Periodic table and trends
Mendeleev arranged the elements in order of increasing atomic mass and, for the most part, similar elements fell into the same column. A few pairs, such as tellurium and iodine, only fitted if he swapped them. Why do elements repeat their chemical behaviour at regular intervals, and what does the position of an element tell you about its electrons? This page answers both questions.
Atomic number and the structure of the table
Compare lithium (Z = 3), sodium (Z = 11) and potassium (Z = 19). All three are soft metals that fizz in water, and each is more vigorous than the last. The same behaviour comes back at regular steps in atomic number. Why?
Mendeleev left gaps for elements not yet discovered and predicted their properties, which is why his table was accepted. He put tellurium (relative atomic mass 127.6) before iodine (126.9) because their properties demanded it, even though that broke his own rule. The puzzle was solved by Henry Moseley, who used X-ray spectra to show that the real ordering is by atomic number (the number of protons), not mass. Tellurium has Z = 52 and iodine Z = 53, so tellurium comes first.
The table is also split into blocks, according to the sublevel being filled by the last electron: the s-block (Groups 1 and 2), the p-block (Groups 13 to 18) and the d-block (Groups 3 to 12, the transition metals).
Mendeleev's table was refined over time. The noble gases, unknown in his day, were discovered later and added as a new group (now Group 18), and Moseley's atomic numbers replaced atomic mass as the ordering rule. The groups you should know by their properties are:
| Group | Name | Typical properties |
|---|---|---|
| 1 | Alkali metals | Solids at room temperature, soft, low melting points, good conductors; very reactive, more so going down, because the single outer electron is further from the nucleus and better shielded, so it is lost easily |
| 2 | Alkaline earth metals | Solids, harder and less reactive than Group 1, good conductors; two outer electrons are lost less easily; reactivity increases going down |
| 17 | Halogens | Non-metals that do not conduct (Cl₂ gas, Br₂ liquid, I₂ solid); very reactive because one more electron completes the outer level; reactivity decreases going down as the incoming electron is further from the nucleus |
| 18 | Noble gases | Monatomic gases, non-conductors, very unreactive because the outer level is full |
At the end of the 19th century, Dmitri Mendeleev arranged the chemical elements in order of increasing atomic mass, which caused certain discrepancies in pairs such as…
At the end of the 19th century, Dmitri Mendeleev arranged the chemical elements in order of increasing atomic mass, which caused certain discrepancies in pairs such as tellurium ($Z = 52$, atomic mass $127.6\text{ u}$) and iodine ($Z = 53$, atomic mass $126.9\text{ u}$). Henry Moseley later resolved this anomaly using X-ray spectroscopy.
- a) Explain the basis of Moseley's law and why the atomic number $Z$, and not the atomic mass, is the correct criterion for periodic ordering.
- b) Justify the relative positions of tellurium and iodine in the current periodic table in terms of their valence electron configurations.
The ground-state electron configurations determine the position of chemical elements in the periodic table. Consider three neutral elements called A, B and C, whose…
The ground-state electron configurations determine the position of chemical elements in the periodic table. Consider three neutral elements called A, B and C, whose valence shells are: A: $3s^2\,3p^1$; B: $[\mathrm{Ar}]\,4s^2$ (complete configuration except for the abbreviated core); C: $3s^2\,3p^5$.
- a) Deduce the period and group (using the IUPAC numbering from 1 to 18) to which each of the three elements belongs.
- b) Classify each element, giving reasons, as a metal, non-metal or metalloid.
Atomic radius and ionic radius
The atomic radius of lithium is 152 pm, but by fluorine, in the same period, it is only 72 pm. Fluorine has more protons and more electrons, yet it is much smaller. How can that be?
Two ideas explain this:
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Down a group, each atom has an extra main energy level, so the outer electrons are further from the nucleus. The inner electrons also shield the outer ones from the nuclear charge. The radius grows even though the nuclear charge grows.
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Across a period, electrons are added to the same main energy level, and these electrons shield each other poorly. The nuclear charge increases, so the effective nuclear charge felt by the outer electrons increases and pulls them closer.
The effective nuclear charge is the pull an outer electron actually feels: $Z_{\text{eff}} = Z - \sigma$, where $Z$ is the nuclear charge (number of protons) and $\sigma$ is the shielding constant, a measure of how much the inner electrons cancel that pull. Down a group, $\sigma$ rises almost as fast as $Z$, so $Z_{\text{eff}}$ stays roughly constant while the outer electrons sit in a higher level, further out.
When an atom forms an ion its size changes sharply. A cation is smaller than its atom because it has lost its outer electrons (usually the whole outer level) and the remaining electrons are pulled in by the same nuclear charge. An anion is larger than its atom: the same nuclear charge now holds more electrons, which repel each other.
