Question

Answer the following questions:
Name two elements of 3d series for which the third ionisation enthalpies are quite high.
Out of KMnO$_4$ and K$_2$MnO$_4$, which one is paramagnetic and why?
Write any one consequence of lanthanoid contraction.
How do you prepare potassium manganate from pyrolusite ore?
Why is the ability of oxygen more than fluorine to stabilise higher oxidation states of transition metals?

Hint:

When studying transition metals, pay attention to their oxidation states and how the ligands around them influence their stability and reactivity.

Answer 1

(i) Two elements of the 3d series for which the third ionization enthalpies are quite high are Zinc (Zn) and Copper (Cu). Both have completely filled d-orbitals, making it more difficult to remove the third electron.
(ii) Out of KMnO$_4$ and K$_2$MnO$_4$, KMnO$_4$ is paramagnetic. This is because in KMnO$_4$, manganese exists in the +7 oxidation state, which leaves it with unpaired electrons in the d-orbitals, making it paramagnetic. K$_2$MnO$_4$, on the other hand, has manganese in the +6 oxidation state, which does not result in unpaired electrons.
(iii) One consequence of lanthanoid contraction is the decrease in ionic radii as we move across the lanthanide series. This causes a similar size among elements, leading to similarities in chemical behavior.
(iv) Potassium manganate can be prepared from pyrolusite ore by reacting it with a suitable oxidizing agent like potassium hydroxide (KOH) under heating. The reaction is as follows:
$\text{MnO}_2 + 4\text{KOH} \longrightarrow \text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O}$
(v) Oxygen stabilizes higher oxidation states of transition metals better than fluorine because oxygen has a smaller size and a higher electronegativity, which helps in accepting electrons from the transition metal. In contrast, fluorine, being more electronegative, tends to form more stable bonds in lower oxidation states.