**CALCULATION OF Zeff:**

**Slater’s Rules**

**Slater’s Rules**

**1) Write the electron configuration for the atom using the following design ( Write the electronic structure of the atom in groupings as follows):
(1s), (2s, 2p), (3s, 3p), (3d), (4s, 4p), (4d), (4f), (5s, 5p), (5d), (5f)…**

**2) Any electrons of principal quantum number higher than the “electron of concern” in electronic configuration contributes no shielding i.e., electrons in higher groups than the electron you are considering do not shield electrons in lower groups.(Approximately correct statement).**

**3) All other electrons in the same group as the “electron of concern” shield to an extent of 0.35 nuclear charge units. **

**4) If the “electron of concern” is an s or p electron: All electrons with one less value of the principal quantum number shield to an extent of 0.85 units of nuclear charge. All electrons with two (or more) less values of the principal quantum number shield to an extent of 1.00 units. i.e, **For

**ns**or

**np**electrons:

- Electrons in the same
*ns, np*group contribute 0.35, except when considering electrons in the*1s*orbital, where 0.30 works better. - Electrons in the
*n-1*group(s) contribute 0.85. - Electrons in the
*n-2*or lower groups contribute 1.00.

**5) ****If the electron of interest is an ***d*** or ***f*** electron: All electrons to the left shield to an extent of 1.00 units of nuclear charge. i.e., **For **nd** and **nf** valence electrons:

- Electrons in the same
*nd*or*nf*group contribute 0.35. - Electrons in groups to the left contribute 1.00.

**5) Sum the shielding amounts from steps 2 through 5 and subtract from the nuclear charge value to obtain the effective nuclear charge(Zeff).**

*Shielding effect* – The decrease in (lessening of)) attractive electrostatic charge from nuclear protons due to presence of *electrons of same shell and of inner shells.* *In an atom electrons are shielded from the pull of nucleus by the electrons of same shell and of inner shells. Poor shielding of more diffused ”d” or “f’ orbitals explain phenomena such as the Lanthanide contraction and inert pair effect or anomalous change in size and IE down the group 13 or anomalous variation in size of 3d series elements. The more shielding, the further the valence electron shell will spread out and the bigger atoms will become. *

The effective nuclear charge( Z* or Zeff) is the net nuclear charge experienced by a given electron. Z* = Z-σ where σ is screening or shielding constant. *Example: Lithium(1s ^{2}2s^{1})- The electron in the 2s orbital is shielded from the full attraction of the protons by the electrons of the 1s orbital. Thus, Z* felt by the 2s electron should be one rather than three. However, lithium’s 2s electron experiences about 1.3 charge units. This can be understood by the fact that the 2s orbital has two maxima in its radial probability function vs radius graph, and the smaller maxima penetrates within the maximum of the inner 1s electron. Although lithium’s 2s electron spends most of its time in the outer lobe of that orbital feeling a nuclear charge of +1, it spends sometime inside the 1s orbital experiencing the full nuclear charge of +3. Thus, Z* is somewhat greater than +1.*

*note 1- In hydrogen atom or H-like single electron species, the net pull on the electron is just exactly equal to the nuclear charge.*

*note 2- 3d electrons experience a much greater Z* than the 4s electron and would be held more tightly. Thus, the 4s electrons are first removed when d-block elements are ionized. *

*note 3-**Since “s” electrons are closer to the nucleus, screening effect or penetration effect follows the order s>p>d>f. *

*note 4- Actual shielding effect is always greater that the screening constant because core electrons are much closer to the nucleus than are valence electrons.*

The periodic table tendency for effective nuclear charge:

- Across a period, Zeff increases and is dominating factor (due to increasing nuclear charge with no accompanying increase in number of shells).
Down the group, Zeff increasesbut increasing number of orbits is dominating factor. The greatly increased size causes a weaker net attraction for the outer electrons and hence decrease in ionization energy. See the values of Zeff here.

