Question

Consider row 4 of the periodic table. The trend is that the$\mathbf{4}\mathit{x}$subshell fills. Then the 3d, then the 4p.

(a) Judging by adherence to and deviation from this trend, whit might be said of the energy difference between the 4sand 3drelative to that between the 3dand 4p?

(b) Is this also true of row 5?

(c) Are these observations in qualitative agreement with Figure 8.13? Explain.

Short answer

(a)$r_1=0.054a_0,r_2=0.3a_0,r_3=1.64a_0,r_4=16a_0$

(b) The resultant answer is yes.

(c) The resultant answer is no.

Step-by-step solution

Given data

$n=1,2,3,4$For the electron in potassium.

Concept of Atomic radii

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

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

Determine the equation

(a)

For the $n=1$electrons, if they orbit the 19 protons in the potassium nucleus and half of the other $n=1$electron, the effective Z that the electron would see would be 18.5, since 19-0.5=18.5. That can then be used in the radii equation, along with being 1:

$$\begin{gathered} r_1=\frac{{\left(1\right)}^2{a}_0}{(18.5)} \\ {r}_1=0.054a_0 \end{gathered}$$

For the $n=2$electrons, if they orbit the 19 protons in the potassium nucleus, both 1 s electrons, and half of the seven others$n=2$ electrons, the effective Z that the electron would see would be 13.5 (since 19-2 -3.5=18.5). that can then be used in the radii equation, along with being 2:

$$\begin{gathered} r_2=\frac{{\left(2\right)}^2{a}_0}{(13.5)} \\ {r}_2=0.296a_0 \end{gathered}$$

For the$n=3$electrons, if they orbit the 19 protons in the potassium nucleus, both 1s electrons, all eight of the n=2 electrons, and half of the seven other electrons, the effective Z that the electron would see would be 5.5 (since 19-2-8-3.5=5.5 ). that can then be used in the radii equation, along with n being 3:

$$\begin{gathered} r_3=\frac{{\left(3\right)}^2{a}_0}{(5.5)} \\ {r}_3=1.636a_0 \end{gathered}$$

For the $▵n=4$electron, if it orbits the 19 protons in the potassium nucleus, both 1s electrons, all eight of the n=2 electrons, and all eight of the n=3 electrons, the effective that the electron would see would be 1 (since 19-2-8-8=1). That can then be used in the radii equation, along with being 4:

$$\begin{gathered} r_4=\frac{{\left(4\right)}^2{a}_0}{I} \\ {r}_4=16a_0 \end{gathered}$$

Determine the radii

(b)

Calculate the radii

$$\begin{gathered} r_1=\left(0.054a_0\right)\left(\frac{0.0529\text{nm}}{a_0}\right) \\ {r}_1=2.86\times {10}^{-3}\text{nm} \\ {r}_2=\left(0.3a_0\right)\left(\frac{0.0529\text{nm}}{a_0}\right) \\ {r}_2=1.59\times {10}^{-2}\text{nm} \end{gathered}$$

Similarly, calculate further:

$$\begin{gathered} r_3=\left(1.64a_0\right)\left(\frac{0.0529\text{nm}}{a_0}\right) \\ {r}_3=8.67\times {10}^{-2}\text{nm} \end{gathered}$$

Determine the equation

(c)

The atomic orbit radius equation can be used to estimate the effective Z that the valence electron sees, using $0.22\text{nm}$for the$r_n$for the n and $0.0529\text{nm}$for the$a_0$ after solving for Z :

$$\begin{gathered} Z=\frac{n^2{a}_0}{r_n} \\ Z=\frac{{\left(4\right)}^2(0.0529\text{nm})}{(0.22\text{nm})} \\ Z=3.85 \end{gathered}$$

So since in the valence electron orbits an effective Z of 1 in the original model, and it would need to orbit an effective $Z$ of $3.85$ based just on the experimental radius, there would need to be $3.8$ more protons be "unscreened"

The valence electron of potassium is in the s-shell, and consequently will have a roughly elliptical orbit. Therefore, it will spend part of its time near the nucleus, inside the orbits of the lower n electrons.