Dot-and-cross diagrams

1. Overview

A dot-and-cross diagram shows how the outer electrons are arranged once atoms have bonded, and which atom each electron came from. You need to draw them for ionic compounds, simple molecules, ions such as NH₄⁺ and molecules with coordinate bonds, including atoms with more than eight outer electrons and species with an unpaired electron. A correct electron count is also the starting point for the shapes of molecules (topic 3.5).

Key Definitions

• Dot-and-cross diagram: a diagram of the electrons in a species, usually the outer shells only, with the electrons from one atom drawn as dots and those from another atom as crosses, showing which atom each electron came from.
• Ionic bonding: the electrostatic attraction between oppositely charged ions (positively charged cations and negatively charged anions).
• Covalent bonding: the electrostatic attraction between the nuclei of two atoms and a shared pair of electrons.
• Coordinate (dative covalent) bond: a covalent bond in which both electrons of the shared pair come from the same atom, which uses one of its lone pairs.
• Bonding pair: a pair of electrons shared between two atoms; each shared pair is one covalent bond.
• Lone pair: a pair of outer-shell electrons on an atom that is not shared with another atom.
• Double and triple covalent bonds: two or three shared pairs of electrons between the same two atoms, as in O=O and N≡N.
• Expanded octet: more than eight electrons in the outer shell of an atom in a compound, possible only for atoms in period 3 and beyond, such as sulfur in SF₆ (12 electrons).
• Free radical: a species with one or more unpaired electrons; every species with an odd total number of electrons is a free radical, such as the hydroxyl radical, •OH.

Content

What a dot-and-cross diagram shows

All electrons are identical. Dots and crosses are bookkeeping: they record which atom each electron came from.

1. Electrons from one kind of atom are dots (•) and from the other kind are crosses (×). For a third kind of atom, use a third symbol such as ○. If the question gives a key, follow it exactly.
2. Show outer-shell electrons only, unless the question asks for all electrons. Then add each inner shell as another circle.
3. Draw each atom's symbol, usually inside a circle for its outer shell. In a covalent bond the two circles overlap, and the shared pair sits in the overlap.
4. Draw paired electrons close together. For eight electrons, a clear layout puts the four pairs top, bottom, left and right.
5. Put every ion inside square brackets with its charge outside the top right corner, for example [Cl]⁻.

Then do two checks:

• Total electrons. Add the outer electrons of all the atoms. Add one for each negative charge; take away one for each positive charge. A molecule or ion must show exactly this number.
• Electrons around each atom. Count every electron inside that atom's circle, shared pairs included. Hydrogen in a covalent bond has 2. A period 2 atom (C, N, O, F) has at most 8. Some period 3 atoms have more than 8, such as 10 in PCl₅ and 12 in SF₆.

Outer electrons = group number for groups 1 and 2, and group number minus 10 for groups 13 to 18. In simple molecules:

Atom (group)	Outer electrons	Covalent bonds that give 8 outer electrons	Lone pairs left
H (1)	1	1 (H holds only 2)	0
B, Al (13)	3	none: 3 bonds give only 6	0
C (14)	4	4	0
N, P (15)	5	3	1
O, S (16)	6	2	2
F, Cl (17)	7	1	3

Ions, coordinate bonds, expanded octets and odd numbers of electrons change these numbers, as the later subsections show.

Ionic compounds: NaCl, MgO and CaF₂

In ionic bonding (topic 3.2), metal atoms transfer outer electrons to non-metal atoms. The diagram shows the ions after the transfer:

• Inside the anion, draw the gained electrons in the metal's symbol. That shows the transfer.
• Draw the cation in either of two correct ways: with no outer electrons (its old outer shell is now empty), or with its new outer shell full, 8 electrons for Na⁺, Mg²⁺ and Ca²⁺ (2 for Li⁺), all in the metal's symbol. Never leave a lost electron on the cation.
• Draw the right number of each ion, all separate: their circles do not overlap and no electrons are shared.

Key: metal electrons ×, non-metal electrons •.

Compound	Atoms	Ions	What to draw
NaCl	Na 2,8,1; Cl 2,8,7	Na⁺ 2,8; Cl⁻ 2,8,8	[Na]⁺ beside [Cl]⁻; Cl⁻ has 7 • and 1 ×, as four pairs
MgO	Mg 2,8,2; O 2,6	Mg²⁺ 2,8; O²⁻ 2,8	[Mg]²⁺ beside [O]²⁻; O²⁻ has 6 • and 2 ×
CaF₂	Ca 2,8,8,2; F 2,7	Ca²⁺ 2,8,8; F⁻ 2,8	[Ca]²⁺ with a separate [F]⁻ on each side; each F⁻ has 7 • and 1 ×

The diagram shows one formula unit. The solid itself is a giant lattice of ions.

