Answers · Ions and ionic bonding

Dots and crosses — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. BQ10 · Practice

Model answer. The potassium ion’s outer shell holds only crosses, so the crosses are potassium’s electrons. Its charge is +, so the potassium atom lost an electron. That electron is the one cross among the chloride ion’s 7 dots: it moved from potassium to chlorine, and it gives the chloride ion its charge of −. Nothing is shared: the ions are separate, each in square brackets with its charge. Dots and crosses are both electrons; the symbols only show which atom each one came from.

A full answer includes

I chose: they were potassium’s electrons, and one of them has moved to chlorine.
I used the potassium ion’s outer shell: it holds only crosses.
I used the single cross among the chloride ion’s dots: it is the electron that moved.
I did not choose chlorine’s electrons: the potassium ion has charge +, so it lost an electron.

Also accept

The crosses were potassium’s electrons; one of them moved to chlorine.

Watch for

“They were chlorine’s electrons, and one of them has moved to potassium.” — The potassium ion has charge +, so potassium lost an electron; it did not gain one. Its outer shell holds only crosses, so the crosses are its own electrons.
“They are shared by both ions, and they hold the two ions together.” — Nothing is shared in an ionic compound. The ions are separate, each in its own brackets, and they are held by the attraction between their opposite charges.
“They are protons, and they give the potassium ion its positive charge.” — Dots and crosses are both electrons in the shells; protons are in the nucleus. The potassium ion is + because it has one electron fewer than protons.

2. BQ11 · Practice

Model answer. M’s ion is 1+, so the M atom lost 1 electron: its outer shell held 1 electron, and the 8 crosses are the full shell underneath. So M is in Group I. X’s ion is 1−, so the X atom gained 1 electron, the cross: it had 7 outer electrons of its own, the dots. So X is in Group VII. Row B. A Group I atom and a Group VII atom combine 1 : 1.

A full answer includes

I chose row B.
I worked out 1 outer electron for an M atom: its ion is 1+, so it lost 1.
I worked out 7 outer electrons for an X atom: the cross came from M.
I did not count the ion’s 8 crosses as the M atom’s outer shell.

Also accept

Row B

Watch for

“Row A.” — The 8 crosses are what the M ion has left. The M atom had 1 more electron, in the shell it lost: 1 outer electron, Group I.
“Row C.” — One of the X ion’s 8 outer electrons is a cross: it came from M. The X atom had 7, the dots: Group VII.
“Row D.” — Both 8s are the ions’ outer shells. Before the electron moved, the atoms had 1 (M) and 7 (X).

3. BQ12 · Practice

Model answer. The fluorine atom had 7 outer electrons, drawn as dots. It gains 1 electron from the potassium atom, drawn as a cross. The fluoride ion’s outer shell holds 7 dots and 1 cross: 8 electrons, a full shell, charge 1−. The potassium atom lost its single electron in the fourth shell, so its outer shell is now the full third shell: 8 crosses, charge 1+.

Checked automatically

Dots in the fluoride ion’s outer shell: 7 (exact value)
- Common wrong answer 8 : One of the 8 came from potassium, so it is a cross.
- Common wrong answer 9 : Only the outer shell: fluorine has 7 outer electrons of its own.
Crosses in the fluoride ion’s outer shell: 1 (exact value)
- Common wrong answer 8 : Potassium gives only its one outer electron.
Crosses in the potassium ion’s outer shell: 8 (exact value)
- Common wrong answer 0 : The potassium ion’s outer shell is not empty: the full third shell is its outer shell now.
- Common wrong answer 1 : Potassium gave its one outer electron to fluorine. Count the shell underneath.

A full answer includes

I wrote 7 dots: fluorine’s own outer electrons.
I wrote 1 cross: the electron from potassium.
I wrote 8 crosses for the potassium ion: its full third shell.
I did not write 0 for the potassium ion: its outer shell is not empty.

Also accept

7 dots and 1 cross; the potassium ion has 8 crosses.

Watch for

“All 8 outer electrons are dots, because they belong to the fluoride ion now.” — The electron from potassium keeps its cross, so you can see where it came from.
“The metal ion has an empty outer shell, so it has no electrons drawn.” — The metal atom loses its whole outer shell. The shell underneath is full, and that is its outer shell now.

4. BQ13 · Practice

Model answer. Lithium is 2,1. Its atom loses its one outer electron, so the lithium ion, Li⁺, keeps only its first shell of 2: two crosses. Bromine has 7 outer electrons and gains lithium’s electron, so the bromide ion, Br⁻, shows 7 dots and 1 cross. The ions are separate, each in square brackets with its charge. Diagram C. A gives the lithium ion 8 electrons, but a lithium atom has only 3. B shows a pair of electrons shared between two atoms, which is a covalent bond, not ions. D draws the moved electron as a dot, so it no longer shows where it came from.

