Classification of elements and Periodic Table

There are 118 known elements. Learning each one separately would be hopeless; learning the pattern they fall into takes an afternoon. Arrange the elements in order of atomic number and their properties come back again and again at regular intervals. This unit explains why the table has the shape it has, and how six properties — atomic radius, ionic radius, ionization energy, electron affinity, electronegativity and metallic character — change in a way you can predict.

4.1 Modern periodic law and modern periodic table

Mendeleev (1869) arranged the elements by increasing atomic mass, but had to reverse some pairs to keep the chemistry right: Ar (39.9) before K (39.1), Co before Ni, Te before I. In 1913 Henry Moseley measured the frequency \nu of the characteristic X-rays emitted by different metals and found

where Z is the atomic number and a, b are constants. Atomic number, not atomic mass, is the fundamental property of an element, and every reversal disappears when the elements are ordered by Z.

The long form (modern) periodic table has 7 horizontal periods and 18 vertical groups. A period number equals the principal quantum number n of the outermost shell being filled, so the number of elements in a period is fixed by how many orbitals that shell offers.

Period	n	Subshells filled	No. of elements	Name
1	1	1s	2	very short
2	2	2s 2p	8	short
3	3	3s 3p	8	short
4	4	4s 3d 4p	18	long
5	5	5s 4d 5p	18	long
6	6	6s 4f 5d 6p	32	very long
7	7	7s 5f 6d 7p	32	very long

Elements in the same group have the same valence configuration, so they behave alike: all of group 1 reacts with water to give a hydroxide and hydrogen, all of group 17 forms an X⁻ ion. Depending on which subshell receives the last electron, an element belongs to one of four blocks.

Block	Last electron enters	Outer configuration	Groups	Character
s	ns	ns¹⁻²	1, 2 (and He)	soft, very reactive metals
p	np	ns² np¹⁻⁶	13–18	metals, metalloids, non-metals
d	(n−1)d	(n−1)d¹⁻¹⁰ ns⁰⁻²	3–12	transition metals
f	(n−2)f	(n−2)f¹⁻¹⁴ (n−1)d⁰⁻¹ ns²	inside group 3	inner transition

4.2 IUPAC classification of elements

The old labels IA, IIA … IB, IIB were used differently in Europe and America, which caused confusion. In 1988 IUPAC replaced them: the groups are simply numbered 1 to 18 from left to right.

IUPAC group	Old (CAS)	Family name	Valence configuration
1	IA	Alkali metals	ns¹
2	IIA	Alkaline earth metals	ns²
3–12	IIIB–IIB	Transition elements	(n−1)d¹⁻¹⁰ns¹⁻²
13	IIIA	Boron family	ns²np¹
14	IVA	Carbon family	ns²np²
15	VA	Nitrogen family	ns²np³
16	VIA	Chalcogens	ns²np⁴
17	VIIA	Halogens	ns²np⁵
18	0	Noble gases	ns²np⁶ (He: 1s²)

IUPAC also fixed the quarrel over who discovered the very heavy elements by giving a systematic name to every element with Z > 100. Write the digits of the atomic number, replace each digit by its Latin/Greek root, join them in order, and add -ium. The symbol is the first letter of each root.

4.3 Nuclear charge and effective nuclear charge

The nucleus carries a charge of +Ze. An outer electron, however, does not feel the full pull of +Ze, because the electrons between it and the nucleus push it outwards. This partial cancellation is the screening or shielding effect, measured by the screening constant \sigma. What the electron actually feels is the effective nuclear charge:

Screening power depends on how close to the nucleus an orbital keeps its electron. Penetration falls in the order s > p > d > f, so the shielding power of the subshells also falls in the order s > p > d > f.

