A Data Booklet may be used.
W, X, Y and Z are four consecutive elements in Period 3, but not necessarily in the order presented.
Which of the following is the correct sequence of the four elements in order of increasing atomic number?
Source: NYJC 2018 Prelim P1 Q14, via EJC 2022 The Periodic Table Topical Revision, Q5
CommonFour elements with atomic numbers from 11 to 17 inclusive have the following properties.
Which of the following is the correct sequence of the four elements in order of increasing atomic number?
Source: CJC 2019 Prelim P1 Q14, via EJC 2022 The Periodic Table Topical Revision, Q7
Which statement does not describe the chemical properties of Period 3 compounds?
Source: NJC 2019 Prelim P1 Q15, via EJC 2022 The Periodic Table Topical Revision, Q11
Source: 2024 DHS J2 Prelim P3 Q2(a)
Elements E and F are Period 3 elements.
Source: SRJC 2011 P2 Q5(a) and (b) (modified), via EJC 2022 JC1 The Periodic Table Tutorial, Q2
Source: VJC 2016 Prelim P3, via EJC 2022 JC1 The Periodic Table Tutorial, Q8
Hydrogen halides can be unstable to heat.
| gas in glass jar | observations |
|---|---|
| hydrogen chloride | no observable change |
| hydrogen bromide | reddish brown vapour formed slowly |
| hydrogen iodide | purple fumes evolved vigorously |
| hydrogen astatide | ? |
Source: ACJC 2018 Prelim P3 (modified), via EJC 2022 JC1 The Periodic Table Tutorial, Q9
Source: 2023 SAJC J2 Prelim P2 Q2(a)(i)
This question is about phosphorus.
Source: 2024 CJC J2 Prelim P2 Q2(a) and (b)
ChallengeMagnesium nitrate and barium nitrate decompose similarly on heating.
Source: RI(JC) H2 Chemistry Tutorial 22 (Group II), Q4(a) and (b)
CY, X, W, Z[1]
The chloride of W dissolves in water and turns blue litmus red: W is not Na (NaCl is neutral).
X is a good conductor of electricity but insoluble in water: X can be Mg or Al. Since Y is Mg, X must be Al.
The oxide of Y has the highest melting point: MgO, so Y is Mg. Z has the highest first ionisation energy and largest ionic radius: Z is P.
Since the four elements in Period 3 must be consecutive, W must be Si. Hence the order is Y, X, W, Z (Mg, Al, Si, P).
CG, F, E, H[1]
MgCl2, AlCl3, SiCl4 and PCl5 dissolve in water to give acidic solutions and hence turn blue litmus red, so E could be Mg, Al, Si or P.
Only Al2O3 is amphoteric, so F is Al. Across a period, the electronegativity increases, so G is the first element among E, F, G and H. Chlorine has the smallest atomic radius, so H is chlorine.
DThe acidity of Period 3 chlorides in water decreases from NaCl to PCl5.[1]
Statement D is wrong. The acidity of Period 3 chlorides in water increases from NaCl to PCl5, and the pH of the resultant solution decreases from pH 7 to pH 1.
NaCl dissolves readily in water to form hydrated ions in a neutral solution. NaCl(s) → Na+(aq) + Cl−(aq), pH = 7[1]
AlCl3 dissolves in water with partial hydrolysis to give an acidic solution. AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl−(aq), [Al(H2O)6]3+(aq) + H2O(l) ⇌ [Al(H2O)5(OH)]2+(aq) + H3O+(aq), pH = 3[1]
PCl5 undergoes complete hydrolysis in water to form a strongly acidic solution. PCl5(s) + 4H2O(l) → H3PO4(aq) + 5HCl(aq), pH = 2[1]
As E has a low melting point, E has a simple covalent structure, hence E could be sulfur or phosphorus. Sulfur is a yellow solid, and the two oxides that sulfur forms (SO2 and SO3) do not have an empirical formula of E2O5, so E is not sulfur.
