The first ionisation enthalpy of boron is lower than that of beryllium because:

Beryllium has a fully filled 2s² subshell, which is extra stable

Boron has a larger atomic radius

Beryllium has a fully filled 2s² subshell, which is extra stable

Boron has a higher nuclear charge

Beryllium is a metal and boron is not

The electron gain enthalpy of chlorine is more negative than that of fluorine because:

Fluorine is so small that the incoming electron faces heavy repulsion

Chlorine is more electronegative

Fluorine is so small that the incoming electron faces heavy repulsion

Chlorine has a higher nuclear charge

Fluorine has a half-filled p subshell

Why is the second ionisation enthalpy of sodium far greater than the first?

Na⁺ has a stable noble-gas configuration, so the next electron comes from a filled shell

Sodium becomes larger after losing one electron

Na⁺ has a stable noble-gas configuration, so the next electron comes from a filled shell

The nuclear charge increases after ionisation

Down a group, effective nuclear charge on the outermost electron changes very little because:

Each new inner shell screens the added protons very effectively

The number of protons does not increase

Each new inner shell screens the added protons very effectively

The Periodic Table and Periodicity

Inorganic chemistry has a reputation for being memory work, and it earns that reputation only when studied in the wrong order. Almost every fact about the p-block, the d-block and reactivity follows from two competing quantities — how strongly the nucleus pulls (effective nuclear charge) and how far away the outer electron sits. Learn to reason with those two, and you can derive trends you would otherwise memorise, and you can also handle the exceptions, which is what the paper actually tests. This chapter is therefore the highest-leverage investment in the whole inorganic syllabus: get it right and the blocks that follow become predictable.

Effective Nuclear Charge — the One Idea Behind Every Trend

An outer electron does not feel the full nuclear charge Z, because the inner electrons screen part of it. What it feels is Z_eff = Z − σ, where σ is the screening constant. Two facts about screening carry the whole chapter: inner-shell electrons screen very effectively, while electrons in the same shell screen each other poorly. So moving across a period adds one proton and one same-shell electron, and since the added electron screens badly, Z_eff rises. Moving down a group adds a whole new shell, so the outer electron is both further away and behind more screening, and Z_eff barely changes while distance grows a great deal.

The trends, all following from Z_eff and distance
Property	Across a period (left → right)	Down a group
Atomic radius	Decreases — stronger pull, same shell	Increases — a new shell each time
Ionisation enthalpy	Increases	Decreases
Electron gain enthalpy	More negative (with exceptions)	Less negative
Electronegativity	Increases; F is the highest of all	Decreases
Metallic character	Decreases	Increases
Oxide character	Basic → amphoteric → acidic	More basic going down

⚠️ A cation is always smaller and an anion always larger than the parent atom

Na is 186 pm but Na⁺ is only 95 pm — a loss of more than half the radius, because removing the 3s electron removes an entire shell and leaves the same nuclear charge pulling on fewer electrons. Cl is 99 pm but Cl⁻ is 181 pm, since the added electron increases repulsion without adding any proton. For an isoelectronic series — species with the same electron count, such as N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺, Al³⁺ — the radius falls steadily as the nuclear charge rises, so N³⁻ is the largest and Al³⁺ the smallest. Questions asking you to rank an isoelectronic set are testing exactly this, and the shortcut is simply: more protons, smaller ion.

🎯 Why the ionisation enthalpy of boron is lower than beryllium's

The general trend says ionisation enthalpy rises across a period, yet Be (899 kJ/mol) is higher than B (801). The reason is that Be has the configuration 2s², a fully filled and therefore stable subshell, so removing an electron from it costs extra; B is 2s²2p¹, and the lone 2p electron is both higher in energy and better screened, so it leaves more easily. The same logic covers the second anomaly, N (1402) being higher than O (1314): nitrogen's 2p³ is half-filled and stabilised by exchange energy, while oxygen's 2p⁴ has one paired-up electron suffering extra repulsion. So both exceptions come from stable configurations rather than from anything special about the elements, and every group-15/group-16 anomaly in the chapter reduces to this.

