Historical development of the periodic table
Today the periodic table hangs on the wall of every chemistry classroom, neat and complete, as if it had always looked like this. It did not: it took more than forty years to find the right ordering criterion, and along the way there were brilliant successes, predictions that came true almost to the gram, and mistakes that nobody could explain until atomic physics arrived.
The first attempts: triads and octaves
In 1829, the German Johann Döbereiner noticed that certain groups of three elements with similar chemical properties — the triads — showed a numerical regularity: the atomic mass of the middle element was approximately the mean of the other two.
Decades later, in 1865, the Englishman John Newlands took a more ambitious step: he arranged all the known elements by increasing atomic mass and observed that, starting from lithium, every eighth element repeated the properties of the first. He called it the "law of octaves", by analogy with the musical scale.
The law of octaves worked well for the lightest elements, but it soon failed: in 1865 only sixty-three elements were known, and whole families — the noble gases, for example — had not yet been discovered. A closed musical scale, in groups of eight with no exceptions, had no room to accommodate what was still waiting to be found.
In $1829$, the German chemist Johann Döbereiner published his 'triads': groups of three elements with very similar chemical properties, such as lithium, sodium and…
In $1829$, the German chemist Johann Döbereiner published his 'triads': groups of three elements with very similar chemical properties, such as lithium, sodium and potassium.
- a) What did Döbereiner observe about the atomic mass of the central element in each triad, compared with the mean of the masses of the other two?
- b) What properties did the elements in the alkali metal triad share? State two.
The atomic masses of chlorine, bromine and iodine are $35,5\ \text{u}$, $79,9\ \text{u}$ and $126,9\ \text{u}$. Work out the mean of the masses of chlorine and iodine…
The atomic masses of chlorine, bromine and iodine are $35,5\ \text{u}$, $79,9\ \text{u}$ and $126,9\ \text{u}$. Work out the mean of the masses of chlorine and iodine and check whether it approximately equals the mass of bromine, as claimed by Döbereiner's law of triads.
In $1865$, the Englishman John Newlands arranged the known elements in order of increasing atomic mass and observed that, starting from lithium, every eighth element…
In $1865$, the Englishman John Newlands arranged the known elements in order of increasing atomic mass and observed that, starting from lithium, every eighth element repeated the properties of the first: he called this the 'law of octaves'. The diagram represents his 'scale' of elements:
- a) Why did he choose this musical name?
- b) Counting lithium as the first, the next seven elements by mass are Be, B, C, N, O, F and Na: which element occupies the 'eighth note' from lithium? Are their properties similar?
Another classic triad of Döbereiner consists of calcium ($40,1\ \text{u}$), strontium ($87,6\ \text{u}$) and barium ($137,3\ \text{u}$). a) Work out the mean of the…
Another classic triad of Döbereiner consists of calcium ($40,1\ \text{u}$), strontium ($87,6\ \text{u}$) and barium ($137,3\ \text{u}$).
- a) Work out the mean of the masses of calcium and barium.
- b) Calculate the percentage deviation of this mean from the actual mass of strontium and evaluate whether the law of triads works in this case.
The end members of the alkali metal triad are lithium ($6,9\ \text{u}$) and potassium ($39,1\ \text{u}$). a) Use Döbereiner's rule to predict the atomic mass of the…
The end members of the alkali metal triad are lithium ($6,9\ \text{u}$) and potassium ($39,1\ \text{u}$).
- a) Use Döbereiner's rule to predict the atomic mass of the central element, sodium.
- b) The actual mass of sodium is $23,0\ \text{u}$: calculate the percentage error of the prediction and comment on the result.
Mendeleev: gaps instead of forcing the order
In 1869, the Russian Dmitri Mendeleev published a table with the 63 known elements, also arranged by increasing atomic mass, but with a decisive idea: when the next element on the list did not fit the properties of a column, he left a gap instead of forcing it into the wrong place.
