Table 1 gives data for three consecutive Period 3 elements. Table 1 Element Atomic radius / nm Ionic radius of M^n+ / nm Na 0.186 0.095 (Na^+) Mg 0.160 0.065 (Mg^2+) Al 0.143 0.050 (Al^3+) (a) Describe and explain the trend in atomic radius from Na to Al, in terms of nuclear cha…

Answer: ((a)) Atomic radius decreases from Na to Al; nuclear charge (proton number) increases while the outer electrons are added to the same shell (n=3) with essentially no extra shielding, so effective nuclear charge increases and pulls the outer electrons in more closely. ((b)(i)) 1s^2 2s^2 2p^6 (the neon configuration, 10 electrons) ((b)(ii)) The three ions are isoelectronic (same number of electrons and shielding) but have increasing nuclear charge (11, 12, 13 protons); the same number of electrons is pulled in more strongly by the greater nuclear charge, so ionic radius decreases from Na+ to Al3+. ((c)(i)) Na, Mg and Al are giant metallic lattices; the number of delocalised outer electrons per atom increases (1, 2, 3) and ionic radius decreases, so the electrostatic attraction between the metal cations and the delocalised electrons gets stronger, raising the melting point. ((c)(ii)) Silicon is a giant covalent (macromolecular) lattice in which every atom is joined to four neighbours by strong covalent bonds; melting requires breaking a huge number of these strong bonds throughout the lattice, needing far more energy than melting a metallic or simple molecul… ((c)(iii)) P4, S8, Cl2 and Ar are simple molecular (or monatomic) structures held together only by weak van der Waals forces between separate molecules or atoms; melting only needs to overcome these weak forces (the strong covalent bonds within each molecule stay intact), so very little en… ((d)) Conductivity rises from Na to Al (more delocalised electrons per atom in the metallic lattice); P4, S8, Cl2 and Ar do not conduct at all because their simple molecular/monatomic structures contain no ions and no delocalised or mobile electrons - every electron is held in a coval…. Part (a): Atomic radius trend from Na to Al Across Period 3, atomic radius decreases from Na (0.186 nm) to Al (0.143 nm). Each successive element has one more proton (Na =11, Mg =12, Al =13), while the extra electron is added to the same o…

AS Chemistry 9701: Period 3 Trends in Radius, Melting Point and Conductivity

Table 1 gives data for three consecutive Period 3 elements.

Table 1

Element	Atomic radius / nm	Ionic radius of $M^{n+}$ / nm
Na	$0.186$	$0.095$ ( $\text{Na}^+$ )
Mg	$0.160$	$0.065$ ( $\text{Mg}^{2+}$ )
Al	$0.143$	$0.050$ ( $\text{Al}^{3+}$ )

(a) Describe and explain the trend in atomic radius from Na to Al, in terms of nuclear charge and electron shielding. [2]

(b) $\text{Na}^+$ , $\text{Mg}^{2+}$ and $\text{Al}^{3+}$ all have the same electron configuration.

(i) State this electron configuration. [1]

(ii) Explain why the ionic radius decreases from $\text{Na}^+$ to $\text{Al}^{3+}$ , even though the three ions have the same number of electrons. [2]

Table 2

Element	Na	Mg	Al	Si	$\text{P}_4$	$\text{S}_8$	$\text{Cl}_2$	Ar
Melting point / $^\circ\text{C}$	$98$	$650$	$660$	$1414$	$44$	$115$	$-101$	$-189$

Explain, in terms of structure and bonding:

(i) why the melting point rises from Na to Al, [2]

(ii) why silicon has by far the highest melting point of any element in the period, [2]

(iii) why the melting point falls sharply after silicon and stays low from $\text{P}_4$ through to Ar. [2]

(d) State whether electrical conductivity generally rises or falls from Na to Al. [1]

Part (a): Atomic radius trend from Na to Al

Across Period 3, atomic radius decreases from Na ( $0.186\ \text{nm}$ ) to Al ( $0.143\ \text{nm}$ ).

Each successive element has one more proton (Na $=11$ , Mg $=12$ , Al $=13$ ), while the extra electron is added to the same outer shell ( $n=3$ ). Because outer-shell electrons shield each other only weakly, the amount of shielding provided by the inner core barely changes across the period, so the effective nuclear charge felt by the outer electrons increases as protons are added. This stronger effective nuclear charge pulls the outer shell in more tightly, so atomic radius decreases from Na to Al.

Part (b): Electron configuration and ionic radius trend

(i)

$\text{Na}^+$ , $\text{Mg}^{2+}$ and $\text{Al}^{3+}$ have all lost their outer $3s$ (and, for $\text{Al}^{3+}$ , $3p$ ) electrons, leaving $10$ electrons in each ion. Their electron configuration is: $1s^2\,2s^2\,2p^6$ (the same configuration as a neon atom).