\sigma can be estimated with Slater's rules. Write the configuration in the groups (1s)(2s,2p)(3s,3p)(3d)(4s,4p)… and, for an electron in an ns or np orbital, add up:

Other electrons located in	Contribution to \sigma each
a group to the right (higher shells)	0.00
the same (ns, np) group	0.35 (but 0.30 in 1s)
the shell just inside, n-1	0.85
shell n-2 and deeper	1.00

Period 3 element	Na	Mg	Al	Si	P	S	Cl
Z	11	12	13	14	15	16	17
Z_{\text{eff}} (Slater)	2.20	2.85	3.50	4.15	4.80	5.45	6.10
Atomic radius / pm	186	160	143	117	110	104	99
IE₁ / kJ mol⁻¹	496	738	578	786	1012	1000	1251

4.4 Periodic trend and periodicity: atomic radii, ionic radii, ionization energy, electron affinity, electronegativity, metallic character

Periodicity is the repetition of similar properties after regular intervals in the periodic table. It happens because the outer-shell configuration repeats: every element in group 1 ends in ns¹, so every element in group 1 behaves alike.

Atomic radii

An atom has no sharp boundary, so radius is always defined operationally.

Type	Defined as	Used for
Covalent radius	half the internuclear distance in X–X of a single bond	non-metals, e.g. Cl 99 pm
Metallic radius	half the internuclear distance of nearest atoms in the metal crystal	metals, e.g. Na 186 pm
van der Waals radius	half the distance of closest approach of two non-bonded atoms	noble gases, e.g. Ar 154 pm

For the same element van der Waals radius > metallic radius > covalent radius.

Across a period the radius decreases: n stays the same while Z_{\text{eff}} rises, pulling the shell in. Down a group it increases: a whole new shell is added, which easily beats the rise in Z_{\text{eff}}.

Ionic radii

A cation is always smaller than its parent atom: Na (186 pm) → Na⁺ (95 pm), because the whole 3s shell is removed and the remaining 10 electrons are pulled in by 11 protons. An anion is always larger: Cl (99 pm) → Cl⁻ (181 pm), because the added electron increases electron–electron repulsion while Z is unchanged.

For an isoelectronic series — ions with the same number of electrons — size falls as nuclear charge rises:

Ion	N³⁻	O²⁻	F⁻	Ne	Na⁺	Mg²⁺	Al³⁺
Electrons	10	10	10	10	10	10	10
Protons	7	8	9	10	11	12	13
Radius / pm	171	140	136	—	95	65	50

Ionization energy

Successive values always increase, IE₁ < IE₂ < IE₃ …, because each electron is pulled from an ion of higher positive charge. A very big jump marks the point at which a noble-gas core is broken: for sodium IE₁ = 496 but IE₂ = 4562 kJ mol⁻¹, which is why sodium forms Na⁺ and never Na²⁺.

Ionization energy increases across a period (Z_{\text{eff}} up, size down) and decreases down a group (size up, shielding up).

Two irregularities are asked almost every year:

• Be (899) > B (801). Beryllium's electron leaves a filled 2s² orbital, while boron's leaves a 2p orbital that is higher in energy and well shielded by 2s².
• N (1402) > O (1314). Nitrogen is 2p³, a half-filled subshell with extra exchange stability and no paired electron. Oxygen is 2p⁴; the two electrons paired in one 2p orbital repel each other, so one is easier to pull off.

The same pair of reasons gives Mg > Al and P > S in period 3.

Electron affinity

Second electron affinity is always endothermic, because the electron must be forced onto an already negative ion: O⁻(g) + e⁻ → O²⁻(g) needs about +780 kJ mol⁻¹. Electron affinity rises across a period and falls down a group, but the first member of each group breaks the rule.

Fluorine (328) has a smaller electron affinity than chlorine (349), and oxygen (141) a smaller one than sulphur (200): in F and O the 2p subshell is so compact that the electrons already there repel the incoming one strongly. Groups 2, 15 and 18 have near-zero or negative values because their subshells are filled or half-filled.

Electronegativity

It increases across a period and decreases down a group. It is not the same as electron affinity: electron affinity belongs to an isolated atom and is an energy in kJ mol⁻¹, whereas electronegativity belongs to a bonded atom and has no units. The electronegativity difference decides whether a bond is non-polar covalent, polar covalent or ionic (Unit 5).