Worked example 1: lithium oxide

Question: Draw a dot-and-cross diagram for lithium oxide, Li₂O, showing outer electrons only. Use × for lithium electrons and • for oxygen electrons.

Step 1: electron arrangements. Li is 2,1 (1 outer electron). O is 2,6 (6 outer electrons).

Step 2: ions. Each Li loses 1 electron to form Li⁺ (2). Each O gains 2 electrons to form O²⁻ (2,8).

Step 3: ratio. Two Li⁺ balance one O²⁻: 2(+1) + (−2) = 0. So the O atom gains one electron from each of two Li atoms.

Step 4: draw. Two separate [Li]⁺ ions with no outer electrons. (The new outer shell of Li⁺ is the first shell, so if you draw it, it holds 2 ×, not 8.) One [O]²⁻ ion with 6 • and 2 × as four pairs: two • • pairs and two • × pairs.

Step 5: check. Electrons lost (2) equal electrons gained (2), and every ion has brackets and a charge.

Answer: [Li]⁺ [O]²⁻ [Li]⁺: two lithium ions with no outer electrons, and one oxide ion with 8 outer electrons (6 • and 2 ×).

Simple molecules from topics 3.4 and 3.5

For any covalent molecule or ion:

1. Choose the key (or copy the question's) and work out the total number of electrons.
2. Put the central atom in the middle. It is usually the atom that forms the most bonds; H forms only one bond, so it is an outer atom.
3. Join each outer atom to it with one shared pair, one electron from each atom.
4. If an atom still needs more bonds (see the table above), add a second or third shared pair. O, N and C often need this.
5. Put each atom's remaining electrons on it as lone pairs, then do both checks.

In the table, the central atom (or, in a two-atom molecule, the first atom named) uses • and the other atoms use ×. Each shared pair in a normal covalent bond has one electron from each atom.

Species	Shared pairs	Lone pairs	Electrons around each atom	Total
H₂	H-H: 1	none	2 on each H	2
Cl₂	Cl-Cl: 1	3 on each Cl	8 on each Cl	14
HCl	H-Cl: 1	3 on Cl	H 2, Cl 8	8
O₂	O=O: 2	2 on each O	8 on each O	12
N₂	N≡N: 3	1 on each N	8 on each N	10
CO₂	each C=O: 2	none on C, 2 on each O	C 8, O 8	16
CH₄	4 C-H	none	C 8, H 2	8
NH₃	3 N-H	1 on N	N 8, H 2	8
H₂O	2 O-H	2 on O	O 8, H 2	8
C₂H₆	C-C: 1; 6 C-H	none	each C 8, H 2	14
C₂H₄	C=C: 2; 4 C-H	none	each C 8, H 2	12
HCN	H-C: 1 (○ •); C≡N: 3	1 on N	H 2, C 8, N 8	10
BF₃	3 B-F	none on B, 3 on each F	B 6, F 8	24

In C₂H₆ and C₂H₄ both carbons use •, so each C-C pair is • •. In HCN, H needs a third symbol, ○.

Reading a row: NH₃. Draw N with its five •. Three H circles overlap the N circle at the left, right and bottom; each overlap holds • ×. The last two • sit together at the top of N as the lone pair.

Reading a row: CO₂. C sits in the middle with an O on each side. Each overlap holds two • and two ×, as two pairs. Each O has two lone pairs (four ×) on its outer side; C has none.

BF₃ is the exception. Boron brings only 3 electrons, so after three B-F bonds it has 6. Do not add a lone pair or a double bond to reach 8. That space is what lets BF₃ accept a lone pair (Worked example 2).

Coordinate bonds: the ammonium ion

A coordinate (dative covalent) bond needs a donor atom with a lone pair (such as N in NH₃) and an acceptor with room for two more electrons (such as H⁺, which has no electrons at all, or B in BF₃).

When ammonia gas meets hydrogen chloride gas, white solid ammonium chloride forms:

NH₃(g) + HCl(g) → NH₄Cl(s)

In effect, HCl transfers H⁺ to NH₃: the H-Cl bond breaks with both of its electrons staying on Cl (making Cl⁻), and the lone pair on N forms a coordinate bond to the H⁺.

Key: N electrons •, H electrons ×, Cl electrons ○.

Ion	What to draw	Electrons around the central atom	Total
NH₄⁺	N with four H around it. Three N-H overlaps hold • ×; the fourth holds • • (the coordinate bond). No lone pair on N. Brackets, charge +	N 8	5 + 4 − 1 = 8
Cl⁻	A separate ion with 7 ○ and 1 × (the electron that came from H), as four pairs. Brackets, charge −	Cl 8	7 + 1 = 8

In a displayed formula a coordinate bond is an arrow from donor to acceptor: [H₃N→H]⁺. Once formed, it is identical to the other three N-H bonds, and the ion is tetrahedral (topic 3.5).

