Lanthanoid contraction definition - In Chemistry, lanthanoid contraction, also called lanthanide contraction, occurs as the atomic size or the ionic radii of the tripositive lanthanide ions steadily decrease from La to Lu because of the electrons entering the inner (n-2) f orbitals and the increasing nuclear charge. This particular gradual decrease in the size with an increasing atomic number is referred to as lanthanide contraction.
Lanthanide contraction happens to all the 14 elements that are present in the lanthanide series. Cerium(Ce), Praseodymium(Pr), Neodymium(Nd), Promethium(Pm), Samarium(Sm), Europium(Eu), Gadolinium(Gd), Terbium(Tb), Dysprosium(Dy), Holmium(Ho), Erbium(Er), Thulium(Tm), Ytterbium(Yb), and Lutetium(Lu) are the total elements that are included in the series. In accordance with the lanthanide contraction of the mentioned elements, as the atomic number increases, the atomic radius of these elements decreases. This is very easy to compare by taking into consideration Ce and Nd in the periodic table. Ce has the atomic number 58 and Nd has the atomic number 60. Now, look at the graph below to find out the atomic radius of the following elements mentioned above.
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About Lanthanide Contraction
Lanthanide contraction is the steady decrease in the size of the ions and atoms of the rare earth elements with increasing atomic numbers from lanthanum (atomic number 57) to lutetium (with an atomic number 71). For every consecutive atom, the nuclear charge can be more positive by a single unit, accompanied by the corresponding increase in the electron count present in the 4f orbitals surrounding the nucleus.
The 4f electrons imperfectly protect each other from the increased positive charge of the nucleus, resulting in a steady rise in the effective nuclear charge attracting every electron as the lanthanide elements progress, resulting in successive decreases in ionic and atomic radii.
Consequences of Lanthanide Contraction
The following points will depict the effect of lanthanide contraction more clearly.:
The size of the atom of the third transition series is approximately similar to that of the atom of the second transition series. For example, the radius of Zr = radius of Hf and the radius of Nb = radius of Ta, and so on.
As there is only a small change in the ionic radii of the Lanthanides, their chemical properties are the same. This makes the element's separation in the pure state difficult.
As the size of the lanthanides decreases from the elements La to Lu, the covalent character of the hydroxides increases, and thus, their basic strength decreases. Therefore, Lu(OH)₃ is said to be least basic, and La (OH)₃ is said to be more basic.
Due to the smaller size and higher nuclear charge, the tendency to produce coordinate. Complexity increases from the element La³⁺to Lu³⁺.
It increases from the elements La to Lu.
Electron's attraction by the nuclear charge is higher, and thus the Ionization energy of the 5d elements is much larger compared to 4d and 3d. In the 5d series, the total elements except Pt and Au contain a filled s-shell.
Elements from Hafnium to rhenium contain similar Ionization energy, and after that, the Ionization energy increases with the number of shared d-electrons such that Gold and Iridium hold the maximum Ionization Energy.
Mercury - the Liquid Metal
At room temperature, mercury is the only metal that remains in its liquid form. The nucleus pulls the 6s valence electrons very close together (due to lanthanide contraction), making the outer s-electrons less involved in metallic bonding.
Lanthanides with a 3+ oxidation state have a higher charge to radius ratio and hence a lower charge to radius ratio. As compared to d-block elements, this decreases the ability of lanthanides to form complexes. Still they form complexes with strong chelating agents like EDTA, β-diketones, oxime, and so on. They do not form Pπ-complexes.
Cause of Lanthanide Contraction
The effect of lanthanide contraction results from the poor shielding of nuclear charge (with the attractive nuclear force on electrons) by 4f electrons; the 6s electrons can be drawn towards the nucleus, hence resulting in the smaller atomic radius.
In the case of single-electron atoms, the average separation of an electron from the nucleus is defined by the subshell it belongs to and decreases with an increased charge on the nucleus, where this, in turn, leads to the decrease in atomic radius. Whereas, in the case of multi-electron atoms, the decrease in the radius brought about by an increase in nuclear charge is partially offset by the increasing electrostatic repulsion among the electrons.
A "shielding effect" operates, in which existing electrons shield the outer electrons from the nuclear charge by causing them to undergo less effective charge on the nucleus as more electrons are applied to the outer shells. The inner electrons' shielding effect decreases in the following order: s > p > d > f.
In general, as a specific subshell is filled in a period, the atomic radius decreases. This particular effect is specifically pronounced in the case of lanthanides, as the 4f subshell that is filled across these elements is not more effective at shielding the outer shell (n=5 and n=6) electrons. Therefore, the shielding effect can be less able to counter the decrease in radius caused by an increasing nuclear charge. This leads to the "lanthanide contraction". And, the ionic radius drops from a range of 103 pm for lanthanum (III) to 86.1 pm for lutetium (III).
The relativistic effects have been blamed for up to 10% of the lanthanide contraction.
Shielding Effect on the Atomic Radii
The lanthanide is a result of the poor shielding effect of the 4f electrons. The shielding effect is a phenomenon where the inner shell electrons shield the outer shell electrons from getting attracted by the charge of the nucleus. Therefore, if the shielding effect is not good, then the outer shell electron is attracted by the positively charged nucleus towards itself, and thus, the radii of that atom decrease with the increase in the positive charge or atomic number. Thus, s-shell has a greater shielding effect as compared to the other subshells while f-shell has the least shielding effect. The d and the p shell are in between where the p-shell has more shielding effect than the d-shell.
This can be seen if the elements with the f-subshell are compared with the d-block elements that do not have any f-subshell in their atom. Such elements are Pd and Pt. The Pd element has 4d electrons whereas Pt has 5d and 4f electrons. These two elements, therefore, have roughly had the same electronic radii. This is due to the shielding effect and the lanthanide contraction. It is because we expect the Pt to have a larger radius as compared to Pd since it has more number of electrons and larger number of protons but not because of the poor shielding effect of the f-subshell present. Hence, this will increase the nucleus pull of the electrons towards the nucleus.
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