Q.How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?

The key idea is that ionization enthalpy depends on nuclear charge, atomic size, and electronic configuration — specifically, the stability of the resulting electron configuration after removal.

Step 1: Sodium ($\text{Na}$, $1s^2 2s^2 2p^6 3s^1$) has a larger atomic radius and lower effective nuclear charge than magnesium ($\text{Mg}$, $1s^2 2s^2 2p^6 3s^2$). Removing the single $3s$ electron from Na is easier, so its first ionization enthalpy is lower.

Step 2: After losing one electron, Na⁺ attains a stable noble gas configuration ($1s^2 2s^2 2p^6$). Removing a second electron from this stable core requires a very large amount of energy. …

Why this happens — the core idea

Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom. The first ionization enthalpy ($IE_1$) depends on how tightly the outermost electron is held. The second ionization enthalpy ($IE_2$) depends on the stability of the resulting ion after the first electron is removed.

Sodium (Na, atomic number 11) and magnesium (Mg, atomic number 12) are neighbours in Period 3. Their electronic configurations are:

• Na: $1s^2 2s^2 2p^6 3s^1$
• Mg: $1s^2 2s^2 2p^6 3s^2$

The key difference: Na has a single electron in the 3s orbital, while Mg has a filled 3s subshell. This small difference leads to a dramatic reversal in ionization enthalpies.

Step-by-step reasoning

1. First ionization enthalpy — Na is lower than Mg

The first ionization enthalpy measures the energy to remove the most loosely bound electron.

• For Na, that electron is the lone $3s^1$ electron. It experiences relatively weak effective nuclear charge because the 3s orbital is shielded by the inner core of 10 electrons ($1s^2 2s^2 2p^6$). The nuclear charge is +11, but the effective nuclear charge felt by the 3s electron is only about +2.2 (due to shielding). So this electron is easy to remove.
• For Mg, the outermost electrons are $3s^2$. The nuclear charge is +12, and the effective nuclear charge felt by each 3s electron is higher (about +3.1) because the same 10 inner electrons shield less effectively when the nuclear charge is larger. Moreover, the two 3s electrons are paired in the same orbital, so there is some electron-electron repulsion, but the dominant factor is the higher effective nuclear charge.

Thus, $IE_1$ of Na (496 kJ/mol) is lower than that of Mg (738 kJ/mol).

2. What happens after the first electron is removed?

• Na → Na⁺: The electron configuration becomes $1s^2 2s^2 2p^6$, which is exactly the noble gas configuration of neon. This is an extremely stable, closed-shell configuration.
• Mg → Mg⁺: The electron configuration becomes $1s^2 2s^2 2p^6 3s^1$. This is not a noble gas configuration — it still has one electron in the 3s orbital.

3. Second ionization enthalpy — Na is much higher than Mg

Now we remove a second electron from the already positively charged ion.

• For Na⁺, the second electron must come from the stable, filled $2p$ subshell (core electron). This electron is much closer to the nucleus, experiences a much higher effective nuclear charge, and is part of a very stable octet. Removing it requires a huge amount of energy — about 4562 kJ/mol.
• For Mg⁺, the second electron is the lone $3s^1$ electron. This is still a valence electron, relatively far from the nucleus, and the ion Mg⁺ has a +1 charge, so the remaining electron is held more tightly than in neutral Mg, but not nearly as tightly as a core electron. The energy required is about 1451 kJ/mol. …