The most famous case is that of eka-aluminium, the gap that Mendeleev left below aluminium. For it he predicted an atomic mass of about $68\ \text{u}$ and a density of $5,9\ \text{g/cm}^3$. A few years later, in 1875, Lecoq de Boisbaudran discovered gallium: mass $69,7\ \text{u}$, density $5,9\ \text{g/cm}^3$.
Mendeleev also had to swap the order of some pairs, such as tellurium ($127,6\ \text{u}$) and iodine ($126,9\ \text{u}$): because of their chemical properties, tellurium had to come first, even though it was heavier. He could not justify why; forty years later, Moseley's work would explain it.
In $1869$, the Russian Dmitri Mendeléiev published the first version of his periodic table with the $63$ elements known at the time. a) By what criterion did he mainly…
In $1869$, the Russian Dmitri Mendeléiev published the first version of his periodic table with the $63$ elements known at the time.
- a) By what criterion did he mainly arrange the elements?
- b) When the next element on the list did not fit the properties of a column, Mendeléiev left a gap instead of forcing it: how did he interpret these gaps?
Mendeléiev predicted the existence of an unknown element which he called eka-aluminium, discovered a few years later and named gallium ($\text{Ga}$). The diagram shows…
Mendeléiev predicted the existence of an unknown element which he called eka-aluminium, discovered a few years later and named gallium ($\text{Ga}$). The diagram shows the gap that Mendeléiev left in his table:
- a) The prefix eka means 'one beyond' in Sanskrit: what position did eka-aluminium occupy relative to aluminium in the table?
- b) What would have happened to the credibility of the periodic table if gallium had never been found?
Tellurium ($M = 127,6\ \text{u}$) is heavier than iodine ($M = 126,9\ \text{u}$); however, Mendeléiev placed tellurium before iodine, contrary to the order of increasing…
Tellurium ($M = 127,6\ \text{u}$) is heavier than iodine ($M = 126,9\ \text{u}$); however, Mendeléiev placed tellurium before iodine, contrary to the order of increasing mass.
- a) What chemical properties led him to make this swap?
- b) What ordering criterion, discovered decades later, justified this swap as being correct?
Mendeléiev predicted an atomic mass of about $68\ \text{u}$ and a density of $5,9\ \text{g/cm}^3$ for eka-aluminium. Gallium, discovered in $1875$ by Lecoq de…
Mendeléiev predicted an atomic mass of about $68\ \text{u}$ and a density of $5,9\ \text{g/cm}^3$ for eka-aluminium. Gallium, discovered in $1875$ by Lecoq de Boisbaudran, has a mass of $69,7\ \text{u}$ and a density of $5,9\ \text{g/cm}^3$.
- a) Calculate the percentage error of the predicted mass.
- b) Evaluate the overall quality of both predictions: what do they reveal about the method used by Mendeléiev?
A student searches for Döbereiner's triads among these elements, with their atomic masses: Mg ($24,3\ \text{u}$), Ca ($40,1\ \text{u}$), Sr ($87,6\ \text{u}$), Ba…
A student searches for Döbereiner's triads among these elements, with their atomic masses: Mg ($24,3\ \text{u}$), Ca ($40,1\ \text{u}$), Sr ($87,6\ \text{u}$), Ba ($137,3\ \text{u}$) and Zn ($65,4\ \text{u}$).
- a) Check, by calculating the relevant means, which sets of three elements consecutive in mass satisfy Döbereiner's rule (the mass of the central element matches the mean of the outer ones with an error of less than $5\ \%$).
- b) Of the triads that satisfy the numerical rule, which ones also group chemically similar elements?
- c) Conclude why the law of triads alone was not sufficient to arrange all the known elements.
In $1871$, Mendeléiev predicted for eka-silicon: atomic mass $72\ \text{u}$, density $5,5\ \text{g/cm}^3$ and an oxide with formula $\text{EO}2$ with a density of…
In $1871$, Mendeléiev predicted for eka-silicon: atomic mass $72\ \text{u}$, density $5,5\ \text{g/cm}^3$ and an oxide with formula $\text{EO}_2$ with a density of $4,7\ \text{g/cm}^3$. Germanium, discovered in $1886$, has mass $72,6\ \text{u}$, density $5,3\ \text{g/cm}^3$, and its oxide $\text{GeO}_2$ has a density of $4,2\ \text{g/cm}^3$.
