Exercises — Silicon and silicates; silicones; zeolites

• Silicon ::: the pure element, symbol $Si$ (metalloid, Group 14).
• Silica ::: the oxide $Si{O}_2$ (quartz, sand, glass).
• Silicate ::: a salt built from the $Si{O}_4^{4-}$ tetrahedron (e.g. olivine).
• Silicone ::: the man-made polymer $[R_{2}SiO{]}_n$ with an Si–O backbone and organic $R$ groups. The trap is the near-identical spelling; anchor each to a physical example.

Silicon contributes $+4$. Each of the four oxygens contributes $-2$. $+4+4(-2)=+4-8=-4\Rightarrow Si{O}_4^{4-}$ This is the only number you memorise; everything else is derived from it.

Carbon forms strong $p\pi$–$p\pi$ double bonds → discrete $O=C=O$ molecules (a gas). Silicon is too big for good sideways $3p$ overlap, so it cannot make those double bonds; instead it satisfies its bonds with single $Si-O-Si$ bridges, giving a giant covalent network. See Allotropy and giant covalent solids.

Rule: a shared (bridging) oxygen belongs to two tetrahedra, so it counts as $\frac{1}{2}$ per Si. With 2 corners shared:

• bridging O: $2\times \frac{1}{2}=1$ net O
• terminal O: $4-2=2$ net O
• total O per Si $=1+2=3$ Charge per Si $=+4+3(-2)=+4-6=-2\Rightarrow Si{O}_3^{2-}$. ✔ (pyroxenes)

• bridging O: $3\times \frac{1}{2}=1.5$
• terminal O: $4-3=1$
• total O per Si $=1.5+1=2.5$ Charge per Si $=+4+2.5(-2)=-1$. Multiply by 2 to clear the fraction: $S{i}_2{O}_5^{2-}$. ✔ (mica, clay)

Every O is shared → each counts $\frac{1}{2}$. O per Si $=4\times \frac{1}{2}=2$. Charge $=+4+2(-2)=+4-4=0$ → neutral $Si{O}_2$. ✔ (quartz, feldspar)

Let each Si share $s$ corners on average. Net O per Si $=(4-s)+\frac{s}{2}=4-\frac{s}{2}$. From the formula: O per Si $=\frac{11}{4}=2.75$. $4-\frac{s}{2}=2.75\Rightarrow \frac{s}{2}=1.25\Rightarrow s=2.5$ So on average 2.5 corners shared — consistent with the "double chain" entry in the parent table. Charge check: per Si, $+4+2.75(-2)=+4-5.5=-1.5$; for 4 Si: $4(-1.5)=-6\Rightarrow S{i}_4{O}_{11}^{6-}$. ✔

Each Si–Cl becomes an Si–OH on hydrolysis, and each Si–OH can condense to form one Si–O–Si link (see Condensation polymerisation).

• $R_{3}SiCl$: 1 reactive bond → can link on one side only → caps a chain end.
• $R_{2}SiC{l}_2$: 2 reactive bonds → links on both sides → a growing line.
• $RSiC{l}_3$: 3 bonds → a branch/cross-link point. Exactly two link-points per unit is the geometric requirement for an unbranched chain. Fewer stops growth; more branches it.

Each $Si$ ($+4$) sitting among 4 half-shared O is neutral, so $(Si{O}_2{)}_5$ contributes $0$. Replacing $S{i}^{4+}$ by $A{l}^{3+}$ loses $+1$ per Al → each $Al{O}_2$ unit carries $-1$. Net framework charge $=3\times (-1)=-3$. To balance, you need $3$ singly-charged $N{a}^{+}$ ions. (This is the basis of ion exchange — see Ion exchange and water softening.)

Step 1 — Rochow (make the monomer): why a dichloride? → two link-points needed for a chain. $2C{H}_{3}Cl+Si\stackrel{Cu,570K}{\to }(C{H}_3{)}_{2}SiC{l}_2$ Step 2 — Hydrolysis: turn reactive Si–Cl into condensable Si–OH. $(C{H}_3{)}_{2}SiC{l}_2+2H_{2}O\to (C{H}_3{)}_{2}Si(OH{)}_2+2HCl$ Step 3 — Condensation polymerisation: two silanols lose water to build the Si–O–Si backbone. $n(C{H}_3{)}_{2}Si(OH{)}_2\to [-(C{H}_3{)}_{2}Si-O-{]}_n+n{H}_{2}O$ Mnemonic: dichloride → diol → dehydrate.

• Add more $RSiC{l}_3$ (functionality 3): each such unit is a branch point, so more of them means more cross-links → a harder, rubbery, resinous network.
• Use less $R_{3}SiCl$ (functionality 1): these are chain-terminators; too many of them keep chains short → runny oil. The linear $R_{2}SiC{l}_2$ supplies the backbone in between. So: raise trichloride, lower monochloride.

• $(Si{O}_2{)}_y$: the neutral silica part of the framework.
• $(Al{O}_2{)}_x$: each Al carries $-1$, so this part carries $-x$ total charge.
• $M_{x/n}$: a cation $M$ of charge $+n$ supplies $n$ units of positive charge each, so to cancel $-x$ you need $x/n$ of them. Charge balance: $\frac{x}{n}\cdot n=x$. ✔
• $z{H}_{2}O$: water molecules occupying the open cages/channels.

Sheet ⇒ 3 corners shared per tetrahedron. O per Si $=(4-3)+\frac{3}{2}=1+1.5=2.5$. Charge per Si $=+4+2.5(-2)=-1$. Two-Si repeat: O $=5$, charge $=-2\Rightarrow S{i}_2{O}_5^{2-}$. Class: sheet silicate. The weak forces between sheets are exactly why talc feels slippery — a real-world payoff of the geometry.

Reason 1 — shape-selectivity: the pores are molecule-sized, so only reactants/products of the right shape fit; this steers the reaction toward specific hydrocarbons. Reason 2 — acidic framework sites: the $Al$-induced $-1$ charges, balanced by protons ($H^{+}$), create strong Brønsted acid sites that actually catalyse C–C bond formation. Rebuttal of "the metal ions react": the loose exchangeable cations mostly balance charge and can be swapped; the catalysis is dominated by pore geometry + acidic sites, not by the cations reacting.

Because Si is large, its $3p$ orbitals overlap sideways poorly, so it cannot form the $p\pi$–$p\pi$ double bonds carbon uses — hence no discrete $O=Si=O$ gas; instead it makes single $Si-O-Si$ bridges into giant solids (a). The $Si-O$ bond is unusually strong ($\approx 452$ kJ/mol) partly because oxygen lone pairs donate into Si's empty $3d$ orbitals ($p\pi$–$d\pi$), which is why nearly all silicon on Earth is bonded to oxygen (b) — see Lewis acids and empty d-orbitals. We can borrow that same tough Si–O backbone but cap the free corners with organic $R$ groups, giving silicones, inorganic inside and organic (water-repelling) outside (c). And if we replace some framework $S{i}^{4+}$ by $A{l}^{3+}$ in a 3D silica network, each swap leaves a $-1$ charge that must be balanced by a loose, exchangeable cation in the cage — precisely the property that lets zeolites soften water by ion exchange (d). Compare with Group 14 trends and Carbon and its allotropes.