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InkFlaskCambridge IGCSE
Contains the answers

Answers · Covalent molecules

Sharing electrons — answer key

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

1. VQ1 · Practice

Model answer. A covalent bond is a pair of electrons shared between two atoms, one electron from each. Both atoms count the pair in their outer shell, so both reach a noble gas electronic configuration. No electron is transferred and no ions form.

A full answer includes

  • I chose: a pair of electrons shared between two atoms.
  • I did not choose an electron transferred: that makes ions, as in ionic bonding.
  • I did not choose the weak attraction between molecules: that is an intermolecular force.

Also accept

  • A pair of electrons shared between two atoms.

Watch for

  • “One atom gives its electron to the other, as in ionic bonding.” — Two non-metal atoms both need electrons, so neither gives one away. They share a pair, and no ions form.
  • “A covalent bond is the attraction between oppositely charged ions.” — That is the ionic bond. Covalent bonds join non-metal atoms, which share electrons instead.
  • “A covalent bond is a weak attraction between neighbouring molecules.” — That describes the forces between molecules. A covalent bond holds two atoms together inside one molecule, and it is strong.

2. VQ2 · Practice

Model answer. In H2 each hydrogen atom gives one electron to the shared pair, so 2 electrons are shared, and each atom counts both of them: 2, like helium. In HI iodine puts 1 of its 7 outer electrons into the shared pair and keeps 6, which sit as 6 ÷ 2 = 3 lone pairs. Row A.

A full answer includes

  • I chose row A.
  • I counted 2 shared electrons in H2.
  • I counted 2 electrons for each H atom: both electrons of the shared pair.
  • I counted 3 lone pairs on iodine.
  • I did not count iodine’s 6 unshared electrons as 6 pairs.

Also accept

  • Row A

Watch for

  • “Row B.” — Each hydrogen atom counts both electrons of the shared pair, its own and its partner’s: 2, like helium.
  • “Row C.” — 6 is the number of unshared electrons on iodine. They sit in pairs: 6 ÷ 2 = 3 lone pairs.
  • “Row D.” — A covalent bond is a pair: each hydrogen atom gives one electron, so 2 electrons are shared, and each atom counts both.

3. VQ3 · Practice

Options as numbered on the pupil sheet: 1 — SiH4; 2 — PH3; 3 — H6S; 4 — HF; 5 — H2S; 6 — PH5.

Model answer. Each hydrogen atom makes one bond, so the number of hydrogen atoms is the number of electrons the other atom needs to reach 8. Fluorine (7 outer) needs 1: HF. Sulfur (6 outer) needs 2: H2S. Phosphorus (5 outer) needs 3: PH3. Silicon (4 outer) needs 4: SiH4.

A full answer includes

  • I matched fluorine to HF: it needs 1 electron.
  • I matched sulfur to H2S: it needs 2.
  • I matched phosphorus to PH3: it needs 3.
  • I matched silicon to SiH4: it needs 4.
  • I did not give an atom one hydrogen atom for each of its outer electrons.

Also accept

  • F: HF. S: H2S. P: PH3. Si: SiH4.

Watch for

  • “Phosphorus has 5 outer electrons, so it bonds to 5 hydrogen atoms (PH5).” — The bonds equal the electrons it needs, 8 − 5 = 3, not the electrons it has: PH3.
  • “Sulfur has 6 outer electrons, so it bonds to 6 hydrogen atoms (H6S).” — Sulfur needs only 8 − 6 = 2 more electrons, so it shares 2 pairs: H2S.

4. VQ4 · Practice

Model answer. Oxygen has 6 outer electrons. It shares 1 with each hydrogen, which leaves 4 unshared: two lone pairs. The pupil drew only one lone pair, so oxygen counts 2 + 2 + 2 = 6 and has lost two of its own electrons. Add the second lone pair; then oxygen counts 2 + 2 + 4 = 8.

Checked automatically

  • (a) Electrons oxygen counts in this diagram: 6 (exact value)
    • Common wrong answer 4 : Each shared pair counts both its electrons for oxygen, not only its own dot: 2 + 2 + 2 = 6.
    • Common wrong answer 8 : That is what oxygen should count. Count what the pupil drew: two shared pairs and one lone pair, 2 + 2 + 2 = 6.