E is phosphorus, as one of its oxides is P4O10, with an empirical formula of E2O5.[1]
P4(s) + 5O2(g) → P4O10(s)[1]
P4O10(s) + 12NaOH(aq) → 4Na3PO4(aq) + 6H2O(l)[1]
P4O10(s) + 6H2O(l) → 4H3PO4(aq)[1]
Resulting pH: 2[1]
MgO is an ionic oxide and is basic, so it can react with acids. P4O10 is a covalent oxide and is acidic, so it can react with bases. The oxide of F is Al2O3, which is an amphoteric oxide, so it can react with both acids and bases.[1]
As Al3+ has a high charge and small ionic radius, it has a high charge density. Al3+ polarises the O2− electron cloud to a large extent, resulting in significant covalent character.[1]
As Al2O3 is an ionic compound with significant covalent character, it possesses the properties of both a metallic and a non-metallic oxide, making it amphoteric.
Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l)[1]
Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2Na[Al(OH)4](aq)[1]
From the Data Booklet: F2 + 2e− ⇌ 2F−, E⦵ = +2.87 V; Cl2 + 2e− ⇌ 2Cl−, E⦵ = +1.36 V; I2 + 2e− ⇌ 2I−, E⦵ = +0.54 V
As E⦵(X2/X−) is less positive down the group, oxidising power: F2 > Cl2 > I2.[1]
F2 is the strongest oxidising agent and oxidises Cr to CrF4, where the oxidation state of Cr increases from 0 to +4. Cl2 is less oxidising than F2 and oxidises Cr to CrCl3, where the oxidation state of Cr increases from 0 to +3.[1]
I2 is the weakest oxidising agent and oxidises Cr to CrI2, where the oxidation state of Cr increases from 0 to +2.[1]
Black fumes would evolve very vigorously for HAt, as it is the least thermally stable of the four HX.[1]
Decomposition of H–X depends on the energy supplied by the red hot wire to break the H–X bond. From Cl to At, the size of the X atom increases, the effectiveness of overlap between H and X atoms decreases, and the bond strength decreases: HAt < HI < HBr < HCl. So thermal stability: HAt < HI < HBr < HCl, and the vigour of formation of X2 fumes: HAt > HI > HBr > HCl.[1]
The nuclear charge of calcium is higher than that of potassium, while the shielding effect by the inner shells of electrons remains relatively constant. Hence, the effective nuclear charge is higher for calcium than for potassium.[1]
The attraction between the valence electrons and the nucleus is stronger in calcium, so its valence electrons are closer to the nucleus, resulting in a smaller atomic radius for calcium.[1]
Compared with Si, P has a smaller atomic radius and a greater nuclear charge, while the shielding effect by the same number of inner electrons is similar. Therefore, the nuclear attraction in P is larger, and hence the first ionisation energy of P is higher than that of Si.[1]
P: 1s2 2s2 2p6 3s2 3px1 3py1 3pz1 and S: 1s2 2s2 2p6 3s2 3px2 3py1 3pz1[1]
In S, the two electrons occupying the same 3p orbital (3px) give rise to inter-electronic repulsion. Thus, less energy is required to remove a paired 3p electron from S than to remove an unpaired 3p electron from P.[1]
Si has a giant molecular structure, with the Si atoms held together by an extensive network of strong covalent bonds. However, P4 and S8 have simple molecular structures, with weak instantaneous dipole–induced dipole interactions between the molecules, which require less energy to overcome. Thus P4 (and S8) have lower melting points than Si.[1]
P4 has fewer electrons than S8, hence its instantaneous dipole–induced dipole interactions are weaker and require less energy to overcome. Thus P4 has a lower melting point than S8.[1]
Mg(NO3)2 decomposes at a lower temperature.[1]
Mg2+ has a much smaller ionic radius than Ba2+ but the same charge, so Mg2+ has a higher charge density and hence greater polarising power than Ba2+.[1]
Mg2+ distorts the electron cloud of the NO3− anion to a greater extent than Ba2+.[1]
Hence, the N–O bond in Mg(NO3)2 is more easily broken, resulting in Mg(NO3)2 decomposing at a lower temperature.[1]