Chapter 3 · Classification of Elements and Periodicity in Properties
3.1 Development of the Periodic Table and the Modern Periodic Law
How chemists learned to arrange the elements: Döbereiner's triads, Newlands' octaves, Lothar Meyer's curve and Mendeleev's bold table with its gaps, then Moseley's atomic numbers, the modern periodic law, the long form of the table, and IUPAC names for elements beyond Z = 100.
By 1860 about sixty elements were known, and new ones were being found every few years with the help of the spectroscope. Each had its own atomic mass, density, melting point, valence and set of compounds. Learning them one at a time was hopeless; what chemists needed was an organising principle — a way of grouping elements so that knowing one member of a family tells you a great deal about the others, and so that the gaps in the pattern point to elements still waiting to be discovered.
This section follows that search. It begins with families of three (triads) and repeating eights (octaves), reaches its turning point in Mendeleev's table of 1869, which was bold enough to leave gaps and to correct accepted atomic masses, and ends with the discovery that the true ordering number is not the atomic mass but the atomic number $Z$. That gives the modern periodic law and the long form of the periodic table used today. Finally, you will learn how the newest, heaviest elements are named before they receive permanent names.
Early Attempts: Triads, Octaves and Atomic Volumes
The first clue came in 1829 from the German chemist Johann Döbereiner. He noticed that some elements with very similar properties come in groups of three, and that the middle member has an atomic mass close to the average of the other two.
📘 Definition 3.1 — Döbereiner's triad
A Döbereiner triad is a set of three chemically similar elements, arranged in order of increasing atomic mass, in which the atomic mass of the middle element is approximately the arithmetic mean of the atomic masses of the other two:
$$A_{\text{middle}}\approx\tfrac12\left(A_{\text{first}}+A_{\text{third}}\right).$$
Examples: lithium, sodium, potassium; calcium, strontium, barium; chlorine, bromine, iodine. The middle element is also intermediate in properties such as reactivity and density.
✏️ Example 1 — Testing triads
Use the atomic masses (in $\mathrm{u}$) $\mathrm{Li}$ $6.94$, $\mathrm{Na}$ $22.99$, $\mathrm{K}$ $39.10$; $\mathrm{Cl}$ $35.45$, $\mathrm{Br}$ $79.90$, $\mathrm{I}$ $126.90$; $\mathrm{F}$ $19.00$. (a) Show that Li, Na, K and Cl, Br, I behave as triads. (b) Is F, Cl, Br a triad? (c) What does (b) suggest about the triad idea?
💡 Show solution▶
(a) Li, Na, K Mean of the outer masses $=\tfrac12(6.94+39.10)=23.02\ \mathrm{u}$; sodium is $22.99\ \mathrm{u}$ — agreement to $0.03\ \mathrm{u}$ ✓.
(a) Cl, Br, I $\tfrac12(35.45+126.90)=81.175\approx81.18\ \mathrm{u}$; bromine is $79.90\ \mathrm{u}$ — within about $1.6\%$ ✓.
(b) $\tfrac12(19.00+79.90)=49.45\ \mathrm{u}$, but chlorine is $35.45\ \mathrm{u}$. F, Cl and Br are certainly chemically similar (all halogens), yet they are not a triad.
(c) The mass relation works only for a few hand-picked sets; similar elements need not form triads, and most elements could not be placed in any triad. The idea was a useful hint that atomic mass and properties are linked, but not a classification of all the elements.
In 1865 the English chemist John Newlands arranged the elements then known in order of increasing atomic mass and noticed a rhythm.
📐 Law 3.1 — Newlands' law of octaves (historical)
When the elements are arranged in order of increasing atomic mass, every eighth element has properties similar to the first, like the eighth note of a musical octave. For example, starting from lithium: Li, Be, B, C, N, O, F, Na — sodium, the eighth, resembles lithium. The relation holds well only up to calcium.
Beyond calcium the octaves break down: to keep the rhythm Newlands had to put dissimilar elements together (iron, a metal, fell in the same row as oxygen and sulphur) and to squeeze two elements into one place. The noble gases were not yet known; when neon and argon were found, the "eighth" element became the ninth, so the rule could not survive. Newlands' work was ridiculed at first, but in 1887 the Royal Society of London honoured him for it.
