Question

Four successive members of the first-row transition elements are listed below with atomic numbers. Which one of them is expected to have the highest $\text{E}_{{{\text{M}}^{+3}}\text{/}{{\text{M}}^{+2}}}^{{}^\circ }$ value?
A. Mn (Z = 25)
B. Fe (Z = 26)
C. Co (Z = 27)
D. Cr (Z = 24)

Answer 1

For this problem, we have to determine the reduction potential of each ion and this can be determined with the help of the ionisation energy. And then we can choose the correct answer among the given options.

Complete step by step answer:
- In the given question, we have to choose the correct element which will have the highest value of the reduction potential or $\text{E}_{{{\text{M}}^{+3}}\text{/}{{\text{M}}^{+2}}}^{{}^\circ }$.
- As we know that ionisation potential of an element is the minimum amount of energy that is required to remove the outermost loosely bound electron.
- If the outermost electron will be stable, more ionisation energy will be required.
- So, Manganese has the electronic configuration $\text{(Ar)3}{{\text{d}}^{5}}\text{ 4}{{\text{s}}^{2}}$ and its ion will have the electronic configuration:
$\text{M}{{\text{n}}^{3+}}\text{ = (Ar)3}{{\text{d}}^{4}}$; $\text{M}{{\text{n}}^{2+}}\text{ = (Ar)3}{{\text{d}}^{5}}$
- So, as we can see that the ionisation potential of +2 state is more because it has to break its half-filled orbital stability.
- So, to reduce from +3 to +2 it will need less reduction potential and have a lower value of $\text{E}_{{{\text{M}}^{+3}}\text{/}{{\text{M}}^{+2}}}^{{}^\circ }$ or $\text{E}_{\text{M}{{\text{n}}^{+3}}\text{/M}{{\text{n}}^{+2}}}^{{}^\circ }$ i.e. 1.57V.

- Now, iron has the electronic configuration $\text{(Ar)3}{{\text{d}}^{6}}\text{ 4}{{\text{s}}^{2}}$ and its ion will have the electronic configuration:
$\text{F}{{\text{e}}^{3+}}\text{ = (Ar)3}{{\text{d}}^{5}}$; $\text{F}{{\text{e}}^{2+}}\text{ = (Ar)3}{{\text{d}}^{6}}$
- So, as we can see that the ionisation potential of +2 state is more because it has to remove the electrons from the full-filled s-orbital.
- So, to reduce from +3 to +2 it will need less reduction potential and have a lower value of $\text{E}_{{{\text{M}}^{+3}}\text{/}{{\text{M}}^{+2}}}^{{}^\circ }$ or $\text{E}_{\text{F}{{\text{e}}^{+3}}\text{/F}{{\text{e}}^{+2}}}^{{}^\circ }$ i.e. 0.77V.

- Now, cobalt has the electronic configuration $\text{(Ar)3}{{\text{d}}^{7}}\text{ 4}{{\text{s}}^{2}}$ and its ion will have the electronic configuration:
$\text{C}{{\text{o}}^{3+}}\text{ = (Ar)3}{{\text{d}}^{6}}$; $\text{C}{{\text{o}}^{2+}}\text{ = (Ar)3}{{\text{d}}^{7}}$
- So, as we can see that the ionisation potential of +3 state will be less because it has to remove the electron which is neither half-filled or full-filled.
- So, to reduce from +3 to +2 it will need the maximum reduction potential and have the highest value of $\text{E}_{{{\text{M}}^{+3}}\text{/}{{\text{M}}^{+2}}}^{{}^\circ }$ or $\text{E}_{\text{C}{{\text{o}}^{+3}}\text{/C}{{\text{o}}^{+2}}}^{{}^\circ }$ i.e. 1.97 because +3 state is more stable than the +2 state.

- Now, chromium has the electronic configuration $\text{(Ar)3}{{\text{d}}^{5}}\text{ 4}{{\text{s}}^{1}}$ and its ion will have the electronic configuration:
$\text{C}{{\text{r}}^{3+}}\text{ = (Ar)3}{{\text{d}}^{3}}$; $\text{C}{{\text{r}}^{2+}}\text{ = (Ar)3}{{\text{d}}^{4}}$
- So, as we can see that the ionisation potential of +2 state will be more because it has a stable configuration than +3.
- So, to reduce from +3 to +2 it will need less reduction potential and have less value of $\text{E}_{{{\text{M}}^{+3}}\text{/}{{\text{M}}^{+2}}}^{{}^\circ }$ or $\text{E}_{\text{C}{{\text{r}}^{+3}}\text{/C}{{\text{r}}^{+2}}}^{{}^\circ }$ i.e. -0.41V.
So, the correct answer is “Option C”.

Note: The successive electrode potential is the value of the electrode potential when first an electron is removed and then the second electron is removed and so on.