Tetrahedral and Octahedral Holes

Take a box of marbles poured into the tightest arrangement you can manage. Shake it until every sphere nestles against its neighbors. You have just built a close-packed lattice. But the marbles do not fill space completely. The gaps between them are not random slivers of empty volume. They are two specific, highly symmetric types of cavity. These are called tetrahedral and octahedral holes. Their number, geometry, and size follow directly from the packing geometry itself. This matters because most of solid-state ionic chemistry is really the chemistry of what sits in those holes. Rock salt, spinel, corundum, fluorite, the perovskites, every close-packed metal alloy and interstitial carbide — all of them are built the same way. Take a close-packed array of the larger ion. This is almost always the anion, because anions are usually bigger. Then populate some fraction of its tetrahedral and octahedral holes with the smaller cation. Which holes get filled, and how many of them, is dictated by charge balance. It also depends on whether the cation actually fits. That is the geometric question that Radius Ratio Rules will answer quantitatively. Get the holes right, and the entire architecture of ionic solids becomes predictable rather than memorized. The big picture is a marriage of two ideas already in your toolkit. Close-Packing of Spheres tells you how identical spheres pack. This can be cubic close-packed (ccp) or hexagonal close-packed (hcp). Both fill 74.05% of space. Simple coordination geometry tells you what shape of cavity is left over. Every close-packed layer sits with each sphere nestled in the pocket of three spheres beneath it. This pocket, capped by an atom above, is either a tetrahedron of four surrounding spheres. Or, when two such pockets from adjacent layers point at each other, it is an octahedron of six surrounding spheres. The bookkeeping of how many of each hole exist per sphere is where the real predictive power lives. It is exact: 2 tetrahedral holes and 1 octahedral hole per sphere. This holds always, regardless of ccp or hcp. This article derives the number, geometry, and radius ratio of tetrahedral and octahedral holes from first principles. It works through three fully quantitative examples: NaCl, a hypothetical MX₂ hole-filling problem, and a competition-style packing-fraction calculation. It also sets up the geometric vocabulary that Zinc Blende and Wurtzite Structures and the Spinel Structure will put to direct structural use.

Key takeaways

In any close-packed array of $N$ spheres there are exactly $2N$ tetrahedral holes and $N$ octahedral holes.
A tetrahedral hole is surrounded by 4 spheres at the corners of a tetrahedron; an octahedral hole is surrounded by 6 spheres at the corners of an octahedron.
The maximum radius of a sphere that fits exactly in a tetrahedral hole is $r/R = 0.225$ ; for an octahedral hole it is $r/R = 0.414$ , where $R$ is the packing-sphere radius.
These ratios come from pure geometry (bond angles and touching-sphere conditions), not from any specific element — they hold for ccp and hcp alike.
Ionic and interstitial structures are built by filling a chosen fraction of these holes: NaCl fills all octahedral holes, zinc blende fills half the tetrahedral holes, spinel fills 1/8 tetrahedral + 1/2 octahedral holes.
Key idea: the number of interstitial sites is fixed by packing topology (2 tetrahedral + 1 octahedral per sphere), while the size of each site type is fixed by the touching-sphere geometry — together they gate which ions can occupy which holes.

Where this lives today

The tetrahedral/octahedral hole framework is not a historical curiosity — it is the working vocabulary of modern materials design. Battery cathode materials such as LiCoO₂ and the layered NMC (Ni-Mn-Co) oxides are built on close-packed oxide sublattices in which Li⁺ ions shuttle in and out of octahedral holes during charge/discharge; the reversibility and rate of that intercalation depends directly on octahedral-hole size and connectivity (edge-sharing octahedral chains provide fast diffusion pathways). Spinel-structured LiMn₂O₄, an important cathode chemistry, is a direct real-world instance of the tetrahedral/octahedral hole-filling stoichiometry worked out in Example 3, with Li⁺ in tetrahedral sites and Mn³⁺/Mn⁴⁺ in octahedral sites. In semiconductor materials science, the choice between zinc-blende (tetrahedral-hole-filled) and wurtzite polytypes for compounds like GaN, ZnO, and SiC — both technologically central, GaN in blue LEDs and power electronics, ZnO in transparent conducting oxides and varistors — is governed by exactly the radius-ratio and stacking arguments introduced here, refined further in Zinc Blende and Wurtzite Structures. In steelmaking and hard-coating technology, interstitial carbides and nitrides (Fe₃C cementite, TiN, WC) place small carbon or nitrogen atoms into octahedral holes of a close-packed or near-close-packed metal lattice, and the resulting lattice strain and hole occupancy directly control hardness and phase stability. Even biomineralization exploits these ideas: some biogenic and synthetic hydroxyapatite/calcium phosphate materials are analyzed using close-packed anion frameworks with cations distributed over tetrahedral and octahedral interstices, informing the design of bone-substitute ceramics.