- a) Calculate the percentage error of each of the three predictions.
- b) Why did he predict an oxide with formula $\text{EO}_2$ and not $\text{EO}$ or $\text{E}_2\text{O}_3$? Think about the silicon column and the formulas of its compounds.
- c) In light of these results, comment on the statement 'the periodic table is a predictive tool'.
Moseley: atomic number replaces mass
Mendeleev's table had a fundamental problem: atomic mass, its ordering criterion, produced inversions (Ar–K, Co–Ni, Te–I) that nobody could explain. The solution came in 1913, from a 26-year-old physicist: Henry Moseley.
Moseley bombarded different elements with electrons and measured the frequency $\nu$ of the characteristic X-rays they emitted. He found a very clean mathematical relationship between that frequency and an increasing whole number, one for each element:
$$\sqrt{\nu} = k\cdot\left(Z - 1\right)$$where $k$ is a constant and $Z$ is the atomic number: the number of protons in the nucleus. Unlike mass, which depends on the mixture of natural isotopes of each element, $Z$ is an exact whole number, with no possible ambiguity.
With this criterion, Mendeleev's inversions stopped being a mystery: argon ($Z=18$) comes before potassium ($Z=19$) even though it is slightly heavier, because atomic mass is the weighted mean of the natural isotopes, and one element can have fewer protons but, on average, more neutrons than the next one.
In $1913$, the young British physicist Henry Moseley studied the X-rays emitted by different elements when bombarded with electrons and found a perfect regularity. The…
In $1913$, the young British physicist Henry Moseley studied the X-rays emitted by different elements when bombarded with electrons and found a perfect regularity. The diagram summarises his experimental set-up:
- a) What physical quantity did Moseley assign to each element as an ordering criterion, instead of atomic mass?
- b) What is this number called today in the periodic table and what exactly does it count in the atom?
Argon has an atomic mass of $39,9\ \text{u}$ and potassium $39,1\ \text{u}$; however, in the modern table argon ($Z = 18$) comes before potassium ($Z = 19$). a) Arrange…
Argon has an atomic mass of $39,9\ \text{u}$ and potassium $39,1\ \text{u}$; however, in the modern table argon ($Z = 18$) comes before potassium ($Z = 19$).
- a) Arrange Ar and K first by mass and then by $Z$: which of the two arrangements matches their properties (inert gas versus highly reactive soft metal)?
- b) Explain how Moseley's work permanently eliminated this type of 'reversal' that had remained unexplained for Mendeléiev.
Moseley measured the frequency $\nu$ of the characteristic X-rays of many elements and found that the square root of the frequency increases linearly with atomic number:…
Moseley measured the frequency $\nu$ of the characteristic X-rays of many elements and found that the square root of the frequency increases linearly with atomic number:
$\sqrt{\nu} = 4,98\cdot10^{7}\cdot\left(Z - 1\right)\ \text{Hz}^{1/2}$
Plotted on a graph, the values for the elements lie on a straight line:
- a) Work out the frequency of the characteristic X-rays of copper ($Z = 29$).
- b) An unknown metallic sample emits X-rays with $\nu = 1,54\cdot10^{18}\ \text{Hz}$: calculate its atomic number and identify the element knowing that $Z = 26$ corresponds to iron.
- c) Explain why this law allowed the table to be arranged by $Z$ without any ambiguity, something impossible with atomic mass.
