A Simple Model: The multielectron atom is unsolvable, but simple models go a long way. Section$7.8$gives energies and orbit radii forone-electron/hydrogenlike atoms. Let us see how useful these are by considering lithium.

(a) Treat one of lithium's$\mathit{n}\mathbf{=}\mathbf{1}$electrons as a single electron in a one-electron atom ofrole="math" localid="1659948261120" $\mathit{Z}\mathbf{=}\mathbf{3}$. Find the energy and orbit radius.

(b) The other$\mathit{n}\mathbf{=}\mathbf{1}$electron being in the same spatial state. must have the same energy and radius, but we must account for the repulsion between these electrons. Assuming they are roughly one orbit diameter apart, what repulsive energy would they share, and if each claims half this energy. what would be the energies of these two electrons?

(c) Approximately what charge does lithium's lone valence electron orbit, and what radius and energy would it have?

(d) Is in reasonable to dismiss the role of the$\mathit{n}\mathbf{=}\mathbf{1}$electrons in chemical reactions?

(e) The actual energies of lithium's electrons are about$\mathbf{-}\mathbf{98}\mathbf{\text{eV}}$(twice, of course) and$\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{4}\mathbf{\text{eV}}$. How good is the model?

(f) Why should the model's prediction for the valence electron's energy differ in the direction it does from the actual value?

Hydrogen like atom, atomic number, $Z=3$, Bohr radius, $a_0=5.29\times {10}^{-11}\text{m}$.

Expression for atomic radii$r_n$is given by,$r_n=\frac{n^2{a}_0}{\text{Z}}$

Where,Z represents number of protons, nrepresents principal quantum number and$a_0$represents Bohr radius.

(a)

The energy of the electron in first orbit

$$\begin{gathered} E_n=(-13.6\text{eV}){\left(\frac{\text{Z}}{n}\right)}^2 \\ {E}_1=(-13.6\text{eV}){\left(\frac{3}{1}\right)}^2 \\ {E}_1=-122\text{eV} \end{gathered}$$

The radius of the orbit of the electron is,

$$\begin{gathered} r_n=\frac{n^2{a}_0}{Z} \\ {r}_1=\frac{{\left(1\right)}^2\left(5.29\times {10}^{-11}\text{m}\right)}{\left(3\right)} \\ {r}_1=1.76\times {10}^{-11}\text{m} \end{gathered}$$

(b)

The electric potential between two electron charges

$$\begin{gathered} U=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r} \\ U=\frac{1}{4\pi \left(8.85\times {10}^{-12}\frac{{\text{c}}^2}{\text{N}\times {\text{m}}^2}\right)}\frac{{\left(-1.6\times {10}^{-19}\text{C}\right)}^2}{\left(3.52\times {10}^{-11}\text{m}\right)} \\ U=6.54\times {10}^{-18}\text{J}=\left(6.54\times {10}^{-18}\text{J}\right)\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right) \\ U=40.87\text{eV} \end{gathered}$$

If half of potential energy added to both of the electron's energy, the new electron energy become

$$\begin{gathered} E_1^{\text{'}}=E_1+\frac{U}{2} \\ {E}_1^{\text{'}}=(-122\text{eV})+\frac{(40.87\text{eV})}{2} \\ {E}_1^{\text{'}}=-101.6\text{eV} \end{gathered}$$

(c)

On lithium has one electron on second orbit, so the charge of valance electron orbit of lithium is equal to the charge of one electron.

The charge of valance electron orbit,$Q_{\text{valence orbit}}=-1.6\times {10}^{-19}\text{C}$

The energy for lithium's valence electron is

$$\begin{gathered} E_n=(-13.6\text{eV}){\left(\frac{Z}{n}\right)}^2 \\ {E}_2=(-13.6\text{eV}){\left(\frac{1}{2}\right)}^2 \\ {E}_2=-3.4\text{eV} \end{gathered}$$

The radius of the orbit for lithium's valence electron is

$$\begin{gathered} r_n=\frac{n^2{a}_0}{\text{Z}} \\ {r}_2=\frac{{\left(2\right)}^2\left(5.29\times {10}^{-11}\text{m}\right)}{\left(1\right)} \\ {r}_2=2.1\times {10}^{-10}\text{m} \end{gathered}$$

(d)

The inner orbit electrons are more closely to the neutron so they likely to experience quite high binding energy. This high binding force ceases them to participate in chemical reaction directly through exchanging or sharing. But do play a part in altering the energy and orbit radius of the valence electron due to their shielding of the nucleus. And that does affect the kind of reactions that the valence electron participates in.

(e)

Actual energy of lithium's 1s electron is $-98\text{eV}$, but the model predicts the energy approx. $-102\text{eV}$, On the other hand lithium's second orbit electron has energy$-5.4\text{eV}$ but the model predicts $-3.4\text{eV}$. Percentage of error in predicting second orbit electron is very large. So, the model is good in predicting energy of smaller orbit's electron, but not good for higher orbit's electron.

(f)

The valence electrons predicted energy being less negative than the actual energy can be attributed to incomplete shielding of the nucleus by the 1s electrons. If the 1s electrons aren't completely efficient at using their charge of to$-2$ shield the charge on the nucleus of $+3$ the valence electron will see a higher effective charge on the nucleus. Consequently, the Z used in equation $E_n=(-13.6\text{eV}){\left(\frac{Z}{n}\right)}^2$to determine the energy of the electron would be higher, leading to a greater negative energy.