Question

The trend of basicity of lanthanide hydroxides:
A. increases across the lanthanide series
B. decreases across the lanthanide series
C. first increases and then decreases
D. first decreases and then increases

Answer 1

To answer this question you must recall the factors affecting the basicity of metal hydroxides and the variation in the properties of lanthanides as we move from left to right. Lanthanides are the first series of f-block elements.

Complete step by step answer:
With the change in electronic configuration of a metal, many metallic properties begin to vary. The basic character of metal hydroxides depends upon the bond formed between the metal and hydroxide group. If the bond is ionic, then basic strength is higher as compared to when the bond formed between metal and oxygen is covalent.
Ionic character of a metal decreases with decrease in its size and vice versa because smaller the atomic radius, the valence electrons are bound more strongly and thus lesser is the ionic character of the metal.
The lanthanide series corresponds to the filling of $4f$ orbital. Thus, lanthanides have three valence electrons and form tri positive ions.
In the lanthanide series, as we move from left to right, the atomic radii and the ionic radii of tri positive lanthanide ions decrease from lanthanum to lutetium. This change in size is observed due to increasing nuclear charge and electrons entering the inner $4f$ orbital which has poor shielding effect. This decrease in size with increasing atomic number is known as lanthanide contraction.
As the size of lanthanides decreases across the series due to lanthanide contraction, we can infer that the metallic character decreases, and the covalent character of the hydroxides increases. The increase in covalent character of the hydroxides results in a decrease in the basic strength.

The correct option is B.

Note:
The lanthanide series contains 15 elements which correspond to the filling of $4f$orbital.
The general electronic configuration of lanthanide elements is $\left( {Xe} \right)4{f^{0 - 14}}5{d^{0 - 1}}6{s^2}$.