Simple Molecules and Covalent Bonds Worked ExamplesIGCSE Chemistry 0620

(a) Two marking points: a covalent bond is a pair of electrons (M1) shared between two atoms (M2). The sharing gives each atom a noble gas electronic configuration.

(b) Each line in a displayed formula is one covalent bond. Count them by type.

(c) Di- means two: a diatomic molecule contains two atoms, e.g. hydrogen $\text{H}_2$, oxygen $\text{O}_2$, nitrogen $\text{N}_2$ or chlorine $\text{Cl}_2$.

Oxygen ($2,6$) has six outer electrons; each hydrogen has one. Each O–H bond is one shared pair: one dot from oxygen and one cross from hydrogen.

That uses $2$ of oxygen's $6$ outer electrons, so 4 remain as two lone (non-bonding) pairs. Oxygen then has $8$ outer electrons and each hydrogen has $2$. This is diagram C.

A omits both lone pairs, so oxygen shows only $2$ of its own electrons. B shows two crosses in each bond, but a hydrogen atom has only one electron. D shows only one lone pair, so oxygen has only $4$ of its $6$ electrons.

(a) Around carbon: one pair in each overlap, so 4 electrons. Carbon should have 8 (a noble gas configuration). Carbon has 4 outer electrons of its own and must share all of them.

(b) Each oxygen shows 7 dots (one in the bond and six in lone pairs). An oxygen atom ($2,6$) has only 6 outer electrons.

(c) Carbon needs to share four electrons and each oxygen needs to share two, so each carbon–oxygen bond is a double bond: two shared pairs, each pair made of one dot (oxygen) and one cross (carbon).

Each oxygen then has two lone pairs (four dots) as well as its two bonding pairs, giving $8$. Carbon has four bonding pairs, giving $8$. Structure: $\text{O=C=O}$.

(a) Each carbon has $4$ outer electrons. In the diagram each carbon has used $2$ (one in each C–H bond), so each has 2 more to share.

Add four dots in the overlap between the carbon atoms — two shared pairs, a C=C double bond. Each carbon now has $4$ pairs $= 8$ electrons; each hydrogen has $2$. There are no lone pairs.

(b) Four C–H bonds ($4 \times 2 = 8$ electrons) plus the double bond ($4$ electrons):

(c) Ethene is a simple molecular substance. The intermolecular forces (between the molecules) are weak, so little energy is needed to overcome them. The strong covalent bonds inside the molecules are not broken when it boils.

(a) Each O–H bond: one shared pair — one dot (oxygen) and one cross (hydrogen).

The O–O bond: one shared pair, one electron from each oxygen atom. Both are oxygen electrons, so the O–O overlap contains two dots.

Each oxygen has $6$ outer electrons and has used $2$ in its two bonds, so it has two lone pairs (four dots). Each oxygen: $2$ bonding pairs $+ 2$ lone pairs $= 8$ electrons. Each hydrogen: $2$.

(b) Two lone pairs on each oxygen: 4 non-bonding pairs.

(c) Total outer electrons in the atoms, compared with the electrons in the diagram:

Simple molecular substances have low melting and boiling points (weak intermolecular forces) and are poor conductors in any state (no ions, no mobile electrons). B fits both. Answer: B.

C has a low melting point too, but it conducts as a solid and a liquid — it is a metal (mercury). Low melting point alone is not enough.

A conducts only when liquid — ionic. D has very high melting and boiling points and does not conduct — giant covalent.

(a) When a simple molecular substance boils, its molecules separate. The forces overcome are the intermolecular forces — the weak attractive forces between molecules.

(b) M1: Boiling overcomes the attractive forces between methane molecules.

M2: Decomposition requires breaking the covalent bonds (C–H) within each molecule.

M3: Intermolecular forces are weak, so little energy is needed and the boiling point is low; covalent bonds are strong, so a lot of energy is needed and decomposition happens only at a high temperature.

(a) Hexane is a simple molecular substance. It contains no ions (M1) and no mobile (delocalised) electrons (M2) — all its electrons are held in covalent bonds within the molecules — so no charged particles can move through the liquid.

