Question

Which of the following lanthanoids
show +2 oxidation state besides
the characteristic oxidation state
+3 of lanthanoids?

• A. $Eu$
• B. $Ce$
• C. $Yb$
• D. $Ho$

Answer 1

Answer: (A), (C)

$Yb$

Solution
Analyzing the given options

Option (A)

Automic number of $Ce\left(Cerium\right)$ is 58 The electronic configuration of Ce is $\left[Xe\right]4f^{1}5d^{1}6s^2$


So,
Losing two electrons doesn't provide it any extra stability, but on loaing three electrons it attains stability.

Thu, it exhibits +3 oxdation no and is not stable in +2 oxidation state.

Option(B)
Atomic number of EU ( Europium) is 63.
The electronic configuration of Eu is $\left[Xe\right]4f^{7}5d^{0}6s^2$

So, it tends to lose 2 electrons from the outer s-orbital and attain stable half-filled configuration of $4f^7$

Thus, Eu shows +2 oxidation state as well.


Option (C)
Atomic number of Yb (Ytterbium) is 70.
The electronic configuration of Yb is:
[Xe]$4f^{14}5d06s2$

So Yb tends to lose its 2 electrons from the
outer $s-orbital$ and attain a stable fully-filled configuration.
$i.e.,\left[Xe\right]4f^{14}$

Thus, due to the extra stability of fully-filledconfigurations, Yb exhibit +2 oxidation states
as well

Option(D)
Automic number of Ho ( Holmium) is 67

The electronic configuration of Ho is
[Xe]$4f^{11}5d^{0}6s^2$, so losing 2 electrons will not provide it any stability. That's why it does not show+2 oxidation state.

Hence , the correct options are (B) and (C).