A full answer includes

  • (a) Method: I counted 2 + 2 in the shared pairs and 2 in the lone pair.
  • (a) Answer: oxygen counts 6.
  • (b) I wrote that a second lone pair of dots on oxygen is missing.
  • (b) I wrote that with it oxygen counts 8.
  • (b) I did not call the diagram right because each hydrogen counts 2.

Also accept

  • (a) 6. (b) One lone pair (two dots) is missing; add it so oxygen has 8.

Watch for

  • “The diagram is right, because each hydrogen has 2.” — Check every atom: the hydrogens are full, but oxygen counts only 6.
  • “A shared pair counts for only one of the two atoms.” — It counts for both atoms. Each atom adds both electrons of every pair it shares to its own count.

5. VQ5 · Practice

Model answer. Each chlorine atom brings 7 outer electrons: 7 + 7 = 14. Each atom needs 1 more, so the atoms share one pair: 2 electrons. The other 14 − 2 = 12 electrons are unshared and sit as 12 ÷ 2 = 6 lone pairs, 3 on each atom. Check: each atom counts 6 + 2 = 8.

Checked automatically

  • Outer-shell electrons in the molecule: 14 (exact value)
    • Common wrong answer 16 : That adds each atom’s count of 8, so the shared pair is counted twice. Add each atom’s own outer electrons: 7 + 7 = 14.
    • Common wrong answer 34 : That counts the inner shells too. Only the outer shells: 7 + 7 = 14.
  • Shared electrons: 2 (exact value)
    • Common wrong answer 1 : That is the number of pairs. One shared pair is 2 electrons.
  • Lone pairs in the molecule: 6 (exact value)
    • Common wrong answer 12 : That is the number of unshared electrons. They form 12 ÷ 2 = 6 pairs.
    • Common wrong answer 3 : That is the lone pairs on one atom. The molecule has two atoms, with 3 each.

A full answer includes

  • (a) Method: I added 7 + 7.
  • (a) Answer: 14 electrons.
  • (b) I wrote 2 electrons: one shared pair.
  • (c) Method: I took 14 − 2 = 12 unshared electrons and halved them.
  • (c) Answer: 6 lone pairs, 3 on each atom.
  • (a) I did not add 8 + 8: that counts the shared pair twice.

Also accept

  • 14; 2; 6.

Watch for

  • “Each atom counts 8, so the molecule has 16 outer electrons.” — Both atoms count the shared pair, but it is only 2 electrons. Count each electron once: 7 + 7 = 14.
  • “Count lone electrons, not lone pairs.” — A lone pair is two electrons. Halve the number of unshared electrons to get the pairs.

6. VQ19 · Practice

Model answer. (a) The description gives carbon 4 crosses in the overlaps and 2 in the lone pair: 6. (b) A carbon atom has only 4 outer electrons, and all 4 go into the four shared pairs with hydrogen, so carbon has 0 lone pairs. Its four shared pairs already give it 4 × 2 = 8. Nitrogen keeps a lone pair in ammonia because it has 5 outer electrons and shares only 3.

Checked automatically

  • (a) Crosses on carbon in the pupil’s description: 6 (exact value)
    • Common wrong answer 4 : Those are the crosses in the overlaps. The lone pair adds 2 more.
    • Common wrong answer 10 : That is what carbon would count, its partners’ dots included. Count only the crosses, carbon’s own electrons.

A full answer includes

  • (a) Method: I added 4 crosses in the overlaps and 2 in the lone pair.
  • (a) Answer: 6 crosses.
  • (b) I wrote that a carbon atom has only 4 outer electrons.
  • (b) I wrote that all 4 are used in the four shared pairs.
  • (b) I wrote that carbon has 0 lone pairs in methane.
  • (b) I did not write that carbon needs a lone pair to reach 8.

Also accept

  • (a) 6. (b) Carbon has only 4 outer electrons, all in the shared pairs: 0 lone pairs.

Watch for

  • “Carbon needs a lone pair, like nitrogen, to reach 8.” — Carbon reaches 8 from its four shared pairs alone: 4 × 2 = 8. Nitrogen keeps a lone pair because it has 5 outer electrons and shares only 3.

