Exercise practice: Classification of Elements and Periodicity in Properties

4.1 How the exercises are grouped

The chapter prints exercises 3.1 to 3.40. The exercise practice page groups them by skill and works a representative question from each group.

4.2 Group A: position and classification

Exercise 3.4 asks why the sixth period holds 32 elements. The period fills 6s, 4f, 5d and 6p orbitals. Two electrons go into 6s, fourteen into 4f, ten into 5d and six into 6p, and the total is 32. The seventh period is also 32 for the same reason.

Exercise 3.5 locates Z = 114. The nearest noble gas before it is radon at Z = 86, so the element falls in period 7. Counting from group 1 of that period, it reaches group 14, which makes it a p-block element below lead. Exercise 3.6 asks for the third period and seventeenth group: that is chlorine, with Z = 17.

Exercise 3.7 asks which elements were named after two laboratories. Lawrencium, Z = 103, honours the Lawrence Berkeley Laboratory, and seaborgium, Z = 106, was named by Glenn Seaborg's group. Exercise 3.8 explains why a group shows similar chemistry: elements of one group share the same outer electronic configuration, so their valence electrons behave in the same way.

4.3 Group B: radii and isoelectronic species

Exercise 3.11 defines isoelectronic species as species with the same number of electrons. F- is isoelectronic with Ne and with Na+ and Mg2+. Argon shares its 18 electrons with Cl-, K+ and Ca2+. Mg2+ shares ten electrons with F- and N3-, and Rb+ shares its electron count with Kr and Br-.

Exercise 3.12 arranges N3-, O2-, F-, Na+, Mg2+ and Al3- in increasing radius. All have ten electrons, so the ion with the highest nuclear charge is smallest. The atomic numbers are 7, 8, 9, 11, 12 and 13, so the order of increasing radius is Al3+ below Mg2+ below Na+ below F- below O2- below N3-.

Exercise 3.13 asks why cations are smaller and anions larger than their parent atoms. A cation has lost electrons, so the same nuclear charge now attracts fewer electrons and the cloud contracts. An anion has gained electrons, so electron repulsion grows and the cloud expands.

4.4 Group C: ionization enthalpy

Exercise 3.14 points out why the definition insists on an isolated gaseous atom in its ground state. Comparison between elements is fair only if every atom is free and unexcited, so the measured energy depends on the atom's structure rather than on its environment or its starting state.

Exercise 3.15 converts the ground state energy of hydrogen into an ionization enthalpy. The energy of one atom is 2.18 x 10^-18 J, and one mole contains the Avogadro number of atoms, so the energy per mole is 2.18 x 10^-18 x 6.022 x 10^23, which is 1.31 x 10^6 J mol^-1 or 1312 kJ mol^-1. The positive sign means energy must be supplied.

Exercise 3.16 gives the second period values, which rise from left to right with two dips, one just after beryllium and one just after nitrogen. Beryllium sits above boron because its electron comes from a filled 2s subshell, while boron loses a 2p electron that is more shielded. Nitrogen sits above oxygen because nitrogen's 2p subshell is half filled and therefore stable, while oxygen's paired electron is easier to remove.

Exercise 3.17 compares sodium and magnesium. Sodium's first value is lower because it loses a single 3s electron, while magnesium holds its 3s pair more tightly. After one electron has gone, the sodium ion has the neon configuration, so its second ionization enthalpy is very high; the magnesium ion still has a 3s electron and loses it more easily.

Exercise 3.19 takes the group 13 values 801, 577, 579, 558 and 589 kJ mol^-1 for B, Al, Ga, In and Tl. The trend is not smooth. Gallium's value is slightly higher than aluminium's because the d electrons of the preceding series shield the outer electron poorly, so the effective nuclear charge on gallium is a little larger than a simple size argument would predict. Thallium rises again because of the poor shielding by its filled f subshell.

4.5 Group D: electron gain enthalpy and character

Exercise 3.20 compares pairs. Between oxygen and fluorine, fluorine has the more negative value because oxygen's half-filled 2p subshell resists an extra electron. Between fluorine and chlorine, chlorine is more negative because fluorine's small 2p subshell causes strong repulsion for the incoming electron.

Exercise 3.21 asks about the second electron gain enthalpy of oxygen. It is positive, because a second electron must be pushed onto an already negative oxide ion and that costs energy.

Exercise 3.22 separates the two ideas. Electron gain enthalpy is an enthalpy change for adding an electron to a gaseous atom and has a unit of energy per mole. Electronegativity is a relative, unitless tendency of an atom in a bond to attract the shared electron pair.

Exercise 3.29 writes the general outer configurations. For the s-block the outer configuration is n s1 or n s2. For the p-block it is n s2 n p1 to n s2 n p6. For the d-block it is (n-1) d1 to (n-1) d10 with n s1 or n s2. For the f-block it is (n-2) f1 to (n-2) f14 with (n-1) d0 or d1 and n s2.

Exercise 3.30 assigns positions. The configuration n s2 n p4 with n = 3 is sulphur, period 3 and group 16. The configuration (n-1) d2 n s2 with n = 4 is titanium, period 4 and group 4. The configuration (n-2) f7 (n-1) d1 n s2 with n = 6 is gadolinium, a lanthanoid in period 6, placed in the f-block.

Exercise 3.38 ranks metallic character: potassium is the most metallic, then magnesium, then aluminium, then boron, because metallic character falls across the period and rises down the group. Exercise 3.39 ranks non-metallic character as fluorine above nitrogen above carbon above boron above silicon. Exercise 3.40 compares oxidizing power among F, Cl, O and N. Fluorine is the strongest oxidiser, chlorine comes next in acidic aqueous conditions, then oxygen, and nitrogen is not an oxidising agent in this comparison, so the order is F greater than Cl greater than O greater than N.