Periodicity: Atomic Radius and Electronegativity

Periodicity is the recurring pattern of physical and chemical properties as you move across a period of the periodic table. These patterns arise directly from electronic structure. In this lesson, we examine two key periodic properties: atomic radius and electronegativity.

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Atomic Radius

The atomic radius is a measure of the size of an atom. Because electron clouds do not have sharp edges, we define atomic radius in practical terms:

• Covalent radius: Half the distance between the nuclei of two bonded atoms of the same element.
• Metallic radius: Half the distance between the nuclei of adjacent atoms in a metallic crystal.
• Van der Waals radius: Half the distance between the nuclei of two non-bonded atoms at their closest approach (used for noble gases and comparing non-bonded atoms).

Trend Across a Period

Atomic radius decreases across a period (left to right). Data for Period 3:

Element	Na	Mg	Al	Si	P	S	Cl	Ar
Z	11	12	13	14	15	16	17	18
Covalent radius (pm)	186	160	143	117	110	104	99	--
Decrease from Na (%)	--	14	23	37	41	44	47	--

(Argon does not form covalent bonds, so its covalent radius is not directly comparable.)

Explanation:

• Each element has one more proton in the nucleus (nuclear charge increases)
• Electrons are added to the same outer shell (same principal energy level)
• The increased nuclear charge pulls the electron cloud closer to the nucleus
• Shielding remains approximately constant (same number of inner shells)
• The effective nuclear charge experienced by outer electrons increases

The result is a stronger pull on the outer electrons, shrinking the atom.

Trend Down a Group

Atomic radius increases down a group because:

• Each element has an additional electron shell
• The outer electrons are further from the nucleus
• Increased shielding from additional inner shells reduces the effective nuclear charge
• These effects outweigh the increased nuclear charge

Group 1	Li	Na	K	Rb	Cs
Metallic radius (pm)	152	186	227	248	265
Number of shells	2	3	4	5	6

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Ionic Radius

When atoms form ions, their radii change significantly:

flowchart LR
    A["Metal atom (large)"] -- "Loses electrons" --> B["Cation (smaller)"]
    C["Non-metal atom (medium)"] -- "Gains electrons" --> D["Anion (larger)"]

Cations (positive ions) are smaller than their parent atoms because:

• Electrons are removed from the outermost shell
• Often, an entire shell is lost (e.g., Na -> Na+ loses its 3s electron, leaving only shells 1 and 2)
• The remaining electrons are pulled closer by the same (or now unbalanced) nuclear charge
• Example: Na (186 pm) -> Na+ (95 pm) -- nearly halved

Anions (negative ions) are larger than their parent atoms because:

• Extra electrons are added, increasing electron-electron repulsion
• The nuclear charge is now insufficient to hold the larger electron cloud as tightly
• Example: Cl (99 pm) -> Cl- (181 pm) -- nearly doubled

Ionic Radii Data for Period 3

Species	Na+	Mg2+	Al3+	Si4+	P3-	S2-	Cl-
Ionic radius (pm)	95	65	50	41	212	184	181
Protons	11	12	13	14	15	16	17
Electrons	10	10	10	10	18	18	18

Notice how the cations (Na+ to Al3+) are much smaller, while the anions (P3- to Cl-) are much larger than their parent atoms.

Isoelectronic Series

An isoelectronic series is a set of species with the same number of electrons. For example: O2-, F-, Ne, Na+, Mg2+, Al3+ all have 10 electrons.

Across this series, the nuclear charge increases (8, 9, 10, 11, 12, 13), but the electron count stays constant. Therefore, the radius decreases as nuclear charge increases:

Species	O2-	F-	Ne	Na+	Mg2+	Al3+
Protons	8	9	10	11	12	13
Electrons	10	10	10	10	10	10
Radius (pm)	140	136	--	95	65	50

Key insight: In an isoelectronic series, the species with the most protons has the smallest radius because it pulls the same number of electrons more tightly.

