Question

a) What will be the complete equations of the following?
$C{r_2}O_7^{2 - } + 2O{H^ - } \to$
$MnO_4^ - + 4{H^ + } + 3{e^ - } \to$
b) Account for the following:
(i) $Zn$ is not considered as a transition element.
ii) Transition metals form a large number of complexes.
iii) The ${E^\circ }$ value for the $M{n^{3 + }}$/$M{n^{2 + }}$ couple is much more positive than that for a $C{r^{3 + }}$/$C{r^{2 + }}$ couple.

Answer 1

As we know that transition elements possess incompletely filled d-orbitals in its ground state or in their most common oxidation states but zinc, cadmium and mercury contain fully filled d-orbitals.

Complete step-by-step answer:
(a). As we know that in an aqueous solution the dichromate ions will be converted to chromate ions and a water molecule will be released in the reaction. Hence the balanced chemical equation for the reaction will be:
(i) $C{r_2}O_7^{2 - } + 2O{H^ - } \to 2CrO_4^{2 - } + {H_2}O$
In the second reaction, when the medium is neutral the permanganate ions are converted to manganese dioxide and the oxidation state is reduced to $+ 4$ from $+ 7$. Hence the complete balanced equation will be as follows:
(ii) $MnO_4^ - + 4{H^ + } + 3{e^ - } \to Mn{O_2} + 2{H_2}O$

(b). (i) Transition element characteristics:
-It should have partially filled d orbitals.
-It shows variable oxidation states.

The valence configuration should have at least an unpaired electron. But zinc has an electronic configuration $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^2}$ which suggests that its d-orbital is fully filled in ground state so it has no tendency to lose and electron or gain an electron and thus it is stable. Therefore, it is not considered as a transition element. It has a fixed oxidation state of $+ 2$ and has no unpaired electron.

(ii). Transition elements have large number of vacant d-orbitals along with the electronic configuration as $n{s^2}\left( {n - 1} \right){d^{1 - 10}}$ and hence have empty d orbitals to accommodate or accept electrons coming from the ligands that are attached to them. the ions formed by these metals have small size and high ionic charges. As a result of this small size, high effective nuclear charge, variable oxidation states of the transition metal ion make the complex compounds more stable.

(iii) Electronic configuration of
$M{n^{3 + }} \to \left[ {Ar} \right]4{s^0}3{d^4} \to 3{d^5}$
$C{r^{3 + }} \to \left[ {Ar} \right]4{s^0}3{d^3} \to 3{d^4}$
$M{n^{2 + }}$ion is stable due to extra stability of its half-filled d-orbital of metal ion whereas the $C{r^{3 + }}$ is less stable due to the four electrons in d-orbital so it is neither completely filled nor half-filled so it has a tendency to gain electron and form the much more stable ion.

Hence the ${E^\circ }$ value for the $M{n^{3 + }}$/$M{n^{2 + }}$ couple is much more positive than that for a $C{r^{3 + }}$/$C{r^{2 + }}$ couple.

Note: One of the notable features of the transition element is the great variety in its oxidation states and small size which results in coloured compounds. They can form various coloured compounds in different oxidation states.