Metallic character

Metallic (electropositive) character is the tendency to lose electrons. It runs opposite to ionization energy: it decreases across a period and increases down a group. The diagonal staircase B, Si, Ge, As, Sb, Te (the metalloids) marks the metal / non-metal boundary.

Property	Across a period (left → right)	Down a group	Main reason
Atomic radius	decreases	increases	Z_{\text{eff}} vs new shell
Ionic radius	decreases (same charge)	increases	same
Ionization energy	increases	decreases	Z_{\text{eff}} vs size and shielding
Electron affinity	increases	decreases	Z_{\text{eff}} vs size
Electronegativity	increases	decreases	Z_{\text{eff}} vs size
Metallic character	decreases	increases	opposite of IE
Valency w.r.t. oxygen	1 → 7	constant	valence electrons

Chapter summary

• Modern periodic law: properties are periodic functions of atomic number, established by Moseley's relation \sqrt{\nu} = a(Z-b).
• The long form has 7 periods (2, 8, 8, 18, 18, 32, 32 elements) and 18 IUPAC groups; blocks s, p, d, f are named after the subshell filled last.
• Elements with Z > 100 are named systematically from the roots nil, un, bi, tri, quad, pent, hex, sept, oct, enn plus -ium.
• Z_{\text{eff}} = Z - \sigma; screening power falls as s > p > d > f. Slater's rules give Z_{\text{eff}} = 2.20 for Na(3s) and 6.10 for Cl(3p).
• Across a period: radius ↓, IE ↑, EA ↑, EN ↑, metallic character ↓. Down a group: radius ↑, IE ↓, EA ↓, EN ↓, metallic character ↑.
• Cations are smaller and anions larger than the parent atom; in an isoelectronic series radius falls as the number of protons rises.
• Anomalies: Be > B and N > O (and Mg > Al, P > S) in ionization energy because of filled 2s and half-filled 2p stability; Cl > F and S > O in electron affinity because the second-period atoms are too compact.

Practice questions

Group A — Multiple choice (1 mark each)

1. An element has the configuration [Ar] 3d¹⁰ 4s² 4p³. It belongs to [1]agroup 13, p-blockbgroup 15, p-blockcgroup 5, d-blockdgroup 15, d-block
2. The correct order of ionic radii is [1]aAl³⁺ > Mg²⁺ > Na⁺ > F⁻bO²⁻ > F⁻ > Na⁺ > Mg²⁺cNa⁺ > F⁻ > O²⁻ > N³⁻dF⁻ > O²⁻ > N³⁻ > Na⁺
3. The first ionization energy of nitrogen is greater than that of oxygen because [1]anitrogen has higher nuclear chargebnitrogen has a half-filled 2p subshellcoxygen is more electronegativedoxygen has a larger radius
4. The IUPAC systematic name of the element with Z = 106 is [1]aUnnilhexiumbUnnilsextiumcUnunhexiumdUnhexium
5. Using Slater's rules, the effective nuclear charge felt by the 3s electron of sodium is [1]a11.00b8.80c2.20d1.00

Answers to Group A

1. (b) — the last electron enters 4p, so p-block; 2 + 3 = 5 valence electrons gives group 15 (arsenic, Z = 33).

2. (b) — all are isoelectronic with 10 electrons, so radius falls as protons rise: 8 < 9 < 11 < 12.

3. (b) — 2p³ is half-filled and has no paired 2p electron, so it is extra stable.

4. (a) — 1, 0, 6 → un + nil + hex + ium.

5. (c) — \sigma = 8(0.85) + 2(1.00) = 8.80, so Z_{\text{eff}} = 11 - 8.80 = 2.20.