Around 1869 the German chemist Lothar Meyer plotted a physical property — the atomic volume, i.e. molar mass divided by density — against atomic mass. The curve rose and fell in a repeating way: the alkali metals sat at the peaks, and elements with similar properties occupied similar positions on the curve. Meyer had shown clearly that properties are periodic, repeating at regular intervals, and he produced a table very like Mendeleev's. Mendeleev published first, and went much further.
✏️ Example 2 — Octaves and atomic volumes
(a) In Newlands' list Li, Be, B, C, N, O, F, Na, Mg, Al, Si, P, S, Cl, K, Ca, which element is the "octave" of beryllium, and which of sodium? (b) Explain why the discovery of argon (atomic mass $39.95$) upset the octaves. (c) The densities of sodium, magnesium and caesium are $0.97$, $1.74$ and $1.93\ \mathrm{g\,cm^{-3}}$ and their molar masses $22.99$, $24.31$ and $132.91\ \mathrm{g\,mol^{-1}}$. Calculate their atomic volumes and say where on Lothar Meyer's curve each lies.
💡 Show solution▶
(a) Counting Be as the first, the eighth is Mg; counting Na as the first, the eighth is K. Both pairs are indeed similar (alkaline earth metals; alkali metals).
(b) Argon lies between chlorine and potassium. Inserting it (and the other noble gases) puts nine places between similar elements, e.g. Na … Ar, K: the regular "eighth" disappears.
Position Na and Cs, alkali metals, sit at peaks (Cs at the highest, far to the right); Mg, the next element after Na, is already on the falling side of the peak. (Check: atomic volume more than halves from Na to Mg although the molar mass barely changes ✓.)
Mendeleev's Periodic Table
The Russian chemist Dmitri Mendeleev (1834–1907) arranged the $63$ elements known in 1869 in order of increasing atomic mass, in rows and columns, so that elements with similar properties — especially similar formulae of their oxides and hydrides — fell in the same vertical column (group).
📐 Law 3.2 — Mendeleev's periodic law (1869)
The properties of the elements are a periodic function of their atomic masses: when the elements are arranged in order of increasing atomic mass, similar properties recur at regular intervals.
What made Mendeleev's table so successful was that he did not follow atomic mass blindly. He trusted chemical similarity more than the measured masses, which were often unreliable at the time. This led him to three bold steps:
He left gaps for undiscovered elements and predicted their properties from those of their neighbours. He called them eka-boron, eka-aluminium and eka-silicon (Sanskrit eka = one: "one place beyond" boron, aluminium, silicon). Scandium (1879), gallium (1875) and germanium (1886) were later found with properties remarkably close to his predictions.
He corrected atomic masses. Where an element did not fit, he argued that its valence, and hence its atomic mass, had been wrongly assigned — beryllium and indium are famous examples (Example 4).
He reversed a few pairs when similarity demanded it: tellurium ($127.6$) was placed before iodine ($126.9$), so that iodine fell with the halogens and tellurium with sulphur and selenium.
How could anyone predict the properties of an element nobody had seen? Mendeleev's method was essentially interpolation: a property of the missing element was estimated from the same property of its neighbours above and below in its group and to the left and right in its row. Interactive 1 lets you repeat his reasoning with modern data.
Interactive 1 — Mendeleev's Prediction Lab: Averaging the Neighbours
The unknown element sits in the centre of a cross formed by its four neighbours in the table. Choose an element and a property; press ▶ Play to light up the neighbours one by one and build the running average, then compare it with Mendeleev's prediction (where he made one) and the value measured today.
average of neighbours—
Mendeleev's prediction—
measured today—
average vs measured—
✏️ Example 3 — Predicting germanium
Germanium's neighbours are silicon (above; $28.09\ \mathrm{u}$, $2.33\ \mathrm{g\,cm^{-3}}$), tin (below; $118.71\ \mathrm{u}$, $7.29\ \mathrm{g\,cm^{-3}}$), gallium (left; $69.72\ \mathrm{u}$, $5.91\ \mathrm{g\,cm^{-3}}$) and arsenic (right; $74.92\ \mathrm{u}$, $5.73\ \mathrm{g\,cm^{-3}}$). (a) Estimate the atomic mass and density of germanium by averaging. (b) Compare with the measured $72.63\ \mathrm{u}$ and $5.32\ \mathrm{g\,cm^{-3}}$. (c) Germanium is in the same group as carbon and silicon. Predict the formulae of its oxide and chloride, and the percentage by mass of germanium in the oxide.