The three classic 'reversals' in the table are Ar ($39,9\ \text{u}$)–K ($39,1\ \text{u}$), Co ($58,9\ \text{u}$)–Ni ($58,7\ \text{u}$) and Te ($127,6\ \text{u}$)–I…
The three classic 'reversals' in the table are Ar ($39,9\ \text{u}$)–K ($39,1\ \text{u}$), Co ($58,9\ \text{u}$)–Ni ($58,7\ \text{u}$) and Te ($127,6\ \text{u}$)–I ($126,9\ \text{u}$): in all three pairs, the heavier element comes first in the table.
- a) Using atomic numbers (Ar: $Z = 18$, K: $Z = 19$; Co: $Z = 27$, Ni: $Z = 28$; Te: $Z = 52$, I: $Z = 53$), verify that the order by $Z$ is the opposite of the order by mass in all three cases.
- b) The atomic mass of an element is the weighted mean of the masses of its naturally occurring isotopes: use this idea to explain how an element with a lower $Z$ can turn out to be heavier than the next one.
- c) Conclude why atomic mass cannot be the ordering criterion of the table, whereas atomic number can.
What the modern table solved
Mendeleev's table, for all its success, had difficulties that today's table does not have: there was no column for the noble gases (discovered later), hydrogen did not fit well in any group, the isotopes of the same element seemed to require different cells, and the rare earths (lanthanides) could not find a place. Ordering by $Z$ and basing the table on electron configurations — which you will see in the next section — solved all four at once.
State the modern periodic law in your own words. a) On what quantity do the properties of the elements periodically depend? b) What does it mean for properties to repeat…
State the modern periodic law in your own words.
- a) On what quantity do the properties of the elements periodically depend?
- b) What does it mean for properties to repeat 'periodically'? Support your answer with an example of two elements with very similar properties located in the same column.
Newlands' law of octaves worked reasonably well with lighter elements, but failed beyond calcium. a) Arrange these eight elements in order of increasing mass: Li (…
Newlands' law of octaves worked reasonably well with lighter elements, but failed beyond calcium.
- a) Arrange these eight elements in order of increasing mass: Li ($6,9\ \text{u}$), B ($10,8\ \text{u}$), Na ($23,0\ \text{u}$), Be ($9,0\ \text{u}$), C ($12,0\ \text{u}$), F ($19,0\ \text{u}$), O ($16,0\ \text{u}$) and N ($14,0\ \text{u}$), and check that the eighth is chemically similar to the first.
- b) Propose a historical reason why the law of octaves could not be generalised: think about how many elements were known in $1865$ and which entire families had yet to be discovered.
In $1886$, Clemens Winkler isolated germanium ($\text{Ge}$) from the mineral argyrodite and verified that its properties matched almost exactly those of the eka-silicon…
In $1886$, Clemens Winkler isolated germanium ($\text{Ge}$) from the mineral argyrodite and verified that its properties matched almost exactly those of the eka-silicon predicted by Mendeléiev fifteen years earlier. The first clue that there was a new element in that mineral came from the analysis of its spectrum.
- a) Why does the appearance of unknown spectral lines indicate the presence of a new element?
- b) How does this episode relate to the idea of the periodic table as a predictive tool?
The melting points of the halogens are: $\text{F}2$: $-220\ \celsius$, $\text{Cl}2$: $-101\ \celsius$, $\text{Br}2$: $-7\ \celsius$ and $\text{I}2$: $114\ \celsius$. a)…
The melting points of the halogens are: $\text{F}_2$: $-220\ \celsius$, $\text{Cl}_2$: $-101\ \celsius$, $\text{Br}_2$: $-7\ \celsius$ and $\text{I}_2$: $114\ \celsius$.
- a) Plot melting point against period number in your notebook and describe the trend down the group.
- b) Astatine, the next halogen, barely exists in nature and has hardly been studied: estimate its melting point graphically and deduce whether it will be a solid at room temperature ($25\ \celsius$).
You can do it on paper and send a photo with your answer.
