# Why second ionisation energy greater than the first?

Why is the second ionisation energy for atoms with more than 1 electron in the outermost sublevel greater than that of the first, if you are removing an electron from the same sublevel ? Why is there a greater attraction for an electron in the same sublevel?

The ionisation energy basically measures how tightly an electron is held. That in turn is determined by a couple of factors, namely:

1. Nucleus-electron attraction. If this increases, the ionisation energy increases.
2. Electron-electron repulsion. If this increases, the ionisation energy decreases.

More formally these can be quantified by $Z$, the nuclear charge, and $\sigma$, the shielding parameter. The effective nuclear charge, which is hopefully a term you have come across, is defined as

$$Z_\mathrm{eff} = Z - \sigma$$

and the larger $Z_\mathrm{eff}$ is, the larger the ionisation energy. You don't necessarily need a formula to prove this, as it should be pretty intuitive. Larger effective nuclear charge $\implies$ electron is held more tightly $\implies$ more energy needed to ionise. But here is the mathematical relationship:

$$\text{IE} \propto \left(\frac{Z_\mathrm{eff}}{n}\right)^2$$

As long as you are staying within the same subshell, i.e. the principal quantum number $n$ is not being changed. So, the ionisation energy is determined by $Z_\mathrm{eff}$.

When you remove an electron, you decrease the electron-electron repulsions; this means that $\sigma$ decreases. As a result $Z_\mathrm{eff}$ decreases and the ionisation energy decreases. (Technically, there would be an increased attractive force ($\propto kq_1q_2/r^2$) from the nucleus, because the electrons on average will now be closer to the nucleus. However, within this inorganic chemistry framework, the nuclear charge $Z$ is simply the number of protons in the nucleus and therefore remains constant.)

• Is the additional culonic attraction after first ionisation accounted for in $Z_{\text{eff}}$, i.e. in $\sigma$, or instead just the decrease in repulsion between negative charges? (Or am I just confused and this is the same thing?) – Linear Christmas Mar 2 '17 at 15:37
• I guess it is two sides of the same coin. Less repulsion means that electrons on average are closer to the nucleus, so experience a stronger Coulombic attraction from the nucleus. I should make that slightly clearer. – orthocresol Mar 2 '17 at 16:33