Coordination number, voids (tetrahedral, octahedral)

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1. Coordination Number (CN)

WHAT decides it: the packing arrangement, not the element.

Structure	CN	Why
Simple cubic	6	6 face-direction neighbours
Body-centred cubic (bcc)	8	body-centre touches 8 corners
Hexagonal / cubic close packed (hcp / ccp/fcc)	12	densest possible for equal spheres

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2. Voids — the leftover holes

When you place one close-packed layer (A) and drop the next layer (B) into its dimples, two kinds of hole appear.

HOW MANY voids per atom (the golden numbers)

Derivation-from-scratch (fcc unit cell), WHY these numbers:

Take an fcc (ccp) unit cell.

Number of atoms $N$ in fcc:

• 8 corners × $\tfrac18$ = 1
• 6 faces × $\tfrac12$ = 3 $N = 1 + 3 = 4$

Octahedral voids:

• 1 at the body centre (fully inside) = 1
• 12 at edge centres, each shared by 4 cells → $12\times\tfrac14 = 3$ $\text{Oct} = 1 + 3 = 4 = N \checkmark$ Why this step? Only these positions are equidistant from 6 atoms.

Tetrahedral voids:

• Split the cube into 8 small cubes; the centre of each small cube is a tetrahedral void, all inside the cell. $\text{Tet} = 8 = 2N \checkmark$ Why this step? Each small-cube centre is surrounded by 4 nearest fcc atoms forming a tetrahedron.

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3. Void SIZE — the radius ratio (derived, not memorised)

We want the biggest sphere of radius $r$ that fits snugly into a hole made by big spheres of radius $R$, touching them.

3a. Octahedral void

Along a face diagonal of that square, spheres touch:

• Big spheres touch along the edge of the square: edge $= 2R$.
• The small sphere sits at the centre; along the diagonal we have $R + 2r + R$.

Diagonal of a square of side $2R$ is $2R\sqrt2$. So: $2R + 2r = 2R\sqrt2$ $r + R = R\sqrt2 \implies r = R(\sqrt2 - 1)$ $\boxed{\dfrac{r}{R} = \sqrt2 - 1 \approx 0.414}$

3b. Tetrahedral void

Let cube edge $= a$.

• Big spheres touch across a face diagonal: $a\sqrt2 = 2R \Rightarrow R = \tfrac{a\sqrt2}{2}$.
• Small + big along body diagonal: $a\sqrt3 = 2R + 2r$.

Divide the second by $2R$: $\frac{a\sqrt3}{2R} = 1 + \frac{r}{R}$ Substitute $2R = a\sqrt2$: $\frac{a\sqrt3}{a\sqrt2} = 1 + \frac{r}{R} \implies \sqrt{\tfrac32} = 1 + \frac{r}{R}$ $\boxed{\dfrac{r}{R} = \sqrt{\tfrac32} - 1 \approx 0.225}$

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4. Worked examples

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5. Common mistakes (Steel-man + fix)

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Recall Feynman: explain to a 12-year-old

Imagine oranges packed tightly in a box. Each orange snugly touches 12 others — that's its "coordination number." Between the oranges there are little empty pockets. Small pockets are hugged by 4 oranges (tetrahedral) and big pockets by 6 oranges (octahedral). If you drop tiny marbles into these pockets, that's exactly how tiny ions hide inside a crystal of big ions! For every orange there's 1 big pocket and 2 small pockets.

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Flashcards

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Connections

• Close packing in solids (hcp, ccp, bcc)
• Packing efficiency and unit cell dimensions
• Radius ratio rule and ionic structures
• NaCl, ZnS, CaF2 crystal structures
• Density of a unit cell
• States of Matter (Quantitative)

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Socho tumne oranges ko ek box mein tight pack kiya. Har orange apne 12 padosi oranges ko touch karta hai — yehi uska coordination number hai (close packing mein hamesha 12, bcc mein 8). Packing ke beech mein kuch khaali jagah bach jaati hai — inhe voids kehte hain. Do type ke hote hain: tetrahedral void (4 spheres ghere hue, chhota hole, CN=4) aur octahedral void (6 spheres ghere hue, bada hole, CN=6).

Sabse important golden rule: agar N atoms close pack karein, toh octahedral voids = N aur tetrahedral voids = 2N. Matlab har ek atom ke liye 1 bada aur 2 chhote hole. Yeh fcc unit cell se derive hota hai — cell mein 4 atoms, body centre + edge centres pe 4 octahedral, aur cube ko 8 mini-cubes mein todo toh 8 tetrahedral. Isko yaad rakho: "Oct = One, Tet = Two".

Ab void ka size — kitna bada chhota ion andar fit hoga? Geometry se nikalta hai: octahedral ke liye $r/R = \sqrt2 - 1 \approx 0.414$, tetrahedral ke liye $r/R = \sqrt{3/2}-1 \approx 0.225$. Yaad rakho octahedral hole bada hai (6 spheres zyada room dete hain), isliye ratio bada. Common galti: log sochte hain "octa" chhota hoga — nahi bhai, ulta hai!

Ye kyun matter karta hai? Ionic crystals mein bade anions close pack karte hain aur chhote cations in voids mein baithte hain. Kaunsa void, kitne fill hue — isse compound ka formula decide hota hai. Jaise anions ccp + saare octahedral voids filled → AB (NaCl); saare tetrahedral voids filled → A₂B (Na₂O). Toh geometry directly chemistry banati hai!