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Electronegativity

Electronegativity is the ability of an atom to attract the bonding pair of electrons in a covalent bond towards itself. The most commonly used scale is the Pauling scale.

Pauling Electronegativity Values

Element	F	O	Cl	N	Br	S	C	H	Na
Pauling EN	4.0	3.4	3.2	3.0	3.0	2.6	2.6	2.2	0.9

Fluorine is the most electronegative element (4.0). Caesium and francium are the least electronegative (approximately 0.7).

Trend Across a Period

Electronegativity increases across a period because:

• Nuclear charge increases
• Atomic radius decreases
• The bonding electrons are closer to the nucleus and experience a stronger pull
• Shielding remains roughly constant

graph LR
    A["Na<br/>EN = 0.9"] --> B["Mg<br/>EN = 1.3"]
    B --> C["Al<br/>EN = 1.6"]
    C --> D["Si<br/>EN = 1.9"]
    D --> E["P<br/>EN = 2.2"]
    E --> F["S<br/>EN = 2.6"]
    F --> G["Cl<br/>EN = 3.2"]

Trend Down a Group

Electronegativity decreases down a group because:

• Atomic radius increases
• More shielding from inner shells
• The bonding pair is further from the nucleus, experiencing less pull
• The increase in nuclear charge is insufficient to compensate

Group 17	F	Cl	Br	I
Pauling EN	4.0	3.2	3.0	2.7

Noble gases are generally not assigned electronegativity values because they do not typically form covalent bonds.

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Relationship Between Electronegativity and Bonding

The difference in electronegativity between two bonded atoms determines the bond type:

Electronegativity Difference	Bond Type	Example
0	Pure covalent (non-polar)	Cl-Cl (0)
0.1 - 0.4	Slightly polar covalent	C-H (0.4)
0.5 - 1.7	Polar covalent	H-Cl (1.0)
> 1.7	Ionic (electron transfer)	Na-Cl (2.3)

These boundaries are approximate. The transition from polar covalent to ionic is gradual, not abrupt. In reality, very few bonds are 100% ionic or 100% covalent -- most have some degree of both characters.

Worked Example 1

Predict the bond type in HCl.

• Electronegativity of H = 2.2
• Electronegativity of Cl = 3.2
• Difference = 3.2 - 2.2 = 1.0

This falls in the polar covalent range. The bonding pair is pulled towards chlorine, giving Cl a partial negative charge (delta-minus) and H a partial positive charge (delta-plus).

Worked Example 2

Predict the bond type in MgO.

• Electronegativity of Mg = 1.3
• Electronegativity of O = 3.4
• Difference = 3.4 - 1.3 = 2.1

This exceeds 1.7, indicating ionic bonding. Magnesium transfers two electrons to oxygen, forming Mg2+ and O2-.

Worked Example 3

Arrange the following bonds in order of increasing polarity: C-H, O-H, N-H, F-H.

Bond	EN Difference
C-H	2.6 - 2.2 = 0.4
N-H	3.0 - 2.2 = 0.8
O-H	3.4 - 2.2 = 1.2
F-H	4.0 - 2.2 = 1.8

Order of increasing polarity: C-H < N-H < O-H < F-H

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Electron Affinity (Brief Note)

The first electron affinity is the energy change when one mole of gaseous atoms gains one mole of electrons:

X(g) + e- -> X-(g)

For most non-metals, this is an exothermic process (negative value) because the incoming electron is attracted by the nuclear charge.

Element	O	F	Cl	Br	I
1st EA (kJ/mol)	-141	-328	-349	-325	-295

Note that chlorine has a more negative first electron affinity than fluorine, despite fluorine being more electronegative. This is because fluorine's very small atomic radius means the incoming electron experiences significant repulsion from the existing tightly packed electrons. Chlorine's slightly larger atom allows the electron to be accommodated with less repulsion.

