Question

Use the data in Table 3.1 to plot the logarithm of ionization energy versus the number of electrons removed for Be. Describe the electronic structure of the Be atom.

Short answer

The electronic structure of the Beryllium element is$1s^{2}2s^2$, and the graphical representation of the Logarithm of ionization energy versus the number of electrons removed from Beryllium (Be) is

Step-by-step solution

Step 1: The graphical representation of the Logarithm of ionization energy versus the number of electrons removed from Beryllium (Be).

Electronic structure of Beryllium element.

The graph of Log$\left({\text{IE}}_{\text{n}}\right)$for n=1, 2, 3, and 4. The graph Logarithm of ionization energy versus the number of electrons removed from Beryllium (Be) shows a dramatic increase between the values of$\text{(n = 2)}$, and$\text{n =3}$which shows two (2) easily removed electrons$\left(e^{-}\right)$.The ionization energies for Beryllium (Be) element show a large gap between$I{E}_2$and$I{E}_3$. So, the Beryllium (Be) element easily loses two electrons$\left(e^{-}\right)$to form.

The Beryllium element has two electrons$\left(e^{-}\right)$, so the Beryllium element shows Helium (He) like stability.

The electronic configuration of the Beryllium element is$1s^{2}2s^2$.