Chemical periodicity of other elements

1. Overview

The Periodic Table is a prediction tool. Elements in the same group (Groups 1, 2 and 13 to 17) have the same number of outer electrons, so their compounds follow the same formula pattern and their properties change gradually down the group. Across each period from Period 2 onwards, the Period 3 pattern repeats: metals on the left, non-metals on the right. This section uses these patterns both ways: to predict how an unfamiliar element behaves, and to deduce an unknown element's position and identity from data.

Key Definitions

• Periodicity: the pattern of physical and chemical properties that repeats across each period of the Periodic Table, because outer-shell electron arrangements repeat.
• Group: a vertical column of the Periodic Table; in Groups 1, 2 and 13 to 17 every element has the same number of outer-shell electrons, so the elements react similarly and form compounds with similar formulae.
• Period: a horizontal row of the Periodic Table; for s-block and p-block elements the period number equals the number of occupied electron shells, and from Period 2 onwards the elements change from metals to non-metals across the row.
• Valence (outer-shell) electrons: the electrons in the outermost shell of an atom, which take part in bonding; there are as many as the group number in Groups 1 and 2, and the group number minus 10 in Groups 13 to 17.
• Electronegativity: the power of an atom to attract electrons to itself.
• Basic oxide: an oxide that reacts with acids to form a salt and water; most metal oxides are basic.
• Acidic oxide: an oxide that reacts with bases to form a salt and water; most non-metal oxides are acidic.
• Amphoteric oxide: an oxide that reacts with both acids and bases to form salts, for example aluminium oxide.
• Hydrolysis: a reaction in which a compound is broken down by water; when a covalent chloride such as SiCl₄ is hydrolysed, hydrogen chloride forms and the solution becomes acidic.

Content

What an element's position tells you

The group gives the number of outer electrons, and the period the number of occupied shells: germanium, in Group 14 and Period 4, has 4s² 4p² outside three inner shells. The number of outer electrons is the group number in Groups 1 and 2, and the group number minus 10 in Groups 13 to 17. The outer electrons set the usual oxidation numbers, and so the formulae:

Group	Usual oxidation number	Period 3 examples
1	+1	NaCl, Na₂O
2	+2	MgCl₂, MgO
13	+3	AlCl₃, Al₂O₃
14	+4	SiCl₄, SiO₂
15	+5 or +3	PCl₅, PCl₃, P₄O₁₀
16	+6 or +4 in oxides; −2 with H	SO₃, SO₂, H₂S
17	−1 with metals and H	NaCl, HCl

Swap the Period 3 element for any element E of the same group to get the pattern: Group 13 gives ECl₃ and E₂O₃, and Group 14 gives ECl₄ and EO₂.

In Period 3 and below, the highest oxidation number usually equals the number of outer electrons, because all of them can be used in bonding. Period 2 is more limited. Nitrogen cannot hold more than eight electrons in its outer shell, so it forms NCl₃ but not NCl₅, and oxygen and fluorine never reach +6 or +7.

Across a period: Period 3 as the template

Period 3, covered in topics 9.1 and 9.2, is the pattern that every later period repeats. Silicon and SiO₂ have giant molecular (giant covalent) structures. The elements:

Element	Structure	Conducts electricity?
Na, Mg, Al	giant metallic	well
Si	giant molecular	slightly (semiconductor)
P, S	simple molecular	no

Their oxides and chlorides:

Element	Oxide	Chloride with water
Na, Mg	ionic, basic	dissolves: pH 7 (NaCl) or just below (MgCl₂)
Al	amphoteric	hydrolysed: acidic
Si	giant molecular, acidic	hydrolysed: strongly acidic
P, S	simple molecular, acidic	hydrolysed: strongly acidic (PCl₅)

As electronegativity rises across a period, oxides and chlorides change from ionic to covalent, and oxides from basic to acidic. The metal/non-metal border runs diagonally, near boron, silicon, germanium, arsenic, antimony and tellurium, so lower periods contain more metals: in Group 14, carbon is a non-metal, silicon and germanium are semiconductors, and tin and lead are metals.

Down a group: what changes

Each step down a group adds one occupied shell of electrons.

