Why do d and f electrons have a poor shielding effect compared to s and p electrons?
It is down to the properties of the wavefunction which corresponds to the orbital (indeed, an orbital and a wavefunction are the same thing). Michael Flynn already answered your question, so I don't really have much to add. I had started a draft with an answer which I didn't manage to complete earlier but I decided to post it anyway. So, a wavefunction has a radial part and an angular component. The analytical expression of the radial component can be plotted to show the radial distribution function of each orbital. On the x axis you have radius and on the y axis is the square of the radial component of the wavefunction multiplied by [math] 4 \pi r^2 [/math]. This is done, so that one can obtain the probability density of finding an electron. As you can see from this figure, the electron density that is defned through those orbitals is concentrated in a different region. The distance which corresponds to the highest probability of finding the electron from a quantum-mechanical point of view corresponds to the classical radius. The s orbital is the most penetrating one due to the fact that its orbital quantum number is equal to zero. This means that the electron which occupies an s orbital only has Coulombic interaction with the nucleus but there is no centrifugal force trying to push it away from the nucleus. So, an electron which is located farther from the nucleus doesn't experience the same electrostatic potential from the nucleus when there already are s electrons. This is what the shielding effect is - a less penetrating electron interacts with a "composite particle" so to say which includes the nucleus and the more penetrating electrons, so the sum positive charge is smaller than the charge of the nucleus itself. The same can be said about the higher-energy orbitals. The rule beginner chemistry classes introduce that from orbitals with an equal sum of principle and orbital quantum numbers, [math] n + l [/math] is not some random abstraction. You can check out an answer I wrote some time ago with some further details on the nature of orbital angular momentum.Suzanka Bett's answer to How does the ℓ quantum number effect the energy of an orbital?
Electron shielding, effects?
If an electron is strongly attracted to its nucleus then it has great resistance to give up that electron. Chemical reactions are about giving up or accepting electrons. Each orbit around the nucleus has a specific number of electrons it needs to make the orbit stable. Unstability happens when these orbits are missing electrons. If it is easier to grab an electron to complete the orbit then that determines it Valence. The same if the orbit shell becomes more stable to give up electrons (opposite side of the Valence chart.) The reactivity scale of a chemical reaction are controlled by the rapidness of an atom or molecule to give up or attract electrons. Strong affinity to the nucleus would be an inhibition to reaction when it is looking to give up an electron not an enhancer. Strong affinity to the nucleus with an incomplete orbit is an enhancer to accept electrons.
What is the effective nuclear charge for the 4s electron in Cr?
The formula to be used is: Z* = Z - S where: Z* = effective nuclear charge Z = atomic number of Cr = 24 S = shielding Let us first calculate S using Slater's rule. *Write the electronic configuration as follows: (1s)^2 (2s 2p)^8 (3s 3p)^8 (3d)^5 (4s)^1 *Electrons in an "s" or "p" orbit in the same shell as the electron for which you're solving contribute 0.35, electrons in an "s" or "p" orbital in the shell one energy level lower contribute 0.85, and electrons in an "s" or "p" orbital in shells two energy levels and lower contribute 1. Therefore, the (4s)^1 do not contribute since we are focusing on the 3d orbital. Therefore, S = (5 x 0.35) + (8 x 0.85) + (10 x 1) = 18.55 Thus, Z* Z - S = 24 - 18.55 = 5.45 * Electrons in a "d" or "f" orbital in the same shell as the electron for which you're calculating contribute 0.35, and electrons in an "d" or "f" orbital in all lower energy levels contribute 1. Electrons in shells higher than the electron for which you're solving do not contribute to shielding. Therefore, the (4s)^1 do not contribute since we are focusing on the 3d orbital. Therefore, S = (4 x 0.35) + (18 x 1) = 19.4 Thus, Z* = 24 - 19.4 = 4.6 A good discussion of this concept can be found in the following site: http://www.ehow.com/how_5977365_calculat... That's it!n_n
Why are d electrons less effective in shielding the nuclear charge than the s or p electrons and are thus more polarizing?
d electrons orbits are not spherical symmetric.
Why does the f-orbital have a poor screening effect?
I could try answering this although I have left studying chemistry years ago .Screening effect — the ability of electrons to reduce the attraction effect from nucleus on the electrons behind the former ones — depends on two factors obviously,No. of electrons in the orbital which is supposed to screen other electrons.distance from the nucleus.Both the factors have effects opposite to each other .f orbital although has large number of electrons but its distance from the nucleus is also quite large compared to other inner orbitals. So this distance supersedes the large negative charge (due to large no. of electrons in f-orbital) resulting in less ability of f orbital to screen the outer electrons .I guess I could give a very crude answer hoping to see much better answers but still I guess it should be satisfying to some extent.
Why do electrons in d and f orbital produce a less shielding effect?
Unlike s orbital due to their shape they are not fully able to cover outer electrons .so outer electron experience more force.
Why is shielding effect constant across a period?
The shielding constants DO depend on the number of electrons. For example, Slater's rules (from Wikipedia) for shielding go as follows: The shielding constant for each group is formed as the sum of the following contributions: ...(1) An amount of 0.35 FROM EACH other electron within the same group except for the [1s] group, where the other electron contributes only 0.30. ...(2) If the group is of the [s p] type, an amount of 0.85 FROM EACH electron with principal quantum number (n) one less and an amount of 1.00 for each electron with an even smaller principal quantum number ...(3) If the group is of the [d] or [f], type, an amount of 1.00 FOR EACH electron inside it. This includes i) electrons with a smaller principal quantum number and ii) electrons with an equal principal quantum number and a smaller azimuthal quantum number (l). (CAPS are mine, for emphasis.) So, six 2p electrons exert a greater sheilding effect on, say, a 3p electron of iron than two 2s electrons. Collectively, the six 2p electrons exert a shielding of 6 × 0.85 = 5.1 e and the two 2s electrons exert a shielding of 2 × 0.85 = 1.7 e, so altogether the n = 2 subshell reduces the effective nuclear charge felt by the 3p electrons by 5.1 + 1.7 = 6.8 e. (The iron atom example is worked for all the electrons in Wikipedia - see the link below.)