Q.The first ionisation enthalpies of Na, Mg, Al and Si are in the order:

The key concept here is the Periodic Trend of First Ionisation Enthalpy.

1. First ionisation enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom. Generally, it increases across a period due to increasing effective nuclear charge and decreasing atomic radius, making it harder to remove an electron.
2. However, there is an exception between Group 2 and Group 13 elements. Group 2 elements have a stable, fully filled $ns^2$ electron configuration, requiring more energy to remove an electron. Group 13 elements have their first electron in a $np^1$ orbital, which is higher in energy and experiences greater shielding, making it easier to remove.
3. For Na ($3s^1$), Mg ($3s^2$), Al ($3s^2 3p^1$), and Si ($3s^2 3p^2$):
   • Na has the lowest ionisation enthalpy in the period.
   • Mg has a higher ionisation enthalpy than Na. …

Concept and Intuition: Periodic Trend of Ionisation Enthalpy

Ionisation enthalpy (or ionisation energy) is the minimum energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. The first ionisation enthalpy ($IE_1$) refers to the removal of the first electron.

The general trend for first ionisation enthalpy across a period (from left to right) is that it increases. This is primarily due to two factors:

1. Increasing Effective Nuclear Charge ($Z_{eff}$): As we move across a period, the number of protons in the nucleus increases, leading to a stronger attraction for the valence electrons. While the number of inner electrons (shielding electrons) remains the same, the valence electrons are in the same principal energy level, so the shielding effect doesn't fully counteract the increased nuclear charge.
2. Decreasing Atomic Radius: The stronger attraction from the nucleus pulls the electron cloud closer, resulting in a smaller atomic radius. Electrons closer to the nucleus are held more tightly and require more energy to remove.

However, there are important exceptions to this general trend, particularly between Group 2 and Group 13 elements, and between Group 15 and Group 16 elements. These exceptions arise from the extra stability associated with half-filled or fully-filled subshells.

For the elements in question (Na, Mg, Al, Si), the key exception occurs between Magnesium (Group 2) and Aluminium (Group 13).

Step-by-Step Solution

Let's analyse the first ionisation enthalpies of Na, Mg, Al, and Si by considering their electronic configurations and positions in the periodic table (all are in Period 3).

1. Electronic Configurations:

   • Na (Sodium, Z=11): $[Ne] 3s^1$ (Group 1)
   • Mg (Magnesium, Z=12): $[Ne] 3s^2$ (Group 2)
   • Al (Aluminium, Z=13): $[Ne] 3s^2 3p^1$ (Group 13)
   • Si (Silicon, Z=14): $[Ne] 3s^2 3p^2$ (Group 14)

2. Comparing Na and Mg:

   • Moving from Na to Mg, the nuclear charge increases from $+11$ to $+12$.
   • Na has one electron in the $3s$ orbital ($3s^1$), which it readily loses to achieve a stable noble gas configuration ($[Ne]$).
   • Mg has a fully-filled $3s$ orbital ($3s^2$), which is a very stable configuration. Removing an electron from this stable, fully-filled orbital requires more energy than removing the single $3s$ electron from Na.
   • Therefore, $IE_1(\text{Na}) < IE_1(\text{Mg})$. This follows the general trend.

3. Comparing Mg and Al (The Exception):

   • Moving from Mg to Al, the nuclear charge increases from $+12$ to $+13$. According to the general trend, $IE_1$ should increase. However, this is where the exception occurs.
   • Mg ($[Ne] 3s^2$): To remove the first electron, we must break the highly stable, fully-filled $3s$ orbital.
   • Al ($[Ne] 3s^2 3p^1$): To remove the first electron, we remove the single electron from the $3p$ orbital.
     • The $3p$ electron is at a slightly higher energy level than the $3s$ electrons.
     • The $3p$ electron experiences greater shielding from the inner $3s^2$ electrons compared to a $3s$ electron. This means the $3p$ electron is less tightly held by the nucleus.
     • Removing the $3p^1$ electron leaves Al with a stable, fully-filled $3s^2$ configuration. …