You are given a series of isoelectronic monatomic chemical species: $\mathrm{N}^{3-}$ ($Z=7$), $\mathrm{O}^{2-}$ ($Z=8$), $\mathrm{F}^-$ ($Z=9$), $\mathrm{Na}^+$…
You are given a series of isoelectronic monatomic chemical species: $\mathrm{N}^{3-}$ ($Z=7$), $\mathrm{O}^{2-}$ ($Z=8$), $\mathrm{F}^-$ ($Z=9$), $\mathrm{Na}^+$ ($Z=11$) and $\mathrm{Mg}^{2+}$ ($Z=12$).
- a) Write the ground-state electron configuration common to all these species and order the ions, giving reasons, from smallest to largest ionic radius.
- b) Explain why the radius of the cation $\mathrm{Na}^+$ is significantly smaller than the atomic radius of the neutral $\mathrm{Na}$ atom ($Z=11$), whereas the radius of the anion $\mathrm{F}^-$ is larger than that of the neutral $\mathrm{F}$ atom ($Z=9$).
The concept of effective nuclear charge ($Z_{\text{ef}} = Z - \sigma$, where $\sigma$ is the shielding constant) underpins the periodic variations in radius and redox…
The concept of effective nuclear charge ($Z_{\text{ef}} = Z - \sigma$, where $\sigma$ is the shielding constant) underpins the periodic variations in radius and redox reactivity.
- a) Using the shielding model, justify why, going down group 1 of the alkali metals (from $\mathrm{Li}$ to $\mathrm{Cs}$), the atomic radius increases significantly despite the enormous increase in the total nuclear charge $Z$ (from $Z=3$ to $Z=55$).
- b) Consider the ions $\mathrm{Na}^+$ ($Z=11$) and $\mathrm{K}^+$ ($Z=19$), with noble gas configurations of 10 and 18 electrons respectively. The ionic potential is defined here as $|z|/r$, where $z$ is the charge number and $r$ is the ionic radius. Justify which ion has the greater ionic potential and, according to a simple electrostatic model, which has the more negative enthalpy of hydration.
Ionisation energy
The first ionisation energy generally increases across a period and decreases down a group, the opposite of the radius trend. The closer an electron is to the nucleus, and the stronger the effective nuclear charge, the harder it is to remove. Down a group the outer electron is further away and better shielded, so it is removed more easily. This is why Group 1 metals become more reactive going down.
Exceptions across a period
There are two dips in the second and third periods:
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Group 2 to Group 13 (Mg to Al): I₁(Mg) = 738 kJ mol⁻¹ but I₁(Al) = 578 kJ mol⁻¹. In magnesium the electron comes from a full 3s sublevel; in aluminium it comes from a 3p sublevel, which is higher in energy and slightly further from the nucleus, so it is easier to remove.
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Group 15 to Group 16 (N to O): nitrogen (2p³, a half-filled sublevel with one electron in each p orbital) has a higher I₁ than oxygen (2p⁴). In oxygen two electrons share one p orbital and repel each other, so one is removed more easily.
Successive ionisation energies make it possible to identify the valence shell electron configuration of a chemical element. For a representative element $X$ in the third…
Successive ionisation energies make it possible to identify the valence shell electron configuration of a chemical element. For a representative element $X$ in the third period of the periodic table, the values of its first four ionisation energies have been measured experimentally: $I_1 = 580\text{ kJ}\times\text{mol}^{-1}$, $I_2 = 1815\text{ kJ}\times\text{mol}^{-1}$, $I_3 = 2740\text{ kJ}\times\text{mol}^{-1}$ and $I_4 = 11600\text{ kJ}\times\text{mol}^{-1}$. The relative magnitude of these energies is illustrated in the following bar chart:
- a) Justify, giving reasons, the most stable chemical valency or oxidation state of $X$ in its compounds with electronegative elements.
- b) Deduce which group of the periodic table $X$ belongs to and identify the specific element.
You can do this on paper and send a photo with your answer.
The first ionisation energy generally increases across a period from left to right. However, occasional anomalies are observed when comparing adjacent elements in the…
The first ionisation energy generally increases across a period from left to right. However, occasional anomalies are observed when comparing adjacent elements in the second and third periods.
- a) The first ionisation energy of magnesium ($Z = 12$) is $738\text{ kJ}\times\text{mol}^{-1}$, whereas that of aluminium ($Z = 13$) is $578\text{ kJ}\times\text{mol}^{-1}$. Justify this discrepancy based on the electron configurations and the type of sublevel from which the electron is removed.
- b) The first ionisation energy of nitrogen ($Z = 7$) is $1402\text{ kJ}\times\text{mol}^{-1}$, higher than that of oxygen ($Z = 8$), which is $1314\text{ kJ}\times\text{mol}^{-1}$. Explain this using Hund's rule and the stability of half-filled sublevels.
