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) Zinc (Zn) and Copper (Cu) exhibit high third ionization enthalpies among 3d series elements. This is attributed to their completely filled d-orbitals, which require substantial energy to remove the third electron.
(ii) KMnO$_4$ is paramagnetic, unlike K$_2$MnO$_4$. In KMnO$_4$, manganese is in the +7 oxidation state, possessing unpaired d-electrons. Conversely, manganese in K$_2$MnO$_4$ is in the +6 oxidation state, which lacks unpaired d-electrons.
(iii) Lanthanoid contraction results in a progressive decrease in ionic radii across the lanthanide series. This leads to elements having similar sizes and consequently, comparable chemical properties.
(iv) Potassium manganate (K$_2$MnO$_4$) can be synthesized from pyrolusite ore (MnO$_2$) by reacting it with potassium hydroxide (KOH) under heating conditions, as shown by the reaction:
$\text{MnO}_2 + 4\text{KOH} \longrightarrow \text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O}$
(v) Oxygen is a more effective stabilizer of higher oxidation states in transition metals compared to fluorine. This is due to oxygen's smaller atomic size and higher electronegativity, facilitating electron acceptance from the transition metal. Fluorine, while more electronegative, tends to form more stable bonds in lower oxidation states.