Ions and Ionic Bonds Worked ExamplesIGCSE Chemistry 0620

(a) Fill the shells $2, 8, 8$ in order. Potassium ($19$ electrons): $2,8,8,1$. Fluorine ($9$ electrons): $2,7$.

(b) A potassium atom loses its one outer electron to form $\text{K}^{+}$ ($2,8,8$). A fluorine atom gains one electron to form $\text{F}^{-}$ ($2,8$). Each ion now has a full outer shell.

(c) (i) A positive ion is a cation. (ii) A negative ion is an anion. (iii) The anions of Group VII elements are halide ions (fluoride, chloride, bromide, iodide).

(d) $\text{F}^{-}$ has $10$ electrons arranged $2,8$ — the same as neon. ($\text{K}^{+}$, with $2,8,8$, matches argon.)

(a) Two marking points: an ionic bond is a strong electrostatic attraction (M1) between oppositely charged ions (M2).

(b) Lithium ($2,1$) transfers its single outer electron (a cross) to fluorine ($2,7$).

Lithium ion: one shell containing two crosses, in square brackets, with charge $+$. Its second shell is now empty, so it is not drawn.

Fluoride ion: inner shell of two dots; outer shell of seven dots and one cross (the electron gained from lithium), arranged in four pairs; square brackets with charge $-$.

Start from the atoms. Each calcium atom loses two electrons, one to each of two fluorine atoms. The formula is $\text{CaF}_2$, so the diagram needs one $\text{Ca}^{2+}$ ion and two $\text{F}^{-}$ ions — the student has the right number of ions.

Error 1: the calcium ion still shows two crosses in its outer shell. Those two electrons have been lost. The outer shell of $\text{Ca}^{2+}$ is the third shell, which is full: it should show eight crosses ($2,8,8$).

Error 2: each fluoride ion shows eight dots. Fluorine provides only seven outer electrons; the eighth came from calcium, so it must be drawn as a cross.

Error 3: the fluoride ions have no charge. Each must show a $-$ charge outside its square brackets.

(b) Correct diagram: $[\text{Ca}]^{2+}$ with an outer shell of eight crosses in four pairs; two separate $[\text{F}]^{-}$ ions, each with seven dots and one cross in four pairs.

Calcium is in Group II ($2,8,8,2$): it loses 2 electrons to form $\text{Ca}^{2+}$. Nitrogen is in Group V ($2,5$): it gains 3 electrons to form $\text{N}^{3-}$. That rules out C and D.

The charges must cancel. The lowest common multiple of $2$ and $3$ is $6$: three $\text{Ca}^{2+}$ give $6+$ and two $\text{N}^{3-}$ give $6-$.

Answer: A. B has the right electron numbers but the formula swaps the numbers of each ion (it would have a charge of $4 - 9 = -5$). D uses nitrogen's group number, $5$, as the number of electrons gained.

X is a metal in Group III: it loses three outer electrons to form $\text{X}^{3+}$. Z is a non-metal in Group VII: it gains one electron to form $\text{Z}^{-}$.

One $\text{X}^{3+}$ needs three $\text{Z}^{-}$ to balance the charges: $\text{XZ}_3$. Answer: B.

D uses the group numbers ($3$ and $7$) directly instead of the ion charges. C swaps the ratio. A would only be right if the charges were equal in size, e.g. Group I with Group VII.

(a) $2,8$ is $10$ electrons. (i) The atom with $10$ electrons is neon: Ne.

(ii) Positive ions with $10$ electrons come from elements with more than $10$ protons that lose electrons: $\text{Na}^{+}$ ($11 - 1$) or $\text{Al}^{3+}$ ($13 - 3$).

(iii) Negative ions with $10$ electrons come from elements with fewer than $10$ protons that gain electrons: $\text{F}^{-}$ ($9 + 1$), $\text{O}^{2-}$ ($8 + 2$) or $\text{N}^{3-}$ ($7 + 3$).

(b) A $2+$ ion has lost two electrons, so the atom had $23 + 2 = 25$ electrons and therefore $25$ protons. Proton number $25$ is manganese (Mn).

(c) A $3-$ ion has gained three electrons, so the atom had $18 - 3 = 15$ protons: phosphorus. The ion is the phosphide ion, $\text{P}^{3-}$.

