Silicon and silicates; silicones; zeolites: Worked examples

The scenario matrix

Before any worked example, let us list every distinct case this topic can throw at you. Each row is a "cell"; every cell is covered by at least one example below.

Cell	Case class	Key variable	Limiting/edge feature	Example
A	Isolated tetrahedron	0 corners shared	the "start" case, max charge	Ex 1
B	Pyro (pair)	1 corner shared	first bridge appears	Ex 2
C	Chain / ring	2 corners shared	repeating unit $(S i O_{3}^{2 -})_{n}$	Ex 3
D	Sheet	3 corners shared	half-integer O count	Ex 4
E	Framework	4 corners shared	degenerate limit → neutral $S i O_{2}$	Ex 5
F	Silicone functionality	Si–Cl count 1,2,3,4	end-cap vs chain vs cross-link vs net	Ex 6
G	Silicone mass/water bookkeeping	condensation stoichiometry	count the $H_{2} O$ released	Ex 7
H	Zeolite charge balance	number of $A l$	cations needed $= x / n$	Ex 8
I	Real-world word problem	water softening	ion-exchange capacity	Ex 9
J	Exam twist / trap	$C O_{2}$ vs $S i O_{2}$ , naming	reasoning, not a number	Ex 10

The single tool behind cells A–E is one counting idea:

Look at the figure: it shows the same tetrahedron gaining bridges one corner at a time, and the charge dropping from $- 4$ to $0$ .

Figure — Silicon and silicates; silicones; zeolites

Cell A — the isolated tetrahedron (0 shared)

Cell B — the pair (1 shared)

Worked example Ex 2 · Pyrosilicate

$S i_{2} O_{7}^{6 -}$ Two tetrahedra share one corner (thortveitite). Confirm the unit is $S i_{2} O_{7}^{6 -}$ .

Forecast: Two Si and one shared O — how many O total, and what charge?

Per-Si count. One bridging O ( $n_{b} = 1$ ), three terminal ( $n_{t} = 3$ ). Why this step? Only the single shared corner is a bridge; the other three corners still hang free.
Per-Si charge. $+ 4 + (3 \cdot 1 + 1 \cdot \frac{1}{2}) (- 2) = + 4 + (3.5) (- 2) = + 4 - 7 = - 3.$ Why this step? Half the bridging O is "owned" here, so $3.5$ oxygens per Si.
Two Si together. Charge $= 2 \times (- 3) = - 6$ . Total O $= 2 \times 3.5 = 7$ . Why this step? The formula unit is two tetrahedra; double everything.
Write it. $S i_{2} O_{7}^{6 -}$ .

Verify: Count atoms directly instead of using halves: 2 Si, and O = 3 terminal + 3 terminal + 1 shared = 7. Charge from atoms: $2 (+ 4) + 7 (- 2) = 8 - 14 = - 6$ . ✔ Matches.

Cell C — the chain (2 shared)

Worked example Ex 3 · Single-chain pyroxene

$(S i O_{3}^{2 -})_{n}$ A pyroxene has each tetrahedron sharing two corners, forming an endless chain. Find the repeating charge per Si.

Forecast: Two bridges now — will the charge halve again?

Per-Si count. $n_{b} = 2$ , $n_{t} = 2$ . Why this step? In an infinite chain each tetrahedron links to a neighbour on the left and on the right → two bridging O.
Per-Si charge. $+ 4 + (2 \cdot 1 + 2 \cdot \frac{1}{2}) (- 2) = + 4 + (3) (- 2) = + 4 - 6 = - 2.$ Why this step? Two whole terminal + two halves = 3 oxygens per Si.
Repeating unit. $S i O_{3}^{2 -}$ , so the chain is $(S i O_{3}^{2 -})_{n}$ .

Verify: The parent's derivation gives exactly $S i O_{3}^{2 -}$ . ✔ Note the O count fell from 4 (isolated) → 3.5 (pair) → 3 (chain): each new bridge removes half an oxygen. This monotone pattern is the sanity check for the whole matrix.

Cell D — the sheet (3 shared, half-integer)

Cell E — the framework limit (4 shared → neutral)

Worked example Ex 5 · Quartz

$S i O_{2}$ (degenerate limit) Every corner is shared. Show the framework is electrically neutral.

Forecast: If more bridges = less charge, does the charge reach zero?