(b) Sodium chloride is ionic. When dissolved, its $\text{Na}^{+}$ and $\text{Cl}^{-}$ ions are free to move and carry the charge.

(c) Keep naked flames (e.g. Bunsen burners) away from the liquids.

(d) To remove traces of the previous liquid. Ions left from the sodium chloride solution could make hexane or ethanol appear to conduct.

(e) Replace the lamp with an ammeter, which can detect a current too small to light a lamp.

Write each formula first: $\text{NH}_3$, $\text{C}_4\text{H}_{10}$, $\text{CO}_2$, $\text{C}_2\text{H}_4$, $\text{CH}_3\text{OH}$, $\text{N}_2$, $\text{H}_2\text{O}$.

(a) Nitrogen, $\text{N}\equiv\text{N}$: each N shares three electrons. (b) Ethene ($2 + 4 = 6$ atoms, C=C). Methanol also has six atoms but no double bond. (c) Butane: $4 + 10 = 14$ atoms.

(d) Non-bonding pairs: $\text{CO}_2$ has $2$ on each O $= 4$; $\text{N}_2$ $2$; $\text{H}_2\text{O}$ $2$; $\text{CH}_3\text{OH}$ $2$; $\text{NH}_3$ $1$; alkanes and ethene $0$. Carbon dioxide has the most.

(e) Butane is an alkane, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$: $3$ C–C bonds and $10$ C–H bonds.

Iodine, $\text{I}_2$: covalent bonding — a shared pair of electrons holding two iodine atoms together in each molecule.

Calcium oxide: ionic bonding — electrostatic attraction between $\text{Ca}^{2+}$ and $\text{O}^{2-}$ ions.

Copper: metallic bonding — attraction between positive copper ions and a sea of delocalised electrons.

Diamond: covalent bonding — shared pairs of electrons between carbon atoms, each bonded to four others.

(b) Iodine is simple molecular. Warming overcomes the weak intermolecular forces between the $\text{I}_2$ molecules; the covalent I–I bonds do not break.

(c) Diamond has a giant covalent structure: to melt it, many strong covalent bonds must be broken, which needs a lot of energy. Iodine is simple molecular: melting only overcomes the weak intermolecular forces between $\text{I}_2$ molecules, which needs little energy.

Outer electrons: H $1$, C $4$, N $5$. H needs $1$ more, C needs $4$ more, N needs $3$ more to reach a full shell.

(a) H–C overlap: one shared pair (one cross from H, one dot from C). Carbon has $3$ electrons left, and nitrogen needs $3$: the C–N overlap holds three shared pairs — a triple bond — made of three dots (C) and three crosses (N).

Nitrogen has used $3$ of its $5$ electrons, so it has one lone pair (two crosses) on the far side. Check: C has $1 + 3 = 4$ pairs $= 8$; N has $3$ pairs $+ 1$ lone pair $= 8$; H has $2$.

(b) Three shared pairs: $3 \times 2 = 6$ electrons.

(c) Compare the atoms' outer electrons with the electrons in the diagram:

(a) Valencies: C forms $4$ bonds, O $2$, H $1$. Two C–H bonds use $2$ of carbon's bonds, leaving $2$ for oxygen: a C=O double bond. Displayed formula: carbon in the centre, single bonds to two H atoms and a double bond to O.

(b) Each C–H: one shared pair. C=O: two shared pairs. Oxygen ($6$ outer electrons) uses $2$ in the double bond and keeps two lone pairs. Carbon: $4$ pairs $= 8$; oxygen: $2$ bonding pairs $+ 2$ lone pairs $= 8$.

Electron check: $4 + (2 \times 1) + 6 = 12$ outer electrons $= 4$ bonding pairs ($8$) $+ 2$ lone pairs ($4$).

(c) Room temperature is above $-19\,^{\circ}\text{C}$, so methanal is a gas. It is simple molecular: only weak intermolecular forces have to be overcome, which needs little energy.

(d) No. Liquid methanal is made of molecules: there are no ions and no mobile electrons.