(a) In an ionic lattice the charges alternate along every row and column, so each ion's nearest neighbours carry the opposite charge. P has $2+$ ions on all sides, so P is $\text{O}^{2-}$. Q has $2-$ ions on all sides, so Q is $\text{Mg}^{2+}$. R has $2+$ ions beside it, so R is $\text{O}^{2-}$.

(b) Magnesium oxide has a giant lattice: a regular arrangement of alternating positive and negative ions ($\text{Mg}^{2+}$ and $\text{O}^{2-}$) held together by strong electrostatic attraction between oppositely charged ions, repeated in three dimensions.

(c) Each magnesium ion inside the layer has four oxide ions around it (above, below, left and right). In three dimensions there is one more above and one below the layer, making six.

(d) The charges $2+$ and $2-$ are equal in size, so the lattice contains equal numbers of each ion to be neutral: ratio $1:1$.

(a) (i) The lamp does not light while the zinc chloride is solid. (ii) The lamp lights once the zinc chloride has melted.

(b) Zinc chloride is ionic ($\text{Zn}^{2+}$ and $\text{Cl}^{-}$ ions). In the solid, the ions are held in fixed positions in the lattice and cannot move, so there is no current. When molten, the ions are free to move and carry charge to the electrodes.

(c) Chloride ions are attracted to the positive electrode, where they form chlorine. Test: hold damp litmus paper in the gas — it is bleached.

(d) Chlorine is toxic, so the gas must not escape into the laboratory.

(e) The lamp lights. In aqueous solution the lattice has broken up and the ions are mobile, so the solution conducts.

(a) Y is the ionic compound: it conducts only when molten, not when solid. (Its brittleness — it shatters — is also typical of an ionic lattice.) X conducts as a solid and changes shape when hammered: it is a metal.

(b) (i) In a metal, the current is carried by delocalised electrons that move through the lattice of positive ions. (ii) In molten Y, the current is carried by ions — positive ions and negative ions moving in opposite directions.

(c) Solid Y is a giant lattice of alternating positive and negative ions. The ions are held in fixed positions by strong electrostatic attraction, so they cannot move. There are no delocalised electrons in an ionic compound, so no charged particles can move and there is no current.

(a) C. It has a high melting point and conducts when molten but not when solid. (It also dissolves in water.) A has a very low melting point (simple molecular); B does not conduct even when molten (giant covalent); D conducts as a solid (metal).

(b) C has a giant ionic lattice. There are strong electrostatic attractions between the oppositely charged ions throughout the lattice, and a lot of energy is needed to overcome them.

(c) Yes. When C dissolves, its ions separate and become free to move through the solution, so they can carry the current.

(a)(i) Each X ion has a charge of $3+$, so each X atom lost three electrons. Losing its outer electrons leaves a full shell, so the atom had three outer electrons: Group III.

(ii) Each Y ion shows six dots (Y's own outer electrons) and two crosses (gained from X), with a charge of $2-$. Six outer electrons means Group VI.

(b) The diagram shows $2$ X ions and $3$ Y ions. Check the charges: $2 \times (+3) + 3 \times (-2) = 0$.

(c) Two X atoms each lose $3$ electrons: $2 \times 3 = 6$ electrons, which is the same as three Y atoms each gaining $2$. Six electrons are transferred.

(d) Group III, Period 3 is aluminium; Group VI, Period 2 is oxygen. The compound is aluminium oxide, $\text{Al}_2\text{O}_3$.

(a) $\text{X}^{2+}$ has lost two electrons: $18 + 2 = 20$ protons, so X is calcium. $\text{Z}^{-}$ has gained one electron: $18 - 1 = 17$ protons, so Z is chlorine.

(b) $\text{Ca}^{2+}$ needs two $\text{Cl}^{-}$ to balance the charges: $\text{CaCl}_2$.

(c) $[\text{Ca}]^{2+}$ with an outer shell of eight crosses (its third shell). Two separate $[\text{Cl}]^{-}$ ions, each with an outer shell of seven dots and one cross — the cross being the electron gained from calcium. Charges shown outside the square brackets.

(d) Both ions have $18$ electrons, $2,8,8$ — the configuration of argon.

(e) In the solid, the ions are held in fixed positions in the giant lattice and cannot move. When dissolved, the ions are free to move and carry charge through the solution.