Review the key concepts, formulae, and examples before starting your quiz.
🔑Concepts
Lanthanoid Contraction is the steady decrease in the atomic and ionic radii (specifically for ions) of lanthanoid elements with an increase in atomic number from Lanthanum () to Lutetium ().
The primary cause is the poor shielding effect of electrons. As the atomic number increases, the nuclear charge increases by unity at each step, and the new electron enters the subshell. Due to the diffused shape of orbitals, they shield the outer electrons from the nucleus very effectively, leading to an increase in the effective nuclear charge ().
One major consequence is the similarity in atomic and ionic radii of the second () and third () transition series. For example, Zirconium () and Hafnium () have almost identical sizes, making their separation difficult.
The basic strength of lanthanoid hydroxides decreases from to . As the size of the ion decreases, the covalent character of the bond increases (according to Fajan's Rule), leading to a decrease in basicity.
The separation of lanthanoid elements in their pure state is difficult due to their very similar chemical properties resulting from the contraction, though it can be achieved through ion-exchange methods.
📐Formulae
💡Examples
Problem 1:
Explain why (Atomic radius pm) and (Atomic radius pm) have almost identical sizes despite being in the period below .
Solution:
This is due to the phenomenon of Lanthanoid Contraction.
Explanation:
Normally, the size increases down a group ( to ). However, before () in the series, there are 14 lanthanoid elements ( to ) where the shell is being filled. The poor shielding of these 14 electrons results in a contraction in size that almost exactly cancels the expected increase in size due to the addition of a new energy shell. Thus, and have nearly identical radii.
Problem 2:
Arrange the following in increasing order of basic strength: , , .
Solution:
Explanation:
According to lanthanoid contraction, the size of the ion decreases from to . As the size of the cation decreases, its polarizing power increases, which increases the covalent character of the bond. A more covalent bond releases ions less easily in solution, making it a weaker base.