Ionic or covalent, ionic radii and radius ratios

Ionic and covalent are two ends of a slope

Intuition

Some bonds are like a tug of war that one side wins outright, so one atom grabs the electrons. Others are an even tug, so the electrons are shared. Most solids sit somewhere along the slope between the two.

Lecture 7 recaps electron counting from Lecture 2. Carbon has 4 valence electrons and 4 valence orbitals, so it can reach an inert gas configuration by 2-centre, 2-electron bonds. Boron has 3 and 4, so it cannot by covalent bonding alone. Isoelectronic elements have the same number of valence electrons, so they are in the same group, but elements in the same group near the “line” can bond differently (glyoxylic acid C$_2$H$_2$O$_3$ against Sn$_2$K$_2$O$_3$).

Ketelaar’s triangle recognises a continuum of bonding for intermediate electronegativities, between ionic, covalent and metallic. The lecture sums up solids by bond type.

• Atoms in molecules: covalent. Directional, complex shapes, low symmetry.
• Molecules in crystals: van der Waals. Non-directional, moderate symmetry, limited by molecular shape.
• Atoms in “solid state” compounds: ionic. Non-directional, high coordination numbers, high packing efficiency and high symmetry. Combinations are possible.

Compare

Largely ionic

Large cations with low valence (groups I, II). Small electronegative anions (O²⁻, F⁻). Example NaCl.

Largely covalent

Small highly polarising cations (B³⁺, Si⁴⁺, P⁵⁺, S⁶⁺). Large polarisable anions (I⁻, S²⁻). Example SiC.

Small formal charges such as Na$^+$ and Cl$^-$ are reasonable. High formal charges such as Ti$^{4+}$ and C$^{4-}$ are unlikely to really exist, though TiC is classified as ionic by convention. A formula $A_mX_p$ follows from charge neutrality, with $m$ and $p$ set so the net charge is zero.

An ion has no sharp edge, so its radius is defined by a convention

Intuition

An ion is a fuzzy cloud, not a hard ball. To give it a size you have to agree where the cloud stops. Different agreements give slightly different sizes.

The lecture uses the electron density map of LiF. The minimum in electron density between the ions defines the effective ionic radius. Pauling and Goldschmidt radii are shown for comparison. Shannon and Prewitt (1969, 1970) gave experimentally derived radii, with $r(\text{O}^{2-}) = 1.40$ Å (similar to Pauling) and $r(\text{O}^{2-}) = 1.26$ Å (the minimum in electron density). Radii depend on coordination number. This is a purely ionic model, but ions are not treated as incompressible spheres.

Exam detail

Trends in effective ionic radii. In the s and p block, $r$ increases down a group. Isoelectronic cations: $r$ decreases as oxidation number increases. Ions of the same charge: $r$ decreases as atomic number increases across a period (for example the lanthanide contraction). For an element, $r$ decreases as oxidation number increases. For an ion, $r$ increases as coordination number increases.

The slide table of Shannon radii (Acta Cryst. 1976, A32, 751) lists ion, coordination number and radius in angstrom. The page does not copy it. Look up values on the slide when a question gives them.

The radius ratio predicts how many neighbours fit

Intuition

A small ion in a big gap rattles around, so the structure picks a tighter gap. A big ion needs more room, so it takes a bigger gap. The ratio of the two sizes tells you which gap fits.

The size of the octahedral (or tetrahedral) hole in an fcc structure determines the size of ion it can hold. Lecture 7 gives $r_\text{oct}/r_\text{sphere} = \sqrt{2} - 1 = 0.414$ and $r_\text{tet}/r_\text{sphere} = 1/\sin(109.47/2) - 1 = 0.225$. Coordination number increases with $r_\text{cation}/r_\text{anion}$.

$r_\text{cation}/r_\text{anion}$	Coordination number	Example structure
< 0.155	2	linear pair
0.155 to 0.225	3	triangle
0.225 to 0.414	4	ZnS (zinc blende), fcc
0.414 to 0.732	6	NaCl (sodium chloride), fcc
0.732 to 1.000	8	CsCl (caesium chloride), simple cubic

Try it: worked example

A compound has cation/anion radius ratio 0.60. Which coordination number and structure type does the table give? (Practice number.)

Show answer and steps

AnswerCoordination number 6, the NaCl structure.

1. Compare 0.60 with the table ranges.
2. It is above 0.414 and below 0.732.
3. That row gives coordination number 6, with NaCl as the example.

Optional aside

Lecture 5 gives the tetrahedral hole ratio as 0.255. Lecture 7 gives 0.225, and its table uses 0.225 as the boundary between 3 and 4. Quote 0.225 and note the mismatch if asked.

Structure maps that include electronegativity predict coordination number more consistently than radius-ratio rules. On the Mooser–Pearson diagram a greater electronegativity difference (x-axis) means more ionic, and higher ionicity gives a higher coordination number. Higher principal quantum numbers (y-axis) mean less s-p hybridisation, more diffuse orbitals, less directional bonding and a higher coordination number.

Covalency also explains deviations from close packing. For AlX$_3$ (AlF$_3$, AlCl$_3$, AlBr$_3$, AlI$_3$) the lecture shows two structure types: close-packed AlX$_6$ octahedra with Al coordination 6 (ionic), and layered Al$_2$X$_6$ dimers with Al coordination 4 (covalent). The slide’s text does not say which halide gets which type, and the slide was not checked against its image. Check the slide before using this.