Question

(i) Scandium $\left( {{\text{Z}} = 21} \right)$ is a transition element but Zinc $\left( {{\text{Z}} = 30} \right)$ is not. Explain.
(ii) Calculate the equivalent weight of ${\text{KMn}}{{\text{O}}_4}$ in acidic medium.
(iii) What do you mean by Lanthanide contraction?

Answer 1

Transition elements are the elements which have vacant d orbital. Manganese is one of the transition elements. Its equivalent weight depends upon the medium it reacts. The elements which have partly filled $\left( {{\text{n}} - 2} \right){\text{f}}$ orbitals are f block elements. Since their inner subshells are successively filled with electrons, these electrons are called inner transition elements.

Complete step by step answer:
(i) It is given that the atomic number of Scandium is $21$ while that of Zinc is $30$.
Transition elements are the elements which show properties which are transitional between s and p block elements.
The outer electronic configuration of Scandium is $3{{\text{d}}^1}4{{\text{s}}^2}$ while that of zinc is $3{{\text{d}}^{10}}4{{\text{s}}^2}$. This indicates that Scandium has incompletely filled d orbitals in its ground state. Zinc does not have incompletely filled d orbitals in its ground state or any of its oxidation state.
(ii) In ${\text{KMn}}{{\text{O}}_4}$, the oxidation state of ${\text{Mn}}$ is $+ 7$. In an acidic medium, the oxidation state of ${\text{Mn}}$ changes from $+ 7$ to $+ 2$ state. So there will be a gain of $5{{\text{e}}^ - }$.
Therefore ${\text{n}} = 5$
Molar mass of ${\text{KMn}}{{\text{O}}_4},{{\text{M}}_{{\text{KMn}}{{\text{O}}_4}}}$ $= 158{\text{gmo}}{{\text{l}}^{ - 1}}$
Equivalent weight of ${\text{KMn}}{{\text{O}}_4},{{\text{W}}_{{\text{eq}}}}$ $= \dfrac{{{{\text{M}}_{{\text{KMn}}{{\text{O}}_4}}}}}{{\text{n}}}$
Equivalent weight of ${\text{KMn}}{{\text{O}}_4}$ $= \dfrac{{158}}{5} = 31.6{\text{g}}$
(iii) There are two series of inner transition elements-Lanthanide series and Actinide series. Lanthanides are the fourteen elements after the element Lanthanum. The shielding effect of $4{\text{f}}$ is very small than d orbitals as $4{\text{f}}$ orbital is much diffuse in nature. On moving across the lanthanide series, ionic radii decrease as we increase the atomic number.

Additional information:
On moving across the lanthanide series, the nuclear charge increases by $+ 1$ at each step and the addition of extra electrons takes place.

Note:
Zinc does not show any of the properties of transition elements like paramagnetism, formation of colored ions etc. Post Lanthanide elements like ${\text{Zr}}$ and ${\text{Hf}}$ show similar ionic radii though they belong to the first and second transition series. Lanthanides show similar physical and chemical properties. So they are hard to separate.