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Question

Answers

A. $\text{V}$

B. $\text{Cr}$

C. $\text{Mn}$

D. $\text{Fe}$

Answer
Verified

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;

- 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.