First ionization energies

Question

For row 5 on the periodic table, as one moves across, it is seen that ionization energy is higher for p than d and d is higher than s. Explain this trend while talking about atomic structure and periodic trends. Explain this thoroughly.

Answer 1

The reason that this is the case is that is as one moves across the periodic table from left to right, the number of valence electrons increases and it becomes more difficult for an atom to want to give them up in order to achieve noble gas configuration.

Think of it this way, for sodium (Na), it needs to give up 1 electron in order to reach the noble gas configuration of Neon (Ne) while for fluorine (F), a halogen, it needs to gain 1 more valence electron to become Neon.

It is much easier for fluorine to gain 1 more electron to become a negative ion than to lose 7 electrons to reach noble gas configuration and this trend is true elements located in the p position of the periodic table.

p has a higher ionization energy than d because elements in the d orbital have fewer valence electrons than those on the p orbital and d is has a higher ionization energy than s because d is able to store more valence electrons in its subshell than the s orbital but not as many as the p orbital, which is often why you see metal d ions like iron or zinc have charges limited to +2 or +3. Whereas the s orbital has ions that have charges of +1 and +2 and ions within the p orbital prefer to become anions.

This is very closely related to the periodic trends of electronegativity and electron affinity because elements that have higher numbers of valence electrons have a higher ionization energy which makes sense because such elements will require a higher energy to get rid of their valence electrons due to the fact that these elements have a strong nuclear attractive force that wants to keep the electrons in place and also due to the fact that these elements are only a couple electrons away from filling their octect to become a noble gas. The definition of electron affinity is like the name says: how attracted the electrons are to their respective atoms. And electronegativity is very similar: with it being how strong the nucleus pulls electrons to keep them around its vicinity.

You can also argue that atomic radius also plays a role since the trend is the exact opposite of that of electron affinity. You can think of it as: the bigger the atomic radius gets, the further away the electrons get away from the nucleus, and the easier it is to just take one off it since the attraction between the electron and the nucleus become weaker in accordance to Coulumb's Law: F = kQq/r^2. As you increase the atomic radius, the attractive force between protons and electrons will decrease by a squared factor, which means it's easier to take an electron away from the nucleus.