Property	Down the group	Why
Atomic radius	increases	one more occupied shell
First ionisation energy and electronegativity	generally decrease	outer electrons are further from the nucleus and more shielded, which outweighs the larger nuclear charge
Metallic character; reactivity of Group 1 and 2 metals	increase	outer electrons are lost more easily
Basic character of oxides and hydroxides	increases	the element is more metallic, its bonding more ionic
Oxidising power of the halogens	decreases	an extra electron is attracted less strongly (topic 11.2)
Melting point	depends on the structure	falls down Group 1 (weaker metallic bonding); rises down Group 17 (stronger id-id forces)

Trends are not always smooth. Data Booklet electronegativities fall down Groups 1, 2 and 17, but not from Period 3 to Period 4 in Groups 13 to 16 (Al 1.5, Ga 1.8; Si 1.9, Ge 2.0; P and As both 2.2; S and Se both 2.6). Aluminium and gallium also share a first ionisation energy of 577 kJ mol⁻¹. The cause is the d-block: from aluminium to gallium the nuclear charge rises by 18, but ten of the added electrons are 3d electrons, which shield poorly. So gallium to bromine hold their outer electrons more strongly than the trend suggests.

Predicting the properties of an element

Start from the familiar element next to it in the same group, then adjust for the trend down the group. For bonding, compare electronegativity differences with compounds whose bonding you know: there are no fixed cut-offs between ionic and covalent bonding.

Worked example 1: predicting the chemistry of germanium

Question: Germanium, Ge, is in Group 14, directly below silicon. Predict: (a) the formulae of its chloride and oxide; (b) the structure and electrical conductivity of the element; (c) the bonding in the chloride and its reaction with water; (d) the acid-base nature of the oxide. (Pauling electronegativity values: Na 0.9, Si 1.9, Ge 2.0, Cl 3.0, O 3.5.)

Step 1: formulae. Group 14 gives 14 − 10 = 4 outer electrons, so oxidation number +4, as in SiCl₄ and SiO₂. Predict GeCl₄ and GeO₂.

Step 2: the element. Like silicon, germanium should form a giant molecular lattice, each atom bonded to four others, with a high melting point (probably lower than silicon's, as larger atoms form longer, weaker bonds). Near the metal/non-metal border, it should be a semiconductor.

Step 3: the chloride. Compare electronegativity differences:

• Ge-Cl: 3.0 − 2.0 = 1.0
• Si-Cl: 3.0 − 1.9 = 1.1
• Na-Cl: 3.0 − 0.9 = 2.1

Ge-Cl is close to Si-Cl and far from Na-Cl, so GeCl₄ should be a covalent, simple molecular liquid like SiCl₄. Water should hydrolyse it, giving a white solid, misty fumes of HCl and a strongly acidic solution (pH about 1 to 2):

GeCl₄(l) + 2H₂O(l) → GeO₂(s) + 4HCl(aq)

Step 4: the oxide. Ge-O (3.5 − 2.0 = 1.5) is like Si-O (3.5 − 1.9 = 1.6). So predict a giant molecular, acidic oxide that, like SiO₂, reacts with hot, concentrated sodium hydroxide:

GeO₂(s) + 2NaOH(aq) → Na₂GeO₃(aq) + H₂O(l)

Oxides become more basic down a group, so GeO₂ may also show some basic behaviour.

Answer: (a) GeCl₄ and GeO₂; (b) a giant molecular semiconductor with a high melting point; (c) GeCl₄ is a covalent liquid that water hydrolyses to GeO₂ and HCl; (d) GeO₂ is mainly acidic and reacts with NaOH, possibly with some basic character.

All of this fits the real element. GeO₂ also reacts with concentrated hydrochloric acid, so it is classed as amphoteric: one step more basic than SiO₂, as the trend suggests.

Worked example 2: predicting the properties of rubidium

Question: Rubidium, Rb, is in Group 1 between potassium and caesium, whose first ionisation energies are 418 and 376 kJ mol⁻¹. Predict: (a) the first ionisation energy of rubidium; (b) its reaction with water, with an equation and the likely pH; (c) the bonding in rubidium chloride and the pH of its solution. (Pauling electronegativity values: Na 0.9, K 0.8, Cl 3.0.)

