Question

The trend for ionization energy is a general increase from left to right across a period. However, phosphorus (P) is found to have a higher first ionization energy value than sulfur (S). Explain this exception to the general trend in terms of electron arrangements and attraction/repulsion.

Answer 1

Step-by-step explanation:

The electrons are attracted to the nucleus and it is necessary to provide energy to start them. So, the ionization energy, I, is the energy needed to start an electron to a gaseous atom, isolated and in a fundamental state.

The trend for ionization energy is a general increase from left to right across a period. But there are exceptions to this periodic trend due to the great stability of atoms with semi-occupied or occupied orbitals, because electrons are more difficult to extract. One of these exceptions is P and S.

P and S are neighboring elements in the Periodic Table. Both belong to period 3 and have the electronic valence configurations (last layer):

• P 3s² 3p³

• S 3s² 3p⁴

In sub-level p there can be a maximum of 6 electrons

Then the Phosphorus has the p half-filled orbitals (3 electrons). So it is stable and therefore to start an electron, more energy is needed.

Another way of thinking is that in the case of sulfur S the electron enters an orbital already previously occupied, half full and stable. Therefore the interelectronic repulsion between the two makes it easier to remove the electron.

Answer 2

₁₅P 1s²2s²2p⁶3s²3p³
₁₆S 1s²2s²2p⁶3s²3p⁴
It is because: 1) phosphorus have 3 unpaired electrons in 3p shell, and that is more stable than 3p⁴ in sulfur, so it ie easier to remove 1 paired electron in sulphur to became 3p³.
2) repulsion in 3p shell in sulfur between 2 paired electrons is higher, so it is easier to remove that electron.