Na vs Mg: Why First IE of Na is Lower but Second IE is Higher
Chemistry · Periodic Classification Of Properties · NEET
The FIRST ionization enthalpy of sodium (Na) is lower than magnesium (Mg) because Mg has a higher effective nuclear charge, so it holds its outer electrons more tightly. But the SECOND ionization enthalpy flips: Na's second IE is much higher than Mg's, because after Na loses one electron it becomes Na+ with a stable noble-gas core [Ne], and pulling an electron out of that stable core is very hard. Memory hook: "Na fights back on the SECOND punch" — its second electron sits in a full noble-gas shell.
First IE: Mg > Na (higher Zeff). Second IE: Na > Mg, because Na+ has the stable noble-gas core [Ne] and breaking it needs huge energy, while Mg+ still has an easy 3s electron to lose.
Your doubts, answered
Why is the FIRST ionization enthalpy of Na lower than Mg?
Na is [Ne]3s1 and Mg is [Ne]3s2. Mg has one more proton in its nucleus, so its effective nuclear charge (Zeff) is higher. The 3s electrons in Mg feel a stronger pull, so more energy is needed to remove the first electron. That is why first IE: Na (496 kJ/mol) < Mg (737 kJ/mol).
Why is the SECOND ionization enthalpy of Na higher than Mg?
After losing one electron, Na becomes Na+ which is [Ne] — a complete, stable noble-gas configuration. Removing a second electron means breaking into this very stable core, which needs huge energy. Mg becomes Mg+ = [Ne]3s1, which still has one loose valence electron that is easy to remove. So second IE: Na > Mg.
What is the exact electronic configuration change for Na and Mg?
Na: [Ne]3s1 -> Na+ [Ne] (noble-gas core) -> Na2+ needs to break the core (very hard). Mg: [Ne]3s2 -> Mg+ [Ne]3s1 (still has valence electron) -> Mg2+ [Ne] (easy, reaches noble-gas). Na hits the stable core one step earlier than Mg.
Is this a case of a 'big jump' in successive ionization enthalpy?
Yes. For Na the big jump comes between IE1 and IE2, because IE2 removes an electron from the noble-gas core. For Mg the big jump comes between IE2 and IE3. The jump always appears right after all valence electrons are gone.
Does higher Zeff always mean higher ionization enthalpy?
For neutral atoms across a period, yes — Mg beats Na in first IE because of higher Zeff. But for successive IE you must also check the electron configuration of the ION being ionized. A noble-gas core (like Na+) raises the IE massively, which is why the trend can flip.
⚠️ The NEET trap ✗ Since Mg has higher effective nuclear charge, Mg must have higher ionization enthalpy than Na at EVERY step (both first and second). ✓ Only the FIRST IE follows Zeff (Mg > Na). The SECOND IE flips: Na+ is [Ne] (noble-gas core), so Na's second IE is HIGHER than Mg's second IE. 🧠 NTA loves the flip. Always check the configuration of the ION you are ionizing, not just the neutral atom.
Real NEET questions
2025
Which statement about first ionization enthalpies is correct? 'The correct order of the first ionization enthalpies of Na, Mg, Al and Si is Si > Al > Mg > Na.'
A · The statement is correct as written: Si > Al > Mg > Na
B · The correct order is Si > Mg > Al > Na (Al dips below Mg) ✓
C · The correct order is Na > Mg > Al > Si
D · The correct order is Si > Al > Na > Mg
Solution: First IE generally rises across period 3, but Al ([Ne]3s2 3p1) has its outer electron in a 3p orbital that is easier to remove than the filled 3s2 of Mg. So Al dips below Mg. The real order is Si > Mg > Al > Na, NOT Si > Al > Mg > Na. This confirms Mg has a higher first IE than Na, the starting point of the Na vs Mg comparison.
2016
In which option does the order NOT agree with the property indicated: (B) B < C < N < O (increasing first ionization enthalpy)?
A · B < C < N < O is correct
B · B < C < N < O is wrong; correct is B < C < O < N ✓
C · O has the highest first IE in the set
D · All second period orders here are correct
Solution: Across a period first IE generally increases with Zeff, but N has a stable half-filled 2p3 configuration, giving N a higher first IE than O. So the correct order is B < C < O < N, meaning option (B) does not agree. This is the same principle — stable configurations override the simple Zeff trend — that makes Na's second IE higher than Mg's.
Solved Periodic Classification Of Properties NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
First IE: Na is about 496 kJ/mol, Mg is about 737 kJ/mol (Mg higher). Second IE: Na is about 4562 kJ/mol, Mg is about 1451 kJ/mol (Na much higher). NEET rarely needs exact numbers, but the direction of each trend is important.
In one line, why does the trend reverse for the second IE?
Because Na+ reaches the stable noble-gas core [Ne] after just one electron loss, while Mg+ still has a loose 3s electron left. Breaking a noble-gas core needs far more energy.
Is this concept directly in NCERT?
Yes. NCERT Chemistry (Classification of Elements and Periodicity) Exercise 3.17 asks exactly this: why first IE of Na is lower than Mg but its second IE is higher. The answer uses the noble-gas-core logic explained here.
How should I remember which flips?
Remember 'one step earlier': Na reaches the noble-gas core one ionization step earlier than Mg. So Na's big jump (and high second IE) comes first, at the second electron.