💡 Show solution▶
(a) mass $\tfrac14(28.09+118.71+69.72+74.92)=\tfrac14(291.44)=72.86\ \mathrm{u}$.
(a) density $\tfrac14(2.33+7.29+5.91+5.73)=\tfrac14(21.26)=5.315\approx5.32\ \mathrm{g\,cm^{-3}}$ (the dropped digit is $5$: round to even, $5.32$).
(b) Mass: $72.86$ vs $72.63$, about $0.3\%$ high. Density: $5.32$ vs $5.32$ — agreement to $0.1\%$. Averaging works because properties change smoothly both down a group and along a period.
(c) Group valence $4$ like C and Si: oxide $\mathrm{GeO_{2}}$, chloride $\mathrm{GeCl_{4}}$. $\%\mathrm{Ge}=\dfrac{72.63}{72.63+2(16.00)}\times100=\dfrac{72.63}{104.63}\times100=69.42\%$.
✏️ Example 4 — Mendeleev corrects an atomic mass
(a) Analysis shows that $38.27\ \mathrm{g}$ of indium combine with $8.00\ \mathrm{g}$ of oxygen. Chemists of the 1860s assumed the oxide was $\mathrm{InO}$. What atomic mass did that give? Mendeleev argued the oxide is $\mathrm{In_{2}O_{3}}$. What atomic mass follows, and where does indium then fit? (b) In beryllium oxide, $9.01\ \mathrm{g}$ of beryllium combine with $16.00\ \mathrm{g}$ of oxygen. Find the atomic mass of beryllium if the oxide is $\mathrm{Be_{2}O_{3}}$ (as was once believed, by analogy with aluminium) and if it is $\mathrm{BeO}$.
💡 Show solution▶
Idea The analysis fixes only the mass ratio; the formula decides how many atoms of each element that ratio represents. Use $O=16.00$.
(a) InO One In per O: $8.00\ \mathrm{g}$ O is $0.500\ \mathrm{mol}$ O, so $38.27\ \mathrm{g}$ is $0.500\ \mathrm{mol}$ In: $A=38.27/0.500=76.54\ \mathrm{u}$ — which would put indium between arsenic and selenium, where it does not fit at all.
(a) In₂O₃ Two In per three O: $0.500\ \mathrm{mol}$ O pairs with $\tfrac23\times0.500=0.3333\ \mathrm{mol}$ In, so $A=38.27/0.3333=114.8\ \mathrm{u}$ (accepted value $114.82$). Indium then falls in the boron–aluminium group, below gallium, with trivalent compounds like $\mathrm{InCl_{3}}$ — exactly as Mendeleev said.
(b) $16.00\ \mathrm{g}$ O $=1.000\ \mathrm{mol}$. $\mathrm{BeO}$: $1.000\ \mathrm{mol}$ Be, $A=9.01\ \mathrm{u}$. $\mathrm{Be_{2}O_{3}}$: $\tfrac23\ \mathrm{mol}$ Be, $A=9.01\times\tfrac32=13.5\ \mathrm{u}$ — which would place Be between C and N, with no room for it. Mendeleev chose $9$, putting Be above Mg, and he was right.
⚠️ Limitations of Mendeleev's table
Anomalous pairs. Ordering by mass fails in a few places: $\mathrm{Ar}$ ($39.95$) before $\mathrm{K}$ ($39.10$), $\mathrm{Co}$ ($58.93$) before $\mathrm{Ni}$ ($58.69$), $\mathrm{Te}$ ($127.60$) before $\mathrm{I}$ ($126.90$). Mendeleev had to reverse them by hand; his law gave no reason.