Electronegativity
Fluorine and oxygen hold on to the shared electrons in a bond far more strongly than sodium or caesium do, and argon, right beside chlorine, barely joins in at all. Some atoms pull on bonding electrons far more strongly than others. How do we measure this, and where does it show up?
Electronegativity increases across a period and decreases down a group, like ionisation energy. Across a period the effective nuclear charge rises and the radius falls, so the nucleus attracts bonding electrons more strongly. Fluorine is the most electronegative element and caesium and francium are among the least.
The noble gases are normally left out of the Pauling scale because they do not usually form bonds, so there is no shared pair to attract.
Back to the start: elements repeat their chemical behaviour because the number of outer electrons repeats as the atomic number rises, and an element's position (period, group, block) tells you its outer electronic structure and so its trends in size, ionisation energy and electronegativity.
Electronegativity quantifies the relative tendency of a bonded atom to attract the electron density of the chemical bond towards itself. a) Explain how electronegativity…
Electronegativity quantifies the relative tendency of a bonded atom to attract the electron density of the chemical bond towards itself.
- a) Explain how electronegativity varies across a period of the periodic table from left to right, justifying the behaviour in terms of effective nuclear charge and atomic radius.
- b) Justify why noble gases generally lack defined electronegativity values on the usual Pauling scale.
Electron affinity measures the energy change when a neutral gaseous atom gains an electron: $\mathrm{X}{\text{(g)}} + \mathrm{e}^- \to \mathrm{X}^-{\text{(g)}}$.…
Electron affinity measures the energy change when a neutral gaseous atom gains an electron: $\mathrm{X}_{\text{(g)}} + \mathrm{e}^- \to \mathrm{X}^-_{\text{(g)}}$. According to the usual thermodynamic convention, an exothermic process has $\Delta H_{\text{ae}} < 0$. The thermodynamic values for the halogens chlorine and fluorine are $\Delta H_{\text{ae}}(\mathrm{Cl}) = -349\text{ kJ}\times\text{mol}^{-1}$ and $\Delta H_{\text{ae}}(\mathrm{F}) = -328\text{ kJ}\times\text{mol}^{-1}$, while for gaseous nitrogen it is $\Delta H_{\text{ae}}(\mathrm{N}) \ge 0$.
- a) Explain why adding an electron to fluorine releases less energy than for chlorine, even though fluorine has higher electronegativity and a smaller atomic radius.
- b) Justify in detail why nitrogen has an unfavourable electron affinity (endothermic or zero) compared with carbon ($\Delta H_{\text{ae}} = -122\text{ kJ}\times\text{mol}^{-1}$).
Four representative chemical elements, labelled consecutively W, X, Y and Z, are analysed. The following experimental data are known: 1. W and X belong to the same…
Four representative chemical elements, labelled consecutively W, X, Y and Z, are analysed. The following experimental data are known:
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W and X belong to the same period; W is an alkaline earth metal and X is a non-metal in group 16.
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Y belongs to the same group as W and to the third period; its atomic radius is smaller than that of W.
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Z has the electron configuration $[\mathrm{Ne}]\,3s^2\,3p^5$.
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Element W has a first ionisation energy of $590\text{ kJ}\times\text{mol}^{-1}$ and an atomic radius of $197\text{ pm}$.
- a) Deduce the identity and complete electron configuration of W, X, Y and Z, given that W belongs to the fourth period.
- b) Arrange the four elements in order of increasing first ionisation energy and justify the order of electronegativity between X and Z.
Topic challenge. How does quantum physics explain energy savings in our streets? Design a technical proposal for converting traditional street lighting to more efficient…
Topic challenge. How does quantum physics explain energy savings in our streets? Design a technical proposal for converting traditional street lighting to more efficient technologies.
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Data collection: Go to the EirGrid (Smart Grid Dashboard, system demand) website and note down the electricity demand of your region at midnight compared with midday.
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Sodium spectrum: Classic street lamps use $\text{Na}$ vapour ($Z=11$), with its characteristic doublet centred at $\lambda = 589.0\text{ nm}$. Calculate the energy of this photon ($E = h \times c / \lambda$) and give the four possible quantum numbers for the valence electron in its ground state and in the excited orbital to which it is promoted.
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Trends and new materials: To replace it with LED luminaires, semiconductors containing gallium ($\text{Ga}$) or indium ($\text{In}$) are used. Write their electron configurations and explain, using effective nuclear charge and shielding, why their ionisation energy and atomic radius differ from those of sodium.
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Reasoned proposal: Write a one-page report justifying the technological replacement, relating the electronic transition, the periodic properties of the proposed material and the potential impact on EirGrid's night-time demand curve.
Assessment will consider:
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Accuracy in quantum-energy calculations and rigorous assignment of quantum numbers.
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Coherent justification of periodic properties according to the quantum-mechanical model.
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Ability to connect the EirGrid data with atomic structure and energy efficiency.