Slater's rules are a simple way to estimate the effective...

Slater's rules are a simple way to estimate the effective nuclear charge experienced by an electron. In this approach, the "shielding constant," $S$, is calculated. The effective nuclear charge is then the difference between $S$ and the atomic number, $Z$. (Note that the results in Table 7.2 and Figure 7.2 were calculated in a slightly different way.) $$Z^{*}=Z-S$$ The shielding constant, $S$, is calculated using the following rules: 1. The electrons of an atom are grouped as follows: $(1 s)$ $(2 s, 2 p)(3 s, 3 p)(3 d)(4 s, 4 p)(4 d),$ and so on. 2. Electrons in higher groups (to the right) do not shield those in the lower groups. 3. For $n s$ and $n p$ valence electrons a) Electrons in the same $n s, n p$ group contribute 0.35 (for $1 s 0.30$ works better). b) Electrons in $n-1$ groups contribute 0.85 c) Electrons in $n-2$ groups (and lower) contribute 1.00 4. For $n d$ and $n f$ electrons, electrons in the same $n d$ or $n f$ group contribute $0.35,$ and those in groups to the left contribute 1.00 As an example, let us calculate $Z *$ for the outermost electron of oxygen: $$\begin{array}{c} S=(2 \times 0.85)+(5 \times 0.35)=3.45 \\ Z^{*}=8-3.45=4.55 \end{array}$$ Here is a calculation for a $d$ electron in $\mathrm{Ni}$ : $$Z^{*}=28-[18 \times 1.00]-[7 \times 0.35]=7.55$$ and for an $s$ electron in $\mathrm{Ni}$ : $$\begin{array}{r} Z^{*}=28-[10 \times 1.00]-[16 \times 0.85] \\ -[1 \times 0.35]=4.05 \end{array}$$ (Here $3 s, 3 p,$ and $3 d$ electrons are in the $(n-1)$ groups.) (a) Calculate $Z^{*}$ for $F$ and $N$ e. Relate the $Z^{*}$ values for O, $\mathrm{F}$, and $\mathrm{Ne}$ to their relative atomic radii and ionization energies. (b) Calculate $Z^{*}$ for one of the $3 d$ electrons of $\mathrm{Mn}$, and compare this with $Z^{*}$ for one of the $4 s$ electrons of the element. Do the $Z^{*}$ values give us some insight into the ionization of Mn to give the cation?

Slater's rules are a simple way to estimate the effective nuclear charge experienced by an electron. In this approach, the "shielding constant," $S$, is calculated. The effective nuclear charge is then the difference between $S$ and the atomic number, $Z$. (Note that the results in Table
7.2 and Figure 7.2 were calculated in a slightly different way.)
$$Z^{*}=Z-S$$
The shielding constant, $S$, is calculated using the following rules:
1. The electrons of an atom are grouped as follows: $(1 s)$ $(2 s, 2 p)(3 s, 3 p)(3 d)(4 s, 4 p)(4 d),$ and so on.
2. Electrons in higher groups (to the right) do not shield those in the lower groups.
3. For $n s$ and $n p$ valence electrons
a) Electrons in the same $n s, n p$ group contribute 0.35 (for $1 s 0.30$ works better).
b) Electrons in $n-1$ groups contribute 0.85
c) Electrons in $n-2$ groups (and lower) contribute 1.00
4. For $n d$ and $n f$ electrons, electrons in the same $n d$ or $n f$ group contribute $0.35,$ and those in groups to the left contribute 1.00 As an example, let us calculate $Z *$ for the outermost electron of oxygen:
$$\begin{array}{c}
S=(2 \times 0.85)+(5 \times 0.35)=3.45 \\
Z^{*}=8-3.45=4.55
\end{array}$$
Here is a calculation for a $d$ electron in $\mathrm{Ni}$ :
$$Z^{*}=28-[18 \times 1.00]-[7 \times 0.35]=7.55$$
and for an $s$ electron in $\mathrm{Ni}$ :
$$\begin{array}{r}
Z^{*}=28-[10 \times 1.00]-[16 \times 0.85] \\
-[1 \times 0.35]=4.05
\end{array}$$
(Here $3 s, 3 p,$ and $3 d$ electrons are in the $(n-1)$ groups.)
(a) Calculate $Z^{*}$ for $F$ and $N$ e. Relate the $Z^{*}$ values for O, $\mathrm{F}$, and $\mathrm{Ne}$ to their relative atomic radii and ionization energies.
(b) Calculate $Z^{*}$ for one of the $3 d$ electrons of $\mathrm{Mn}$, and compare this with $Z^{*}$ for one of the $4 s$ electrons of the element. Do the $Z^{*}$ values give us some insight into the ionization of Mn to give the cation?

