Question

(i) With reference to structural variability and chemical reactivity, write the difference between lanthanoids and actinoids.
(ii) Name a member of the lanthanoid series which is well known to exhibit $4+$ oxidation state.
(iii) Complete the following equation:
  \[MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to \]
(iv) Out of $M{{n}^{+3}}$ and $C{{r}^{+3}}$, which is more paramagnetic and why?
(atomic number : \[Mn=25,Cr=24\])

Answer 1

Hint:The lanthanoids and actinoids are known to have the oxidation states ranging from three to seven. The oxidation state of these elements are positive as they readily lose electrons in order to form their respective ions which are more stable in nature.

Complete step-by-step answer: We will be answering the parts and subparts of the question in a stepwise manner.
(i) In the first part we are asked to differentiate between the lanthanoids and actinoids which are present in the lower part of the periodic table separately which is also called the f-block. The difference in both these categories on the basis of their structural variability and the chemical reactivity is listed below.
The lanthanides are non-radioactive elements whereas the actinides are radioactive. Which means they have a heavier nucleus which is unstable in nature, and so it breaks down in small fragments until the nucleus becomes stable.
The next differentiation is based on the ionisation enthalpy of actinoids and lanthanoids. The ionisation energy of the actinides are much less than the ionisation enthalpy of the lanthanides. We know that the action tends to acquire $+3$ oxidation state which is most common, and $+7$ oxidation state too, whereas the lanthanoids show the oxidation states from $+3$ to $+7$ with no specific preferences.
Actinides are known to have more reactivity towards the complexes having higher magnetic properties as compared to the lanthanides.
(ii) The lanthanide which is known to exhibit an oxidation state of $+4$ is known as the Cerium.
(iii) Now, we are supposed to complete the equation,
\[MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to \]
There are eight protons in the reactant side so the product should have half the number of water molecules to balance out the reaction. So, four water molecules would contain four oxygens which balances the number of oxygen on the reactant side too. So, the equation becomes,
\[MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O\]
Which is the required reaction.
(iv) $M{{n}^{2+}}$ is more paramagnetic than $C{{r}^{3+}}$, as it has more number of unpaired electrons and we know that more the number of unpaired electrons more paramagnetic the substance would be.

Note:The lanthanoids and actinoids differ from each other in terms of radioactivity, and ionisation energies along with the reactivity towards magnetic compounds.
The paramagnetic nature of any substance is established because of the availability of the unpaired electrons in their orbitals.