Ask the tutor: How does the choice between filling tetrahedral vs. octahedral holes actually get decided when both radius-ratio windows seem to overlap, as in ZnS?

The physical picture

Start with a single close-packed layer of spheres of radius $R$ , each touching six neighbors, arranged so that every sphere sits in a triangular pocket formed by three others. This is layer A. A second layer, B, is laid down so that each of its spheres drops into alternate triangular pockets of layer A — call these the "up-pointing" pockets it occupies, leaving the other set, the "down-pointing" pockets, empty directly beneath B-layer spheres. This is the single packing rule that generates both ccp (stacking ABCABC…) and hcp (stacking ABABAB…); the local coordination environment around any interior sphere, and hence the holes around it, is identical in both cases. That local identity is precisely why the hole-counting result is stacking-independent. Look at what happens under each B-sphere. Three A-spheres form a triangle beneath it, and the B-sphere caps that triangle from above: four spheres total, arranged as a tetrahedron. This is a tetrahedral hole — small, and lined by four nearest neighbors whose centers define a regular tetrahedron with the void at its centroid. Now look at the other set of pockets in layer A — the ones B did not fill. These remain open below, and when the next layer (C in ccp, or A again in hcp) is added above B, a sphere of that next layer sits over this empty pocket. The result is three A-spheres below, forming a triangle, and three spheres of the layer above forming a second triangle rotated by 60° relative to the first, sandwiching a void between them. Six spheres in total, arranged as a regular octahedron: this is the octahedral hole. The essential geometric fact is that both hole types are interstices, not lattice sites of the packing spheres themselves — they are the leftover space, and there is more leftover space than atoms. Counting carefully (done rigorously in the derivation below) shows each sphere in the lattice is associated with exactly 2 tetrahedral holes and 1 octahedral hole, and this ratio, 2 : 1, is topological — it does not care whether the stacking is ABAB, ABCABC, or any random hcp/ccp mixture, because it only depends on the local touching arrangement, which is identical in every close-packed structure.

Figure

A single close-packed bilayer showing a tetrahedral hole (capped triangle) and an octahedral hole (two opposed triangles)

To find how large a foreign sphere can be and still fit inside each hole, picture the packing spheres as rigid and in contact with their neighbors, and ask for the radius of the largest sphere that can be inserted into the hole without pushing them apart. This is a pure trigonometry problem once the coordination geometry is fixed — worked out explicitly below — and it yields the famous radius-ratio limits: $r/R = 0.225$ for a tetrahedral hole and $r/R = 0.414$ for an octahedral hole, where $R$ is the host-sphere radius and $r$ is the guest. These numbers are the geometric backbone of Radius Ratio Rules: an ion too big for a tetrahedral hole is forced into an octahedral one, and an ion too big for even that is forced into a lower-coordination or non-close-packed structure entirely.

Figure

Cross-sections defining the touching-sphere condition for each hole type

The core ideas

The two governing relationships of this topic are a counting law and two geometric ratio laws.

N_{\text{tet}} = 2N, \qquad N_{\text{oct}} = N

where: $N$ is the number of spheres in the close-packed array (ccp or hcp alike); $N_{\text{tet}}$ and $N_{\text{oct}}$ are the numbers of tetrahedral and octahedral holes respectively. Reading: every close-packed sphere "owns," on a per-sphere basis, exactly two tetrahedral voids and one octahedral void — a fixed topological ratio 2:1 independent of the packing sequence (cubic or hexagonal), of the unit cell chosen, or of how many layers are stacked, provided edge effects are ignored (i.e., in the bulk/infinite-lattice limit).

