III) This is because of the comparable energies of 5f, 6d and 7s orbitals so all of them can participate.

" /> III) This is because of the comparable energies of 5f, 6d and 7s orbitals so all of them can participate.

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Question:

How would you account for the following: (i) Many of the transition elements are known to form interstitial compounds. (ii) The metallic radii of the third (5d) series of transition metals are virtually the same as those of the corresponding group members of the seconds (4d) series. (iii) Lanthanoids form primarily +3 ions, while the actinoids usually have higher oxidation states in their compounds, +4 or even +6 being typical.

Solution:

I) Transition metals exist in hcc geometry in which there is vacant spaces. In these vacant spaces the elements like C, H, O form compounds called interstitial compounds.
II) Due to lanthanide contraction, the metallic radii of the third (5d) series of transition metals are virtually the same as those of the corresponding group members of the seconds (4d) series. "The steady decrease in the values of the radii of M3+ cations in the lanthanide series is called as Lanthanide contraction."
III) This is because of the comparable energies of 5f, 6d and 7s orbitals so all of them can participate.