A full answer includes

I chose diagram C.
I checked: the lithium ion keeps only 2 electrons, its first shell.
I checked: the bromide ion shows 7 dots and 1 cross, the electron from lithium.
I checked: the ions are separate, in square brackets, with charges + and −.
I did not choose A: a lithium atom has only 3 electrons, so its ion cannot show 8.

Also accept

Diagram C

Watch for

“Diagram A.” — A lithium atom has only 3 electrons. After it loses 1, its ion keeps a shell of 2, not 8.
“Diagram B.” — B shows a pair of electrons shared between two atoms: a covalent bond. In an ionic bond the electron moves, and the ions are drawn apart, each in brackets with its charge.
“Diagram D.” — The bromide ion gained its extra electron from lithium, so that electron stays a cross: 7 dots and 1 cross, not 8 dots.

5. BQ14 · Practice

Model answer. The potassium atom (Group I) loses its 1 outer electron, leaving 2,8,8: K⁺. The sulfur atom (Group VI) has 6 outer electrons and gains 2 to make 8: S²⁻ (2,8,8). The numbers lost and gained are not the same: each potassium atom gives only 1 electron, so two potassium atoms are needed for each sulfur atom, K₂S.

Checked automatically

Electrons lost by each potassium atom: 1 (exact value)
- Common wrong answer 2 : That is what each sulfur atom gains. Potassium has only 1 outer electron (2,8,8,1).
- Common wrong answer 8 : The 8 are the full shell left behind. Potassium loses only its 1 outer electron.
Electrons gained by each sulfur atom: 2 (exact value)
- Common wrong answer 6 : 6 is the number sulfur already has. It needs 2 more to make 8.
- Common wrong answer 8 : Its outer shell holds 8 in total; it already has 6.
- Common wrong answer 1 : One more electron gives sulfur only 7 in its outer shell. It needs 2 to make 8.

A full answer includes

I wrote 1 lost by each potassium atom, giving K⁺.
I wrote 2 gained by each sulfur atom, giving S²⁻.
I did not write 6 for sulfur: 6 is what it has, not what it gains.

Also accept

1 and 2.

Watch for

“Potassium loses 2 electrons, because sulfur gains 2.” — Potassium has only 1 outer electron (2,8,8,1), so each atom loses only 1. Two potassium atoms give sulfur its 2.
“Sulfur gains 6 electrons.” — 6 is the number it already has in its outer shell. It needs 2 more to make 8.

6. BQ15 · Practice

Model answer. An aluminium atom (2,8,3) loses its 3 outer electrons, so the aluminium ion, Al³⁺, is 2,8: its outer shell is now the full shell of 8 crosses. A fluorine atom (2,7) has room for only 1 more electron, so each of the 3 electrons goes to a different fluorine atom: three fluoride ions, F⁻, each with 7 dots and 1 cross. Electrons lost: 3; electrons gained: 3 × 1 = 3. Diagram D. A puts all 3 electrons on one fluorine atom, which would give its outer shell 10. B still shows the 3 electrons on the aluminium ion although they are already on the fluoride ions, so each is drawn twice. C draws the moved electrons as dots, so it no longer shows where they came from.

A full answer includes

I chose diagram D.
I checked: there are three fluoride ions, because each fluorine atom takes only 1 electron.
I checked: the aluminium ion’s outer shell is the full shell of 8 crosses underneath.
I checked: each fluoride ion shows 7 dots and 1 cross, the electron from aluminium.
I did not choose A: one fluorine atom cannot take all 3 electrons.

Also accept

Diagram D

Watch for

“Diagram A.” — One fluorine atom takes only 1 electron before its outer shell is full; 3 more would give it 10. Aluminium’s 3 electrons need three fluorine atoms.
“Diagram B.” — Aluminium has lost all 3 outer electrons: they are the crosses on the fluoride ions. Its ion’s outer shell is the full shell of 8 underneath.
“Diagram C.” — The moved electrons came from aluminium, so each stays a cross: 7 dots and 1 cross in every fluoride ion.

7. BQ16 · Practice

Model answer. Each aluminium atom loses its 3 outer electrons: Al³⁺. Each oxygen atom gains 2 to make 8: O²⁻. The electrons lost must equal the electrons gained. The smallest number that both 3 and 2 divide into is 6, so 2 aluminium atoms lose 2 × 3 = 6 electrons and 3 oxygen atoms gain 3 × 2 = 6. So there are 2 Al³⁺ for every 3 O²⁻: Al₂O₃. Check: 2 × (+3) + 3 × (−2) = 0.