7. VQ6 · Practice

Model answer. Oxygen needs 2 electrons and each fluorine needs 1, so oxygen shares one pair with each fluorine atom. Oxygen keeps 6 − 2 = 4 dots as two lone pairs and counts 2 + 2 + 4 = 8; each fluorine keeps 7 − 1 = 6 crosses as three lone pairs and counts 2 + 6 = 8. Diagram C. In A each fluorine counts 4 + 6 = 10 and oxygen 12; in B each fluorine counts only 2 + 4 = 6; in D oxygen counts 2 + 2 + 6 = 10.

A full answer includes

  • I chose diagram C.
  • I checked oxygen in C: 2 + 2 in the shared pairs and 4 in two lone pairs make 8.
  • I checked each fluorine in C: 2 in the shared pair and 6 in three lone pairs make 8.
  • I did not choose diagram A: fluorine shares one pair with oxygen, not two.

Also accept

  • Diagram C

Watch for

  • “Diagram A.” — Fluorine needs only 1 electron, so it shares one pair with oxygen, not two. With two pairs each fluorine counts 4 + 6 = 10.
  • “Diagram B.” — Count a fluorine atom: 2 shared + 4 in two lone pairs = 6. Fluorine has 7 outer electrons and shares 1, so it keeps 6: three lone pairs.
  • “Diagram D.” — Oxygen has 6 outer electrons and shares 2, so it keeps 4: two lone pairs. With three lone pairs oxygen counts 2 + 2 + 6 = 10.

8. VQ7 · Practice

Model answer. An atom makes one covalent bond for each electron it needs to fill its outer shell. Neon’s outer shell already holds 8, a noble gas electronic configuration, so it needs 0 electrons and makes 0 bonds. That is why neon is made of single atoms.

A full answer includes

  • I chose: its outer shell is full, so it needs no more electrons.
  • I can say the rule: an atom makes one bond for each electron it needs.
  • I can say that neon needs 0 electrons.
  • I did not choose “it needs 8 more”: neon already has 8.

Also accept

  • Its outer shell is full, so it needs no more electrons.

Watch for

  • “Neon needs 8 more electrons, too many to share.” — Neon’s outer shell already has 8. Count what an atom needs to reach a full shell, not what it has.
  • “Neon atoms are too small to share electrons.” — Size does not stop sharing: hydrogen atoms are smaller and still share a pair. Neon makes no bonds because it needs no more electrons.
  • “Neon gives its outer electrons away to other atoms.” — Neon neither gives nor takes electrons. Its full outer shell is already a noble gas configuration.

9. VQ8 · Practice

Model answer. In the pupil’s diagram each nitrogen atom counts 2 × 2 = 4 shared electrons and 2 in its lone pair: 6, so it is 2 short of 8. Of its own electrons the pupil drew only 1 + 1 in the shared pairs and 2 in the lone pair: 4 of its 5, so one is missing. Each nitrogen atom needs 3 electrons, so the atoms must share 3 pairs, a triple bond, N≡N. Then each atom keeps 5 − 3 = 2 electrons as one lone pair and counts 6 + 2 = 8.

Checked automatically

  • (a) Electrons each nitrogen atom counts in the pupil’s diagram: 6 (exact value)
    • Common wrong answer 4 : That counts only the shared pairs. Add the lone pair: 2 × 2 + 2 = 6.
    • Common wrong answer 8 : That is what a full shell needs. The pupil’s diagram gives 2 × 2 + 2 = 6.
  • (b) Own outer electrons of each atom in the pupil’s diagram: 4 (exact value)
    • Common wrong answer 5 : Count them: 1 in each of the two shared pairs and 2 in the lone pair make 4, so one is missing.
    • Common wrong answer 6 : Only one electron of each shared pair is the atom’s own: 1 + 1 + 2 = 4.
  • (c) Shared pairs in a correct N2 molecule: 3 (exact value)
    • Common wrong answer 2 : With two pairs each atom counts only 6. A third pair gives 6 + 2 = 8.
    • Common wrong answer 6 : That is the number of shared electrons. A pair is 2 electrons: 6 electrons are 3 pairs.

A full answer includes

  • (a) Method: I counted 2 × 2 in the shared pairs and 2 in the lone pair.
  • (a) Answer: 6 electrons.
  • (b) Method: I counted 1 own electron in each shared pair and 2 in the lone pair.
  • (b) Answer: 4 of its 5.
  • (c) I wrote 3 shared pairs: a triple bond.
  • (c) I did not write 6: that is the number of shared electrons, not pairs.

