Atomic Structure and the Periodic Table Worked ExamplesIGCSE Chemistry 0620

(a) X is an electron (in a shell). Y is a proton (positive, in the nucleus). Z is a neutron (no charge, in the nucleus). W is the nucleus.

(b) The nucleus contains protons and neutrons — both are needed.

(c)(i) Proton number = number of protons = 4. (ii) Nucleon number = protons + neutrons $= 4 + 5 =$ 9. Proton number 4 is beryllium, Be. (Check: 4 electrons, arranged 2,2.)

(d) proton: charge $+1$, mass $1$; neutron: charge $0$, mass $1$; electron: charge $-1$, mass $\frac{1}{1840}$ (negligible).

Fill the shells in order: up to 2 in the first shell and up to 8 in each of the next two shells (for proton numbers 1 to 20).

(a)(i) Nitrogen, proton number 7: 7 electrons → 2,5.

(ii) Aluminium has 13 electrons; $\text{Al}^{3+}$ has lost 3: $13 - 3 = 10$ electrons → 2,8. (iii) Sulfur has 16 electrons; $\text{S}^{2-}$ has gained 2: $16 + 2 = 18$ electrons → 2,8,8.

(iv) Calcium, proton number 20: 20 electrons → 2,8,8,2.

(b) Two outer-shell electrons → Group II. Four occupied shells → Period 4.

Work out each ion. $\text{K}^+$: protons $19$, neutrons $39 - 19 = 20$, electrons $19 - 1 = 18$.

$\text{Cl}^-$: protons $17$, neutrons $37 - 17 = 20$, electrons $17 + 1 = 18$.

Statement 1: both have 18 electrons, 2,8,8 — correct. Statement 2: both have 20 neutrons — correct. Statement 3: 19 ≠ 17 — wrong. Answer: B.

Protons = the lower number (proton number). Neutrons = nucleon number − proton number. Electrons = protons for an atom; subtract the positive charge or add the negative charge for an ion.

${}^{88}_{38}\text{Sr}$ (atom): protons 38; neutrons $88 - 38 = 50$; electrons 38.

${}^{65}_{29}\text{Cu}^{2+}$: protons 29; neutrons $65 - 29 = 36$; electrons $29 - 2 = 27$.

${}^{34}_{16}\text{S}^{2-}$: protons 16; neutrons $34 - 16 = 18$; electrons $16 + 2 = 18$. ${}^{81}_{35}\text{Br}^{-}$: protons 35; neutrons $81 - 35 = 46$; electrons $35 + 1 = 36$.

(a) 26 protons → iron, Fe. Nucleon number $= 26 + 30 = 56$. Electrons: $26 - 23 = 3$ fewer than protons, so the charge is $3+$.

(b) A $2-$ ion has gained 2 electrons, so protons $= 36 - 2 = 34$ → selenium, Se (the ion is $\text{Se}^{2-}$).

(c) A $2+$ ion has lost 2 electrons, so protons $= 28 + 2 = 30$ → zinc, Zn (the ion is $\text{Zn}^{2+}$).

(d) The Group VII element in Period 4 is bromine, Br. Period 4 → 4 occupied shells. Group VII → 7 outer-shell electrons.

Use two rules: number of occupied shells = period number; number of outer-shell electrons = group number (Groups I to VII). Group VIII noble gases have a full outer shell.

(a) Period 3, Group VI → R (sulfur, 2,8,6).

(b) The Group VIII element → V (argon, 2,8,8).

(c) T is in Period 4 (4 shells) and Group II (2 outer electrons): 2,8,8,2 (calcium, 20 electrons).

(d) W and Z are both in Group VII, so both have 7 outer-shell electrons: W and Z.

(e) Z is in Period 5, so it has 5 occupied shells.

For a neutral atom, electrons = protons = the lower number; neutrons = upper number − lower number.

A: neutrons $1 - 1 = 0$, electrons 1 — one fewer neutron. B: neutrons 2, electrons 2 — equal. D: neutrons 8, electrons 8 — equal.

C: neutrons $7 - 3 = 4$, electrons 3 — exactly one more. Answer: C.

(a) Ethanol, $\text{C}_2\text{H}_5\text{OH}$, has 2 C, 6 H and 1 O. In a molecule, electrons = total protons (it has no charge).

(b) $\text{NH}_4^{\,+}$: protons from 1 N and 4 H.

(c) 10 electrons is the same as a neon atom (proton number 10).

(d) $\text{CO}_3^{\,2-}$: protons $= 6 + 3(8) = 30$. A $2-$ charge means 2 extra electrons.

Compare protons with electrons. W: 11 p, 10 e. X: 9 p, 10 e. Y: 10 p, 10 e. Z: 12 p, 12 e.

(a) Y — 10 protons (neon) and 10 electrons, so it is a neutral atom with a full outer shell.

(b) W: 11 protons is sodium; one electron fewer than protons → $\text{Na}^+$. X: 9 protons is fluorine; one electron more than protons → $\text{F}^-$.

(c) Z — a neutral atom (12 p, 12 e) with 2 outer-shell electrons (magnesium). (d) Nucleon number of X $= 9 + 10 =$ 19.

(e) W and Z: sodium (11) and magnesium (12) are both in Period 3 — their atoms have three occupied shells (Na 2,8,1; Mg 2,8,2). For an ion, use the atom of the element, not the ion's own number of shells.

(a)(i) Two electrons with no charge → 2 protons → He. (ii) A positive ion with 2 electrons has lost its outer electrons: lithium (2,1) loses one → $\text{Li}^+$; or beryllium (2,2) loses two → $\text{Be}^{2+}$.

(b) 2,8,8 is 18 electrons. (i) The atom with 18 electrons is Ar.

(ii) Metals just after argon lose electrons: potassium (2,8,8,1) → $\text{K}^+$, or calcium (2,8,8,2) → $\text{Ca}^{2+}$.

(iii) Non-metals just before argon gain electrons: chlorine (2,8,7) → $\text{Cl}^-$, or phosphorus (2,8,5) → $\text{P}^{3-}$.

(a) $\text{X}^{2+}$ has lost 2 electrons: protons $= 18 + 2 = 20$ → calcium. Nucleon number $= 20 + 20 = 40$.

$\text{Y}^{3-}$ has gained 3 electrons: protons $= 18 - 3 = 15$ → phosphorus. Nucleon number $= 15 + 16 = 31$.

(b) Calcium atom (20 electrons): 2,8,8,2 → Group II, Period 4. Phosphorus atom (15 electrons): 2,8,5 → Group V, Period 3.

(c) For ${}^{137}_{56}\text{Ba}^{2+}$:

Order: nucleon number (137) > neutrons (81) > protons (56) > electrons (54).

1 False. An atom of ${}^{1}_{1}\text{H}$ has 1 proton and $1 - 1 = 0$ neutrons.

2 False. $\text{Na}^+$, $\text{Mg}^{2+}$ and $\text{Al}^{3+}$ all have the configuration 2,8, but sodium, magnesium and aluminium are in Period 3 (their atoms have three shells).

3 True. The number of occupied shells in an atom equals the period number.

4 False. Group VII atoms have seven outer-shell electrons, not seven in total — for example chlorine has 17 electrons (2,8,7).

5 True. Protons and neutrons (relative mass 1 each) are in the nucleus; electrons have a relative mass of only $\frac{1}{1840}$.