\left(\frac{r}{R}\right)_{\text{tet}} = \sqrt{\tfrac{3}{2}} - 1 = 0.225

where: $r$ is the radius of the largest sphere that fits without strain inside a tetrahedral hole; $R$ is the radius of the four touching host spheres forming the tetrahedron. Reading: a guest ion filling a tetrahedral hole can be at most 22.5% the size of the host spheres — tetrahedral holes are the small interstices, appropriate for small, highly charged, or low-coordination-preferring cations (Zn²⁺, Be²⁺, many M⁴⁺ in spinels).

\left(\frac{r}{R}\right)_{\text{oct}} = \sqrt{2} - 1 = 0.414

where: $r$ is the radius of the largest sphere fitting exactly inside an octahedral hole; $R$ is the radius of the six touching host spheres forming the octahedron. Reading: octahedral holes are roughly twice as capacious (0.414 vs 0.225) as tetrahedral holes, which is why most 6-coordinate ionic rock-salt-type structures (NaCl, MgO, most transition-metal monoxides) exploit octahedral rather than tetrahedral sites — bigger cations need the bigger hole.

Derivation ACounting tetrahedral and octahedral holes per sphere

Goal. Show that an infinite close-packed lattice of $N$ spheres contains exactly $2N$ tetrahedral holes and $N$ octahedral holes, using the ccp (fcc) unit cell as the concrete bookkeeping device (the same count follows for hcp by an equivalent local argument).

Step 1 — spheres per conventional ccp unit cell. The cubic close-packed structure has the face-centered cubic (fcc) unit cell: 8 corner spheres shared among 8 cells each ( $8 \times \tfrac18 = 1$ ), and 6 face spheres shared between 2 cells each ( $6 \times \tfrac12 = 3$ ). Total:

N_{\text{cell}} = 1 + 3 = 4 \text{ spheres per unit cell.}

Step 2 — locating octahedral holes. In the fcc cell, octahedral voids occur at the body center (1 site, entirely inside the cell, multiplicity 1) and at the midpoint of every edge (12 edges, each edge shared among 4 adjacent cells, so $12 \times \tfrac14 = 3$ ). Total octahedral holes per cell:

N_{\text{oct,cell}} = 1 + 3 = 4.

This exactly equals $N_{\text{cell}} = 4$ , confirming the ratio $N_{\text{oct}} : N_{\text{sphere}} = 1:1$ .

Step 3 — locating tetrahedral holes. Tetrahedral voids sit fully inside the cell at the eight positions $(\tfrac14,\tfrac14,\tfrac14)$ -type coordinates (one in each of the 8 minor octants of the cube), none shared with neighboring cells. Total:

N_{\text{tet,cell}} = 8.

Since $N_{\text{cell}} = 4$ , the ratio is

N_{\text{tet}} : N_{\text{sphere}} = 8 : 4 = 2:1.

Step 4 — generalizing to any $N$ , and to hcp. Because both counts scale linearly with the number of unit cells, for $N$ total spheres:

N_{\text{tet}} = 2N, \qquad N_{\text{oct}} = N.

The hcp lattice has an identical local coordination environment around each interior sphere (12 nearest neighbors, six of each hole type arranged locally the same way as in ccp — the two structures differ only in the stacking sequence of identical close-packed layers, not in the local touching geometry). A direct count in the hcp hexagonal cell (2 spheres/cell, 2 tetrahedral holes/cell inside plus contributions, 1 octahedral hole/cell net after sharing) reproduces the same $2:1$ ratio. The result is therefore a statement about local packing topology, not about long-range stacking order. Limiting check: for a single isolated sphere ( $N=1$ ) embedded deep in a bulk lattice, the formula predicts 2 tetrahedral and 1 octahedral hole "belonging" to it — consistent with the direct local picture (each interior sphere is capped above and below by two tetrahedral pockets, and sits adjacent to one octahedral pocket per layer-pair), confirming the bookkeeping is not an artifact of the particular unit-cell choice.

