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The atomic and ionic radii (${{\text{M}}^{{\text{3 + }}}}{\text{ions}}$) of lanthanides elements decreases with increase in atomic number. This effect is called:A.Lanthanide contractionB.f-f transitionC.Actinoid contractionD.Shielding effect

Last updated date: 10th Aug 2024
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Answer
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Hint:
The atomic radii and ionic radii of lanthanides depend upon the effective nuclear charge. With increasing the effective nuclear charge radii of atoms or ions decreases and vice-versa.

Complete step by step answer:
Lanthanides are f-block elements the general configuration of the lanthanides is$\left[ {Xe} \right]4{f^{1 - 14}}5{d^{0 - 1}}6{s^2}$. Here the last electrons enter to the inner f-orbital. According to the periodic properties from left to right of a period size of the atoms decreases as the electrons enter the same shell. This can be explained on the basis of effective nuclear charge. With increasing the atomic number of the atoms, the effective nuclear charge increases and vice-versa.
The Increase of nuclear charge can be explained on the basis of the screening effect of the orbitals. The orbitals with higher screening power, screen the nuclear charge more than the lower screening power orbitals. The order of screening effect of the orbitals $s > p > d > f$.
For lanthanide elements the last electrons enter to the f-orbitals. Due to the lower screening power of the f-orbital, with increasing the number of electrons of f-orbitals nuclear charge increases, which results the outermost$6{s^2}$ electrons experiences more attraction force by the nucleus and the radii of that corresponding atom or ion(${{\text{M}}^{{\text{ + 3}}}}$) get decreased. This overall effect is called lanthanide contraction effect.

The correct answer is A. lanthanide contraction.

Note:
For lanthanide electrons entering to the inner f orbital which increases the effective nuclear charge due to the lower screening power of f-orbital electrons.