefore, only $\mathrm{Lu}^{3+}$ exhibits zero effective magnetic moment.