Go deeper · Derivation B · 2 minRadius ratios for tetrahedral and octahedral holes

Derivation BRadius ratios for tetrahedral and octahedral holes

Goal. Derive $r/R = 0.225$ (tetrahedral) and $r/R = 0.414$ (octahedral) from the condition that host spheres of radius $R$ touch each other and touch a central guest sphere of radius $r$ .

Step 1 — octahedral hole setup. Place the octahedral hole's six host spheres at the vertices of a regular octahedron, with the guest sphere at the center. Consider the cross-section through four of the six host spheres lying in a plane through the center — they sit at the corners of a square, with the guest sphere at the square's center, and the square's edge length equals $2R$ (adjacent host spheres touch each other along the octahedron edge, which is exactly this in-plane square edge for the standard octahedral arrangement derived from ccp geometry).

Step 2 — octahedral distance condition. The center-to-corner distance of a square of edge $2R$ is the half-diagonal:

d = \frac{2R\sqrt{2}}{2} = R\sqrt{2}.

This distance $d$ must equal $R + r$ , the sum of host and guest radii, since the host sphere (at the corner) touches the guest sphere (at the center):

R + r = R\sqrt{2}.

Step 3 — solve for $r/R$ .

r = R\sqrt{2} - R = R(\sqrt{2}-1)

\frac{r}{R} = \sqrt{2} - 1 = 0.4142\ldots

Step 4 — tetrahedral hole setup. Now place the guest sphere at the centroid of a regular tetrahedron whose four vertices are host-sphere centers, each host sphere of radius $R$ touching its neighbors. The bond angle subtended at the centroid between any two vertices of a regular tetrahedron is the tetrahedral angle,

\theta = 109.47^\circ,

so the half-angle from the centroid-to-vertex axis to the tetrahedron edge, in the standard derivation, uses $\theta/2 = 54.735^\circ$ .

Step 5 — tetrahedral distance condition. Let $d$ be the centroid-to-vertex distance (this equals $R+r$ , guest touching host) and let $a = 2R$ be the host–host contact distance (edge of the tetrahedron, since adjacent host spheres touch). For a regular tetrahedron the centroid-to-vertex distance relates to the edge length $a$ by

d = a\sqrt{\tfrac{3}{8}} = a\,\frac{\sqrt{6}}{4}.

Substituting $a = 2R$ :

R + r = 2R\cdot\frac{\sqrt{6}}{4} = R\frac{\sqrt{6}}{2}.

Step 6 — solve for $r/R$ .

r = R\left(\frac{\sqrt{6}}{2} - 1\right)

\frac{r}{R} = \frac{\sqrt{6}}{2} - 1 = 1.2247 - 1 = 0.2247 \approx 0.225.

(Equivalently, this is often written as $r/R = \sqrt{3/2}-1$ , the same numerical value, since $\sqrt{6}/2 = \sqrt{6/4}=\sqrt{3/2}$ .) Limiting check: since an octahedral hole (coordination number 6) is geometrically "roomier" than a tetrahedral hole (coordination number 4) for a fixed host sphere size, we expect $(r/R)_{\text{oct}} > (r/R)_{\text{tet}}$ — and indeed $0.414 > 0.225$ , consistent with the general radius-ratio trend that higher coordination numbers correlate with larger permissible guest/host ratios (CN 8, the cubic hole, gives $r/R = 0.732$ , continuing the same monotonic trend, a fact used directly in Radius Ratio Rules).

Worked examples

Example 1 (foundational): counting holes in a mole of close-packed spheres, and checking NaCl. Take $N = 6.022\times10^{23}$ close-packed spheres (one mole), representing the Cl⁻ sublattice of rock salt, which is ccp. By the counting law:

N_{\text{tet}} = 2N = 1.204\times10^{24}, \qquad N_{\text{oct}} = N = 6.022\times10^{23}.

