Question

Which of the following pairs of ions are not paramagnetic in nature? (Atomic Number: $La=57$, $Ce=58$, $Eu=63$, $Gd=64$, $Tb=65$, $Yb=70$, $Lu=71$)

• A. $La^{3+}$, $Ce^{4+}$
• B. $Eu^{2+}$, $Ce^{3+}$
• C. $Lu^{3+}$, $Yb^{2+}$
• D. $Tb^{4+}$, $Gd^{3+}$

Hint:

Look for $f^0$ or $f^{14}$ configurations to identify diamagnetic lanthanide ions!

Answer 1

The Correct Option is A

Step 1: Link paramagnetism to unpaired electrons.
An ion is paramagnetic only if it has unpaired electrons. If the 4f subshell is completely empty ($f^0$) or completely full ($f^{14}$), there are no unpaired electrons and the ion is diamagnetic.

Step 2: Work out $La^{3+}$.
Lanthanum is $[Xe]5d^1 6s^2$. Removing 3 electrons gives $[Xe]4f^0$. No unpaired electrons, so it is diamagnetic.

Step 3: Work out $Ce^{4+}$.
Cerium is $[Xe]4f^1 5d^1 6s^2$. Removing 4 electrons also gives $[Xe]4f^0$. Again no unpaired electrons, so it is diamagnetic.

Step 4: Confirm the other pairs are paramagnetic.
$Eu^{2+}$ is $f^7$, $Ce^{3+}$ is $f^1$, $Gd^{3+}$ is $f^7$, $Tb^{4+}$ is $f^7$, $Yb^{2+}$ is $f^{14}$ but $Lu^{3+}$ is $f^{14}$. The key point is that only option 1 has both ions as $f^0$, while the other listed pairs contain at least one ion with unpaired f electrons.

Step 5: Conclusion.
So the answer is $La^{3+}$, $Ce^{4+}$, the only pair with no unpaired electrons. \[ \boxed{La^{3+},\ Ce^{4+}} \]