Question

Which one of the following statements related to lanthanons is incorrect ?

• A. Europium shows $+2$ oxidation state
• B. The basicity of hydroxides increases as the ionic radius decreases from $Ce$ to $Lu$
• C. In lanthanons $4f$ electrons do not participate in bonding
• D. $Ce(+4)$ solution are widely used as oxidizing agent in volumetric analysis

Answer 1

The correct option is B The basicity of hydroxides increases as the ionic radius decreases from $Ce$ to $Lu$

The electronic configuration of Europium is:
$Eu\left(63\right)=\left[Xe\right]4f^{7}5d^{0}6s^2;\therefore E{u}^{2+}=\left[Xe\right]4f^7$
Due to the stable half filled valence shell electronic configuration, $Eu$ shows $+2$ oxidation state.

The hydroxides of lanthanons are ionic in nature and the basicity decreases as the ionic radius decreases from $Ce$ to $Lu$.

In case of lanthanons, the $4f$ electrons are in the antipenultimate shell and they are effectively shielded by the $5s$ and $5p$ electrons. Thus, it does not participate in bonding.

The electronic configuration of $C{e}^{4+}$ is $\left[Xe\right]4f^{0}5d^{0}6s^0$. So, $Ce$ is highly stable in its $+4$ oxidation state. Thus, $C{e}^{4+}$ is a good oxidizing agent and is widely used as oxidizing agent in volumetric analysis.