Key Concepts

Effective Nuclear Charge

Effective nuclear charge (Z*) represents the net positive charge experienced by an electron in an atom. It takes into account the actual nuclear charge as well as the shielding effect of other electrons, thereby influencing many atomic properties such as size and ionization energy.

Shielding and Shielding Constant

Shielding refers to the reduction in the full nuclear charge on an electron due to the repulsion of other electrons. The shielding constant (S) is a numerical value calculated by summing the contributions from all other electrons, with each electron contributing according to its position in the electron configuration. This constant is subtracted from the atomic number to obtain the effective nuclear charge.

Slater's Rules

Slater's rules provide a systematic method to estimate the effective nuclear charge by assigning specific screening values to electrons based on their grouping in shells and subshells. The rules differentiate between electrons in the same group, the immediately lower group, and others, making it easier to predict trends in atomic properties like atomic size and ionization energy.

Electron Configuration Grouping

Electrons are organized into groups based on their principal quantum number and orbital type (such as ns, np, nd, and nf). This grouping is essential for applying Slater's rules, as electrons in different groups contribute differently to the shielding effect, affecting the calculation of the effective nuclear charge.

Periodic Trends (Atomic Radii and Ionization Energies)

Effective nuclear charge is closely related to periodic trends such as atomic radii and ionization energies. Generally, a higher effective nuclear charge leads to a smaller atomic radius and a higher ionization energy, since electrons are drawn closer to the nucleus and require more energy to remove.

Transcript

00:01 Hello everyone, let us see the following question.

00:03 Slater's rule are a simple way to estimate the effective nuclear charge experienced by an electron.

00:09 In this approach, shielding constant s is calculated.

00:13 The effective nuclear charge is then difference between s and the atomic number z.

00:17 So, it is been calculated.

00:20 The s effective nuclear charge in slater's rule as z star, which is nothing but atomic number minus the s, that is the shielding constant.

00:31 So, here the various notations are given.

00:33 This is effective nuclear charge.

00:43 This is atomic number and this is shielding constant.

01:00 And here we are talking about slater's rule.

01:06 So, slater's rule, the electrons experience repulsion will be given as the following.

01:12 So, first of all, for the electrons of the given orbitals of the given shell, experience 0 .35 of effect shielding constant, as that of 1s orbital experience 0 .30.

01:29 So, electrons of n minus 1 level from the given electron experience 0 .85, and that of n minus 2 experiences above 1.

01:40 So, using this information, we have a question where we have to calculate the z star of f and neon.

01:49 So let us calculate it for fluorine first.

01:52 So the flooring, it is written as 1s2, 2 s2 and 2p 6.

01:59 So we have 6.

02:01 Totally there are 8 electrons here.

02:04 Out of that, 7 electrons are effectively pulled by the nucleus.

02:08 So, the atomic number z here would be 9.

02:13 So, it will be 9 minus s.

02:16 So now we have to calculate the s here.

02:20 So, s here would be equal to, there are six electron experiencing totally 0 .35, so 0 .5 is written as e5.

02:32 Sorry, electronic configuration is 5.

02:34 So totally there are 7 electrons.

02:36 That means six electrons experiencing of the given shell, the given level is 0 .35, plus two electrons in the penalty made shell, they experience 0 .85.

02:50 On substituting, we get the value here as 2 .1 plus 1 .7, that would be around 3 .8.

03:01 So, effective nuclear chart z star here is 9 minus 3 .3 .3.

03:06 0 .8 that would be around 5 .2.

03:10 So for f it is 5 .2.

03:13 So let's calculate it for neon...

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