In NaCl, Na⁺ ions occupy all the octahedral holes and none of the tetrahedral holes. Check the stoichiometry: number of Na⁺ = $N_{\text{oct}} = N$ = number of Cl⁻, giving a 1:1 ratio — exactly matching the formula NaCl. This also predicts (and matches experiment) that both Na⁺ and Cl⁻ in rock salt have coordination number 6, since an octahedral hole is bounded by 6 host spheres, and by symmetry of the structure Cl⁻ around each Na⁺ also number 6. Sanity/limiting check: if instead only half the octahedral holes were filled (as happens in some defect or ordered-vacancy rock-salt-related phases), stoichiometry would become $M_{0.5}X$ , i.e. $MX_2$ inverted — a useful reminder that fractional occupancy of the fixed hole count, not the hole count itself, is what varies between structure types.

Example 2 (applied): zinc blende (ZnS) — filling tetrahedral holes, and verifying radius fit. Zinc blende has S²⁻ ions in ccp arrangement, with Zn²⁺ filling exactly half of the tetrahedral holes (alternating ones, chosen so that no two filled tetrahedral holes are adjacent, giving each Zn²⁺ four S²⁻ neighbors and vice versa). Stoichiometry check: with $N$ sulfide ions, $N_{\text{tet}} = 2N$ tetrahedral holes exist; filling half gives $N$ filled tetrahedral holes, hence $N$ Zn²⁺ ions for $N$ S²⁻ ions — a 1:1 ratio, matching ZnS. Both ions end up 4-coordinate, consistent with the tetrahedral hole's coordination number. Now check the radius-ratio prediction against real ionic radii (Shannon, 6-coordinate values scaled appropriately; here we use the commonly tabulated effective ionic radii for 4-coordination):

r(\text{Zn}^{2+}) \approx 0.60\ \text{Å (4-coordinate)}, \qquad r(\text{S}^{2-}) \approx 1.84\ \text{Å (4-coordinate)}.

\frac{r_{\text{cation}}}{r_{\text{anion}}} = \frac{0.60}{1.84} = 0.326.

This value, 0.326, sits above the strict tetrahedral limit of 0.225 but below the octahedral limit of 0.414 — it is in fact right in the "gray zone" where either 4- or 6-coordination is geometrically plausible, which is exactly why ZnS is famous for exhibiting both the tetrahedral zinc-blende form and the (hexagonally-stacked, still tetrahedral) wurtzite form, rather than switching to an octahedral rock-salt structure; this borderline behavior is precisely the subtlety that Radius Ratio Rules and Zinc Blende and Wurtzite Structures examine in more depth. The key point here is only the mechanical bookkeeping: half of $2N$ tetrahedral holes filled reproduces 1:1 stoichiometry with 4:4 coordination, exactly as geometry demands.

Example 3 (exam/competition level): spinel stoichiometry and packing-fraction cross-check. Spinel, MgAl₂O₄, has O²⁻ in ccp arrangement. Of the available holes per formula unit's worth of oxide sublattice, the normal spinel structure places divalent cations (Mg²⁺) in 1/8 of the tetrahedral holes and trivalent cations (Al³⁺) in 1/2 of the octahedral holes. Let the oxide sublattice contain $N$ O²⁻ ions. Then:

N_{\text{tet}} = 2N, \qquad N_{\text{oct}} = N.

Filled tetrahedral sites (Mg²⁺):

n_{\text{Mg}} = \frac{1}{8}\times 2N = \frac{N}{4}.

Filled octahedral sites (Al³⁺):

n_{\text{Al}} = \frac{1}{2}\times N = \frac{N}{2}.

Check charge balance. Total positive charge:

Q_+ = 2\left(\frac{N}{4}\right) + 3\left(\frac{N}{2}\right) = \frac{N}{2} + \frac{3N}{2} = 2N.

Total negative charge from O²⁻:

Q_- = 2N.

$Q_+ = Q_- = 2N$ : charge balance holds exactly, an essential and non-trivial check since the fractions 1/8 and 1/2 were not chosen arbitrarily — they are the unique fractions consistent with electroneutrality for a 2:3:4 (Mg:Al:O ratio via $\tfrac14 : \tfrac12 : 1$ , i.e. $1:2:4$ ) spinel stoichiometry MgAl₂O₄. As a further consistency check, verify the mole ratio directly:

