Question

How would you account for the following:
(i) Of the ${ d }^{ 4 }$ species, ${ Cr }^{ +2 }$ is strongly reduced while manganese (III) is strongly oxidizing.
(ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents, it is easily oxidized.
(iii) The ${ d }^{ 1 }$ configuration is very unstable in ions.

Answer 1

Hint: d-block elements also known as transition elements. Transition elements are elements which have varying valency because their d and f block elements are empty.

Complete step by step solution:
(i) The electronic configuration of ${ Cr }{ =[Ar]3d }^{ 5 }{ 4s }^{ 1 }$
${ Cr }^{ +2 }$ = ${ Cr }{ =[Ar]3d }^{ 4 }$
 ${ Cr }^{ +2 }$ is strongly reduced in nature. It has a ${ d }^{ 4 }$ electronic configuration. While ${ Cr }^{ +2 }$ acts as a reducing agent, it gets oxidized to ${ Cr }^{ +3 }$ (${ d }^{ 3 }$ electronic configuration). This ${ d }^{ 3 }$ configuration can be written as ${ t }_{ 2g }^{ 3 }$ configuration, which is a more stable configuration.
In the case of ${ Mn }^{ +3 }$ (${ d }^{ 4 }$ electronic configuration), it acts as an oxidizing agent and gets reduced to ${ Mn }^{ +2 }$. This has an exactly half-filled orbital which is very stable.

(ii) The electronic configuration of Co(II) and Co(III) are given below;
 ${ Co(II)=[Ar]{ 4s }^{ 0 }{ 3d }^{ 7 } }$
${ Co(III)=[Ar]{ 4s }^{ 0 }{ 3d }^{ 6 } }$
Cobalt (II) is stable in aqueous solution. However, in the presence of strong field complexing agents, it is oxidized to Co (III). In spite of the fact that the third ionization enthalpy for cobalt is high, the higher measure of CFSE (Crystal Field Stabilization Energy) released in the presence of strong-field ligands overcomes the ionization energy. In Co(III) species, six lone pairs of electrons from ligands are obliged by ${ sp }^{ 3 }{ d }^{ 2 }$ hybridization which is not possible in Co(II).

(iii) The ions in ${ d }^{ 1 }$ configuration tend to release one more electron to get into a stable ${ d }^{ 0 }$ configuration. Also, the lattice or hydration energy is more sufficient to remove the only electron present in the d-orbital of these ions. Therefore, they act as reducing agents.

Note: The possibility to make a mistake is that in (i) the ${ E }^{ \circ }$ value of ${ Cr }^{ +3 }{ /Cr }^{ +2 }$ is negative whereas ${ Mn }^{ +3 }{ /Mn }^{ +2 }$ is positive. Cr is not acting as an oxidizing agent while ${ Cr }^{ +2 }$ can lose an electron to form ${ Cr }^{ +3 }$ and it acts as a reducing agent.