The second electron affinity (e.g., O- -> O2-) is always endothermic because you are adding an electron to an already negatively charged ion, which requires energy to overcome the repulsion.

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Common Misconceptions

Misconception 1: "Electronegativity and electron affinity are the same thing."

Correction: Electronegativity measures the attraction for bonding electrons within a covalent bond (a relative scale). Electron affinity measures the energy change when a free gaseous atom gains an electron (a measurable quantity in kJ/mol). They are related but distinct properties.

Misconception 2: "Atomic radius increases across a period because more electrons are added."

Correction: More electrons ARE added, but they go into the SAME shell. The increasing nuclear charge pulls all the electrons in that shell closer to the nucleus, decreasing the radius. It is the increasing effective nuclear charge, not the number of electrons, that determines the trend.

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A-Level Deep Dive: Periodicity: Atomic Radius and Electronegativity

Spec mapping

Edexcel 9CH0 specification Topic 1, sub-topic 1.5 covers the periodic trends in atomic radius and electronegativity across periods (e.g. period 3, Na to Ar) and down groups; the explanation of these trends in terms of nuclear charge, atomic radius, shielding and (for electronegativity) the Pauling scale and its use in predicting bond polarity; and the comparative size of cations and anions relative to their parent atoms (refer to the official Pearson Edexcel specification document for exact wording). Examined in Paper 1 (Advanced Inorganic and Physical Chemistry) as part of the structural foundation of inorganic chemistry, with synoptic links into Paper 2 (bond polarity → reaction mechanisms in organic chemistry) and Paper 3 (data interpretation of trend tables).

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Worked example with full mark scheme

Question (7 marks):

(a) Define electronegativity and explain why fluorine is the most electronegative element on the Pauling scale. (3) (b) Predict, with reasoning, which has the larger radius: Na or Na⁺. State the approximate ratio of the two radii. (2) (c) Compare the atomic radius of N (period 2, Group 5) with that of P (period 3, Group 5). Explain the trend. (2)

Solution with mark scheme:

(a) B1 — electronegativity is the ability of an atom to attract a bonding pair of electrons in a covalent bond. B1 — measured on the Pauling scale; F = 4.0 (highest). B1 — fluorine has the highest electronegativity because it has a small atomic radius (so the bonding pair is close to the nucleus), high effective nuclear charge (Z = 9 with only one inner shell of two electrons shielding) and a nearly complete 2p subshell (one electron away from the noble-gas configuration of Ne, providing strong attraction for an additional electron). Common error: candidates write "F is most electronegative" without giving the small radius + high Zₑff justification.

(b) B1 — Na is larger than Na⁺. B1 — Na has electron configuration [Ne] 3s¹; removing the 3s electron to form Na⁺ gives [Ne], which has only two occupied shells rather than three. Atomic radius drops from ~186 pm to ~102 pm — Na⁺ is about half the size of Na. Common error: candidates assume the radius difference is small; in fact removing an entire valence shell typically halves the radius.

(c) B1 — P has a larger atomic radius than N. B1 — P has an additional shell (3 occupied shells vs N's 2), so its outermost electrons are further from the nucleus despite the higher nuclear charge of P. Shielding by inner electrons compensates for the increased nuclear charge.

Total: 7 marks (B3 + B2 + B2).

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Specimen question modelled on the Edexcel 9CH0 Paper 1 format

Question (5 marks): Use Pauling electronegativity values (H = 2.1, F = 4.0, Cl = 3.0, C = 2.5, O = 3.5, N = 3.0) to answer the following.

(a) Predict the bond polarity of (i) C–H, (ii) C–F, (iii) O–H. State the direction of the dipole. (3) (b) Explain why the C–F bond is more polar than the C–Cl bond, despite both halogens being more electronegative than carbon. (2)

Mark scheme decomposition by AO:

Part	AO1	AO2	AO3	Marks
(a)	1	2	0	3
(b)	0	2	0	2
Total	1	4	0	5