\text{Mg} : \text{Al} : \text{O} = \frac{N}{4} : \frac{N}{2} : N = 1 : 2 : 4,

which is exactly the formula MgAl₂O₄. This worked example is the direct numerical bridge to the Spinel Structure, where the distribution question (which specific 1/8 of tetrahedral and 1/2 of octahedral sites, normal vs. inverse spinel) becomes the central chemistry. Limiting/boundary check: if instead all tetrahedral holes were filled with a divalent cation and no octahedral holes were used, the formula would become $M_2O$ (antifluorite-type stoichiometry, as in Na₂O or Li₂O — cations in all tetrahedral holes of a ccp oxide lattice), a genuinely different and experimentally realized limiting structure, confirming that the fractional-occupancy bookkeeping developed here spans real, distinct compound families rather than being an abstract exercise.

Limiting cases, scope & common traps

Situation	What happens	Why
$r/R < 0.225$	Cation "rattles" in tetrahedral hole; may distort to lower coordination or cause structural instability	Guest smaller than the hole's exact-fit size; contact with host spheres is lost
$0.225 < r/R < 0.414$	Tetrahedral (CN 4) coordination geometrically favored, though octahedral sometimes competitive near the upper end	Guest just fits the tetrahedral hole without straining it
$0.414 < r/R < 0.732$	Octahedral (CN 6) coordination favored	Guest fits the larger octahedral hole; too big for tetrahedral without prying host spheres apart
$r/R > 0.732$	Cubic/8-coordinate or higher packing favored; close packing of the anion sublattice itself may break down	Even the octahedral hole is too small; anion array often adopts simple cubic (CsCl-type) packing instead
hcp vs ccp choice	Hole count and size identical; only long-range connectivity/sharing pattern of octahedra (edge- vs face-sharing chains) differs	Local coordination environment around each sphere is stacking-independent
Only a fraction of holes filled	Determines the empirical formula stoichiometry directly	Charge balance forces specific, often non-obvious, fractional occupancies (e.g., 1/8, 1/2 in spinel)

The chemist behind it

The modern language of close packing and interstitial holes owes its systematic form to William Barlow and, in its most influential 20th-century codification for ionic solids, to Linus Pauling, whose 1929 paper "The Principles Determining the Structure of Complex Ionic Crystals" laid out the radius-ratio rules built directly on tetrahedral/octahedral hole geometry — turning what had been a crystallographic curiosity into a predictive, teachable framework for the entire field of solid-state inorganic chemistry.

Origins · 1 minLinus Carl Pauling and how this was discovered

Linus Carl Pauling

1901–1994 · Structural chemistry and the birth of predictive crystal-structure rules, 1920s–1930s

Pauling trained as a chemical engineer at Oregon Agricultural College before moving to Caltech, where exposure to the newly emerging technique of X-ray diffraction — and to Sommerfeld's quantum theory during a European fellowship — gave him the tools to connect atomic-scale bonding preferences to observed crystal geometries. His 1929 "Pauling's Rules" paper formalized the idea that ionic radius ratios, computed from exactly the touching-sphere geometry of tetrahedral, octahedral, and cubic holes derived here, could predict coordination number and hence crystal structure from first principles, before a structure was even solved. This radius-ratio approach, alongside his subsequent work on chemical bonding (for which he won the 1954 Nobel Prize in Chemistry) and his campaign against nuclear weapons testing (1962 Nobel Peace Prize), made him one of the only people in history to win two unshared Nobel Prizes.

History

- 1883: William Barlow proposes close-packing models for crystal structures, anticipating the geometric description of interstitial voids.

- 1913: W. H. and W. L. Bragg solve the structure of NaCl by X-ray diffraction, providing the first direct experimental confirmation of an ionic close-packed-derived lattice with filled octahedral holes.

- 1926: Victor Goldschmidt tabulates systematic ionic radii and correlates them with observed coordination numbers in minerals.

- 1929: Linus Pauling publishes his rules connecting radius ratios (built on tetrahedral/octahedral hole geometry) to predicted coordination number and crystal structure.

- 1930s–present: Radius-ratio and hole-filling arguments become standard tools for rationalizing spinel, perovskite, and defect structures across mineralogy, ceramics, and materials science.

Are tetrahedral and octahedral holes actual "atoms" or lattice positions?

