Chemistry · Ch 7 — Modern Periodic Table

Periodic Trends in Valency

7.5.3.1

Periodic Trends in Valency

For MAIN-GROUP elements specifically, valency is usually equal either to the number of valence (outermost-shell) electrons the atom has, or to the DIFFERENCE between 8 and that number of valence electrons — whichever gives the value actually observed in real compounds (Table 7.6 illustrates both possibilities using hydride and oxide formulas as concrete examples). For instance, group 15 elements (5 valence electrons) show a valency of 3 in hydrides like NH3 (using the '8 minus 5' rule) but a valency of 5 in oxides like N2O5 (using the 'equal to valence electrons' rule directly); group 16 similarly shows both 2 and 6, and group 17 shows both 1 and 7. As Table 7.6 makes clear, valency STAYS THE SAME all the way down any one group (since the general outer electronic configuration, and hence the valence-electron count, does …

Table 7.6Periodic trends in valency of main group elements (Table 7.6)

Table 7.6 — for groups 1, 2 and 13-18: general outer configuration, number of valence electrons, valency, an example hydride formula, and an example oxide formula, illustrated with period-2, period-3 and period-4 elements:

Group 1 ( $ns^1$ , 1 valence e⁻, valency 1): hydrides LiH / NaH / KH; oxides Li2O / Na2O / K2O.

Group 2 ( $ns^2$ , 2 valence e⁻, valency 2): hydrides BeH2 / MgH2 / CaH2; oxides BeO / MgO / CaO.

Group 13 ( $ns^2np^1$ , 3 valence e⁻, valency 3): hydrides B2H6 / AlH3 / GaH3; oxides B2O3 / Al2O3 / Ga2O3.

Group 14 ( $ns^2np^2$ , 4 valence e⁻, valency 4): hydrides CH4 / SiH4 / GeH4; oxides CO2 / SiO2 / GeO2.

Group 15 ( $ns^2np^3$ , 5 valence e⁻, valency 3 and 5): hydrides NH3 / PH3 / AsH3; oxides N2O3 and N2O5 / P4O6 and P2O5 / As2O3 and As2O5.

Group 16 ( $ns^2np^4$ , 6 valence e⁻, valency 2 and 6): hydrides H2O / H2S / H2Se; oxides OF2 (fluorine's oxide, listed here since O itself does not form an 'oxide') / SO2 and SO3 / SeO2 and SeO3.

Group 17 ( $ns^2np^5$ , 7 valence e⁻, valency 1 and 7): hydrides HF / HCl / HBr; oxides Cl2O7 / Br2O.

Group 18 ( $ns^2np^6$ , 8 valence e⁻, valency 0): no simple hydride or oxide listed. …

Misc Problem 7.8Predicting empirical formulas of Ge-S and Ge-Br compounds from valency

Worked out. Worked example: Ge, S and Br belong to groups 14, 16 and 17 respectively. Their outer configurations and valencies: Ge ( $ns^2np^2$ , 4 valence e⁻, valency 4); S ( $ns^2np^4$ , 6 valence e⁻, valency 8-6=2); Br ( $ns^2np^5$ , 7 valence e⁻, valency 8-7=1). (i) Ge and S: S is more electronegative than Ge, so by cross-multiplying valencies (Ge valency 4 with S, S valency 2 with Ge) the formula is Ge2S4, which reduces to the empirical formula GeS2. (ii) Ge and Br: Br is more electronegative than Ge; cross-multiplying valencies (Ge valency 4, Br valency 1) gives GeBr4 directly as the …

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