Q.The first ionization enthalpy ($\Delta_i H$) values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ mol$^{-1}$. Predict whether the first $\Delta_i H$ value for Al will be more close to 575 or 760 kJ mol$^{-1}$? Justify your answer.

Concept: Ionization enthalpy trends and electronic configuration effects in Period 3.

Ionization enthalpy generally increases across a period as nuclear charge rises and atomic radius decreases. However, aluminium shows an anomaly due to its electronic configuration.

Step 1 – Expected trend:

Moving from Na ($496$) → Mg ($737$) → Al → Si ($786$), we expect a steady increase. A simple interpolation would suggest Al lies between Mg and Si, around $760$ kJ mol$^{-1}$.

Step 2 – Electronic configurations:

• Mg: $[\text{Ne}] 3s^2$ (stable, fully-filled $s$-subshell)
• Al: $[\text{Ne}] 3s^2 3p^1$ (one electron in higher-energy $3p$ orbital)
• Si: $[\text{Ne}] 3s^2 3p^2$

Step 3 – The anomaly: …

Why ionization enthalpy matters and what drives the trend

Ionization enthalpy measures the energy needed to remove the most loosely bound electron from an isolated gaseous atom. Across a period, we expect this value to increase because:

• Nuclear charge rises (more protons pulling on electrons),
• Atomic radius shrinks (electrons sit closer to the nucleus),
• Shielding remains roughly constant (electrons are added to the same shell).

So the general story for Period 3 is: Na < Mg < Al < Si < … But nature loves exceptions, and aluminium is one of them.

Step-by-step reasoning

1. Examine the electronic configurations

Write out what we're actually ionizing:

The jump from Mg to Al involves a subshell change: we move from removing a 3s electron to removing a 3p electron.

2. Compare the 3s and 3p orbitals

A 3p electron sits in a higher-energy orbital than a 3s electron in the same shell. More importantly:

• The 3p orbital has a node at the nucleus and spends more time farther out,
• The filled 3s² subshell shields the 3p electron from the full nuclear charge,
• Even though Al has one more proton than Mg (13 vs. 12), the 3p electron feels a smaller effective nuclear charge than Mg's 3s electrons.

3. Predict the anomaly

Because the 3p¹ electron in Al is:

• farther from the nucleus on average,
• more shielded by the inner 3s² electrons,

it is easier to remove than we'd expect from the trend. So $\Delta_i H(\text{Al})$ should be less than $\Delta_i H(\text{Mg}) = 737$ kJ mol⁻¹, not more.

4. Choose between 575 and 760 kJ mol⁻¹

The value 760 kJ mol⁻¹ is higher than Mg's 737, which would continue the increasing trend — but we've just argued that Al breaks this trend.

The value 575 kJ mol⁻¹ sits between Na (496) and Mg (737), reflecting that Al's ionization is easier than Mg's but harder than Na's. This makes physical sense: Al has more nuclear charge than Na, so it's not as easy as Na, but the subshell jump makes it easier than Mg. …