Compare Atomic Radius of Na, Mg, Al — Across Period 3
Question
Arrange Na, Mg, and Al in decreasing order of atomic radius. Explain the trend with reference to nuclear charge and shielding.
Solution — Step by Step
Step 1: Locate These Elements in the Periodic Table
| Element | Symbol | Atomic Number (Z) | Period | Group | Outer configuration |
|---|---|---|---|---|---|
| Sodium | Na | 11 | 3 | 1 | 3s¹ |
| Magnesium | Mg | 12 | 3 | 2 | 3s² |
| Aluminium | Al | 13 | 3 | 13 | 3s² 3p¹ |
All three are in Period 3 — same outermost shell (n = 3). This is the key: since they all have electrons in the same shell, the comparison becomes purely about how hard the nucleus pulls on those outer electrons.
Step 2: Analyse Nuclear Charge and Shielding
The effective nuclear charge (Z_eff) determines how tightly the outer electrons are held:
Z_eff = Z − σ
The shielding constant σ comes from the inner electrons. For Period 3 elements, the inner core is [Ne] = 2 + 8 = 10 electrons. All three elements have the same inner core, so σ is similar for all.
| Element | Z | Inner core (σ ≈) | Z_eff ≈ |
|---|---|---|---|
| Na | 11 | 10 | 1 |
| Mg | 12 | 10 | 2 |
| Al | 13 | 10 | 3 |
Z_eff increases from Na → Mg → Al.
Step 3: Apply to Atomic Radius
Higher Z_eff = stronger pull on outer electrons = electron cloud contracts = smaller atom.
Since all three elements have outer electrons in the n=3 shell and shielding is roughly the same, the increasing nuclear charge directly causes the outer electron cloud to shrink:
Na (Z_eff ≈ 1) → largest Mg (Z_eff ≈ 2) → medium Al (Z_eff ≈ 3) → smallest
Step 4: Final Answer
Decreasing order of atomic radius: Na > Mg > Al
Actual Atomic Radii (Covalent/Metallic)
Na = 186 pm | Mg = 160 pm | Al = 143 pm
(pm = picometres = 10⁻¹² m)
Each step, nuclear charge increases by 1 while the electron shell stays the same → consistent decrease in radius.
Why This Works
The atomic radius trend across a period is a direct consequence of Z_eff increasing while the principal quantum number (n) of the outermost shell stays constant.
Think of it this way: you have three athletes (outer electrons) all running on the same track (n=3 shell). Each step from left to right adds one more person pulling them toward the centre (one more proton). The athlete is pulled inward with each additional proton — the orbital contracts.
The shielding from inner electrons (same [Ne] core) barely changes across Na → Mg → Al, so the full effect of each additional proton is felt by the outer electrons.
Alternative Method — Group vs Period Thinking
A common source of confusion is mixing period trends with group trends. Let's clarify:
- Down Group 1: Na > Li (radius increases going down — new shell)
- Across Period 3: Na > Mg > Al (radius decreases — same shell, more protons)
If asked to compare Na and K, the group trend dominates: K > Na (K has n=4 shell; Na has n=3). If asked to compare Na, Mg, Al, the period trend dominates: Na > Mg > Al.
If the question mixes group and period comparisons (e.g., Na, K, Mg, Al), go element by element:
- Identify which are in the same period vs different periods
- Within same period: apply Z_eff argument
- Across periods: the element with more shells is always larger
Common Mistake
⚠️ Common Mistake
Mistake: Writing the order as Al > Mg > Na (reversed) or Mg > Na > Al.
Why it happens: Students sometimes think "more electrons = larger atom" — which is true when comparing elements in different periods (more electrons means more shells). But for elements in the same period, more electrons come with more protons, and the extra nuclear charge wins over the added electrons.
Remember: Same period = increasing nuclear charge pulls the cloud inward = radius decreases left to right.
Quick check: Na is in Group 1 (farthest left) → largest in Period 3. Al is in Group 13 → smaller. Always verify your answer by checking group positions.
Extended Comparison — All Period 3 Metals
For a broader picture, Period 3 atomic radii follow:
Na (186 pm) > Mg (160 pm) > Al (143 pm) > Si (117 pm) > P (110 pm) > S (104 pm) > Cl (99 pm) > Ar (≈ 88 pm, van der Waals)
Every element in Period 3 is smaller than the one before it — a smooth, consistent decrease driven by the same Z_eff argument applied repeatedly across the period.