Question

# The atomic numbers of vanadium, chromium, manganese and iron are respectively 23, 24, 25, 26. Which one of these may be expected to have the highest second ionization energy?A. $\text{V}$ B. $\text{Cr}$ C. $\text{Mn}$ D. $\text{Fe}$

Hint: Ionization energy is the amount of energy released when electrons are lost by an element. It is represented as $\text{A}\to {{\text{A}}^{+\text{n}}}+\text{n}{{\text{e}}^{-}}$. To find the order of second ionization energies, just write the configurations of the metal atom and then make them lose one electron, see the stability. As, more the stability of an ion, more will be its second ionization energy.

Let us first write the configuration of elements in their ground state only;
- Vanadium has 23 as its atomic number. Its configuration will be $\text{1}{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{6}}\text{3}{{\text{s}}^{2}}\text{3}{{\text{p}}^{6}}\text{4}{{\text{s}}^{2}}\text{3}{{\text{d}}^{3}}$ .

- Chromium has 24 as its atomic number. After the addition of one electron to ‘$\text{3d}$’, the configuration would have been $\text{1}{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{6}}\text{3}{{\text{s}}^{2}}\text{3}{{\text{p}}^{6}}\text{4}{{\text{s}}^{2}}\text{3}{{\text{d}}^{4}}$ but due to extra stability of $\text{3}{{\text{d}}^{5}}$ configuration, the electron from $4\text{s}$ subshell jumps to $3\text{d}$ subshell. Hence, the configuration will be $\text{1}{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{6}}\text{3}{{\text{s}}^{2}}\text{3}{{\text{p}}^{6}}\text{4}{{\text{s}}^{1}}\text{3}{{\text{d}}^{5}}$ .

- Manganese has 25 as its atomic number. Its configuration will be $\text{1}{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{6}}\text{3}{{\text{s}}^{2}}\text{3}{{\text{p}}^{6}}\text{4}{{\text{s}}^{2}}\text{3}{{\text{d}}^{5}}$ .

- Iron has 26 as its atomic number. Its configuration will be $\text{1}{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{6}}\text{3}{{\text{s}}^{2}}\text{3}{{\text{p}}^{6}}\text{4}{{\text{s}}^{2}}\text{3}{{\text{d}}^{6}}$ .
Now, after losing one electron (removal of electron will take place from $4\text{s}$ ).

-Vanadium will become ${{\text{V}}^{+1}}$ , whose configuration will be $\text{1}{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{6}}\text{3}{{\text{s}}^{2}}\text{3}{{\text{p}}^{6}}\text{4}{{\text{s}}^{1}}\text{3}{{\text{d}}^{3}}$ .
-Chromium will become $\text{C}{{\text{r}}^{+1}}$ , whose configuration will be $\text{1}{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{6}}\text{3}{{\text{s}}^{2}}\text{3}{{\text{p}}^{6}}\text{4}{{\text{s}}^{0}}\text{3}{{\text{d}}^{5}}$ .

-Iron will become $\text{F}{{\text{e}}^{+1}}$ , whose configuration will be $\text{1}{{\text{s}}^{2}}\text{2}{{\text{s}}^{2}}\text{2}{{\text{p}}^{6}}\text{3}{{\text{s}}^{2}}\text{3}{{\text{p}}^{6}}\text{4}{{\text{s}}^{1}}\text{3}{{\text{d}}^{6}}$ .

Now, for ions to lose a second electron, the electrons will be removed from $4\text{s}$ subshell from all other elements except for Cr. In Cr, the next electron will be removed from $\text{3d}$ subshell . The subshell in Cr has the configuration of $\text{3}{{\text{d}}^{5}}$. This $\text{3}{{\text{d}}^{5}}$ configuration is very stable as it is half-filled. This half-filled configuration looks like $\begin{matrix} \uparrow & \uparrow & \uparrow & \uparrow & \uparrow \\ \end{matrix}$ . This is because half-filled and full-filled subshells are highly stable. When the second electron will be lost from $\text{3}{{\text{d}}^{5}}$subshell, the stability of the subshell will be lost. That’s why, $\text{C}{{\text{r}}^{+}}$ would not like to lose second electron easily to form $\text{C}{{\text{r}}^{+2}}$. Hence, chromium has the highest second ionization energy.
The correct answer to this question is chromium has the highest second ionization energy;
So, the correct answer is “Option B”.

Note: The correct electronic configuration of Chromium should be written correctly. As, $\text{Cr}$ have exceptional electronic configuration.
- Correct electronic configuration is $\left[ \text{Ar} \right]3{{\text{d}}^{5}}\text{4}{{\text{s}}^{1}}$
- Incorrect electronic configuration is $\left[ \text{Ar} \right]3{{\text{d}}^{4}}\text{4}{{\text{s}}^{2}}$. Such type of electronic configuration is accepted by elements whose configurations are about to reach stability or half-filled or full-filled configurations.