*Group 1: Li = 1.28, Na = 2.51, K = 3.50, Rb = 4.98, Cs = 6.36
Group 14: C = 3.14, Si, 4.29, Ge = 6.78, Sn = 9.10, Pb = 11.61
Group 17: F = 5.10, Cl = 6.12, Br = 9.03, I = 11.61
*

*Element Zeff Rmax IP kJ mol^-1
Li 1.28 164.1 520.2
Na 2.51 179.4 495.8
K 3.50 230.0 418.8
Rb 4.98 249.0 403.0
Cs 6.36 282.0 375.7*

Zeff for a 2p electron in Nitrogen- (1s2) (2s2, 2p3); Zeff = Z – [(0.35 x 4) + (0.85 x 2)] = 7 – 3.10 = 3.90.Zeff for a 3p electron in Silicon- (1s2) (2s2, 2p6)(3s2, 3p2); Zeff = Z – [(0.35 x 3) + (0.85 x 8) + (1 x 2) ] = 14 – 9.85 = 4.15.

**Example 1: ****Calculate Zeff **for a valence electron in fluorine

**(1**** s**2)(2s2,2p5

**)**

**Rule 2 does not apply; 0.35 · 6 + 0.85 · 2 = 3.8**

**Zeff = 9 – 3.8 = 5.2 for a valence electron.**

Zeff for 4s electron & for 3d electron in Zinc-(1s2) (2s2, 2p6)(3s2, 3p6)(3d10)(4s2);For 4s electron,Zeff = Z – [(0.35 x 1) + (0.85 x 18) + (1 x 10)] = 30 – 25.65 = 4.35;For 3d electron, Zeff= Z – [ (0.35 x 9) + (1 x 18)]= 30 – 21.15 = 8.85

*Example 3: Effective nuclear charge felt by a 4p electron of bromine
*

**First write out the electronic structure in the format of the first rule.**

*(1s) ^{2} (2s, 2p)^{8} (3s, 3p)^{8} (3d)^{10} (4s, 4p)^{7}*

**Then write out an equation for the screening constant according to the appropriate Rule – 3 or 4.**

*Here, Rule 3 applies.
There are 6 other electrons in the same ns, np group.
There are 18 electrons in the n-1 groups. (3s, 3p and 3d)
There are 10 electrons in the n-2 and lower groups. (1s, 2s and 2p)
*σ = 6(0.35)+18(0.85)+10(1.00)

σ= 27.4

**Then subtract the screening constant from Z, the atomic number of the element.**

*Bromine’s atomic number is 35.*

35-27.4=7.6=Z_{eff}

What if we consider the effective nuclear charge of a more inner orbital?

*What is the effective nuclear charge felt by a 3d electron of Bromine?*

**Use the same electronic structure written out before.**

*(1s) ^{2} (2s, 2p)^{8} (3s, 3p)^{8} (3d)^{10} (4s, 4p)^{7}*

**Then write out an equation for the screening constant according to the appropriate Rule – 3 or 4.**

*Here, Rule 4 applies.
There are 9 other electrons in the same nd group.
There are 18 electrons to the left of the 3d group.
*σ = 9(0.35)+18(1.00)

σ = 21.15

**Then subtract the screening constant from Z.**

*Bromine’s atomic number is 35.*

35-21.15=13.85=Z_{eff}

Zeff for a 6s electron in tungsten (74)-(1s2) (2s2, 2p6)(3s2, 3p6)(4s2, 4p6) (3d10) (4f14) (5s2, 5p6)(5d4), (6s2); Zeff = Z –[(0.35 x 1) + (0.85 x 12) + (1 x 60) ] = 74 – 70.55 =3.45

**Example 2: Calculate Zeff f**or a 6s electron in Platinum.

**(1**** s**2)(2s2,2p6)(3s2,3p6) (3d10) (4s2,4p6) (4d10) (4f14) (5s2,5p6) (5d8) (6s2

**)**

**Rule 2 does not apply; 0.35 · 1 + 0.85 · 16 + 60 · 1.00 = 73.95**

**Zeff **= 78 – 73.95 = 4.15 for a valence electron.

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