Inorganic Chemistry
Periodic Trends — Size, Ionisation & Electronegativity
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Periodic-trend questions look like a list of exceptions to memorise. They are not. Almost every trend, and almost every exception, comes from two things: how strongly the nucleus pulls on the outer electrons, and how stable a half-filled or filled subshell is.
What you will be able to do
- Explain trends in atomic and ionic radius using effective nuclear charge and shielding
- Order isoelectronic species by size
- Explain the ionisation-enthalpy exceptions (Be > B, N > O, Mg > Al, P > S) and identify a group from successive ionisation enthalpies
- Explain why Cl has a more negative electron gain enthalpy than F, and why noble gases are positive
- Predict the nature of oxides, diagonal relationships and the inert pair effect
The big picture
Two ideas explain nearly every trend.
In one line: Across a period effective nuclear charge rises, so atoms shrink and hold electrons more tightly; down a group new shells are added, so atoms grow and hold electrons less tightly. Half-filled and filled subshells explain the exceptions.
| Property | Across a period (→) | Down a group (↓) |
|---|---|---|
| Atomic radius | Decreases | Increases |
| Ionisation enthalpy | Increases (with exceptions) | Decreases |
| Electron gain enthalpy | More negative | Less negative (with exceptions) |
| Electronegativity | Increases | Decreases |
| Metallic character | Decreases | Increases |
| Basic nature of oxides | Decreases (basic → acidic) | Increases |
Think of it like this: Picture the nucleus as a magnet and the outer electrons as paper clips. Across a period the magnet gets stronger while the clips stay at about the same distance — they are pulled in closer and held tighter. Down a group the clips are placed on a new, outer ring and inner electrons get in the way — so the grip weakens.
Effective nuclear charge
The pull the outer electron actually feels.
Zeff = Z − σZ = nuclear charge; σ = shielding (screening) constant from the other electrons
- Inner electrons shield the outer ones; electrons in the same shell shield each other only weakly.
- Shielding power: s > p > d > f. Poorly shielding d and f electrons explain the d-block and lanthanoid contractions.
- Across a period Z rises by 1 each step but σ rises by much less, so Zeff increases.
- Down a group Zeff on the outer electron changes only slightly, but the new shell is much farther out.
Atomic radius
How big is an atom?
- Covalent radius: half the distance between two bonded identical atoms (non-metals). Metallic radius: half the distance between neighbouring atoms in the metal lattice. van der Waals radius: half the distance between non-bonded atoms in neighbouring molecules — the largest of the three.
- Across period 2: Li (152 pm) > Be > B > C > N > O > F (64 pm). Noble gases are quoted as van der Waals radii, so they look larger than the halogen before them.
- Down a group: radius increases: Li < Na < K < Rb < Cs.
Key idea: The Ga anomaly: Ga (135 pm) is slightly smaller than Al (143 pm). Ga follows the ten 3d elements, whose d electrons shield the extra nuclear charge poorly, so Ga’s outer electrons feel a larger pull (the d-block contraction).
Ionic radius
Gaining or losing electrons changes size dramatically.
- Cation < parent atom: losing electrons (often a whole shell) leaves the same nuclear charge pulling fewer electrons. Na (186 pm) → Na⁺ (102 pm).
- Anion > parent atom: extra electrons repel each other while nuclear charge is unchanged. Cl (99 pm) → Cl⁻ (181 pm).
- Higher positive charge → smaller: Fe³⁺ < Fe²⁺ < Fe.
Key idea: Isoelectronic species (same number of electrons) — the one with more protons is smaller. For 10 electrons: N³⁻ (171) > O²⁻ (140) > F⁻ (133) > Na⁺ (102) > Mg²⁺ (72) > Al³⁺ (54 pm).
Solved example 01 — Isoelectronic ordering
Arrange S²⁻, Cl⁻, K⁺ and Ca²⁺ in increasing order of size.
- All have 18 electrons.
- Protons: S 16, Cl 17, K 19, Ca 20.
- More protons pull the same 18 electrons in more tightly.
Answer: Ca²⁺ < K⁺ < Cl⁻ < S²⁻.
Check yourself
- C1. Which is larger: Mg or Mg²⁺? Why?
- C2. Arrange O²⁻, F⁻ and Na⁺ in decreasing size.
