Question: 88. The pair of lanthanoid ions which are diamagnetic is
(1) \mathrm{Ce}^{4+} and \mathrm{Yb}^{2+}
(2) \mathrm{Ce}^{3+} and \mathrm{Eu}^{2+}
(3) \mathrm{Gd}^{3+} and \mathrm{Eu}^{3+}
(4) \mathrm{Pm}^{3+} and \mathrm{Sm}^{3+}
Answer: Option (1)
Explanation:
Diamagnetic species are those which have no unpaired electrons in their electronic configuration.
\mathrm{Ce}^{4+} has atomic number 58. Its electronic configuration is
\mathrm{Ce}: [\mathrm{Xe}]\,4f^{1}5d^{1}6s^{2}After losing four electrons, \mathrm{Ce}^{4+} becomes
[\mathrm{Xe}]\,4f^{0}Since there are no electrons in the 4f subshell,
it has no unpaired electrons and is diamagnetic.
\mathrm{Yb}^{2+} has atomic number 70. Its electronic configuration is
\mathrm{Yb}: [\mathrm{Xe}]\,4f^{14}6s^{2}After losing two electrons, \mathrm{Yb}^{2+} becomes
[\mathrm{Xe}]\,4f^{14}The 4f subshell is completely filled, so there are no unpaired electrons and the ion is diamagnetic.
Other given ions contain unpaired 4f electrons and are therefore paramagnetic.
Hence, the correct pair of diamagnetic lanthanoid ions is \mathrm{Ce}^{4+}
and \mathrm{Yb}^{2+}.