The Anomalies That Are Actually Examined

Exceptions worth knowing with their reasons
Observation	Reason
Electron gain enthalpy of Cl is more negative than F	F is so small that the incoming electron faces heavy repulsion in a compact 2p shell
Ga has almost the same radius as Al	d-block contraction — the intervening 3d electrons screen poorly
Hf is nearly identical in size to Zr	Lanthanoid contraction — 4f electrons screen even worse
Second ionisation enthalpy of Na is enormous	Na⁺ already has a noble-gas core, so the next electron comes from a filled shell
Noble gases have positive electron gain enthalpy	A filled shell must be disrupted, which costs energy
Be and Mg have positive electron gain enthalpy	Their ns² subshells are already fully filled

🎯 Why the lanthanoid contraction matters far beyond the lanthanoids

Across the fourteen lanthanoids, fourteen protons are added but the electrons go into the deeply buried 4f orbitals, which screen abysmally. The result is that by the time you reach hafnium, the accumulated contraction has cancelled out the size increase you would expect from adding a whole shell — so Hf (159 pm) is essentially the same size as Zr (160 pm) directly above it. This is not a curiosity: it is why zirconium and hafnium are chemically almost inseparable and had to be told apart by neutron absorption, why the 5d elements are unusually dense (Os and Ir are the densest of all elements), and why the second and third transition series resemble each other far more than either resembles the first.

Two more periodic ideas complete the picture. Diagonal relationships arise because moving right increases electronegativity while moving down decreases it, so the two cancel along a diagonal: Li resembles Mg, Be resembles Al, and B resembles Si. That is why lithium, unlike the other alkali metals, forms a nitride and a carbonate that decomposes on heating — magnesium-like behaviour. Amphoteric oxides sit at the metal–non-metal boundary and react with both acid and base; Al₂O₃, ZnO, PbO and SnO are the ones named in the syllabus, and the boundary itself is the reason they behave both ways.