Answers at the end.
Ionisation enthalpy
The energy needed to pull off the outermost electron.
M(g) → M⁺(g) + e⁻ ΔiH₁Always positive (energy needed); second ΔiH₂ > first, because the electron leaves a positive ion
- Across a period it increases (higher Zeff, smaller atom). Down a group it decreases (larger atom, more shielding).
- Noble gases have the highest values in each period; alkali metals the lowest. He has the highest first ionisation enthalpy of all (2372 kJ mol⁻¹); Cs among the lowest.
| Pair | Values (kJ mol⁻¹) | Reason for the exception |
|---|---|---|
| Be > B | 899 > 801 | B loses a 2p electron, which is higher in energy and shielded by 2s |
| N > O | 1402 > 1314 | N has a stable half-filled 2p³; O loses a paired electron (repulsion helps) |
| Mg > Al | 738 > 577 | Al loses a 3p electron |
| P > S | 1012 > 1000 | P is half-filled 3p³ |
Successive ionisation enthalpies identify the group
A sudden very large jump appears when the next electron must come from a filled inner shell. The number of electrons removed before the jump = the number of valence electrons.
Solved example 02 — Finding the group
An element has ΔiH₁ = 738, ΔiH₂ = 1451 and ΔiH₃ = 7733 kJ mol⁻¹. Which group is it in?
- ΔiH₂ → ΔiH₃ is a five-fold jump; ΔiH₁ → ΔiH₂ is only doubled.
- So two electrons come off easily, and the third comes from the inner shell.
Answer: Group 2 (these are the values for Mg).
Electron gain enthalpy
Energy change when a gaseous atom gains an electron.
X(g) + e⁻ → X⁻(g) ΔegHNegative when energy is released; halogens have the most negative values
- Generally becomes more negative across a period and less negative down a group.
- Noble gases: positive — the electron would have to enter a new shell.
- Be, Mg (filled s²) and N (half-filled p³) have values near zero or positive.
- Second electron gain enthalpy is always positive: adding an electron to a negative ion needs energy (O⁻ → O²⁻ needs about +780 kJ mol⁻¹). O²⁻ exists in solids only because the lattice energy pays for it.
Key idea: Cl is more negative than F (−349 vs −328 kJ mol⁻¹). F is so small that the incoming electron is crowded into the compact 2p subshell and repelled by the electrons already there. The same reason makes S more negative than O. Order for halogens: Cl > F > Br > I.
Check yourself
- C3. Why does neon have a positive electron gain enthalpy?
- C4. Which has the more negative electron gain enthalpy: O or S?
Answers at the end.
Electronegativity
How strongly a bonded atom pulls the shared pair toward itself.
- Not a property of an isolated atom — it describes an atom in a bond, so it has no unit. On the Pauling scale F = 4.0 (highest), O = 3.5, N = Cl = 3.0, C = 2.5, H = 2.1, Cs ≈ 0.7.
- Increases across a period, decreases down a group.
- A larger difference between two atoms → more polar bond; a difference above about 1.7 suggests largely ionic character.
- Electronegativity rises with oxidation state of the same element (Mn in KMnO₄ is far more electronegative than in MnO).
| Ionisation enthalpy | Electron gain enthalpy | Electronegativity |
|---|---|---|
| Isolated gaseous atom | Isolated gaseous atom | Atom in a molecule |
| Measured (kJ mol⁻¹) | Measured (kJ mol⁻¹) | Relative scale, no unit |
Oxides and metallic character
Metals on the left, non-metals on the right — and their oxides follow.
| Period 3 oxide | Na₂O | MgO | Al₂O₃ | SiO₂ | P₄O₁₀ | SO₃ | Cl₂O₇ |
|---|---|---|---|---|---|---|---|
| Nature | Strongly basic | Basic | Amphoteric | Weakly acidic | Acidic | Strongly acidic | Very strongly acidic |
- Amphoteric oxides react with both acids and bases: BeO, Al₂O₃, ZnO, Ga₂O₃, SnO, SnO₂, PbO, PbO₂.
- Neutral oxides: CO, NO, N₂O.
- For one element, a higher oxidation state gives a more acidic oxide: MnO (basic) → Mn₂O₇ (acidic); CrO (basic) → Cr₂O₃ (amphoteric) → CrO₃ (acidic).