Isotopes. Isotopes of an element have different masses but identical chemistry. If mass decided position, each isotope would need its own place.
Hydrogen resembles both the alkali metals (forms $\mathrm{H^{+}}$, valence 1) and the halogens (forms $\mathrm{H^{-}}$, diatomic gas), so it has no unique place.
Chemically very different elements share a group (alkali metals with copper, silver and gold in his group I), and there was no explanation of why properties repeat.
The Modern Periodic Law and the Long Form of the Table
The explanation came from physics. In 1913 the English physicist Henry Moseley measured the frequencies $\nu$ of the characteristic X-rays emitted by different elements. A plot of $\sqrt{\nu}$ against atomic mass was irregular, but a plot of $\sqrt{\nu}$ against the position number of the element in the table was a perfectly straight line. That position number is the atomic number $Z$, the charge on the nucleus — the number of protons (Section 2.1). Atomic number, not atomic mass, is the fundamental property of an element; and since $Z$ also equals the number of electrons in the neutral atom, it fixes the electronic configuration, which decides chemical behaviour.
📐 Law 3.3 — The modern periodic law
The physical and chemical properties of the elements are periodic functions of their atomic numbers. Arranged in order of increasing atomic number, elements with similar properties recur at regular intervals, because similar outer electronic configurations recur.
Ordering by $Z$ removes every anomalous pair at a stroke: $\mathrm{Ar}$ ($Z=18$) comes before $\mathrm{K}$ ($Z=19$), $\mathrm{Co}$ ($27$) before $\mathrm{Ni}$ ($28$), $\mathrm{Te}$ ($52$) before $\mathrm{I}$ ($53$). All isotopes of an element share one $Z$ and therefore one place. The table built on this law, and recommended by IUPAC, is the long form of the periodic table.
📘 Definition 3.2 — Periods and groups of the long form
In the long form of the periodic table:
a period is a horizontal row. There are 7 periods, containing $2,\ 8,\ 8,\ 18,\ 18,\ 32$ and $32$ elements;
a group is a vertical column. There are 18 groups, numbered 1 to 18 (IUPAC). Elements in a group have the same outer electronic configuration and similar properties: group 1 the alkali metals, group 2 the alkaline earth metals, group 17 the halogens, group 18 the noble gases;
the 14 elements after lanthanum (cerium, $Z=58$, to lutetium, $Z=71$), the lanthanoids, and the 14 after actinium (thorium, $Z=90$, to lawrencium, $Z=103$), the actinoids, are placed in two separate rows at the bottom, to keep the table a convenient width. They belong to periods 6 and 7.
The long form in outline. Numbers on the right give the number of elements in each period. Section 3.2 explains the colours — the s, p, d and f blocks — from electronic configurations.
✏️ Example 5 — Atomic number versus atomic mass
(a) Using atomic masses (Ar $39.95$, K $39.10$; Co $58.93$, Ni $58.69$; Te $127.60$, I $126.90$) and atomic numbers (Ar 18, K 19; Co 27, Ni 28; Te 52, I 53), explain why these pairs were a problem for Mendeleev but not for the modern law. (b) Chlorine-35 and chlorine-37 differ in mass by two units. Where do they go in each kind of table? (c) How many elements are there in periods 1 to 7 altogether, and why is the long form drawn with the lanthanoids and actinoids below?
💡 Show solution▶
(a) In each pair the first element is heavier, so ordering by mass would put K before Ar (an alkali metal among the noble gases), Ni before Co and I before Te (iodine with S and Se). Ordering by $Z$ gives Ar, K; Co, Ni; Te, I — the chemically correct order, with no exceptions. (The masses are "out of order" because the heavier element happens to have more neutrons in its common isotopes.)
(b) Both have $Z=17$ and the same configuration, so in the modern table they share chlorine's single place. A table based strictly on mass would need two places for one element — a contradiction.
(c) $2+8+8+18+18+32+32=118$. Periods 6 and 7 have 32 elements each; written in a single row the table would be 32 columns wide. Taking out 14 lanthanoids and 14 actinoids leaves 18 columns.
