Why intervening d and f orbital has poor shielding effect?
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, 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 4πr24πr2. 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.
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.
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