Question

(a) How do you prepare:
i. ${K_2}Mn{O_4}$from $Mn{O_2}$
ii. $N{a_2}C{r_2}{O_7}$from$N{a_2}Cr{O_4}$
(b) Account for the following:
i. $M{n^{2 + }}$is more stable than $F{e^{2 + }}$towards oxidation to $+ 3$ state.
ii. The enthalpy of atomization is lowest for Zn in 3d series of the transition elements.
Actinoid elements show a wide range of oxidation states.

Answer 1

We all know that the d-block elements are the one which has incompletely filled d-orbital in its ground state. One of the notable features of transition elements is that they possess a great variety in their oxidation states and may exhibit all oxidation states from $+ 2$ to$+ 7$.

Complete step by step answer:
a. As we have already discussed, the d-block elements also called the transition elements consist of incompletely filled d-orbitals and show a great variety in their oxidation station. Most of them are sufficiently electropositive to dissolve in mineral acid.
Now let us discuss their oxides which are generally formed when these metals react with oxygen at high temperature. All these metals except Scandium form the metal oxides. So we can easily prepare their oxides using oxygen.
i. Let us discuss the preparation of ${K_2}Mn{O_4}$ from $Mn{O_2}$.
$KMn{O_4}$ is prepared by the reaction of $Mn{O_2}$ with alkali metal hydroxides and any oxidising agent such as $KN{O_3}$. This reaction produces dark green coloured ${K_2}Mn{O_4}$ which disproportionates into a neutral or acidic solution to form permanganate. We can understand it in a better way with the help of reaction:
$2Mn{O_2} + 4KOH + {O_2} \to 2{K_2}Mn{O_4} + 2{H_2}O$
ii. Similarly, we can prepare sodium dichromate ($N{a_2}C{r_2}{O_7}$) from sodium chromate ($N{a_2}Cr{O_4}$). In this preparation, $N{a_2}Cr{O_4}$which is yellow in colour is acidified with ${H_2}S{O_4}$which results in solution using which we can crystallise the orange coloured $N{a_2}C{r_2}{O_7}$ crystals. We can understand this with the help of reaction:
$2N{a_2}Cr{O_4} + 2{H^ + } \to N{a_2}C{r_2}{O_7} + 2N{a^ + } + {H_2}O$

b. Now we can discuss some more properties of transition elements.
(i) We know that $M{n^{2 + }}$possesses the electronic configuration as $[Ar]3{d^5}$, so it is clear that its 3d-orbitals is half filled and thus more stable due to which its third ionization enthalpy is much higher and thus it will be difficult to remove the third electron whereas the electronic configuration of $F{e^{2 + }}$is $[Ar]3{d^6}$which is neither half-filled or fully filled and thus unstable so it can loss an electron to attain half- filled configuration for stability and now it will be easy to remove the third electron.
(ii) We will now discuss about the enthalpy of atomization that depends on number of unpaired electrons and as we all know that Zinc is the most stable element of the group as it has an atomic number 30 and the d-orbital is fully filled and hence stable due to which it does not contain any unpaired electron. Therefore, the interatomic bonding in Zinc is weak and thus its enthalpy of atomization is lowest as compared with other members.
(iii) Now comes the Actinides that includes 14 elements and they all show a wide range of oxidation states exhibiting the oxidation states from $+ 3$ to $+ 6$. This can be explained by the very small gap of energy between their subshells including 5d, 6d and 7s. Among these, $+ 3$ oxidation state is most stable.

Note:
It is also interesting to know that when an electron from a lower energy gets excited to higher energy d-orbital, they absorb light in the visible that is why these elements impart colour of different variety. These elements can also make complex compounds due to their smaller size and availability of d-orbital.