No — they are geometric voids, regions of empty space between the close-packed spheres, not positions occupied by the packing atoms themselves. They only become chemically relevant when a different, usually smaller, ion or atom is inserted into them, as happens throughout ionic and interstitial solid-state chemistry. The host close-packed array (typically the larger anions) defines the hole geometry; the guest species (typically smaller cations, or small atoms like C, N, H in metallic interstitial compounds) occupies some fraction of the holes, and that occupied fraction directly sets the compound's stoichiometry.

Why is the octahedral hole radius ratio (0.414) almost double the tetrahedral one (0.225)?

It follows directly from the different coordination geometry, not from any arbitrary convention. An octahedral hole is bounded by 6 spheres arranged with 90°-type contacts in cross-section, giving a centroid-to-vertex distance of $R\sqrt2$ ; a tetrahedral hole is bounded by only 4 spheres at the tighter 109.47° tetrahedral angle, giving a centroid-to-vertex distance of only $R\sqrt{6}/2 \approx 1.2247R$ . The smaller multiplier for the tetrahedral case directly compresses the available central space, so $(r/R)_{\text{tet}} = 0.225$ comes out smaller than $(r/R)_{\text{oct}} = 0.414$ .

Does every close-packed structure have to fill its holes with a different element?

No. Many pure metals and elemental close-packed structures leave the holes completely empty (e.g., ccp copper, hcp magnesium) — the holes only matter once you ask what else can be inserted. Interstitial alloys (steel, where carbon occupies octahedral holes of γ-iron) and ionic solids (where cations fill anion-array holes) are the two major classes where hole occupancy becomes the chemically defining feature, while pure metallic close packing is simply the baseline geometric reference structure.

How do I quickly tell, without drawing the whole lattice, how many holes of each type a given number of ions creates?

Use the counting law directly: multiply the number of close-packed (host) spheres by 2 for tetrahedral holes and by 1 for octahedral holes. This ratio, 2 tetrahedral holes and 1 octahedral hole per host sphere, is completely general for any close-packed lattice (ccp or hcp), so you never need to redraw the unit cell — you only need to know how many host ions are present and what fraction of each hole type a given real structure fills, which is tabulated for named structure types (rock salt: all octahedral; zinc blende: half tetrahedral; fluorite: all tetrahedral; spinel: 1/8 tetrahedral + 1/2 octahedral).

Can an ion be too big for BOTH tetrahedral and octahedral holes, and what happens then?

Yes — when $r/R$ exceeds about 0.732 (the cubic-hole limit) or when even the octahedral limit of 0.414 is badly exceeded, the ion cannot be accommodated in a close-packed anion framework at all without severe lattice strain. In practice the compound instead adopts a structure with a non-close-packed anion arrangement, such as the CsCl structure (simple cubic anion array, cation in the central cubic hole, $r/R$ up to 1) or the fluorite/antifluorite structures with different coordination geometry, illustrating why radius-ratio breakdowns are a genuine structural fork in the road, not just a numerical curiosity.

Is the 74.05% packing fraction of close packing relevant to the size of the holes?

Yes, they are two faces of the same geometric fact. If spheres packed with 100% efficiency (impossible for spheres, but as a thought limit) there would be no interstitial space at all, and $r/R \to 0$ for every hole type. The actual packing efficiency of 74.05% leaves 25.95% of the volume as voids, and that fixed leftover volume is exactly partitioned, by the touching-sphere geometry, into the specific tetrahedral (0.225) and octahedral (0.414) size classes derived here — the packing fraction and the hole sizes are two consequences of the same fixed sphere-touching geometry, not independent facts.

Why does the tetrahedral hole geometry use the angle 109.47° specifically?

109.47° (more precisely $\arccos(-1/3)$ ) is the angle between any two vertex-to-centroid lines of a regular tetrahedron — it is a purely geometric consequence of four points arranged with maximal, equal mutual spacing in three dimensions, and it is the same angle that governs sp³ hybridized bond angles in molecules like methane. Its appearance in the tetrahedral-hole radius-ratio derivation is not a coincidence of ionic solids specifically; it is the same "four equally spaced directions in 3D space" geometry that recurs throughout chemistry wherever true tetrahedral coordination or hybridization occurs.