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The correct order of ionic radii of Ce, La, Pm and Yb in +3 oxidation state is

  • A.\(\mathrm{La^{3+} \lt Pm^{3+} \lt Ce^{3+} \lt Yb^{3+}}\)
  • B.\(\mathrm{La^{3+} \lt Ce^{3+} \lt Pm^{3+} \lt Yb^{3+}}\)
  • C.\(\mathrm{Yb^{3+} \lt Ce^{3+} \lt Pm^{3+} \lt La^{3+}}\)
  • D.\(\mathrm{Yb^{3+} \lt Pm^{3+} \lt Ce^{3+} \lt La^{3+}}\)
  • E.\(\mathrm{La^{3+} \gt Pm^{3+} \gt Ce^{3+} \gt Yb^{3+}}\)

Answer: D

The correct order is \(\mathrm{Yb^{3+} \lt Pm^{3+} \lt Ce^{3+} \lt La^{3+}}\).

Across the lanthanoid series the ionic radius of the M3+ ions decreases steadily as atomic number increases. The added 4f electrons shield the growing nuclear charge poorly, so the outer electrons are pulled in: the lanthanoid contraction.

In order of atomic number, La (57), Ce (58), Pm (61), Yb (70), so La3+ is largest and Yb3+ smallest, with Ce3+ before Pm3+. The options starting with La3+ as the smallest run the trend backwards, and the '>' option puts Pm3+ above Ce3+.

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