Naming Elements with Atomic Numbers above 100
The heaviest elements are made a few atoms at a time in particle accelerators, and their discovery takes years to confirm. Traditionally the discoverers proposed a name, which led to disputes when two laboratories claimed the same element. Since 1978 IUPAC has used a temporary systematic name and three-letter symbol built directly from the atomic number, used until a permanent name is approved.
📐 Rule 3.4 — IUPAC systematic names for $Z\gt100$
Write the root for each digit of $Z$ in order, then add the ending -ium. The symbol is formed from the first letters of the roots (the first letter capitalised).
digit
0
1
2
3
4
5
6
7
8
9
root
nil
un
bi
tri
quad
pent
hex
sept
oct
enn
letter
n
u
b
t
q
p
h
s
o
e
Two spelling rules: the final i of bi and tri is dropped before -ium (bium, trium, not biium, triium); and the final n of enn is dropped before nil (ennil, not ennnil).
✏️ Example 6 — Systematic names and symbols
(a) Give the IUPAC systematic name and symbol for elements $104$, $109$, $112$ and $190$. (b) Which elements have the systematic symbols $\mathrm{Uuq}$ and $\mathrm{Ubn}$? (c) Element $118$ is now called oganesson ($\mathrm{Og}$). What was its systematic name, and in which period and group is it placed?
(c) 1-1-8: ununoctium, Uuo. With $2+8+8+18+18+32+32=118$, it is the last element of period 7: period 7, group 18, below radon.
Interactive 2 — Systematic Name Generator for $Z\gt100$
Choose an atomic number. Each digit is turned into its root, the roots are joined, the spelling rules are applied (watch for bi/tri + ium and enn + nil), and the symbol is read off from the first letters. For $Z=101$–$118$ the permanent name is shown as well. Press ▶ Play to step through the numbers.
114
systematic name—
symbol—
permanent name—
place in the table—
Exercises
I
Early classifications. Atomic masses in $\mathrm{u}$: Ca $40.08$, Sr $87.62$, Ba $137.33$, S $32.06$, Se $78.97$, Te $127.60$, N $14.01$, P $30.97$, As $74.92$.
1 Show that calcium, strontium and barium form a Döbereiner triad.
2 Do sulphur, selenium and tellurium form a triad?
3 Do nitrogen, phosphorus and arsenic form a triad? What does your answer show?
4 State Newlands' law of octaves and give two reasons why it failed beyond calcium.
5 In the list Li, Be, B, C, N, O, F, Na, Mg, Al, Si, P, S, Cl, K, Ca, which element is the "octave" of carbon? Of oxygen?
6 Calculate the atomic volumes of potassium ($M=39.10\ \mathrm{g\,mol^{-1}}$, $\rho=0.89\ \mathrm{g\,cm^{-3}}$) and aluminium ($26.98\ \mathrm{g\,mol^{-1}}$, $2.70\ \mathrm{g\,cm^{-3}}$). Which lies near a peak of Lothar Meyer's curve?
7 What did Lothar Meyer plot, and what did his curve show?
8 Give one strength and one weakness of Döbereiner's triads.
💡 Show solutions▶
1 $\tfrac12(40.08+137.33)=88.7\ \mathrm{u}$ against Sr $87.62\ \mathrm{u}$ — within about $1.2\%$; the three are chemically similar (alkaline earth metals), so they form a triad.
2 $\tfrac12(32.06+127.60)=79.83\ \mathrm{u}$ against Se $78.97\ \mathrm{u}$ (about $1.1\%$): yes.
3 $\tfrac12(14.01+74.92)=44.47\ \mathrm{u}$ against P $30.97\ \mathrm{u}$: no. Chemically similar elements need not obey the mass rule, so triads cannot be a general classification.
4 Arranged by increasing atomic mass, every eighth element resembles the first. It failed because (i) after Ca it forced unlike elements together (e.g. Fe with O and S) and put two elements in one place; (ii) the noble gases, found later, make the similar element the ninth, not the eighth.
