Question

Fourteen elements following Lanthanum are called Lanthanides:
A. What is Lanthanoid contraction? Give reason for it.
B.${\rm{KMn}}{{\rm{O}}_{\rm{4}}}$ is a purple coloured crystal and it acts as an oxidant. How will you prepare ${\rm{KMn}}{{\rm{O}}_{\rm{4}}}$ from ${\rm{Mn}}{{\rm{O}}_{\rm{2}}}$?

Answer 1

The atomic size in Lanthanide is reduced and we know that as Lanthanoid contraction. We can prepare ${\rm{KMn}}{{\rm{O}}_{\rm{4}}}$ from ${\rm{Mn}}{{\rm{O}}_{\rm{2}}}$by using oxidizing the later in different ways.

Step by step answer: (A) We know that we have many general trends across a period and in a group that results from the arrangement of elements in our modern periodic table. For example, atomic radius or electropositive nature.
For atomic radius, we have a general trend that it increases down a group and decreases in a period on moving towards right. This can be understood by taking into account the attraction between the added electrons and nucleus, repulsion by other electrons and the shielding effect of inner electrons as well.
We can say that the atomic radius will be larger if repulsion is more
On the other hand, atomic radius will be smaller if nuclear attraction is more
Now, we know that the inner electrons shield outer ones from nuclear charge and this attraction is not very strong. So, strong shielding would result in lesser attraction and larger size.
Now, we have to consider that the extent of shielding is dependent on the nature of the electrons. It is maximum for $s -$ electrons, lesser for $p -$ electrons and so on. Let’s consider the given elements. From the general electronic configuration of Lanthanoids that can be written to be $4{f^n}$ where $n$ can vary from $1$ to $14$; we can infer that for these elements, electrons are being added up in the $4f -$ orbital and thus they belong to the $f -$ block of the periodic table. Now, as we know that the shielding by $f -$ electrons is lesser than by $d -$ electrons. So, we can say that this results in a quite regular decrease in atomic size of Lanthanoids with an increase in atomic number and this is what we call, Lanthanoid contraction.
(B) We know that potassium permanganate is very significant as an oxidant and has an intense color.
Here, we will look at the preparation method of potassium permanganate by using ${\rm{Mn}}{{\rm{O}}_{\rm{2}}}$ as starting material. As it is evident that in ${\rm{Mn}}{{\rm{O}}_{\rm{2}}}$, ${\rm{Mn}}$ is in $+ 4$ oxidation state and we need it to be in $+ 7$ in ${\rm{KMn}}{{\rm{O}}_4}$. So, we will use oxidation of ${\rm{Mn}}{{\rm{O}}_2}$ to do the same. We can prepare ${\rm{KMn}}{{\rm{O}}_{\rm{4}}}$ from ${\rm{Mn}}{{\rm{O}}_{\rm{2}}}$by oxidative fusion of the later followed by disproportion reaction. For fusion, we can use an alkali metal hydroxide such as ${\rm{KOH}}$ with oxidizing ${\rm{KN}}{{\rm{O}}_{\rm{3}}}$. We can write the chemical equation as follows:
${\rm{2Mn}}{{\rm{O}}_{\rm{2}}} + {\rm{4KOH}} + {\rm{ }}{{\rm{O}}_{\rm{2}}} \to {\rm{2}}{{\rm{K}}_{\rm{2}}}{\rm{Mn}}{{\rm{O}}_{\rm{4}}} + {\rm{2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}$
As we can see the oxidative fusion of ${\rm{Mn}}{{\rm{O}}_2}$results in the formation of potassium manganate which has ${\rm{Mn}}$ in $+ 6$ oxidation state.
Finally, we will use the disproportion reaction of ${{\rm{K}}_{\rm{2}}}{\rm{Mn}}{{\rm{O}}_{\rm{4}}}$ by using a neutral or acidic medium. We can write the chemical equation as follows:
${\rm{3MnO}}_4^{2 - } + {\rm{4}}{{\rm{H}}^ + } \to 2{\rm{MnO}}_4^ - + {\rm{Mn}}{{\rm{O}}_{\rm{2}}} + {\rm{2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}$

Note: We have to keep in mind and provide the reaction conditions as well that includes the medium of the reaction being acidic, basic or neutral.The oxidation and reduction reactions involved must be observed carefully that which atom loses the oxidation number.