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Question
What are the consequences of lanthanoid contraction?
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Solution
(1) Resemblance of second and third transition series - This significantly affects the relative properties of the elements coming before and after the lanthanides in the periodic table. It is clear from the following table that there is a regular increase in size from Sc to Y and Y to La.
Table Atomic radii of d-block elements (in pm)
| Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn |
| 164 | 147 | 135 | 129 | 137 | 126 | 125 | 125 | 128 | 137 |
| Y | Zr | Nb | Mo | Tc | Ru | Rh | Pd | Ag | Cd |
| 180 | 160 | 146 | 139 | 136 | 134 | 134 | 137 | 144 | 154 |
| La* | Hf | Ta | W | Re | Os | Ir | Pt | Au | Hg |
| 187 | 158 | 146 | 139 | 137 | 135 | 136 | 138 | 144 | 157 |
Similarly, we can expect a general increase in size in other groups, though after the lanthanoids, the increase in radii from the second to the third transition series is almost negligible.
\[\ce{Ti -> Zr -> Hf}\]
\[\ce{V -> Nb -> Ta}\]
Pairs of primary elements, such as Zr – Hf, Nb – Ta, Mo – W, etc., have similar (almost) sizes, and the properties of these elements are also similar. Hence, as a result of lanthanoid contraction, the elements of the second and third transition series have greater similarity with each other than the elements of the first and second transition series.
(2) Similarity among lanthanides - Due to very small variations in the radii of lanthanoids, their chemical properties are almost similar. Hence, it is very difficult to separate the elements in a pure state. Modern methods based on repeated fractional crystallization or ion exchange techniques separate them based on very small differences in the sizes of their trivalent ions. These methods separate them based on very small differences in the properties of the elements, such as solubility, complex ion formation, hydration etc.
(3) Basicity differences - Due to lanthanoid contraction, the size of lanthanoid ions decreases regularly with an increase in atomic number. As a result of a decrease in size, the covalent character between lanthanoid ions and OH ions increases from La3+ to Lu3+. Hence, the basic strength of hydroxides decreases with an increase in atomic number. Thus, La(OH)3 is the most basic, while Lu(OH)3 is the least basic.
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