why does electrical conductivity decrease across period 3

The upward trend In the whole of period 3, the outer electrons are in 3-level orbitals. These are all the same sort of distances from the nucleus, and are screened by the same electrons in the first and second levels. The major difference is the increasing number of protons in the nucleus as you go from sodium across to argon. That causes greater attraction between the nucleus and the electrons and so increases the ionisation energies. In fact the increasing nuclear charge also drags the outer electrons in closer to the nucleus. That increases ionisation energies still more as you go across the period. The fall at aluminium You might expect the aluminium value to be more than the magnesium value because of the extra proton. Offsetting that is the fact that aluminium's outer electron is in a 3p orbital rather than a 3s. The 3p electron is slightly more distant from the nucleus than the 3s, and partially screened by the 3s electrons as well as the inner electrons.


Both of these factors offset the effect of the extra proton. The fall at sulphur As you go from phosphorus to sulphur, something extra must be offsetting the effect of the extra proton
The screening is identical in phosphorus and sulphur (from the inner electrons and, to some extent, from the 3s electrons), and the electron is being removed from an identical orbital. pair. The repulsion between the two electrons in the same orbital means that the electron is easier to remove than it would otherwise be. Atomic radius The trend The diagram shows how the atomic radius changes as you go across Period 3. metallic radii for Na, Mg and Al; covalent radii for Si, P, S and Cl; the van der Waals radius for Ar because it doesn't form any strong bonds. It is fair to compare metallic and covalent radii because they are both being measured in tightly bonded circumstances. It isn't fair to compare these with a van der Waals radius, though.


The general trend towards smaller atoms across the period is NOT broken at argon. You aren't comparing like with like. The only safe thing to do is to ignore argon in the discussion which follows. whether the electron is alone in an orbital or one of a pair. The upward trend: In the whole of period 3, the outer electrons are in 3-level orbitals. These electrons are at approximately the same distance from the nucleus, and are screened by corresponding electrons in orbitals with principal atomic numbers n=1 and n=2. The determining factor in the increase in energy is the increasing number of protons in the nucleus from sodium across to argon. This creates greater attraction between the nucleus and the electrons and thus increases the ionization energies. The increasing nuclear charge also pulls the outer electrons toward the nucleus, further increasing ionization energies across the period. The decrease at aluminum : The value for aluminum might be expected to be greater than that of magnesium due to the extra proton.


However, this effect is offset by the fact that the outer electron of aluminum occupies a 3p orbital rather than a 3s orbital. The 3p electron is slightly farther from the nucleus than the 3s electron, and partially screened by the 3s electrons as well as the inner electrons. Both of these factors offset the effect of the extra proton. The decrease at sulfur: In this case something other than the transition from a 3s orbital to a 3p orbital must offset the effect of an extra proton. The screening (from the inner electrons and, to some extent, from the , and the electron is removed from an identical orbital. The difference is that in the case of sulfur, the electron being removed is one of the pair. The repulsion between the two electrons in the same orbital creates a higher-energy environment, making the electron easier to remove than predicted.

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