Using Periodicity to Predict and Identify Elements

AS · 11 min

The point of learning Period 3 in detail is that the same patterns apply everywhere in the Periodic Table. Cambridge tests this directly: questions describe an element you have never studied (germanium, gallium, selenium, rubidium) and ask you to predict its properties, or give you data about an unknown element and ask you to place it in the Periodic Table and identify it. There is nothing new to memorise here; the skill is applying the trends you already know, systematically and with reasons.

The trends you can use

Two directions matter: across a period and down a group.

Key result

Across a period (left to right)

Nuclear charge increases; atomic radius decreases.
Ionisation energy and electronegativity increase.
Elements change from metals to non-metals.
Structures change: giant metallic, then giant covalent, then simple molecular, then monatomic.
Oxides change from basic (ionic) through amphoteric to acidic (covalent).
Chlorides change from ionic (dissolve, neutral) to covalent (hydrolysed, acidic).
Highest oxidation number rises to equal the number of outer-shell electrons.

Down a group

Atomic radius increases (an extra shell each period).
Ionisation energy and electronegativity decrease.
Metallic character increases: elements become more metallic, oxides more basic.
Metals in Groups 1 and 2 become more reactive; non-metals in Group 17 become less reactive.
Elements in the same group have the same number of outer electrons, so they form compounds with the same formulas (for example $\ce{SiCl4}$ and $\ce{GeCl4}$ ; $\ce{Al2O3}$ and $\ce{Ga2O3}$ ).

The single most useful idea is the last one: same group, same outer electrons, same formula type, similar chemistry, with properties shifting steadily as you go down.

The metal–non-metal boundary

The elements along the diagonal from boron to astatine (B, Si, Ge, As, Sb, Te) are metalloids: they have some properties of metals and some of non-metals. Typical metalloid features are a giant covalent structure, semiconductor behaviour, and oxides that are weakly acidic or amphoteric. Because metallic character increases down a group and decreases across a period, the boundary runs diagonally.

group	top of group	further down
13	$\ce{B2O3}$ acidic	$\ce{Al2O3}$ , $\ce{Ga2O3}$ amphoteric
14	$\ce{CO2}$ , $\ce{SiO2}$ acidic	$\ce{GeO2}$ , $\ce{SnO2}$ amphoteric
2	$\ce{BeO}$ amphoteric	$\ce{MgO}$ , $\ce{CaO}$ , $\ce{SrO}$ , $\ce{BaO}$ basic

A method for predictions

Method

Predicting the properties of an unfamiliar element

Locate the element: group (number of outer electrons) and period.
Find the Period 3 element (or familiar element) in the same group. Its compounds give you the formulas of the new element's compounds.
Decide on the structure of the element: metals (left) are giant metallic; Group 14 elements near the top are giant covalent; Groups 15 to 17 are generally molecular, becoming more metallic down the group.
Apply the group trend to decide how properties shift: more metallic, larger, lower ionisation energy, more basic oxide as you go down.
Predict reactions with water, oxygen or chlorine by analogy, and write balanced equations using the formulas from step 2.
Justify every prediction with a reason ("because it is in the same group as ... and has ... outer electrons").

A method for identifying an unknown element

Method

Deducing the position and identity of an unknown element

Metal or non-metal? Good electrical conductivity in the solid means metallic. Basic oxide means metal; acidic oxide means non-metal; amphoteric suggests Al, Be, Zn or a metalloid.
Structure? Very high melting point and non-conductor (or semiconductor): giant covalent. Low melting point: simple molecular.
Group? Use the formula of the oxide or chloride (oxidation number = outer electrons), or a big jump in successive ionisation energies (number of electrons before the jump = group number for Groups 1 to 18).
Period? Use relative atomic mass, atomic radius, or reactivity compared with a known group member.
Confirm by checking every piece of data against your answer; one mismatch means you need to reconsider.

Worked examples

Predicting the properties of rubidium

Rubidium is in Group 1, Period 5. Predict (a) the formula and nature of its oxide, (b) how it reacts with water, compared with sodium, and (c) the bonding in its chloride. Write equations where relevant.

