6.3 · Periodic properties

Effective nuclear charge and periodic trends

One idea, effective nuclear charge, explains most periodic trends.

By the end you should be able to:

Explain shielding, effective nuclear charge and successive ionization energies
Explain and rank periodic trends in radius, ionization energy and electron affinity

Formula

Shielding and effective nuclear charge

Inner (core) electrons shield valence electrons from the full nuclear charge. In the simple model each core electron cancels one proton, and electrons in the same shell do not shield each other:

Z_{\text{eff}} \approx Z - (\text{number of core electrons})

Atom	$Z$	Core electrons	$Z_{\text{eff}}$
$\ce{Na}$	11	10	1
$\ce{Mg}$	12	10	2
$\ce{Cl}$	17	10	7
$\ce{K}$	19	18	1

$Z_{\text{eff}}$ rises by about one per element across a period and stays about the same down a group, where instead $n$ (and so the size of the valence orbitals) increases. These two facts explain the trends below.

Key idea

Atomic and ionic radius

Across a period radius decreases: $Z_{\text{eff}}$ grows while electrons are added to the same shell (Na 186 pm, Mg 160, Al 143, … Cl 99 pm).
Down a group radius increases: the valence electrons are in a larger shell (Li 152 pm, Na 186, K 227 pm).
Cations are smaller than their atoms (usually a whole shell is emptied). Anions are larger (the extra electron adds repulsion without adding protons).
Isoelectronic ions: the same electrons are held by more protons, so a higher nuclear charge gives a smaller ion:

\ce{O^2-}\ (140\ \text{pm}) > \ce{F-}\ (133) > \ce{Na+}\ (102) > \ce{Mg^2+}\ (72)

Key idea

Ionization energy, its anomalies and successive IEs

The first ionization energy removes the outermost electron from a gaseous atom: $\ce{X(g) -> X+(g) + e-}$ . It runs opposite to radius: it increases across a period and decreases down a group. Two dips interrupt the rise across a period:

Group 2 → 13 (Be 899 > B 801 kJ/mol; Mg 738 > Al 578): the new electron is in a p orbital, higher in energy and partly shielded by the filled s subshell.
Group 15 → 16 (N 1402 > O 1314 kJ/mol; P 1012 > S 1000): the fourth p electron has to pair, and the pair repulsion makes it easier to remove.

Successive ionization energies always increase ( $IE_1 < IE_2 < IE_3 \dots$ ) but jump sharply once a core electron must be removed. The number of electrons removed before the big jump is the number of valence electrons, which identifies the group.

Key idea

Electron affinity

Electron affinity (EA) is the energy change when a gaseous atom gains an electron: $\ce{X(g) + e- -> X-(g)}$ . Many tables list the energy released as a positive number (Cl: 349 kJ/mol); with the energy-change sign convention the same value is written as $-349$ kJ/mol.

More energy is released toward the upper right of the table (the halogens), where $Z_{\text{eff}}$ is high and the added electron completes a p subshell.
Cl (349) releases more than F (328 kJ/mol): the extra electron is crowded into fluorine's small $2p$ shell and repelled.
Noble gases, group 2 and nitrogen do not form stable gaseous anions: the electron would have to enter a new shell or subshell, or pair up in a half-filled p subshell.

Method

Ranking a periodic trend

Place each species on the periodic table: the period gives the valence shell $n$ , the group the number of valence electrons.
Same group: larger $n$ means a larger radius, lower ionization energy and (usually) less energy released on gaining an electron.
Same period: larger $Z_{\text{eff}}$ means a smaller radius, higher ionization energy and more energy released on gaining an electron.
For ionization energy, check the two exceptions: group 13 lies below group 2, and group 16 below group 15.
Ions: for an isoelectronic series, rank by the number of protons (more protons, smaller ion). Otherwise compare the number of occupied shells first.
When one element lies lower-left of the other, both effects point the same way (K is larger than S and has a lower IE).

Common mistake

Trend traps

Wrong: atoms get bigger across a period because they have more electrons. Right: they get smaller: the added electrons go into the same shell and barely shield each other while $Z$ increases.
Wrong: O has a higher first IE than N because it is further right. Right: N (1402) > O (1314 kJ/mol), because O loses a paired $2p$ electron.
Wrong: F releases the most energy on gaining an electron. Right: Cl does (349 vs 328 kJ/mol).
Wrong: $\ce{Na+}$ and $\ce{F-}$ are the same size because they are isoelectronic. Right: Na⁺ has 11 protons and F⁻ has 9, so Na⁺ is much smaller (102 vs 133 pm).
Wrong: $Z_{\text{eff}}$ for the valence electron of K is 19. Right: 18 core electrons shield it, so $Z_{\text{eff}} \approx 1$ .

Worked example

Worked example: reading successive ionization energies

A period-3 element has successive ionization energies of 738, 1451, 7733 and 10 543 kJ/mol. Which element is it?

$IE_2 / IE_1 = 1451/738 \approx 2.0$ , but $IE_3 / IE_2 = 7733/1451 \approx 5.3$ : the big jump comes after the second electron.
Two valence electrons means group 2: the element is magnesium, $[\mathrm{Ne}]\,3s^2$ . The third electron would come from the $2p$ core, which feels $Z_{\text{eff}} \approx 12 - 2 = 10$ instead of about 2.

Rank Na, Mg, Al and Si by first IE: Na (496) < Al (578) < Mg (738) < Si (786 kJ/mol). The general rise across the period is broken at Al, whose $3p$ electron is easier to remove than magnesium's $3s$ electron.

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