Step 1: ionisation energy. Rubidium sits between potassium and caesium, so its first ionisation energy, $IE$, should lie between 418 and 376 kJ mol⁻¹. The simplest estimate is the midpoint: $$IE \approx \frac{418 + 376}{2} = 397\ \text{kJ mol}^{-1}$$ This is only an estimate: any value near 400 kJ mol⁻¹ is a sensible prediction.

Step 2: reaction with water. Rubidium's one outer electron (5s¹) is further from the nucleus and more shielded than potassium's 4s electron, so it is lost more easily. Rubidium reacts like potassium, but more vigorously:

2Rb(s) + 2H₂O(l) → 2RbOH(aq) + H₂(g)

RbOH is soluble and fully dissociated, like KOH, so the solution is strongly alkaline, pH about 13 to 14.

Step 3: the chloride. Rubidium's electronegativity (not in the Data Booklet) should be no higher than potassium's 0.8, so the difference with chlorine is at least 3.0 − 0.8 = 2.2, more than in NaCl (2.1). RbCl is ionic and dissolves without reacting: pH 7.

Answer: (a) about 400 kJ mol⁻¹ (midpoint estimate 397 kJ mol⁻¹); (b) a very vigorous reaction giving hydrogen and RbOH(aq), pH about 13 to 14; (c) ionic, pH 7.

Check: the Data Booklet gives 403 kJ mol⁻¹, 1.5% above the midpoint, because the fall from potassium to rubidium (15 kJ mol⁻¹) is smaller than the fall from rubidium to caesium (27 kJ mol⁻¹). Steps down a group are rarely equal.

Deducing an unknown element from data

Data often narrow the choice rather than fix it, hence a possible position. Take the clues one at a time, note what each shows, then check that every clue fits your answer.

Clue	What it tells you
One jump in successive ionisation energies far larger than the rest, after the nth	n outer electrons, so the group (topic 1.4)
Element: malleable solid, conducts well	a metal (graphite also conducts, but it is not malleable)
Element: very high melting point, conducts slightly or not at all	giant molecular; a slight conductor is a semiconductor near the border
Element: low melting point, does not conduct	simple molecular non-metal, or a noble gas
Oxide reacts with acids only	basic oxide: a metal
Oxide reacts with acids and with alkalis	amphoteric: usually a metal near the border, such as Al
Oxide reacts with alkalis only, or dissolves to give an acidic solution	acidic oxide: a non-metal, or a border element such as Si
Chloride has a high melting point, conducts when molten, gives pH 7 or just below in water	ionic chloride: usually Group 1 or 2
Chloride is volatile, fumes in moist air, gives an acidic solution	covalent and hydrolysed: right of its period or near the border (AlCl₃ too)
Formula of an oxide or chloride	the oxidation number, so a likely group (check it: many elements have several)
Reacting masses	$A_{\text{r}}$, then the identity, chosen within the group you deduced

Worked example 3: identifying an unknown element

Question: Element X is a malleable solid that conducts electricity. Its first four ionisation energies are 590, 1150, 4940 and 6480 kJ mol⁻¹. When 0.522 g of X is heated in oxygen until its mass stops changing, 0.730 g of a white oxide forms. The oxide reacts with dilute hydrochloric acid but not with aqueous sodium hydroxide. Deduce the group of X, the formula of its oxide, its relative atomic mass, and its identity and period. (First ionisation energy of Mg = 736 kJ mol⁻¹.)

Step 1: group. The increases are 560, 3790 and 1540 kJ mol⁻¹. The 3rd ionisation energy is 4.3 times the 2nd, while each of the others is less than twice the one before. The 3rd electron comes from an inner shell, so X has 2 outer electrons: Group 2.

Step 2: nature. X is malleable and conducts as a solid, so it is a metal, and its oxide reacts only with the acid, so it is basic. Both fit Group 2.

Step 3: formula. Oxidation number +2 with oxide O²⁻ gives XO.

Step 4: relative atomic mass. $$m(\text{O}) = 0.730 - 0.522 = 0.208\ \text{g}$$ $$n(\text{O}) = \frac{0.208}{16.0} = 0.0130\ \text{mol}$$ In XO the ratio is 1 : 1, so $n(\text{X}) = 0.0130$ mol. $$M(\text{X}) = \frac{0.522}{0.0130} = 40.2\ \text{g mol}^{-1}$$ So $A_{\text{r}}(\text{X}) = 40.2$.