Solution

(a) Group 1, one outer electron, oxidation number $+1$ : oxide $\ce{Rb2O}$ . It is ionic (very large electronegativity difference) and therefore basic: $\ce{Rb2O(s) + H2O(l) -> 2RbOH(aq)}$ , giving a solution of pH 13–14.

(b) $\ce{2Rb(s) + 2H2O(l) -> 2RbOH(aq) + H2(g)}$ . The reaction is much more vigorous than with sodium (rubidium may ignite and explode). Down Group 1 the outer electron is further from the nucleus and more shielded, so the first ionisation energy is lower and the electron is lost more easily.

Predicting the chemistry of selenium dioxide

Selenium is in Group 16, below sulfur. Its dioxide is $\ce{SeO2}$ . Predict whether $\ce{SeO2}$ is acidic or basic, and write equations for its reactions with water and with sodium hydroxide.

Solution

Selenium is a non-metal in the same group as sulfur (six outer electrons, oxidation number $+4$ in $\ce{SeO2}$ , like $\ce{SO2}$ ). Its electronegativity is high enough that Se–O bonding is covalent, so $\ce{SeO2}$ is an acidic oxide, by analogy with $\ce{SO2}$ .

With water, forming selenous acid: $\ce{SeO2(s) + H2O(l) -> H2SeO3(aq)}$

With sodium hydroxide, forming sodium selenite: $\ce{SeO2(s) + 2NaOH(aq) -> Na2SeO3(aq) + H2O(l)}$

Compare $\ce{SO2 + H2O <=> H2SO3}$ and $\ce{SO2 + 2NaOH -> Na2SO3 + H2O}$ : the same formula types, with Se in place of S.

Identifying a Period 3 element from ionisation energies

The first four ionisation energies of a Period 3 element X are 578, 1817, 2745 and 11 577 kJ mol⁻¹. Identify X, and predict the formula and nature of its oxide and the pH of a solution of its chloride.

Solution

The big jump is between the third and fourth ionisation energies (2745 to 11 577), so the fourth electron is removed from an inner shell. X has three outer electrons: Group 13. In Period 3 this is aluminium.

Oxide: $\ce{Al2O3}$ (oxidation number $+3$ ), amphoteric: reacts with $\ce{HCl}$ and with $\ce{NaOH}$ .

Chloride: $\ce{AlCl3}$ , covalent; hydrolysed in water to give an acidic solution, pH about 3.

Exam-style: identifying an element from its chloride

Element Z is in Period 4. It is a shiny grey solid with a high melting point; it conducts electricity slightly, and its conductivity increases on warming. Its chloride, $\ce{ZCl4}$ , is a colourless liquid that fumes in moist air and contains 66.2% chlorine by mass. Identify Z, state its group, and predict the equation for the reaction of its chloride with water. ( $A_r$ : Cl 35.5)

Solution

Mass calculation. In $\ce{ZCl4}$ , $4 \times 35.5 = 142.0$ is 66.2% of $M_r$ :

M_r = \frac{142.0}{0.662} = 214.5, \qquad A_r(\ce{Z}) = 214.5 - 142.0 = 72.5

This matches germanium ( $A_r = 72.6$ ).

Consistency check. A chloride $\ce{ZCl4}$ means four outer electrons: Group 14, the same group as silicon. A semiconductor with a high melting point fits a giant covalent structure like silicon's. A liquid chloride that fumes in moist air fits a covalent, simple molecular chloride like $\ce{SiCl4}$ .

Prediction, by analogy with $\ce{SiCl4}$ :

\ce{GeCl4(l) + 2H2O(l) -> GeO2(s) + 4HCl(aq)}

Steamy fumes of $\ce{HCl}$ and a strongly acidic solution (pH 1–2).

Exam-hard: predicting gallium chemistry

Gallium is directly below aluminium in Group 13. Predict:

(a) the formula of gallium oxide and its behaviour with dilute hydrochloric acid and with sodium hydroxide, writing equations;

(b) whether gallium chloride is likely to be ionic or covalent, and the likely pH of its aqueous solution;

Solution

(a) Group 13, three outer electrons, oxidation number $+3$ : $\ce{Ga2O3}$ . By analogy with $\ce{Al2O3}$ it is amphoteric.