5 Carbon is 4th, so its octave is the 11th: Si. Oxygen (6th) → 13th: S.
6 K: $39.10/0.89=43.9\ \mathrm{cm^{3}\,mol^{-1}}$; Al: $26.98/2.70=9.99\approx10.0\ \mathrm{cm^{3}\,mol^{-1}}$. Potassium, an alkali metal, lies near a peak; aluminium near a trough.
7 Atomic volume (molar mass ÷ density) against atomic mass. The curve rose and fell repeatedly, with alkali metals at the peaks and similar elements at similar positions — properties are periodic.
8 Strength: first evidence that atomic mass is linked to chemical similarity. Weakness: only a few triads exist; most elements could not be fitted, and some similar sets (F, Cl, Br) fail the rule.
II
Mendeleev's periodic table.
1 State Mendeleev's periodic law.
2 Which elements did Mendeleev call eka-boron, eka-aluminium and eka-silicon? What does "eka" mean?
3 Estimate the atomic mass of gallium from its neighbours Al ($26.98$), In ($114.82$), Zn ($65.38$) and Ge ($72.63$). Compare with $69.72\ \mathrm{u}$.
4 Estimate the density of gallium from Al ($2.70$), In ($7.31$), Zn ($7.14$) and Ge ($5.32\ \mathrm{g\,cm^{-3}}$). By what percentage does the estimate differ from $5.91\ \mathrm{g\,cm^{-3}}$?
5 Mendeleev predicted the oxide $\mathrm{E_{2}O_{3}}$ for eka-aluminium. Write the formula of its chloride, and calculate the percentage by mass of gallium in $\mathrm{Ga_{2}O_{3}}$ ($\mathrm{Ga}=69.72$, $\mathrm{O}=16.00$).
6 Name the three pairs of elements Mendeleev placed in reverse order of atomic mass, and explain why he did so.
7 Why is the position of hydrogen a problem in any periodic table?
8 $9.00\ \mathrm{g}$ of a metal M combine with $8.00\ \mathrm{g}$ of oxygen. Find the atomic mass of M if its oxide is (a) $\mathrm{MO}$, (b) $\mathrm{M_{2}O_{3}}$. Which value identifies a known element?
💡 Show solutions▶
1 The properties of the elements are a periodic function of their atomic masses.
2 Scandium, gallium and germanium. "Eka" (Sanskrit, one) means one place beyond: eka-silicon is the element one place below silicon.
3 $\tfrac14(26.98+114.82+65.38+72.63)=\tfrac14(279.81)=69.95\ \mathrm{u}$ — only $0.3\%$ above $69.72\ \mathrm{u}$.
6 Ar/K, Co/Ni, Te/I. In each pair the heavier element is chemically like the group before the lighter one (Te like S and Se; I like the halogens), so Mendeleev let properties override mass.
7 Hydrogen has one valence electron and forms $\mathrm{H^{+}}$ like the alkali metals, but it also needs one electron to complete its shell, forms $\mathrm{H^{-}}$ and is a diatomic gas like the halogens.
8 $8.00\ \mathrm{g}$ O $=0.500\ \mathrm{mol}$. (a) MO: $0.500\ \mathrm{mol}$ M, $A=9.00/0.500=18.0$. (b) $\mathrm{M_{2}O_{3}}$: $\tfrac23\times0.500=0.3333\ \mathrm{mol}$ M, $A=9.00/0.3333=27.0$ — aluminium. (No metal has atomic mass $18$; the oxide is $\mathrm{Al_{2}O_{3}}$.)
III
The modern periodic law and the long form.
1 State the modern periodic law.
2 What did Moseley plot, and what did he discover?
3 How many elements are there in each of the seven periods? How many altogether?
4 Explain how the modern law resolves the position of tellurium and iodine.
5 Give the range of atomic numbers of the lanthanoids and of the actinoids. How many elements are in each series, and in which periods do they belong?
6 Without using a table, find the period and group of the elements with $Z=18$ and $Z=19$ by counting the elements in each period.
