Question bank — Coordination number, voids (tetrahedral, octahedral)

False — as Figure 1 shows, the octahedral walls sit farther apart, so octahedral $r/R=0.414$ beats tetrahedral $0.225$; the octahedral hole is larger.

False — CN is set purely by the packing geometry; all spheres scale together, so ccp is always CN 12 regardless of atom size.

False — per $N$ packed atoms (full atoms per cell) there are $N$ octahedral but $2N$ tetrahedral voids, so tetrahedral voids are twice as many.

False — a void is empty space; it only matters as a seat where a small sphere of radius $r$ may sit, contributing nothing until occupied.

False — 12 is Kepler's hard limit ($6+3+3$) for touching identical spheres; you physically cannot touch a 13th equal sphere.

True — octahedral voids $=N$ equals the ccp atom count $N$, so filled sites : packed atoms $=1:1$ (rock-salt type).

False — all 8 tetrahedral voids sit at the mini-cube centres (each mini-cube = one-eighth of the cell) fully inside the cell, so none are shared.

False — $0.9$ lands in $0.732$–$1.000$, cubic (CN 8); such a large sphere would rattle badly in the tiny tetrahedral hole.

Wrong count — fcc has $2N=8$ tetrahedral voids (one at each mini-cube centre), not 4; only the octahedral voids number $N=4$.

Wrong positions — face centres hold atoms; octahedral voids live at the body centre (1) and the 12 edge centres ($12\times \frac{1}{4}=3$), giving 4.

It is only the lower (snug-touch) bound derived in Figure 1; any $r/R$ from $0.414$ up to $0.732$ still sits octahedrally, just with some rattle.

Missing the subtraction — the mini-cube body diagonal gives $\sqrt{3/2}=1+r/R$, so $r/R=\sqrt{3/2}-1\approx 0.225$, not $\sqrt{3/2}$.

Wrong — CN (coordination number) counts touching spheres (its 6 vertices), so an ion in an octahedral void has CN 6.

Wrong ratio — tetrahedral voids $=2N$ while anions $=N$, so cations : anions $=2:1$, formula $A_{2}B$ (antifluorite).

The larger ions can't fit inside gaps left by smaller ones; geometry forces the small species (radius $r$) into the leftover pockets of the large-sphere (radius $R$) framework.

The lower bound is the snug-touch fit (Figure 1); between it and the next bound the cation rattles but still fits, only jumping to a higher-CN hole once $r/R$ exceeds the top of the range.

Below the snug-fit $r/R$ the ion cannot touch all surrounding anions at once, so it sits loosely and destabilises the structure, favouring a lower-CN hole.

Simple cubic touches only its 6 axis neighbours; ccp adds 3 nestled spheres above and 3 below the layer, pushing contacts to $6+3+3=12$.

Each mini-cube (one-eighth of the cell) has a centre equidistant from its 4 nearest fcc atoms, which form a tetrahedron — the defining arrangement of a tetrahedral void.

Touching means maximum stable contact and lattice energy; a non-touching (rattling) cation gives a less stable arrangement, so the limiting $r/R$ is defined at contact, exactly as Figure 1 sets it.

Yes — the empty pocket is a geometric feature of the close packing; occupancy is a separate chemical choice from existence.

8 — it is surrounded by 8 anions at the corners of a cube, the highest CN in the radius-ratio scheme.

Cations $=\frac{1}{2}(2N)=N$ against anions $N$, giving $1:1$ (this is the ZnS/zinc-blende arrangement).

This boundary is shared: it is the upper snug-limit for tetrahedral and the lower snug-limit for octahedral (both derived in Figure 1), so the octahedral (CN 6) fit begins here.

The "cation" is now an equal sphere, so it should join the close packing itself (CN 12) rather than hide in any void.

No — it is too small even for the trigonal hole; it would rattle in all voids and no stable coordination geometry from the table fits.