(ii)

The three ions are isoelectronic, the same number of electrons and the same amount of shielding, but they have different nuclear charges: $11+$ ( $\text{Na}^+$ ), $12+$ ( $\text{Mg}^{2+}$ ) and $13+$ ( $\text{Al}^{3+}$ ). Since the same $10$ electrons are attracted by an increasingly large nuclear charge, the effective nuclear charge experienced per electron increases from $\text{Na}^+$ to $\text{Al}^{3+}$ , pulling the electron shell in more tightly. This is why the ionic radius decreases steadily: $0.095\ \text{nm} \to 0.065\ \text{nm} \to 0.050\ \text{nm}$ .

Part (c): Melting point trend and structure

(i) Na → Al

Sodium, magnesium and aluminium are all giant metallic lattices: positive metal ions arranged in a regular lattice, surrounded by a “sea” of delocalised outer-shell electrons. Going from Na to Al:

the number of delocalised electrons contributed by each atom increases ( $1 \to 2 \to 3$ ), and
the ionic radius decreases, so the ions pack closer to the delocalised electrons.

Both effects strengthen the electrostatic attraction between the metal cations and the delocalised electrons, so more energy is needed to break the metallic bonding, hence the melting point rises steadily from Na ( $98\,^\circ\text{C}$ ) to Al ( $660\,^\circ\text{C}$ ).

(ii) Silicon’s very high melting point

Silicon has a giant covalent (macromolecular) lattice, essentially the same structure as diamond: every Si atom forms four strong covalent bonds to neighbouring Si atoms, extending throughout the entire crystal. There are no individual molecules, the whole lattice behaves as one giant network of strong bonds. Melting silicon means breaking a vast number of these strong covalent bonds, which requires far more energy than disrupting even the strongest metallic lattice (Al), so silicon’s melting point ( $1414\,^\circ\text{C}$ ) is by far the highest in the period.

(iii) The sharp fall after silicon

$\text{P}_4$ , $\text{S}_8$ , $\text{Cl}_2$ and Ar are all simple molecular structures (Ar exists as separate atoms). Within each $\text{P}_4$ or $\text{S}_8$ molecule (and within each $\text{Cl}_2$ molecule), the atoms are held together by strong covalent bonds, but these bonds do not need to break on melting. Only the much weaker van der Waals (London dispersion) forces between separate molecules or atoms need to be overcome to melt the solid. Since van der Waals forces are far weaker than the covalent bonds within silicon’s lattice or the metallic bonding in Na/Mg/Al, only a small amount of energy is needed, so the melting point falls sharply after silicon and stays low all the way to argon (with small variations between $\text{P}_4$ , $\text{S}_8$ , $\text{Cl}_2$ and Ar reflecting differences in molecule/atom size and number of electrons).

Part (d): Electrical conductivity trend

Electrical conductivity requires mobile charge carriers, either delocalised electrons or free ions.

Na, Mg and Al are giant metallic lattices with an increasing number of delocalised outer-shell electrons per atom ( $1 \to 2 \to 3$ ), so conductivity rises across these three metals, reaching its highest value at aluminium.

Silicon’s giant covalent lattice holds every outer electron in a localised Si–Si covalent bond rather than as a delocalised “sea” of electrons, so silicon conducts only very weakly compared with the metals (it behaves as a semiconductor rather than a true conductor).

$\text{P}_4$ , $\text{S}_8$ , $\text{Cl}_2$ and Ar are simple molecular (or monatomic) structures. There are no ions present, and every electron is either held in a covalent bond within a molecule or present as a non-bonding pair. None are free to move through the structure. With no mobile charge carriers of any kind, none of these elements conducts electricity at all.

Final answers

(a) Atomic radius decreases Na→Al: increasing nuclear charge with constant shielding (same outer shell) increases effective nuclear charge.
(b)(i) $1s^2\,2s^2\,2p^6$ (neon configuration).
(b)(ii) Isoelectronic ions with increasing nuclear charge (11→12→13) pull the same 10 electrons in more tightly, so ionic radius decreases.
(c)(i) Increasing number of delocalised electrons and decreasing ionic radius strengthen metallic bonding from Na to Al.
(c)(ii) Silicon’s giant covalent lattice requires breaking many strong covalent bonds to melt, far more energy than any metallic lattice.
(c)(iii) $\text{P}_4$ , $\text{S}_8$ , $\text{Cl}_2$ , Ar are simple molecular/atomic; only weak van der Waals forces between molecules break on melting, needing little energy.
(d) Conductivity rises Na→Al (more delocalised electrons per atom); $\text{P}_4$ , $\text{S}_8$ , $\text{Cl}_2$ and Ar do not conduct at all because they are simple molecular/atomic, with no delocalised electrons or ions present.