Coordinate bonds in Al₂Cl₆

In AlCl₃, aluminium has only 6 outer electrons, like boron in BF₃. Two AlCl₃ molecules can join to form Al₂Cl₆: a Cl atom in each unit uses a lone pair to form a coordinate bond to the Al of the other unit, so each Al reaches 8.

Key: Al electrons •, Cl electrons ×. Total = 2(3) + 6(7) = 48.

• Layout. The two Al atoms sit side by side, and their circles do not overlap: there is no Al-Al bond. One bridging Cl sits above the gap and one below; each overlaps both Al circles. Two terminal Cl atoms sit on the outer side of each Al.
• Each terminal Cl: one shared pair with its Al (• ×) and 3 lone pairs.
• Top bridging Cl: a normal pair with the left Al (• ×), a coordinate pair to the right Al (× ×) and 2 lone pairs.
• Bottom bridging Cl: the mirror image: a normal pair with the right Al and a coordinate pair (× ×) to the left Al.
• Check: each Al has 3 normal pairs and 1 coordinate pair (8 electrons); every Cl has 8.

Both electrons of each coordinate pair come from Cl, so the arrows in a displayed formula point from the bridging Cl atoms to Al.

Worked example 2: the tetrafluoroborate ion, BF₄⁻

Question: Boron trifluoride reacts with a fluoride ion: BF₃ + F⁻ → BF₄⁻. Draw a dot-and-cross diagram of BF₄⁻, outer electrons only. Use • for boron electrons, × for fluorine electrons and ○ for the extra electron that gives the ion its charge.

Step 1: total. B has 3 outer electrons and each F has 7. The charge is 1−, so add one: 3 + 4(7) + 1 = 32.

Step 2: the BF₃ part. Three B-F pairs, each • ×. Boron has 6 outer electrons, so it can accept one more pair.

Step 3: the coordinate bond. F⁻ has 8 outer electrons (7 × and 1 ○) in four pairs. It donates one pair to boron, so that B-F pair contains no •.

Step 4: check. B: 4 shared pairs = 8. Each F: 1 shared pair + 3 lone pairs = 8. Total: 4 shared pairs (8) + 12 lone pairs (24) = 32, matching Step 1.

Answer: B in the centre with four F atoms, each F with three lone pairs. Three B-F overlaps hold • ×; the fourth holds two electrons from that F (× × or × ○) and no •. The ○ appears once, and the whole ion sits in one pair of square brackets with − outside.

Once formed, the four B-F bonds are identical and the ion is tetrahedral. Here the donor is an ion (F⁻) and the acceptor a neutral molecule (BF₃); for NH₄⁺ it is the other way round.

Expanded octets: SO₂, PCl₅, PF₅ and SF₆

A period 2 atom can never have more than 8 outer electrons, because its outer shell has only one 2s and three 2p orbitals. Atoms from period 3 onwards are larger, and their outer shell also has empty d orbitals, so a central atom can hold more than 8 electrons, such as 10 or 12. That is why PCl₅ exists but NCl₅ does not.

Draw them as before: each atom brings its own electrons, and you do not stop at 8. Central atom •, outer atoms ×.

Species	Shared pairs	Lone pairs on central atom	Electrons around central atom	Outer atoms	Total
SO₂	two S=O, each 2 pairs	1	10	2 lone pairs on each O	18
PCl₅	five P-Cl	0	10	3 lone pairs on each Cl	40
PF₅	five P-F	0	10	3 lone pairs on each F	40
SF₆	six S-F	0	12	3 lone pairs on each F	48

In SO₂, sulfur uses 4 of its 6 electrons in the two double bonds and keeps 2 as a lone pair. Some books show SO₂ with one double bond and one coordinate bond, leaving S with 8. At this level SO₂ is treated as an expanded-octet molecule, so draw two S=O double bonds unless a question says otherwise.

Worked example 3: sulfur tetrafluoride, SF₄

Question: Draw a dot-and-cross diagram of SF₄, outer electrons only, and state the number of electrons around the sulfur atom.

Step 1: key and total. • for S (group 16, 6 outer electrons) and × for F (group 17, 7). Total: 6 + 4(7) = 34.

Step 2: bonds. Each F needs one bond: four S-F pairs, each • ×, using 4 sulfur electrons.

Step 3: lone pairs. Sulfur's last 2 • form a lone pair. Each F keeps 6 × as 3 lone pairs.

Step 4: check. Around S: 4 shared pairs (8) + 1 lone pair (2) = 10. Each F: 8. Total: 8 + 2 + 4(6) = 34, matching Step 1. Ten is allowed because S is in period 3.