7 How do the modern law and the long form deal with isotopes?
8 How many groups does the long form have? Give the family names of groups 1, 2, 17 and 18.
💡 Show solutions▶
1 The physical and chemical properties of the elements are periodic functions of their atomic numbers.
2 The square root of the frequency of the characteristic X-rays of each element, $\sqrt{\nu}$. Against atomic mass the plot was irregular; against the position number $Z$ it was a straight line, showing that $Z$ (the nuclear charge) is the fundamental property.
3 $2,8,8,18,18,32,32$; total $118$.
4 Te has $Z=52$ and I has $Z=53$, so in order of $Z$ tellurium comes first — exactly where chemistry needs it, with no special exception.
5 Lanthanoids $Z=58$–$71$ (Ce–Lu), actinoids $Z=90$–$103$ (Th–Lr); 14 each; periods 6 and 7 respectively.
6 Periods 1–3 hold $2+8+8=18$ elements, so $Z=18$ is the last of period 3: period 3, group 18 (argon). $Z=19$ starts period 4: period 4, group 1 (potassium).
7 All isotopes of an element have the same $Z$, so they occupy the same single place; their different masses do not matter.
8 18 groups. Group 1: alkali metals; group 2: alkaline earth metals; group 17: halogens; group 18: noble gases.
IV
IUPAC systematic names.
1 Write the systematic names and symbols of elements $101$, $105$, $113$ and $117$.
2 Do the same for $120$, $126$ and $139$.
3 Do the same for $150$ and $202$.
4 Give the atomic numbers of the elements with systematic symbols $\mathrm{Uno}$, $\mathrm{Une}$ and $\mathrm{Uue}$.
5 Correct the spelling of "ununbiium" ($Z=112$) and "unennnilium" ($Z=190$), stating the rule used in each case.
6 Element $119$ has not yet been made. Give its systematic name and predict its period and group.
7 Element $117$ is now called tennessine ($\mathrm{Ts}$). In which group is it placed, and which family-style ending does its name share with chlorine and iodine?
8 Why did IUPAC introduce systematic names for new elements?
5ununbium — the final i of "bi" is dropped before "-ium". unennilium — the final n of "enn" is dropped before "nil".
6 Ununennium, Uue. Period 7 ends at $118$, so $119$ would begin period 8 in group 1, below francium.
7 Group 17, the halogens (below astatine); the ending -ine, used for group-17 elements.
8 Superheavy elements are made a few atoms at a time and their discovery is slow to confirm; laboratories sometimes claimed the same element and proposed different names. A temporary name generated from $Z$ avoids disputes until a permanent name is approved.
🎯 Exam corner BoardJEE MainNEET
E1.NEET The IUPAC systematic symbol of the element with atomic number $109$ is (A) $\mathrm{Unn}$ (B) $\mathrm{Uue}$ (C) $\mathrm{Une}$ (D) $\mathrm{Uen}$
💡 Show solution▶
Roots 1 → un (u), 0 → nil (n), 9 → enn (e): unnilennium.
Answer(C) Une. (Uue is 119; Uen is 190.)
E2.JEE MainNumerical answer. What is the atomic number of the element whose IUPAC systematic name is unbiennium?
💡 Show solution▶
Split un-bi-enn-ium → 1, 2, 9.
Answer129.
E3.Board · 3 marks State Mendeleev's periodic law and the modern periodic law. Give two defects of Mendeleev's table that the modern law removes.
💡 Show solution▶
1 mark Mendeleev: properties of elements are a periodic function of their atomic masses.
1 mark Modern: properties of elements are a periodic function of their atomic numbers.
1 mark (i) Anomalous pairs (Ar/K, Co/Ni, Te/I) fall into correct order by $Z$ with no exceptions; (ii) isotopes, having the same $Z$, need only one place.
E4.JEE MainNEETAssertion–reason.Assertion (A): In the modern periodic table cobalt ($58.93\ \mathrm{u}$) is placed before nickel ($58.69\ \mathrm{u}$). Reason (R): In the modern periodic table elements are arranged in order of increasing atomic number. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
💡 Show solution▶
Check Co has $Z=27$ and Ni $Z=28$, so ordering by $Z$ places Co first even though it is heavier. A is true and R explains it.
Answer(A).