Per-Si count. $n_{b} = 4$ , $n_{t} = 0$ . Why this step? In a 3D framework all four oxygens bridge to other silicons — nothing dangles.
Per-Si oxygens. $4 \cdot \frac{1}{2} = 2$ O per Si. Why this step? Every O is shared by exactly two Si, so on average one Si owns two whole oxygens.
Per-Si charge. $+ 4 + (2) (- 2) = + 4 - 4 = 0.$
Formula. $S i O_{2}$ — neutral giant covalent solid.

Verify: This is the terminus of the sequence $- 4, - 3, - 2, - 1, 0$ as bridges go $0, 1, 2, 3, 4$ . ✔ The charge hits exactly zero at full sharing — the limiting case, which is why quartz needs no counter-ions (contrast the zeolite below). See how Allotropy and giant covalent solids treats this as a network solid.

Cell F — silicone functionality (Si–Cl count)

Worked example Ex 6 · Predict the product from monomer functionality

Match each chlorosilane to what it builds, and predict what a $9 : 1$ mix of $(C H_{3})_{2} S i C l_{2}$ to $C H_{3} S i C l_{3}$ gives.

Forecast: Which monomer stops growth, which grows chains, which cross-links?

Read functionality = number of Si–Cl bonds.
- $R_{3} S i C l$ → 1 reactive bond → end-cap (stops a chain).
- $R_{2} S i C l_{2}$ → 2 → linear chain.
- $R S i C l_{3}$ → 3 → cross-link point.
- $S i C l_{4}$ → 4 → dense 3D network. Why this step? Each Si–Cl becomes a Si–O–Si link after hydrolysis+condensation, so the count of links = number of directions the chain can grow. This is exactly the corner-sharing idea from silicates, now man-made — see Condensation polymerisation.
Interpret the $9 : 1$ mix. Mostly $R_{2} S i C l_{2}$ → long linear chains; the $10%$ $R S i C l_{3}$ injects occasional 3-way junctions → light cross-linking → a soft rubber, not an oil and not a hard resin. Why this step? More $R S i C l_{3}$ ⇒ more junctions ⇒ harder/rubbery; almost none ⇒ oil. A tenth is "somewhat rubbery".

Verify (functionality vs role): counts $1, 2, 3, 4$ map to roles end / chain / cross / network — the mnemonic "1 ends, 2 lines, 3 cross, 4 net." ✔ Adding cross-linker to a chain former must raise rigidity, which it does.

Cell G — silicone stoichiometry (count the water)

Worked example Ex 7 · How much water is released?

Polymerise $n$ molecules of $(C H_{3})_{2} S i (O H)_{2}$ into $[- (C H_{3})_{2} S i - O -]_{n}$ . If $n = 100$ , how many $H_{2} O$ are lost, and check the atom balance.

Forecast: Each new Si–O–Si bond loses one water — how many bonds form in a chain of 100?

Write the condensation. $n (C H_{3})_{2} S i (O H)_{2} \to [(C H_{3})_{2} S i O]_{n} + n H_{2} O$ . Why this step? Two silanols ( $- O H$ each) join, releasing one $H_{2} O$ per new bridge. In a ring/long chain of $n$ units there are $n$ bridges formed net.
Plug $n = 100$ . Water released $= n = 100$ molecules. Why this step? The stoichiometric coefficient of $H_{2} O$ equals $n$ .
Atom check per repeat unit. Reactant $(C H_{3})_{2} S i (O H)_{2} = C_{2} H_{8} O_{2} S i$ . Product repeat $(C H_{3})_{2} S i O = C_{2} H_{6} O S i$ . Difference $= H_{2} O$ . ✔ Why this step? Losing one O and two H per unit is precisely one water per unit.

Verify: For $n = 100$ : total H in reactants $= 100 \times 8 = 800$ ; in product $100 \times 6 = 600$ ; plus water $100 \times 2 = 200$ → $600 + 200 = 800$ . ✔ O: reactant $100 \times 2 = 200$ = product $100 \times 1 = 100$ + water $100 \times 1 = 100$ → $200$ . ✔

Cell H — zeolite charge balance

Worked example Ex 8 · Cations in

$N a_{x / n} [(A l O_{2})_{x} (S i O_{2})_{y}]$ A zeolite framework contains $12$ aluminium and $36$ silicon per formula unit, balanced by $C a^{2 +}$ . How many calcium ions?

Forecast: Each Al gives $- 1$ to the framework — how many $C a^{2 +}$ neutralise 12 of them?