- Metallic character decreases across a period and increases down a group; reducing power of metals follows the same pattern.
Solved example 03 — Amphoteric behaviour
Show that Al₂O₃ is amphoteric.
- With acid: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O (acts as a base).
- With alkali: Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄] (acts as an acid).
Answer: It reacts with both acids and bases.
Special behaviour
Second-period anomalies, diagonal pairs and the inert pair effect.
- First member of each group is anomalous (Li, Be, B, C, N, O, F): small size, high electronegativity, and no d orbitals. Its maximum covalency is 4 (B forms [BF₄]⁻, Al forms [AlF₆]³⁻). It forms pπ–pπ multiple bonds easily (C=C, N≡N, O=O) where heavier members prefer single bonds.
- Diagonal relationship: Li–Mg, Be–Al, B–Si resemble each other because their charge-to-radius ratios (and electronegativities) are similar. Example: Li and Mg both form nitrides on heating in N₂; BeO and Al₂O₃ are both amphoteric; B and Si both form acidic oxides and covalent hydrides.
- Inert pair effect: in groups 13–15, the ns² pair becomes more reluctant to take part in bonding down the group. So Tl⁺ is more stable than Tl³⁺, Pb²⁺ more stable than Pb⁴⁺ (PbO₂ is a strong oxidant), and Bi³⁺ more stable than Bi⁵⁺.
Key idea: Why the inert pair effect appears: the poor shielding by the 4f and 5d electrons pulls the 6s pair closer to the nucleus, making it harder to remove; the energy released by forming two extra bonds is not enough to pay for this.
Exam trap gallery
Where marks are lost in periodic trends.
| # | What students write | What is actually correct | Why |
|---|---|---|---|
| 1 | F has the most negative electron gain enthalpy. | Cl (−349) is more negative than F (−328 kJ mol⁻¹). | F’s small 2p subshell repels the incoming electron. |
| 2 | O has a higher ionisation enthalpy than N. | N (1402) > O (1314 kJ mol⁻¹). | N is half-filled 2p³; O loses a paired electron. |
| 3 | Al is smaller than Ga. | Ga (135 pm) is slightly smaller than Al (143 pm). | Poor shielding by 3d electrons in Ga. |
| 4 | Electronegativity and electron gain enthalpy are the same thing. | ΔegH is for an isolated atom (kJ mol⁻¹); electronegativity is for a bonded atom (no unit). | Different definitions, different scales. |
| 5 | Na⁺ is larger than F⁻ because Na is larger than F. | F⁻ (133 pm) > Na⁺ (102 pm). | Isoelectronic: more protons → smaller. |
| 6 | All second electron gain enthalpies are negative. | They are always positive. | An electron is being added to a negative ion. |
| 7 | Pb⁴⁺ is more stable than Pb²⁺, like C⁴⁺ chemistry. | Pb²⁺ is more stable (inert pair effect). | The 6s² pair resists bonding. |
| 8 | Al₂O₃ is basic because Al is a metal. | Al₂O₃ is amphoteric. | It dissolves in both acid and alkali. |
Memory hook: “Left-down big, right-up tight” — radius large at bottom-left; IE, EN high at top-right. · “Be-B, N-O, Mg-Al, P-S” — the four IE exceptions. · “Chlorine wins the electron” — most negative ΔegH. · “Li-Mg, Be-Al, B-Si go diagonal.”
Practice MCQs
Levels: Easy, Moderate, Tough.