Also accept

  • 6; 4; 3.

Watch for

  • “The diagram is right: N2 has a double bond, like O2.” — Nitrogen needs 3 electrons, not 2 like oxygen. With two pairs each nitrogen counts only 6, so the atoms share a third pair.
  • “A triple bond is 3 shared electrons.” — A triple bond is three shared pairs, 6 electrons; each pair has one electron from each atom.

10. VQ18 · Practice

Model answer. Carbon needs 4 electrons and each sulfur needs 2. One pair with each sulfur would leave every atom short, so carbon shares two pairs with each sulfur: two double bonds, S=C=S. Carbon uses all 4 of its electrons in bonds, so it has no lone pairs, and counts 4 × 2 = 8. Each sulfur keeps 6 − 2 = 4 electrons as two lone pairs and counts 4 + 4 = 8. Diagram A. In B carbon counts 2 + 2 + 2 = 6; in C each sulfur counts 4 + 6 = 10; in D carbon has 6 crosses and counts 10.

A full answer includes

  • I chose diagram A.
  • I checked carbon in A: 2 pairs with each sulfur give it 4 × 2 = 8.
  • I checked each sulfur in A: 4 shared + 4 in two lone pairs = 8.
  • I did not choose single bonds (diagram B): carbon would count only 6.
  • I did not give carbon a lone pair (diagram D): all 4 of its electrons are in the shared pairs.

Also accept

  • Diagram A

Watch for

  • “Diagram B.” — Count carbon’s electrons: one pair to each sulfur plus a lone pair gives only 2 + 2 + 2 = 6. Carbon must share two pairs with each sulfur.
  • “Diagram C.” — Sulfur has 6 outer electrons and puts 2 into the double bond, so it keeps 4: two lone pairs, not three.
  • “Diagram D.” — Carbon has only 4 outer electrons, and all 4 are in the shared pairs. It has no electrons left for a lone pair.

11. VQ9 · Practice

Model answer. In the pupil’s diagram each carbon has 3 shared pairs (two with hydrogen, one with carbon): 3 × 2 = 6 electrons, 2 short of 8. Each carbon must share one more pair with the other carbon, so the carbons share 2 pairs: a double bond, C=C. A correct molecule has 4 C–H pairs + 2 C=C pairs = 6 shared pairs, and each carbon counts 2 + 2 + 4 = 8, with no lone pairs.

Checked automatically

  • Electrons each carbon counts in the pupil’s diagram: 6 (exact value)
    • Common wrong answer 3 : Each shared pair counts both its electrons for carbon: 3 pairs are 6 electrons.
    • Common wrong answer 8 : With one pair between the carbons each carbon has only 3 pairs: 6 electrons.
  • Shared pairs between the carbon atoms: 2 (exact value)
    • Common wrong answer 1 : With one pair each carbon counts only 6. They share a second pair.
    • Common wrong answer 4 : Each carbon needs 4 in total, but 2 come from its hydrogen atoms.
  • Shared pairs in a correct molecule: 6 (exact value)
    • Common wrong answer 5 : Count both pairs of the double bond: 4 + 2 = 6.
    • Common wrong answer 3 : Count the pairs of both carbon atoms: each has 2 with hydrogen.

A full answer includes

  • (a) Method: I counted 3 pairs around each carbon, 3 × 2.
  • (a) Answer: 6 electrons.
  • (b) I wrote 2 pairs between the carbon atoms: a double bond.
  • (c) Method: I added 4 C–H pairs and 2 C=C pairs.
  • (c) Answer: 6 shared pairs.
  • (c) I did not count the double bond as one pair.

Also accept

  • 6; 2; 6.

Watch for

  • “Each carbon shares one pair with the other carbon, as in methane’s single bonds.” — Then each carbon counts only 6. A second pair between the carbons fills both shells.
  • “A double bond is one shared pair counted twice.” — A double bond is two separate shared pairs: 4 electrons, two from each carbon.