E5.NEET Which statement is not correct? (A) Mendeleev arranged the elements in order of increasing atomic mass. (B) Germanium is the element Mendeleev called eka-silicon. (C) Mendeleev's original table already contained a group for the noble gases. (D) Mendeleev changed the accepted atomic mass of beryllium.
💡 Show solution▶
Check (A), (B), (D) are true (beryllium: $13.5\to9$). The noble gases were discovered in the 1890s and were added afterwards as a new group.
Answer(C).
E6.Board · Case study A student lists three soft, reactive metals that all form chlorides $\mathrm{MCl}$: lithium ($6.94\ \mathrm{u}$), sodium ($22.99\ \mathrm{u}$) and potassium ($39.10\ \mathrm{u}$); and three similar non-metals: chlorine ($35.45$), bromine ($79.90$) and iodine ($126.90$). (i) Show that each set obeys Döbereiner's rule. (2 marks) (ii) Why was the triad idea abandoned? (1 mark) (iii) In Newlands' arrangement, which element is the "octave" of lithium? (1 mark)
💡 Show solution▶
(i) $\tfrac12(6.94+39.10)=23.02\approx22.99$ (Na) ✓; $\tfrac12(35.45+126.90)=81.18\approx79.90$ (Br) ✓. In each case the middle mass is close to the mean of the other two.
(ii) Only a few such triads could be found; most elements could not be placed in any triad.
(iii) Li, Be, B, C, N, O, F, Na — sodium.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. How are the newest elements made, how do we know they exist, and how did four of them earn their names in 2016?
Why it matters. Elements 113 to 118 completed the seventh period. They were made at laboratories in Japan (RIKEN), Russia (Dubna) and the USA (Lawrence Livermore, Oak Ridge) by smashing lighter nuclei together, often a few atoms in months of work, and each atom lives for less than a second. Their names — nihonium, moscovium, tennessine and oganesson — honour Japan, Moscow, Tennessee and the physicist Yuri Oganessian. The same accelerator techniques are used to make medical isotopes for cancer treatment.
What to do.
Write the systematic name and symbol of each element from 113 to 118, and the permanent name and symbol approved in 2016.
Oganesson-294 was made by firing calcium-48 nuclei at californium-249 ($Z=98$), with three neutrons released. Check that the atomic numbers and mass numbers balance.
How many neutrons are there in $\mathrm{{}^{294}Og}$, $\mathrm{{}^{294}Ts}$ and $\mathrm{{}^{289}Fl}$?
Using the long form of the table, predict the group of each element 113–118 and one property it might be expected to share with the element above it. Find out (from IUPAC or a national laboratory website) why scientists doubt that oganesson really behaves like a noble gas.
Tools: Rule 3.4 and Interactive 2; the IUPAC website or a recent review article on superheavy elements.
Hand in: A table (Z, systematic name, permanent name, group, neutrons in one isotope) and a half-page note on how a new element is confirmed.
Going further. Look up the "island of stability" and the attempts to make elements 119 and 120: what are their systematic names, and where would they go in the table?
Instructor note: 113 ununtrium Uut / nihonium Nh; 114 ununquadium Uuq / flerovium Fl (named 2012); 115 ununpentium Uup / moscovium Mc; 116 ununhexium Uuh / livermorium Lv (2012); 117 ununseptium Uus / tennessine Ts; 118 ununoctium Uuo / oganesson Og. Balance: $Z$: $20+98=118$; $A$: $48+249=297=294+3$ ✓. Neutrons: $294-118=176$, $294-117=177$, $289-114=175$. Groups 13–18. Relativistic effects on the outer electrons are expected to make Og more reactive than a typical noble gas (some calculations suggest a semiconductor-like solid). 119 ununennium (group 1, period 8), 120 unbinilium (group 2).
🧪 Self-check quiz
0/5 correct
Q1. The modern periodic law states that properties of elements are periodic functions of their
Q2. The element Mendeleev called eka-aluminium is
Q3. The IUPAC systematic symbol for element $110$ is
Q4. How many elements are there in the fourth period?
Q5. Which pair would be in the wrong order if elements were arranged strictly by atomic mass?