Answer: S in the centre with four S-F pairs (• ×), one lone pair on S (• •) and three lone pairs on each F (× ×). There are 10 electrons around sulfur, an expanded octet.

Unlike SF₆, SF₄ has a lone pair on S, and that lone pair affects its shape (topic 3.5).

Species with an odd number of electrons

If the total is odd, the electrons cannot all be paired. At least one is unpaired, so the species is a free radical (topic 1.3). Draw the unpaired electron on its own, clearly apart from any pair.

Species	Key	Shared pairs	Lone pairs and unpaired electron	Electrons around each atom	Total
•OH	O •, H ×	O-H: 1 (• ×)	O: 2 lone pairs and 1 single •; H: none	O 7, H 2	6 + 1 = 7
CH₃•	C •, H ×	3 C-H (• ×)	C: 1 single •, no lone pair	C 7, H 2	4 + 3 = 7

Placing the single electron in •OH. After the O-H bond forms, O has 5 of its own electrons left. It can make only two lone pairs from 5, so one electron stays unpaired on O, which has 7 electrons. H already has its 2 and takes no more.

The methyl radical, CH₃•, takes part in free-radical substitution (topic 14.1); the dot in its formula is the unpaired electron on carbon.

Key Equations

Total outer electrons in a molecule or ion ($N_{\text{total}}$ = electrons the diagram must show; $n$ = outer electrons of each atom; $q$ = charge on the species, with its sign): $$N_{\text{total}} = \sum n - q$$ For NH₄⁺: $N_{\text{total}} = 5 + 4(1) - (+1) = 8$. For BF₄⁻: $N_{\text{total}} = 3 + 4(7) - (-1) = 32$. An odd total means at least one unpaired electron.

Electrons around one atom ($b$ = shared pairs that include the atom, a double bond counting as 2 and a triple bond as 3; $l$ = lone pairs on it; $u$ = unpaired electrons on it): $$N_{\text{atom}} = 2b + 2l + u$$ $N_{\text{atom}}$ is 2 for H and at most 8 for a period 2 atom; a period 3 atom can have more than 8, such as 10 or 12.

Ionic compounds: electrons lost by the metal atoms = electrons gained by the non-metal atoms, so the charges add up to zero. For CaF₂: (+2) + 2(−1) = 0.

These are counts of electrons, so they have no units.

Common Mistakes to Avoid

• Mistake: using the wrong symbol for an atom's electrons. Fix: write the key beside your diagram first. If a question says "use • for phosphorus" in PCl₅, all five phosphorus electrons are •, including the one in each P-Cl pair.
• Mistake: leaving out lone pairs on the outer atoms. Fix: a halogen atom with one bond has 3 lone pairs. In PCl₅ that is 15 lone pairs (30 electrons) on chlorine, so 40 electrons in all.
• Mistake: drawing a coordinate bond with one electron from each atom, or the wrong way round. Fix: both electrons come from the donor. In Al₂Cl₆ each coordinate pair is two Cl electrons (× ×) given to Al, and there is no Al-Al bond.
• Mistake: drawing ammonium chloride as one molecule with N joined to five atoms. Fix: draw [NH₄]⁺ (four N-H bonds, one coordinate, 8 electrons round N) and a separate [Cl]⁻, each with brackets and a charge.
• Mistake: forgetting an anion's extra electron, or the brackets and charge. Fix: count with $N_{\text{total}} = \sum n - q$. BF₄⁻ needs 32 electrons, not 31, inside one pair of brackets with − outside.
• Mistake: sharing electrons in an ionic compound, or drawing the wrong number of ions. Fix: ions never overlap. CaF₂ needs one [Ca]²⁺ and two separate [F]⁻, each with 7 • and 1 ×.
• Mistake: forcing every atom to 8. Fix: count what each atom brings. B in BF₃ has 6, S in SF₆ has 12 and O in •OH has 7.
• Mistake: drawing a double bond as a coordinate bond, or the reverse. Fix: in a double bond each atom gives two electrons (each S=O in SO₂ holds two • and two ×); in a coordinate bond one atom gives both. If the question names the bond type, draw that type.

Exam Tips

• Draw or complete: if part of the diagram is printed, keep its layout and key and add only the missing electrons. Show outer electrons only means no inner shells.
• Before moving on, run both checks: the total ($N_{\text{total}}$) and each atom's count. Any atom other than H that does not have 8 needs a reason: three bonds on B or Al, an expanded octet, or an unpaired electron.
• To describe a coordinate bond, use the words "lone pair" and "both electrons from the same atom", and name the donor and acceptor, for example "the lone pair on N forms a coordinate bond to H⁺".
• If the question tells you about the bonding (a double bond, a coordinate bond, an unpaired electron on a named atom), build that feature in first, then add the rest.