Q01. Atomic radius generally decreases across a period because: (Easy)
- (a) shielding increases
- (b) effective nuclear charge increases
- (c) number of shells increases
- (d) mass increases
Q02. The element with the highest electronegativity is: (Easy)
- (a) O
- (b) Cl
- (c) F
- (d) N
Q03. The smallest ion among these is: (Easy)
- (a) Na⁺
- (b) Mg²⁺
- (c) Al³⁺
- (d) F⁻
Q04. Which has a positive electron gain enthalpy? (Easy)
- (a) Cl
- (b) O
- (c) Ne
- (d) S
Q05. Which oxide is amphoteric? (Easy)
- (a) Na₂O
- (b) Al₂O₃
- (c) SO₃
- (d) P₄O₁₀
Q06. Correct order of first ionisation enthalpy: (Moderate)
- (a) B > Be
- (b) O > N
- (c) N > O
- (d) Al > Mg
Q07. The most negative electron gain enthalpy is shown by: (Moderate)
- (a) F
- (b) Cl
- (c) Br
- (d) I
Q08. Correct order of size: (Moderate)
- (a) O²⁻ < F⁻ < Na⁺
- (b) Na⁺ < F⁻ < O²⁻
- (c) F⁻ < O²⁻ < Na⁺
- (d) Na⁺ < O²⁻ < F⁻
Q09. Ionisation enthalpies 496, 4562, 6910 kJ mol⁻¹ suggest the element is in group: (Moderate)
- (a) 1
- (b) 2
- (c) 13
- (d) 14
Q10. Diagonal relationship exists between: (Moderate)
- (a) Li and Na
- (b) Be and Al
- (c) B and Al
- (d) Na and Mg
Q11. Maximum covalency of boron is 4 because: (Moderate)
- (a) it is small
- (b) it has no d orbitals
- (c) it is a metal
- (d) it is electron-rich
Q12. Which pair has almost the same atomic radius? (Moderate)
- (a) Al and Ga
- (b) Na and K
- (c) B and Al
- (d) Li and Be
Q13. The more stable oxidation state of thallium is: (Tough)
- (a) +1
- (b) +3
- (c) +2
- (d) +5
Q14. Second electron gain enthalpy of oxygen is: (Tough)
- (a) negative
- (b) positive
- (c) zero
- (d) equal to the first
Q15. Which is the most acidic oxide? (Tough)
- (a) Na₂O
- (b) MgO
- (c) SiO₂
- (d) Cl₂O₇
Answer key with explanations
- (b) effective nuclear charge increases — Zeff rises.
- (c) F — Pauling value 4.0.
- (c) Al³⁺ — Isoelectronic; most protons.
- (c) Ne — Noble gas.
- (b) Al₂O₃ — Reacts with acid and base.
- (c) N > O — Half-filled 2p³.
- (b) Cl — F is too small.
- (b) Na⁺ < F⁻ < O²⁻ — Fewer protons → larger.
- (a) 1 — Big jump after the first.
- (b) Be and Al — Similar charge/radius ratio.
- (b) it has no d orbitals — Only 2s and 2p available.
- (a) Al and Ga — Ga 135, Al 143 pm.
- (a) +1 — Inert pair effect.
- (b) positive — Adding e⁻ to O⁻.
- (d) Cl₂O₇ — Cl in +7, far right.
Check-yourself answers
- C1. Mg. Mg²⁺ has lost its whole 3s shell, and 12 protons pull 10 electrons.
- C2. O²⁻ > F⁻ > Na⁺.
- C3. Its next electron would enter a new shell (3s), far from the nucleus and well shielded, so energy must be supplied.
- C4. S (−200 vs −141 kJ mol⁻¹): O is small and its 2p subshell is crowded.
One-page revision
Read this page the night before the exam.
Zeff and size
- Zeff = Z − σ; rises across a period
- Radius: ↓ across, ↑ down
- Shielding s > p > d > f
- Ga (135) < Al (143) — d contraction
Ions
- Cation < atom < anion
- Isoelectronic: more protons → smaller
- N³⁻ > O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺
- Fe³⁺ < Fe²⁺
Ionisation enthalpy
- ↑ across, ↓ down; He highest
- Exceptions: Be > B, N > O, Mg > Al, P > S
- Big jump = inner shell reached
- ΔiH₂ > ΔiH₁ always
Electron gain enthalpy
- Cl > F > Br > I (most negative)
- S more negative than O
- Noble gases, Be, Mg, N ≈ 0 or +
- Second ΔegH always positive
Electronegativity and oxides
- F 4.0 > O 3.5 > N = Cl 3.0
- Na₂O basic → Al₂O₃ amphoteric → Cl₂O₇ acidic
- Amphoteric: BeO, Al₂O₃, ZnO, SnO, PbO
- Neutral: CO, NO, N₂O
Special behaviour
- 2nd-period: no d orbitals, max covalency 4
- Diagonal: Li–Mg, Be–Al, B–Si
- Inert pair: Tl⁺, Pb²⁺, Bi³⁺ stable
- PbO₂ strong oxidant