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The lanthanide ion having four unpaired electrons is (Given: Atomic numbers of Ce = 58, Nd = 60, Tb = 65 and Ho = 67)
Detailed Solution
$Nd^{3+}=4f^3$: n = 3. $Ce^{3+}=4f^1$: n = 1. $Tb^{3+}=4f^8$: n = 6. $Ho^{3+}=4f^{10}$: n = 4. So $Ho^{3+}$ has four unpaired electrons.