Key takeaways

Z_eff = Z − σ. Same-shell electrons screen poorly, inner-shell electrons screen well — that drives every trend.
Across a period radius falls and ionisation enthalpy rises; down a group the reverse.
Cations are much smaller and anions much larger than their parent atoms.
In an isoelectronic series, more protons means a smaller ion: N³⁻ > O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺.
Be > B and N > O in ionisation enthalpy, because filled 2s² and half-filled 2p³ are extra stable.
Cl has a more negative electron gain enthalpy than F, because F is too small and the incoming electron is crowded.
The lanthanoid contraction makes Hf the same size as Zr and the 5d elements unusually dense.
Diagonal relationships: Li–Mg, Be–Al, B–Si, because the period and group trends cancel along a diagonal.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The correct order of ionic radii for the isoelectronic species N³⁻, O²⁻, F⁻, Na⁺ is:
1. Na⁺ > F⁻ > O²⁻ > N³⁻
2. N³⁻ > O²⁻ > F⁻ > Na⁺
3. All four are equal
4. F⁻ > N³⁻ > Na⁺ > O²⁻
Show answer
Answer: B — N³⁻ > O²⁻ > F⁻ > Na⁺
All four have ten electrons, so the only variable is nuclear charge: 7, 8, 9 and 11 protons respectively. More protons pull the same ten electrons in tighter, so the radius falls steadily from N³⁻ to Na⁺. The third option is the trap for candidates who assume equal electron counts must mean equal size, which ignores the nucleus entirely.
The first ionisation enthalpy of boron is lower than that of beryllium because:
1. Boron has a larger atomic radius
2. Beryllium has a fully filled 2s² subshell, which is extra stable
3. Boron has a higher nuclear charge
4. Beryllium is a metal and boron is not
Show answer
Answer: B — Beryllium has a fully filled 2s² subshell, which is extra stable
Beryllium is 1s²2s², and removing an electron from a completely filled subshell costs extra energy, while boron's single 2p electron is higher in energy and better screened, so it leaves more easily. Boron does have the higher nuclear charge, which is exactly why the general trend would predict the opposite — the configuration effect overrides it, and that is what makes this an examinable anomaly.
The electron gain enthalpy of chlorine is more negative than that of fluorine because:
1. Chlorine is more electronegative
2. Fluorine is so small that the incoming electron faces heavy repulsion
3. Chlorine has a higher nuclear charge
4. Fluorine has a half-filled p subshell
Show answer
Answer: B — Fluorine is so small that the incoming electron faces heavy repulsion
The 2p shell of fluorine is very compact, so an added electron is squeezed into a small volume already crowded with electrons and suffers strong repulsion, which offsets much of the energy released. Chlorine's larger 3p shell accommodates the electron comfortably. Note that fluorine is still the more electronegative element — electronegativity and electron gain enthalpy are different quantities, which is precisely the confusion the first option exploits.
Hafnium has almost the same atomic radius as zirconium because of:
1. The d-block contraction
2. The lanthanoid contraction
3. Their identical nuclear charge
4. Both being in the same period
Show answer
Answer: B — The lanthanoid contraction
Between Zr and Hf lie the fourteen lanthanoids, whose added electrons enter deeply buried 4f orbitals that screen very poorly, so the accumulated increase in effective nuclear charge cancels the size gain expected from an extra shell. The d-block contraction is the analogous but weaker effect responsible for gallium matching aluminium, and Zr and Hf are in different periods with very different nuclear charges.
Lithium resembles magnesium in forming a nitride and a thermally unstable carbonate. This is an example of:
1. A diagonal relationship
2. The inert pair effect
3. Isoelectronic behaviour
4. Amphoterism
Show answer
Answer: A — A diagonal relationship
Moving right across a period raises electronegativity and moving down a group lowers it, so the two changes cancel along a diagonal and Li ends up resembling Mg, Be resembling Al and B resembling Si. This is why lithium is the odd one out among alkali metals. The inert pair effect concerns heavy p-block elements preferring an oxidation state two lower, which is a different phenomenon entirely.
Which oxide is amphoteric?
1. Na₂O
2. Al₂O₃
3. SO₃
4. MgO
Show answer
Answer: B — Al₂O₃
Al₂O₃ dissolves in acid to give Al³⁺ and in alkali to give the aluminate ion, so it reacts with both — the definition of amphoteric. Na₂O and MgO are firmly basic because they sit on the metallic side of the period, and SO₃ is firmly acidic because it sits on the non-metallic side. Amphoteric oxides cluster at the metal–non-metal boundary, which is exactly where aluminium lies.
Why is the second ionisation enthalpy of sodium far greater than the first?
1. Sodium becomes larger after losing one electron
2. Na⁺ has a stable noble-gas configuration, so the next electron comes from a filled shell
3. The nuclear charge increases after ionisation
4. Sodium is a metal
Show answer
Answer: B — Na⁺ has a stable noble-gas configuration, so the next electron comes from a filled shell
Removing sodium's single 3s electron leaves Na⁺ with the neon configuration, so the second electron must be pulled out of a complete 2p⁶ shell that is both closer to the nucleus and far more stable — hence the huge jump. The nuclear charge is unchanged by ionisation, since protons are untouched; what changes is that fewer electrons now share the same pull, making the ion smaller and its electrons more tightly held.
Down a group, effective nuclear charge on the outermost electron changes very little because:
1. The number of protons does not increase
2. Each new inner shell screens the added protons very effectively
3. Atomic radius decreases
4. Electronegativity increases
Show answer
Answer: B — Each new inner shell screens the added protons very effectively
Protons certainly do increase down a group, but each step also adds a complete inner shell, and inner-shell electrons screen almost perfectly, so the net pull on the outermost electron stays nearly flat while the distance to it grows sharply. That is why ionisation enthalpy and electronegativity both fall down a group even though the nucleus is getting more positive.