Group B — Short answer (5 marks each)

1. State the modern periodic law. Explain, in terms of orbital filling, why the second period contains 8 elements while the fourth contains 18. [5]
2. Define effective nuclear charge. Using Slater's rules calculate Z_{\text{eff}} for a 3p electron of phosphorus (Z=15) and of sulphur (Z=16), and comment on the result. [5]
3. Define ionization energy. Why is the second ionization energy of sodium (4562 kJ mol⁻¹) nearly nine times the first (496 kJ mol⁻¹)? [5]
4. Why is the electron affinity of fluorine smaller than that of chlorine even though fluorine is the more electronegative element? Distinguish clearly between electron affinity and electronegativity. [5]
5. Arrange with reasons: (i) Li, Be, B, C in increasing atomic radius; (ii) N³⁻, O²⁻, F⁻, Na⁺ in decreasing ionic radius. [5]

Answers to Group B

1. Properties are periodic functions of atomic number. Period 2 fills 2s and 2p only: 1 + 3 = 4 orbitals × 2 electrons = 8 elements. Period 4 fills 4s, then 3d, then 4p: 1 + 5 + 3 = 9 orbitals × 2 = 18 elements.

2. Z_{\text{eff}} = Z - \sigma. Phosphorus (1s²)(2s²2p⁶)(3s²3p³): \sigma = 4(0.35)+8(0.85)+2(1.00) = 1.40+6.80+2.00 = 10.20, so Z_{\text{eff}} = 15-10.20 = 4.80. Sulphur (3s²3p⁴): \sigma = 5(0.35)+6.80+2.00 = 10.55, so Z_{\text{eff}} = 16-10.55 = 5.45. Adding one proton raises Z_{\text{eff}} by only 0.65 because the new electron screens 0.35 of it — this is why atoms shrink only gradually across a period.

3. IE₁ removes the 3s¹ electron from a neutral atom. IE₂ must break into the filled 2s²2p⁶ neon core of Na⁺, where the electron is in a lower shell, feels a much larger Z_{\text{eff}} and must be pulled away from a positive ion. Hence the huge jump, and hence sodium's valency of 1.

4. Adding an electron to the very small fluorine atom forces it into a compact 2p subshell where inter-electronic repulsion is severe; chlorine's 3p subshell is larger, so more energy is released (F 328, Cl 349 kJ mol⁻¹). Electron affinity is an energy (kJ mol⁻¹) belonging to an isolated gaseous atom; electronegativity is a dimensionless relative number describing a bonded atom's pull on a shared pair.

5. (i) C (77) < B (88) < Be (111) < Li (152) pm: across period 2, n is fixed but Z_{\text{eff}} rises, so the shell contracts. (ii) N³⁻ (171) > O²⁻ (140) > F⁻ (136) > Na⁺ (95) pm: isoelectronic, so radius falls as protons rise.

Group C — Long answer (8 marks each)

1. (a) Classify the elements into s-, p-, d- and f-blocks, giving the general outer electronic configuration and one example of each. [4](b) Discuss the variation of atomic radius and first ionization energy along period 3, and explain the irregularities observed at aluminium and sulphur. [4]
2. (a) Define ionization energy, electron affinity and electronegativity, and state how each varies across a period and down a group, with the reason. [5](b) An element X has atomic number 26. Write its ground-state electronic configuration and deduce its period, group and block. [3]

Answers to Group C

1.(b) Radius falls steadily 186 → 99 pm because Z_{\text{eff}} climbs from 2.20 to 6.10 while n = 3 throughout. IE₁ rises overall 496 → 1251 kJ mol⁻¹ but dips twice. Al (578) < Mg (738): aluminium's electron comes from 3p, which is higher in energy and shielded by the filled 3s². S (1000) < P (1012): phosphorus is 3p³, half-filled and stable, whereas sulphur is 3p⁴ with one repelling pair.

2.(b) X is iron: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s². The outermost shell is n = 4, so the period is 4. The last electron enters 3d, so it is a d-block element, and for groups 3–10 the group number = (number of d electrons) + (number of s electrons) = 6 + 2 = group 8. Common oxidation states +2 and +3.