Question

Consider the following table of three lanthanoid ions X, Y, and Z and their properties.

Atomic numbers of Ce, Eu, and Lu are 58, 63, and 71, respectively. Given these atomic numbers, the lanthanoid ions X, Y, and Z, respectively, are:

• A. $\text{Lu}^{3+}$, $\text{Eu}^{2+}$, and $\text{Ce}^{4+}$
• B. $\text{Lu}^{3+}$, $\text{Ce}^{4+}$, and $\text{Eu}^{2+}$
• C. $\text{Eu}^{2+}$, $\text{Ce}^{4+}$, and $\text{Lu}^{3+}$
• D. $\text{Eu}^{2+}$, $\text{Lu}^{3+}$, and $\text{Ce}^{4+}$

Hint:

Lanthanoids have a preferred oxidation state of $+3$. Any $+2$ lanthanoid ion (like $\text{Eu}^{2+}$, $\text{Yb}^{2+}$) will act as a reducing agent, while any $+4$ ion (like $\text{Ce}^{4+}$) will act as an oxidizing agent.

Answer 1

The Correct Option is A

Step 1: Set up the electron configurations.
The three ions come from Ce (Z = 58), Eu (Z = 63), and Lu (Z = 71). The key property is how many 4f electrons each ion has, because that decides whether the ion is magnetic and how it behaves chemically. So we write the f-electron count for each possible ion.

Step 2: Work out $\text{Ce}^{4+}$.
Cerium losing four electrons reaches $\text{Ce}^{4+}$, which has an empty 4f shell ($4f^0$). With no unpaired electrons it is diamagnetic, and the empty stable shell makes it a good oxidising ion. So whichever entry in the table is diamagnetic and oxidising must be $\text{Ce}^{4+}$.

Step 3: Work out $\text{Lu}^{3+}$.
Lutetium as $\text{Lu}^{3+}$ has a completely full 4f shell ($4f^{14}$). A full shell also means no unpaired electrons, so it too is diamagnetic, but it is chemically very stable and not an oxidiser. So the diamagnetic but stable, ordinary $+3$ entry is $\text{Lu}^{3+}$.

Step 4: Work out $\text{Eu}^{2+}$.
Europium as $\text{Eu}^{2+}$ has a half filled 4f shell ($4f^7$). Seven unpaired electrons make it strongly paramagnetic, and the $+2$ state makes it a reducing ion. So the paramagnetic, reducing entry must be $\text{Eu}^{2+}$.

Step 5: Match X, Y, Z to the table.
Reading the properties in the table in order, the first ion matches the stable diamagnetic $+3$ behaviour (Lu), the second matches the diamagnetic oxidising behaviour (Ce), and the third matches the paramagnetic reducing behaviour (Eu).

Step 6: State the answer.
So X, Y, Z line up as lutetium(III), cerium(IV), and europium(II) respectively.

\[ \boxed{\text{X} = \text{Lu}^{3+},\ \text{Y} = \text{Ce}^{4+},\ \text{Z} = \text{Eu}^{2+}} \]