Giant Covalent Structures Worked ExamplesIGCSE Chemistry 0620

(a) A is diamond: a three-dimensional network in which each carbon atom is bonded to four others. B, made of flat layers, is graphite.

(b) (i) In diamond each carbon forms four covalent bonds (arranged tetrahedrally). (ii) In graphite each carbon forms three covalent bonds.

(c) Graphite is made of layers of carbon atoms arranged in hexagonal rings. Within a layer each carbon is covalently bonded to three others; there are weak forces between the layers. The fourth outer electron of each carbon is delocalised.

(d) Both are giant covalent structures.

(a) Graphite has layers with only weak forces between them, so the layers can slide over each other easily.

(b) It conducts electricity and has a very high melting point (it is not melted by the heat of the process). It conducts because each carbon atom has one delocalised electron that can move through the layers.

(c) Diamond is very hard. Each carbon atom is joined to four others by strong covalent bonds in a giant three-dimensional network, so a great deal of energy is needed to break it.

Silicon(IV) oxide has a giant covalent structure. Silicon (Group IV) forms four covalent bonds and oxygen forms two, so every Si is bonded to 4 O and every O links 2 Si. Answer: C.

A is the tempting error: silicon and oxygen are both non-metals, joined by covalent bonds, not ions. B describes carbon dioxide, not silicon(IV) oxide. D describes graphite; silicon(IV) oxide does not conduct.

The formula $\text{SiO}_2$ is a ratio (1 Si : 2 O), not the formula of a molecule: each Si has 4 O neighbours, but each O is shared between 2 Si, so $4 \div 2 = 2$ O per Si.

(a) Count the lines from each atom. (i) Each P atom has 4 bonds. (ii) Each Q atom has 2 bonds.

(b) Silicon is in Group IV, with 4 outer electrons: it shares them to form four covalent bonds. Oxygen is in Group VI and needs two more electrons: it forms two bonds. So P is silicon and Q is oxygen.

(c) Each Si is bonded to 4 O atoms, but each O atom is shared between 2 Si atoms.

(d) A giant covalent structure (in reality three-dimensional, with the four bonds on each Si arranged tetrahedrally).

(a) Any two of: very high melting point; very hard; do not conduct electricity; insoluble in water.

(b) Both have a giant covalent structure: every atom is joined to its neighbours by strong covalent bonds in a three-dimensional network — each C bonded to four C in diamond, each Si bonded to four O in silicon(IV) oxide.

Melting or scratching them means breaking many strong covalent bonds, which needs a lot of energy. All the outer electrons are held in bonds, so there are no delocalised electrons (and no ions) to carry a current.

(c) Graphite. It conducts electricity (diamond does not), or it is soft/slippery (diamond is very hard).

(a) (i) Graphite: delocalised electrons. (ii) Copper: delocalised electrons (in the sea of electrons around the positive ions).

(b) In diamond each carbon uses all four outer electrons in covalent bonds, so there are no delocalised (mobile) electrons. In graphite each carbon bonds to only three others, leaving one electron per atom delocalised; these can move along the layers.

(c) Loose grains make poor contact with each other and with the clips, so a fair test needs a single solid piece joining the two clips. (Silicon(IV) oxide still does not conduct: it has no mobile electrons or ions.)

(d) Replace the lamp with an ammeter, which detects currents far too small to light a lamp.

Sort the list first. Giant covalent: diamond, graphite, silicon(IV) oxide. Simple molecular: carbon dioxide, iodine, methane. Giant ionic: sodium chloride. Metallic: copper.

(a) A compound contains two or more elements. Diamond and graphite are elements, so the answer is silicon(IV) oxide.

(b) Graphite (copper conducts but is metallic, not giant covalent). (c) Diamond.

(d) Iodine, $\text{I}_2$ — carbon dioxide and methane are simple molecular but are compounds. (e) Diamond.

Graphite: each C bonds to 3 others, and its fourth electron is delocalised, so graphite conducts. Diamond: each C bonds to 4 others, with no delocalised electrons, so diamond does not conduct. Answer: A.

B swaps the bond numbers. C has the right bond numbers but wrongly says diamond conducts. D has the wrong number for graphite and the wrong conductor.

(a) Like diamond: a very high melting point, very hard, and does not conduct electricity (also insoluble in water).

(b) Silicon carbide has a giant covalent structure. Melting or scratching it requires breaking many strong covalent bonds, needing a lot of energy.

Each Si and each C uses all four of its outer electrons in bonds, so there are no delocalised electrons and no ions to carry a current.

(c) In cutting, grinding or drilling tools (e.g. grinding wheels or sandpaper), because it is very hard.

(a) X: a covalent bond (a shared pair of electrons between two carbon atoms). Y: weak forces between the layers.

(b) Only the weak forces Y hold the layers together, so the layers slide over each other easily. The strong covalent bonds stay intact within each layer.

(c) Each carbon forms only three covalent bonds; its fourth outer electron is delocalised and free to move along the layer.

(d) To melt graphite, the strong covalent bonds within the layers must be broken, not just the weak forces between layers. There are very many of these bonds in a giant structure, so a lot of energy is needed.

(a) Carbon dioxide is simple molecular: separate small $\text{CO}_2$ molecules, each C double-bonded to two O atoms. Silicon(IV) oxide is giant covalent: a continuous network in which each Si is bonded to four O and each O to two Si.

(b) When carbon dioxide changes state, only the weak intermolecular forces between molecules are overcome — little energy is needed, so the temperature is low. The covalent bonds inside the molecules do not break.

To melt silicon(IV) oxide, many strong covalent bonds throughout the giant structure must be broken — a lot of energy is needed, so the temperature is very high.

(c) Both contain only covalent bonds: there are no ions and no delocalised electrons to carry a current.

(a) $M_r(\text{SiO}_2) = 28 + (2 \times 16) = 60$. Moles of $\text{SiO}_2$:

Each Si atom forms 4 Si–O bonds, and each Si–O bond involves exactly one Si atom, so there are 4 mol of Si–O bonds per mole of $\text{SiO}_2$ — there is no double counting.

(b) Multiply by the Avogadro constant.

(c) Moles of C $= \dfrac{3.0}{12} = 0.25$ mol. Each C has 4 bonds, but each bond is shared by 2 atoms, so there are $\dfrac{4}{2} = 2$ mol of C–C bonds per mole of C.

(d) In graphite each C forms 3 covalent bonds, each shared by 2 atoms: