![]() Chlorine: Chlorine, Cl 2, is a much smaller molecule with comparatively weak van der Waals attractions, and thus chlorine will have a lower melting and boiling point than sulfur or phosphorus.The molecules are bigger than phosphorus molecules, and thus the van der Waals attractions are stronger, leading to a higher melting and boiling point. Sulfur: Elemental sulfur forms S 8 rings of atoms.Melting phosphorus breaks no covalent bonds instead, it disrupts the much weaker van der Waals forces between the molecules. Phosphorus: Elemental phosphorus adopts the tetrahedral P 4 arrangement.The magnitudes of the melting and boiling points are governed entirely by the sizes of the molecules, which are shown again for reference: Their melting or boiling points are lower than those of the first four members of the period which have complex structures. Phosphorus, sulfur, chlorine and argon are simple molecular substances with only van der Waals attractions between the molecules. The repulsion between the two electrons in the same orbital creates a higher-energy environment, making the electron easier to remove than predicted. The difference is that in the case of sulfur, the electron being removed is one of the 3p x 2 pair. The screening (from the inner electrons and, to some extent, from the 3s electrons) is identical in phosphorus and sulfur, and the electron is removed from an identical orbital. 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. Both of these factors offset the effect of the extra proton. 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. However, this effect is offset by the fact that the outer electron of aluminum occupies a 3p orbital rather than a 3s orbital. The decrease at aluminum: The value for aluminum might be expected to be greater than that of magnesium due to the extra proton. The increasing nuclear charge also pulls the outer electrons toward the nucleus, further increasing ionization energies across the period. This creates greater attraction between the nucleus and the electrons and thus increases the ionization energies. The determining factor in the increase in energy is the increasing number of protons in the nucleus from sodium across to argon. 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 upward trend: In the whole of period 3, the outer electrons are in 3-level orbitals. ![]() ![]() whether the electron is alone in an orbital or one of a pair.the amount of screening by inner electrons.the distance of the outer electron from the nucleus.\)įirst ionization energy is dependent on four factors:
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