Question

What is Lanthanoid contraction? Write in increasing order of basic nature of oxides \( (Ln_2O_3) \) and hydroxides \( [Ln(OH)_3] \) of Lanthanoids.

Hint:

Poor shielding by 4f electrons shrinks the ions from La to Lu; smaller ion means weaker base, so La(OH)3 is most basic.

Answer 1

Step 1: Definition of Lanthanoid contraction.
The steady and regular decrease in the atomic and ionic radii of the lanthanoid elements with increasing atomic number, as we move from lanthanum \( (La) \) to lutetium \( (Lu) \), is called lanthanoid contraction.

Step 2: Cause.
As the atomic number increases across the series, the added electrons enter the inner \( 4f \) subshell. The \( 4f \) electrons shield the outer electrons from the nucleus very poorly (imperfect shielding). Therefore the effective nuclear charge felt by the outer electrons increases steadily, pulling them closer and causing a gradual decrease in size.

Step 3: Link size to basic strength.
As the ionic size of \( Ln^{3+} \) decreases from \( La^{3+} \) to \( Lu^{3+} \), the covalent character of the \( Ln-OH \) bond increases and the basic strength of the oxides and hydroxides decreases. Thus \( La(OH)_3 \) is the most basic and \( Lu(OH)_3 \) is the least basic.

Step 4: Increasing order of basic nature.
Hydroxides: \[ Lu(OH)_3 < Yb(OH)_3 < \ldots < Ce(OH)_3 < La(OH)_3 \]
Oxides: \[ Lu_2O_3 < Yb_2O_3 < \ldots < Ce_2O_3 < La_2O_3 \]
That is, basic character increases from Lu to La (with decreasing atomic number).