Q28. For four elements the ${\mathrm{IP}}_2$ curve is shown.  In the graph the elements represented by $A, B, C$ and $D$ are

Given Data: It is provided in question a graph with four elements marked on it, depicting their ionisation potential.

Concept Used:  Electronic configuration

Detailed Solution:

The second ionisation potential is presented as IP2 is equal to the second ionisation energy. 

Mathematically, writing this relation for Mg, Al, Si and P we have:

For Mg:

 $$\begin{gathered} \mathrm{Mg}\left(\mathrm{Z}=12\right)\stackrel{1 \mathrm{electron} \mathrm{loss}}{\to }{\mathrm{Mg}}^{+}\to 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^1 \left({\mathrm{IP}}_1\right) \\ {\mathrm{Mg}}^{+}\stackrel{1 \mathrm{electron} \mathrm{loss}}{\to }{\mathrm{Mg}}^{2+}\to 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^6 \left({\mathrm{IP}}_2\right) \end{gathered}$$

For Al:

$$\begin{gathered} \mathrm{Al}\left(\mathrm{Z}=13\right)\stackrel{1 \mathrm{electron} \mathrm{loss}}{\to }{\mathrm{Al}}^{+}\to 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^2 \left({\mathrm{IP}}_1\right) \\ {\mathrm{Al}}^{+}\stackrel{1 \mathrm{electron} \mathrm{loss}}{\to }{\mathrm{Al}}^{2+}\to 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^1 \left({\mathrm{IP}}_2\right) \end{gathered}$$

For Si:

$$\begin{gathered} \mathrm{Si}\left(\mathrm{Z}=14\right)\stackrel{1 \mathrm{electron} \mathrm{loss}}{\to }{\mathrm{Si}}^{+}\to 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^1 \left({\mathrm{IP}}_1\right) \\ {\mathrm{Si}}^{+}\stackrel{1 \mathrm{electron} \mathrm{loss}}{\to }{\mathrm{Si}}^{2+}\to 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^2 \left({\mathrm{IP}}_2\right) \end{gathered}$$

For P:

$$\begin{gathered} P\left(\mathrm{Z}=15\right)\stackrel{1 \mathrm{electron} \mathrm{loss}}{\to }{P}^{+}\to 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^2 \left({\mathrm{IP}}_1\right) \\ {P}^{+}\stackrel{1 \mathrm{electron} \mathrm{loss}}{\to }{P}^{2+}\to 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^2 3{\mathrm{p}}^1 \left({\mathrm{IP}}_2\right) \end{gathered}$$

So, the element A is Mg as its second ionisation energy is less due to its instability and Al is more stable after the second ionisation, so having a higher value of $I{P}_2$. Similarly, Si would have a smaller value of second ionisation energy due to its instability and P would have a higher value for the same as it becomes stable.

Hence, the correct labelling would be $A\to Mg, B\to Al, C\to Si$, and $D\to P,$ making option B the correct answer.