$\ce{Ga2O3(s) + 6HCl(aq) -> 2GaCl3(aq) + 3H2O(l)}$

$\ce{Ga2O3(s) + 2NaOH(aq) + 3H2O(l) -> 2NaGa(OH)4(aq)}$

(b) Gallium's electronegativity is similar to aluminium's, and $\ce{Ga^3+}$ is small and highly charged, so $\ce{GaCl3}$ is expected to be covalent (like $\ce{AlCl3}$ , it forms $\ce{Ga2Cl6}$ dimers). It would be hydrolysed in water, the hydrated $\ce{Ga^3+}$ ion releasing $\ce{H+}$ , giving an acidic solution, pH about 3.

(c) Gallium is expected to be at least as metallic as aluminium: down a group the atomic radius increases and ionisation energies decrease, so metallic character increases. (In reality gallium's ionisation energies are unusually close to aluminium's, because the filled $3d$ sub-shell shields the outer electrons poorly; examiners accept "more metallic" with this reason, or "similar" if explained.)

Watch out

Predictions must be justified. " $\ce{SeO2}$ is acidic" scores less than " $\ce{SeO2}$ is acidic, like $\ce{SO2}$ , because selenium is a non-metal in the same group and its oxide is covalent".
Do not use Period 3 numbers for a different period. A Period 4 element has a larger radius and lower ionisation energy than its Period 3 partner.
Do not assume every element at the bottom of a group is a metal with a basic oxide. Iodine is still a non-metal; trends shift properties gradually.
Formulas follow the group. A Group 14 element forms $\ce{XO2}$ and $\ce{XCl4}$ ; a Group 13 element forms $\ce{X2O3}$ and $\ce{XCl3}$ .

Exam tip

Typical command words: predict (give an answer with a brief reason), suggest (apply knowledge to an unfamiliar situation; any sensible, justified answer can score), deduce (reach a conclusion from the data given; show the link).
In "identify the element" questions, every piece of data is there for a reason. Examiners expect you to use them all and show how each one fits.
Equations by analogy are often worth a mark each. Copy the formula pattern of the Period 3 compound exactly and rebalance.
In multiple-choice questions on unknown elements, eliminate options by testing one property at a time (conductivity, melting point, oxide nature, chloride pH).

Summary

Same group, same number of outer electrons: same formulas and similar chemistry.
Down a group: larger atoms, lower ionisation energy and electronegativity, more metallic, more basic oxides.
Across a period: metals to non-metals, basic to acidic oxides, ionic to covalent chlorides.
Metalloids lie on the diagonal from B to At: giant covalent, semiconducting, weakly acidic or amphoteric oxides.
To identify an element: decide metal or non-metal, structure, group (formula or ionisation energy jump), period (mass or reactivity), then check every clue.
Always justify predictions with the trend and a reason.

Practice

Question

Germanium is in Group 14. Predict the formula of its oxide and its chloride.
Predict, with a reason, whether strontium oxide, $\ce{SrO}$ , is acidic, basic or amphoteric, and write its equation with water.
Write an equation for the reaction of $0.855\ \text{g}$ of rubidium with water and calculate the volume of hydrogen at room conditions. ( $A_r$ : Rb 85.5)
An element has a melting point of 1683 K, is a semiconductor, and forms an oxide that reacts with hot concentrated sodium hydroxide but not with acids. Identify the type of structure of the element and suggest which group it is in.
Arsenic is in Group 15, below phosphorus. Predict the formula of its highest chloride and of its highest oxide, and predict whether the oxide is acidic or basic.
An oxide of a Group 13 element, $\ce{M2O3}$ , contains 52.9% M by mass. Identify M. ( $A_r$ : O 16.0)
An oxide $\ce{XO2}$ contains 30.6% oxygen by mass. Calculate $A_r$ of X and identify it. Predict the nature of the oxide.
Explain why the oxides of Group 14 become more basic down the group from $\ce{CO2}$ to $\ce{PbO}$ .
Element Q: solid at room temperature; does not conduct; melting point 387 K; burns in oxygen to form a colourless gas whose solution in water turns universal indicator red; forms a chloride that fumes in moist air. Its first ionisation energy is lower than that of the element before it in the same period. Identify Q and justify each piece of evidence.
Astatine is at the bottom of Group 17. Predict its physical state at room temperature, its colour, and how its reactivity as an oxidising agent compares with iodine. Explain your predictions using trends in Group 17.