12. VQ10 · Practice

Model answer. Carbon is bonded to four atoms, and each bond gives it 2, so 4 × 2 = 8 already: every bond to carbon is single. Oxygen is bonded to two atoms, which gives it the 2 it needs, so its bonds are single too. Methanol has 3 C–H pairs, 1 C–O pair and 1 O–H pair: 5 shared pairs, 10 electrons. Its atoms bring 4 (C) + 6 (O) + 4 × 1 (H) = 14 outer electrons. So 14 − 10 = 4 are unshared: 2 lone pairs, both on oxygen, which shares only 2 of its 6. Check: carbon counts 4 × 2 = 8, oxygen 2 + 2 + 4 = 8, each hydrogen 2.

Checked automatically

  • Shared pairs in the molecule: 5 (exact value)
    • Common wrong answer 4 : Count the O–H pair too: 3 C–H + 1 C–O + 1 O–H = 5.
  • Outer-shell electrons from all six atoms: 14 (exact value)
    • Common wrong answer 24 : That adds every atom’s full count, so each shared pair is counted twice. Add each atom’s own outer electrons: 4 + 6 + 4 = 14.
    • Common wrong answer 11 : There are four hydrogen atoms, not one: 4 + 6 + 4 = 14.
  • Lone pairs in the molecule: 2 (exact value)
    • Common wrong answer 4 : That is the number of unshared electrons: 4 ÷ 2 = 2 pairs.
    • Common wrong answer 0 : Oxygen shares only 2 of its 6 outer electrons, so it keeps 4: two lone pairs.

A full answer includes

  • (a) I wrote 5 shared pairs: 3 C–H, 1 C–O and 1 O–H.
  • (a) I did not put a double bond between C and O: carbon already has four bonded atoms.
  • (b) Method: I added 4 + 6 + 4 × 1.
  • (b) Answer: 14 electrons.
  • (c) Method: I took 14 − 10 = 4 unshared electrons and halved them.
  • (c) Answer: 2 lone pairs, both on oxygen.
  • (b) I did not add the atoms’ full counts of 8 and 2.

Also accept

  • 5; 14; 2.

Watch for

  • “Carbon and oxygen share a double bond in methanol.” — Count carbon’s bonds: three hydrogens and one oxygen already make 4. All the bonds are single.
  • “Count lone electrons, not lone pairs.” — A lone pair is two electrons. Halve the number of unshared electrons to get the pairs.

13. VT1 · Apply

Model answer. (a) Phosphorus has 5 outer electrons and needs 3; each chlorine needs 1. So phosphorus shares one pair with each of three chlorine atoms: PCl3. (b) Three shared pairs; phosphorus keeps 5 − 3 = 2 electrons, one lone pair; each chlorine keeps 6, three lone pairs. (c) Every atom counts 8 in its third shell: 2,8,8, the configuration of argon. (d) Each nitrogen atom needs 3 electrons. It already shares one pair with each of its two hydrogen atoms, so it needs only one more pair: a single bond to the other nitrogen, N–N. Each nitrogen uses 3 of its 5 outer electrons in bonds and keeps 2 as one lone pair, and counts 3 × 2 + 2 = 8.

A full answer includes

  • (a) I wrote PCl3.
  • (a) I did not give phosphorus five chlorine atoms.
  • (b) I described three shared pairs, one between phosphorus and each chlorine.
  • (b) I put one lone pair on phosphorus.
  • (b) I put three lone pairs on each chlorine.
  • (c) I wrote argon’s configuration, 2,8,8, for every atom.
  • (d) I wrote a single bond between the nitrogen atoms.
  • (d) I put one lone pair on each nitrogen atom.
  • (d) I did not write a triple bond, as in N2.

Also accept

  • (a) PCl3. (b) 3 shared pairs, 1 lone pair on P, 3 on each Cl. (c) Argon. (d) A single N–N bond; one lone pair on each N.

Watch for

  • “Phosphorus has 5 outer electrons, so it bonds to 5 chlorine atoms.” — The number of bonds is the electrons it needs, 8 − 5 = 3, as for nitrogen.
  • “Count lone electrons, not lone pairs.” — A lone pair is two electrons. Halve the number of unshared electrons to get the pairs.
  • “The two nitrogen atoms in hydrazine share three pairs, as in N2.” — Each nitrogen already shares two pairs with hydrogen, so it needs only one more: a single N–N bond. With three, each nitrogen would count 12.