Step 5: identity. Potassium (39.1), argon (39.9) and calcium (40.1) all have $A_{\text{r}}$ close to 40. Argon is a noble gas, not a conducting solid, and potassium is in Group 1: only calcium is a Group 2 metal. Its first ionisation energy (590 kJ mol⁻¹) is below magnesium's (736 kJ mol⁻¹), so X lies below magnesium, which also fits calcium.

Answer: X is calcium, Ca: Group 2, Period 4, with the basic oxide CaO; $A_{\text{r}} = 40.2$ (3 s.f.), close to calcium's 40.1.

Why the group evidence decides. The masses alone cannot choose between three elements near 40, and measured $A_{\text{r}}$ values are often a few percent out. So decide the group first, then choose within it. The masses can also test a group: a Group 1 metal would form X₂O, and the same masses would give $A_{\text{r}} = 20.1$, which fits no Group 1 metal (lithium 6.9, sodium 23.0).

Key Equations

Outer electrons from the group number: outer electrons = group number in Groups 1 and 2; outer electrons = group number − 10 in Groups 13 to 17.

Formula of a compound: the oxidation numbers add up to zero. For gallium (+3) and oxygen (−2): $$2(+3) + 3(-2) = 0 \quad\Rightarrow\quad \text{Ga}_2\text{O}_3$$

Relative atomic mass from reacting masses ($m$ = mass in g; $M$ = molar mass in g mol⁻¹; $n$ = amount in mol, with $n(\text{X})$ from the formula of the compound): $$n = \frac{m}{M} \qquad M(\text{X}) = \frac{m(\text{X})}{n(\text{X})}$$ $A_{\text{r}}(\text{X})$ is the number part of $M(\text{X})$ in g mol⁻¹; it has no units.

Estimate for an element between two others in a group (a prediction, not a data value): $$\text{value} \approx \frac{\text{value above} + \text{value below}}{2}$$

Common Mistakes to Avoid

• Mistake: basing a deduction on one clue. Fix: test every clue against the others. A high-melting oxide could be giant ionic (MgO) or giant molecular (SiO₂); if the molten oxide conducts, it is ionic.
• Mistake: saying a covalent chloride "dissolves", or describing a test for chloride ions instead of its reaction with water. Fix: describe hydrolysis: water breaks the Ge-Cl bonds in GeCl₄, forming GeO₂ and HCl (pH about 1 to 2), whereas ionic RbCl just dissolves (pH 7).
• Mistake: assuming every property falls evenly down a group. Fix: check the Data Booklet: germanium (2.0) is more electronegative than silicon (1.9), not less.
• Mistake: a formula whose oxidation numbers do not balance. Fix: Ga (+3) with O (−2) gives Ga₂O₃, not GaO₃ or Ga₃O₂.
• Mistake: giving the oxide as the product when a Group 1 metal reacts with water. Fix: water gives the hydroxide and hydrogen: 2Rb(s) + 2H₂O(l) → 2RbOH(aq) + H₂(g).
• Mistake: choosing the element whose $A_{\text{r}}$ is nearest your answer without checking its group, or changing your data to fit. Fix: keep your value, decide the group, then choose within it. An $A_{\text{r}}$ near 40 fits potassium (39.1), argon (39.9) and calcium (40.1), but only calcium is in Group 2.

Exam Tips

• Predict or suggest: give the property and a reason from the element's position, for example "GeCl₄ is covalent, like SiCl₄, because germanium is directly below silicon and the electronegativity differences are similar". For an unfamiliar element, more than one answer can be sensible, so the reason matters.
• Deduce or identify: write one line per clue (the clue, then what it shows) and finish with the element, its group and its period. If asked for a name, give a name, not a formula, and include the oxidation number when the element has more than one, such as germanium(IV) chloride.
• Construct an equation for an unfamiliar element by adapting its neighbour's equation, then check the balancing and state symbols.
• Explain a physical property through structure: "conducts when molten, so it contains mobile ions, so it is ionic".
• Calculate $A_{\text{r}}$ on separate lines (mass, amount, then $A_{\text{r}}$), give it to 3 s.f., and justify the identity with the group evidence.