Answers

Four outer electrons, oxidation number $+4$ : $\ce{GeO2}$ and $\ce{GeCl4}$ (like $\ce{SiO2}$ and $\ce{SiCl4}$ ).
Basic. Strontium is a Group 2 metal below magnesium and calcium with low electronegativity, so $\ce{SrO}$ is ionic and contains $\ce{O^2-}$ ions. $\ce{SrO(s) + H2O(l) -> Sr(OH)2(aq)}$ , giving an alkaline solution.
$\ce{2Rb(s) + 2H2O(l) -> 2RbOH(aq) + H2(g)}$ . $n(\ce{Rb}) = 0.855 / 85.5 = 0.0100\ \text{mol}$ ; $n(\ce{H2}) = 0.00500\ \text{mol}$ ; $V = 0.00500 \times 24.0 = 0.120\ \text{dm}^3 = 120\ \text{cm}^3$ .
Very high melting point and semiconductor: giant covalent structure. An acidic oxide (reacts with alkali only) also points to a non-metal or metalloid. These properties match Group 14 (the data are those of silicon).
Five outer electrons, so the highest oxidation number is $+5$ : $\ce{AsCl5}$ (by analogy with $\ce{PCl5}$ ) and $\ce{As4O10}$ (or empirical formula $\ce{As2O5}$ , by analogy with $\ce{P4O10}$ ). The oxide is expected to be acidic (non-metal/metalloid oxide, like $\ce{P4O10}$ ), though less strongly so than phosphorus's because arsenic is lower in the group.
If 52.9% is M, then $\dfrac{2A_r}{2A_r + 48.0} = 0.529$ , so $2A_r = 0.529(2A_r + 48.0)$ , $0.942A_r = 25.39$ , $A_r = 27.0$ : aluminium.
$M_r = \dfrac{32.0}{0.306} = 104.6$ ; $A_r(\ce{X}) = 104.6 - 32.0 = 72.6$ : germanium. $\ce{GeO2}$ is expected to be amphoteric (weakly acidic), between acidic $\ce{SiO2}$ and amphoteric $\ce{SnO2}$ .
Down Group 14 the atoms get larger and the electronegativity decreases, so the elements become more metallic (C and Si non-metals, Ge metalloid, Sn and Pb metals). The bonding in the oxides changes from covalent (acidic oxides) towards ionic (basic oxides), so the oxides become more basic: $\ce{CO2}$ and $\ce{SiO2}$ acidic, $\ce{GeO2}$ and $\ce{SnO2}$ amphoteric, $\ce{PbO}$ amphoteric with more basic character.
Q is sulfur. Solid non-conductor with a low melting point: simple molecular non-metal ( $\ce{S8}$ ). Burns to give a colourless gas forming an acidic solution: $\ce{SO2}$ , an acidic oxide. Chloride fumes in moist air: covalent chloride hydrolysed by water. First ionisation energy lower than the previous element (phosphorus): the $3p^4$ electron is removed from a paired orbital, where spin-pair repulsion makes it easier to remove.
Down Group 17, id–id forces increase as the molecules have more electrons, so volatility decreases: chlorine is a gas, bromine a liquid, iodine a solid. Astatine is predicted to be a solid, and darker than iodine (black). The colours darken down the group. Its oxidising power is predicted to be weaker than iodine's, because its atoms are larger, the incoming electron is further from the nucleus and more shielded, so it is attracted less strongly.

The trends you can use

The metal–non-metal boundary

A method for predictions

A method for identifying an unknown element

Worked examples

Practice

Console