Framework charge. Replacing one $S i^{4 +}$ by $A l^{3 +}$ costs $+ 1$ of charge, so each $A l O_{2}$ unit is $- 1$ . With $x = 12$ Al, framework charge $= - 12$ . Why this step? $S i$ contributes $+ 4$ , $A l$ only $+ 3$ ; the missing $+ 1$ leaves a $- 1$ hole per Al. (This is why zeolites need loose exchangeable cations — see Ion exchange and water softening.)
Balance with $C a^{2 +}$ . Each $C a^{2 +}$ supplies $+ 2$ . Number needed $= \frac{12}{2} = 6$ . Why this step? General rule $x / n$ with $x = 12$ Al, $n = 2$ charge → $12/2 = 6$ cations.
Answer. $6 C a^{2 +}$ ions per formula unit.

Verify: Total cation charge $= 6 \times (+ 2) = + 12$ ; framework $= - 12$ ; net $= 0$ . ✔ Neutral overall, as any real solid must be.

Cell I — real-world word problem (water softening)

Worked example Ex 9 · Softening capacity

A sodium zeolite has framework charge $- 6$ per formula unit (i.e. $6 N a^{+}$ inside). Hard water passes through and every possible $N a^{+}$ is swapped for $C a^{2 +}$ . How many $C a^{2 +}$ ions are removed per formula unit?

Forecast: $N a^{+}$ is $+ 1$ , $C a^{2 +}$ is $+ 2$ — does 6 sodium out mean 6 calcium in?

Charge must stay balanced. Sites hold $- 6$ of framework charge. That is fixed. Why this step? Ion exchange cannot change the framework; only the counter-ions swap, and the total positive charge they carry must still be $+ 6$ .
Set up the swap. $2 N a^{+} ⇌ C a^{2 +}$ keeps charge: two $+ 1$ ions ( $+ 2$ ) leave, one $+ 2$ ion enters. Why this step? Charge conservation forces a $2 : 1$ ratio because $C a$ is doubly charged.
Compute. $6 N a^{+}$ released ⇒ $6/2 = 3 C a^{2 +}$ captured. Why this step? Each captured $C a^{2 +}$ displaces two $N a^{+}$ .

Verify: Positive charge before $= 6 (+ 1) = + 6$ ; after $= 3 (+ 2) = + 6$ . ✔ Water is softened because 3 hardness-causing $C a^{2 +}$ ions were removed and harmless $N a^{+}$ released.

Cell J — the exam twist (reasoning, no number)

Worked example Ex 10 · Why is

$C O_{2}$ a gas but $S i O_{2}$ a rock? Both are Group-14 dioxides. Explain the different states.

Forecast: Same formula $X O_{2}$ , same group — why such different physical states?

Carbon's option. Small C forms strong $p π - p π$ double bonds → discrete linear $O = C = O$ molecules held together only by weak forces → gas. Why this step? Effective sideways 2p–2p overlap lets carbon "finish" its bonding within one small molecule.
Silicon's constraint. Large Si has diffuse 3p orbitals → poor sideways $p π - p π$ overlap → no $S i = O$ double bonds. Si must instead satisfy its bonding with four single Si–O–Si bridges → a giant covalent framework → high-melting solid (quartz/sand). Contrast the Group 14 trends and Carbon and its allotropes. Why this step? Unable to double-bond, silicon reaches full coordination only by networking — exactly the Cell E framework limit.
Naming trap check. silicon = element Si; silica = $S i O_{2}$ ; silicate = $S i O_{4}^{4 -}$ salts; silicone = $[R_{2} S i O]_{n}$ polymer. Why this step? The examiner often hides a naming error in an otherwise-correct sentence.

Verify: Prediction (Si → network solid) matches observation: quartz melts near $170 0^{\circ} C$ , $C O_{2}$ sublimes at $- 7 8^{\circ} C$ . ✔ Reasoning consistent with Lewis acids and empty d-orbitals (O donates into Si's empty $d$ , strengthening Si–O).

Recall Rapid self-test across every cell

Charge on isolated $S i O_{4}$ tetrahedron? ::: $- 4$ Charge per Si in a single chain (2 shared)? ::: $- 2$ , i.e. $S i O_{3}^{2 -}$ Smallest whole formula of a sheet silicate (3 shared)? ::: $S i_{2} O_{5}^{2 -}$ Why is $S i O_{2}$ neutral? ::: 4 shared → 2 O per Si → $+ 4 - 4 = 0$ Monomer that makes linear silicone chains? ::: $(C H_{3})_{2} S i C l_{2}$ Water molecules released polymerising $n$ silanols? ::: $n$ How many $C a^{2 +}$ balance 12 Al in a zeolite? ::: $6$ $6 N a^{+}$ exchanged → how many $C a^{2 +}$ captured? ::: $3$