The densities of the alkali metals are: Li: $0,53\ \text{g/cm}^3$, Na: $0,97\ \text{g/cm}^3$, K: $0,86\ \text{g/cm}^3$, Rb: $1,53\ \text{g/cm}^3$ and Cs:…
The densities of the alkali metals are: Li: $0,53\ \text{g/cm}^3$, Na: $0,97\ \text{g/cm}^3$, K: $0,86\ \text{g/cm}^3$, Rb: $1,53\ \text{g/cm}^3$ and Cs: $1,87\ \text{g/cm}^3$. The diagram shows these values:
- a) Do they follow a strictly increasing trend going down the group? Point out the exception.
- b) Estimate a reasonable value for the density of francium, the period 7 alkali metal, and justify your estimate based on the general trend.
Mendeléiev's table had four difficulties that the modern table does not have: ① there was no column for noble gases; ② hydrogen did not fit well into any group; ③…
Mendeléiev's table had four difficulties that the modern table does not have: ① there was no column for noble gases; ② hydrogen did not fit well into any group; ③ isotopes of the same element seemed to require several positions; ④ rare earth elements (lanthanides) had no clear place.
- a) Explain the historical or chemical origin of each of the four difficulties.
- b) Explain how the modern table resolves them, being ordered by $Z$ and constructed from electron configurations.
- c) When Ramsay discovered the noble gases in the $1890$s, the table was able to add an entire column without breaking down: deduce why this discovery strengthened the table rather than destroying it.
Imagine it is $1865$ and you only know these ten elements, already arranged by mass: Li ($6,9\ \text{u}$), Be ($9,0\ \text{u}$), B ($10,8\ \text{u}$), C (…
Imagine it is $1865$ and you only know these ten elements, already arranged by mass: Li ($6,9\ \text{u}$), Be ($9,0\ \text{u}$), B ($10,8\ \text{u}$), C ($12,0\ \text{u}$), N ($14,0\ \text{u}$), O ($16,0\ \text{u}$), F ($19,0\ \text{u}$), Na ($23,0\ \text{u}$), Mg ($24,3\ \text{u}$) and Si ($28,1\ \text{u}$). Their properties: Li and Na are soft metals that react with water; Be and Mg are harder and less reactive metals; B and Si are metalloids; C, N, O and F are non-metals.
- a) Verify Newlands' law of octaves: does every eighth element resemble the first in its 'scale'? Identify the three pairs that confirm it.
- b) Argon ($Z = 18$, $39,9\ \text{u}$) was not discovered until $1894$: deduce where Newlands would have placed it and what problem an element that resembles no other would have caused for his musical scale.
- c) Explain why Mendeléiev's table, which left gaps, was more flexible than Newlands' closed scale.
The melting points of the alkali metals are: Li: $181\ \celsius$, Na: $98\ \celsius$, K: $64\ \celsius$, Rb: $39\ \celsius$ and Cs: $28\ \celsius$. a) Plot the trend in…
The melting points of the alkali metals are: Li: $181\ \celsius$, Na: $98\ \celsius$, K: $64\ \celsius$, Rb: $39\ \celsius$ and Cs: $28\ \celsius$.
- a) Plot the trend in your notebook and estimate the melting point of francium, the period 7 alkali metal.
- b) Also predict, justifying your answer from the group's behaviour: its state at room temperature ($25\ \celsius$), its hardness and its reactivity with water.
- c) Francium is so rare and radioactive that its measured properties are barely known: explain why the periodic table allows predictive science to be done even with elements that are almost impossible to study in the laboratory.
You can do it on paper and send a photo with your answer.
Write a short report titled 'From triads to atomic number: how the periodic table was built'. a) Summarise the contributions of Döbereiner, Newlands, Mendeléiev and…
Write a short report titled 'From triads to atomic number: how the periodic table was built'.
- a) Summarise the contributions of Döbereiner, Newlands, Mendeléiev and Moseley, indicating the ordering criterion proposed by each.
- b) Point out what problem remained unsolved after each scientist and how the next one addressed it.
- c) Conclude by explaining why it is said that a good scientific theory not only organizes what is known, but predicts the unknown: illustrate this with the case of eka-aluminium.