Identify the trend across Period 3

The elements are: $\mathrm{Na},\ \mathrm{Mg},\ \mathrm{Al},\ \mathrm{Si}$

In general, first ionization enthalpy increases across a period because effective nuclear charge increases.

Recall the important exception

Although the general trend is increasing, there is a known exception between $\mathrm{Mg}$ and $\mathrm{Al}$.

• $\mathrm{Mg}$ has electronic configuration: $1s^2 2s^2 2p^6 3s^2$
• $\mathrm{Al}$ has electronic configuration: $1s^2 2s^2 2p^6 3s^2 3p^1$

The electron removed from $\mathrm{Al}$ is a $3p$ electron, which is:

• higher in energy than $3s$
• less penetrating
• more easily removed

Therefore, $I_1(\mathrm{Al}) < I_1(\mathrm{Mg})$

Use the given values

Given: $I_1(\mathrm{Na}) = 496\ \mathrm{kJ\,mol^{-1}}$ $I_1(\mathrm{Mg}) = 737\ \mathrm{kJ\,mol^{-1}}$ $I_1(\mathrm{Si}) = 786\ \mathrm{kJ\,mol^{-1}}$

So the ionization enthalpy of $\mathrm{Al}$ should be less than 737 but greater than that of $\mathrm{Na}$.

Check the options

• A: 487
  Too low; even lower than Na, so not possible.

• B: 768
  Greater than Mg, contradicts the known exception.

• C: 577
  Less than Mg and reasonable for Al.

• D: 856
  Much too high.

Final answer

Hence, $I_1(\mathrm{Al}) = 577\ \mathrm{kJ\,mol^{-1}}$

So the correct option is C.