# Question 2fc5b

Feb 10, 2017

You can do it like this:

#### Explanation:

The effective nuclear charge $\textsf{{Z}_{\text{eff}}}$, experienced by an electron is less than the nuclear charge $\textsf{Z}$ because of the shielding effect of the other electrons:

$\textsf{{Z}_{e f f} = Z - s}$

Where $\textsf{s}$ is the shielding constant.

We can calculate $\textsf{s}$ by applying Slater's Rules:

Electrons are grouped together in increasing order of $\textsf{n}$ and $\textsf{l}$ values. $\textsf{n}$ tells you the number of the energy level and $\textsf{l}$ denotes the sub - shell such as $\textsf{s , p , d}$. Because $\textsf{s}$ and $\textsf{p}$ electrons are close in energy they are grouped together:

$\textsf{\left[1 s\right] \left[2 s , 2 p\right] \left[3 s , 3 p\right] \left[3 d\right] \left[4 s , 4 p\right] \left[4 d\right] \left[4 f\right] \left[5 s , 5 p\right]}$ and so on.

You can calculate the screening constant as follows:

Add 0.35 for each other electron within the same group, except for $\textsf{1 s}$ when this is 0.3

If the group is of the $\textsf{\left[\text{ns,np}\right]}$ type you add an amount of 0.85 for each electron with a principal quantum number of $\textsf{\left(n - 1\right)}$ and an amount of 1.00 for each electron with a quantum number of $\textsf{\left(n - 2\right)}$ or less.

For $\textsf{\left[d\right]}$ and $\textsf{\left[f\right]}$ electrons just add 1.00 for each electron which is closer to the nucleus.

Lets apply this to sf(""_25Mn)#:

The electron structure is:

$\textsf{\left[1 {s}^{2}\right] \left[2 {s}^{2} 2 {p}^{6}\right] \left[3 {s}^{2} 3 {p}^{6}\right] \left[3 {d}^{5}\right] \left[4 {s}^{2}\right]}$

For a 3d electron:

Within the 3d group there are 4 other electrons giving 4 x 0.35 = 1.40 units.

There are 18 other electrons closer to the nucleus giving 18 x 1.00 =18.0 units.

$\therefore$$\textsf{s = 1.40 + 18.0 = 19.4}$ units.

$\therefore$$\textsf{{Z}_{e f f} = 25 - 19.4 = 5.6}$ units.

For a 4s electron:

Within the 4s group there is 1 other electron giving 1 x 0.35 = 0.35 units.

In the (n - 1) group, which are the 3s, 3p and 3d, electrons there are 13 electrons giving 13 x 0.85 = 11.05 units

This leaves 10 (n - 2) electrons. This gives 10 x 1 = 10 units.

$\therefore$$\textsf{s = 0.35 + 11.05 + 10 = 21.4}$ units.

$\therefore$$\textsf{{Z}_{e f f} = 25 - 21.4 = 3.6}$ units.

This shows that the 4s electrons experience a smaller effective nuclear charge than the 3d electrons so are more easily lost

This is consistent with observations.