Checked automatically

Aluminium ions: 2 (exact value)
- Common wrong answer 3 : That is the number of oxide ions. Each Al³⁺ gives 3 and each O²⁻ takes 2: 2 × 3 = 3 × 2 = 6.
- Common wrong answer 1 : One aluminium atom loses 3, but oxide ions take 2 each, and 3 cannot be shared out in 2s. Use two aluminium atoms.
Oxide ions: 3 (exact value)
- Common wrong answer 2 : Two Al³⁺ lose 6 electrons; each oxide ion takes 2, so 3 oxide ions are needed.
- Common wrong answer 1 : One oxide ion takes only 2 electrons. The aluminium atoms lose 6.
Electrons moved in total: 6 (exact value)
- Common wrong answer 5 : 5 is 3 + 2. Multiply instead: 2 aluminium atoms × 3 electrons each = 6.
- Common wrong answer 3 : That is what one aluminium atom loses. Two aluminium atoms lose 6.
- Common wrong answer 12 : Count each electron once: the 6 lost by aluminium are the same 6 gained by oxygen.

A full answer includes

I wrote 2 aluminium ions: 2 × 3 = 6 electrons lost.
I wrote 3 oxide ions: 3 × 2 = 6 electrons gained.
I wrote 6 electrons moved in total.
I did not stop at 1 aluminium ion: its 3 electrons cannot be shared out in 2s.

Also accept

2 aluminium ions, 3 oxide ions, 6 electrons.

Watch for

“Aluminium oxide is AlO, one of each.” — Al³⁺ and O²⁻ do not cancel: +3 − 2 = +1. Use 2 Al³⁺ and 3 O²⁻.
“There are 3 aluminium ions and 2 oxide ions.” — That swaps the numbers. 3 Al³⁺ would lose 9 electrons, but 2 O²⁻ take only 4.

8. BQ17 · Practice

Model answer. A Group II atom loses its 2 outer electrons, so the barium ion is Ba²⁺. A Group VII atom gains only 1 electron, to fill its outer shell, so each fluoride ion is F⁻. Two fluorine atoms are needed to take barium’s 2 electrons: 2 F⁻ for each Ba²⁺, BaF₂. Each fluoride ion’s outer shell has its own 7 dots and 1 cross from barium. Row C.

A full answer includes

I chose row C.
I worked out 2+ for the barium ion: a Group II atom loses 2 electrons.
I worked out 2 fluoride ions: each fluorine atom takes only 1 electron.
I worked out 7 dots and 1 cross in each fluoride ion’s outer shell.
I did not give one fluoride ion both of barium’s electrons.

Also accept

Row C

Watch for

“Row A.” — Barium is in Group II: its atom loses 2 outer electrons, so its ion is 2+, not 1+.
“Row B.” — A fluorine atom has 7 outer electrons and can take only 1 more. Two fluorine atoms are needed to take barium’s 2 electrons.
“Row D.” — The electron that came from barium keeps its cross, so each fluoride ion shows 7 dots and 1 cross.

9. BT2 · Apply

Model answer. (a) A Group I atom has 1 outer electron and loses it: Cs⁺. A Group VII atom has 7 and gains 1: I⁻. One electron lost and one gained, so the ions combine 1 : 1: CsI. (b) Each lithium atom (Group I) loses its 1 outer electron: Li⁺. The nitrogen atom has 5 outer electrons and gains 3 to make 8: N³⁻ (2,8). One lithium atom gives only 1 electron, so 3 lithium atoms are needed: 3 Li⁺ for each N³⁻, Li₃N. Electrons lost: 3 × 1 = 3; electrons gained: 1 × 3 = 3. They are equal.

A full answer includes

(a) I predicted Cs⁺ for the caesium ion.
(a) I predicted I⁻ for the iodide ion.
(a) I gave the ratio 1 : 1.
(a) I explained it: one electron lost by each caesium atom, one gained by each iodine atom.
(b) I wrote Li⁺ for the lithium ion.
(b) I wrote N³⁻ for the nitride ion.
(b) I wrote 3 lithium ions for each nitride ion.
(b) I showed the electrons lost: 3 × 1 = 3.
(b) I showed the electrons gained: 1 × 3 = 3.
I did not write N⁵⁻: nitrogen gains 3 electrons, not 5.

Also accept

(a) Cs⁺, I⁻, 1 : 1 (one electron lost and one gained). (b) Li⁺, N³⁻, 3 lithium ions for each nitride ion; 3 lost = 3 gained.

Watch for

“Nitrogen forms N⁵⁻, because it has 5 outer electrons.” — The 5 are what it has. It gains 3 more to reach 8, so its charge is 3−.
“Lithium nitride is LiN, one of each.” — One lithium atom gives only 1 electron, and nitrogen needs 3. Three lithium atoms are needed.