Chemistry · Periodic Classification Of Properties · NEET
First ionization enthalpy usually rises across a period, but there are two dips: Be is higher than B, and N is higher than O. Beryllium has a full and stable 2s2 shell, so its electron is hard to pull out; boron's lone 2p1 electron comes out more easily. Nitrogen has a stable half-filled 2p3 set, so it holds its electrons tighter than oxygen. Memory hook: "Full and half-filled shells are strong and stubborn" — Be (full 2s) and N (half-full 2p) both refuse to give up an electron easily.
First ionization enthalpy rises across period 2, but dips at B (red: Be > B, full 2s2) and at O (green: N > O, half-filled 2p3). These two swaps are the anomalies NEET tests.
Your doubts, answered
Why is the first ionization enthalpy of Be greater than B, when B is further right in the period?
Beryllium (Z=4) has the configuration 1s2 2s2. Its outer electron sits in a completely filled 2s orbital, which is a stable, low-energy arrangement, so a lot of energy is needed to remove it. Boron (Z=5) is 1s2 2s2 2p1. Its outermost electron is a single 2p1 electron. The 2p orbital is higher in energy and less tightly held than a filled 2s, so boron's electron comes out more easily. That is why IE1 of Be is greater than B, breaking the normal left-to-right increase.
Why does nitrogen have a higher first ionization enthalpy than oxygen?
Nitrogen (Z=7) is 1s2 2s2 2p3. The three 2p electrons sit one in each of the px, py, pz orbitals, giving a half-filled 2p3 set. This symmetric, half-filled arrangement is extra stable, so nitrogen holds its electrons tightly. Oxygen (Z=8) is 1s2 2s2 2p4. Its fourth 2p electron must pair up in an already-occupied orbital. That pairing brings electron-electron repulsion, so this electron is easier to remove. Result: IE1 of N is greater than O, even though O has more protons.
What does 'half-filled' and 'fully-filled stability' actually mean here?
When a subshell is exactly half-filled (like p3, one electron in each p orbital) or completely filled (like s2 or p6), the electron cloud is symmetric and the exchange energy is high. This makes the atom more stable and lowers its energy. To pull an electron out of such an atom, you must break this stable arrangement, which costs extra energy. That extra cost is why Be (full 2s2) and N (half-filled 2p3) have unusually high first ionization enthalpies compared to their neighbours.
Does the same anomaly repeat in the third period (Mg>Al and P>S)?
Yes. The same two reasons apply. Magnesium (3s2, full s) has a higher IE1 than aluminium (3s2 3p1, lone p electron), so Mg is greater than Al. Phosphorus (3p3, half-filled) has a higher IE1 than sulphur (3p4, paired electron), so P is greater than S. NEET often tests the Na, Mg, Al, Si order: the correct one is Si > Mg > Al > Na, because Al dips below Mg.
Why is nitrogen's ionization enthalpy still less than fluorine's?
Fluorine (2p5) is further right, with a larger effective nuclear charge pulling the electrons in tightly, so IE1 of F is very high. Nitrogen's half-filled stability only lifts it above its immediate neighbour oxygen; it does not beat fluorine. So the order in that region is O < N < F. Remember: the anomaly is local — it swaps only the pair Be/B and the pair N/O, nothing more.
⚠️ The NEET trap ✗ First ionization enthalpy always increases smoothly across a period, so the order for period 2 is Li < Be < B < C < N < O < F < Ne. ✓ There are two dips. Be > B (full 2s2 is stable) and N > O (half-filled 2p3 is stable). The real order is Li < B < Be < C < O < N < F < Ne. 🧠 NEET loves hiding the two swaps. Whenever you see B, Be, N, O in an ionization order, check: did they put Be above B, and N above O? If not, that option is the wrong one.
Real NEET questions
2019
For the second period elements the correct increasing order of first ionisation enthalpy is:
A · Li < Be < B < C < N < O < F < Ne
B · Li < B < Be < C < O < N < F < Ne ✓
C · Li < B < Be < C < N < O < F < Ne
D · Li < Be < B < C < O < N < F < Ne
Solution: First ionisation enthalpy generally increases across a period, but two anomalies appear. Be > B because boron's single 2p1 electron is easier to remove than an electron from beryllium's filled 2s2. N > O because nitrogen's half-filled 2p3 is extra stable. Applying both swaps gives Li < B < Be < C < O < N < F < Ne, which is option B.
2024
Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N.
A · Li < B < Be < C < N ✓
B · Li < Be < C < B < N
C · Li < Be < N < B < C
D · Li < Be < B < C < N
Solution: IE1 rises across period 2, but there is an anomaly at boron. Be has a stable, fully filled 2s2, so it is harder to ionize than B, whose lone 2p1 electron leaves more easily. Hence B < Be. There is no N/O pair here, so the rest increases normally: Li < B < Be < C < N. That is option A.
ReNEET 2026
Assertion A: The first ionization enthalpy of O is lower than that of N and F. Reason R: The loss of an electron from O leads to a stable half-filled p orbital. Choose the most appropriate answer:
A · Both A and R are correct and R is the correct explanation of A ✓
B · Both A and R are correct but R is NOT the correct explanation of A
C · A is correct but R is not correct
D · A is not correct but R is correct
Solution: O (2p4) has a lower first ionization enthalpy than both N (2p3) and F (2p5), so the Assertion is correct. Removing one electron from oxygen gives O+ (2p3), the extra-stable half-filled configuration, which makes that electron easier to remove. So the Reason correctly explains the Assertion. Answer: A.
Solved Periodic Classification Of Properties NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What are the only two exceptions in first ionization enthalpy across period 2?
Be > B and N > O. Both come from stable electron arrangements: Be has a full 2s2 shell, and N has a half-filled 2p3 set. These are the swaps NEET tests most.
Is it enough to just memorise Be > B and N > O for NEET?
You can memorise the order, but knowing the reason (full 2s2 stability and half-filled 2p3 stability) lets you answer assertion-reason and third-period questions like Mg > Al and P > S too. The reason wins more marks.
Does the same rule apply to Al and S in period 3?
Yes. Mg > Al (full 3s2 vs lone 3p1) and P > S (half-filled 3p3 vs paired 3p4). It is the same logic, so learn it once.
Why is oxygen's ionization enthalpy still higher than nitrogen's is a common wrong idea — is it true?
No. It is the opposite. Nitrogen's ionization enthalpy is HIGHER than oxygen's because of half-filled 2p3 stability. The correct order is O < N.
