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Enduring Understanding 1.C: Periodicity

  • Atoms display periodicity, or repeating periodic trends. This can be explained by the electronic structure of the atom - the arrangement of its electrons into shells and subshells.
  • Atoms in the same group in the periodic table will similar electronic structures and will tend to have similar chemical properties. Examples:
  • The alkali metals sodium [Na] and potassium [K] have similar chemical properties. Their electronic structures are similar, both having a single 's' valence electron:
  • Na: [Ne] 3s1
  • K: [Ar] 4s1
  • The halogens fluorine [F] and chlorine [Cl] both have seven valence electrons, one away from a filled shell:
  • F: [He] 2s2 2p5
  • Cl: [Ne] 3s2 3p5
  • The electronic structure of an atom can (usually) be deduced from its position in the periodic table. Its chemical properties can then be deduced from its electronic structure.
  • Atoms with a few valence electrons in their outer shell will tend to lose those electrons to form cations. E.g. Na (electron structure [Ne] 3s1) → Na+ (e.s. [Ne])
  • Atoms with nearly filled outer shells will tend to gain electrons and form anions.
    E.g. F (e.s. [He] 2s2 2p5) → F- (e.s. [He] 2s2 2p6, a filled shell)

  • These properties of alkali metals (tending to lose an electron) and halogens (tending to gain an electron) can be explained by their differing ionization energies and electron affinities, which result from the different effective nuclear charges electrons in the different atoms feel.
  • In alkali metals, the single valence electron is shielded from most of the nuclear charge by the lower energy core electrons in lower shells. It feels very little effective nuclear charge, and therefore has low ionization energy and the electron can easily be lost to form a cation.
  • Halogens, by contrast, have a higher nuclear charge, which is less well shielded by all the valence electrons. The valence electrons therefore feel a much higher effective nuclear charge, resulting in a high ionization energy and electron affinity, causing these atoms to tend to gain electrons and form anions.

  • Because effective nuclear charge increases across a row in the periodic table, atomic radius decreases across a row (i.e. N > O > F). In ions with the same electron configuration, ones with the lower total nuclear charge will be larger.
    • (i.e. Cl- > Ar > K+, all 1s22s22p63s23p6)

  • (Very Brief and simplified) overview of chemical characteristics of groups:

  • Group 1 (Alkali Metals). Li, Na, K...
  • Ionic charge +1
  • Oxide formula: M2O; Halide formula: MX (salt)
  • Group 2 (Alkali Earth Metals). Ca, Mg, Ba...
  • Ionic charge +2
  • Oxide formula: MO; Halide formula: MX2 (salt)
  • Group 13. Al, Ga
  • Ionic charge +3
  • Oxide formula: M2O3. Halide formula MX3 (salt)
  • Group 14. C, Si, Sn, Pb
  • Pb, Sn Ionic charge +2 and +4.
  • C and Si tend to form covalent bonds.
  • Oxide formula: MO2; Halide formula MX2, MX4 (covalent)
  • Group 15 (Pnictogens). N, P, As...
  • Ionic Charge: -3 (Bi is +3)
  • Oxide formula: M2O5. Halide formula MX3, MX5 (covalent)
  • Group 16 (Chalcogens). O, S...
  • Ionic Charge -2
  • Oxide formula MO3. Halide formula MX2, MX6 (covalent)
  • Group 17 (Halogens). F, Cl, Br...
  • Ionic Charge -1
  • Oxide formula: M2O7; Halide formula MX (covalent)
  • Group 18 (Noble Gases) He, Ne, Ar, Kr...
  • Generally inert, do not form ions.

  • Remember that analogous compounds from the same group often can have similar properties and uses. For example, SiO2 and SnO2 can both be used as ceramic materials.
  • Sample Questions:
  • What is the most likely valence electron configuration of an element M that would form a chloride salt with the formula MCl?
  • A) 1s2
  • B) 1s2 2s2 2p2
  • C) 1s2 2s2 2p6 3s1
  • Answer (C), 1s2 2s2 2p6 3s1. To form a monochloride salt, an element should have an atomic electron configuration with a filled shell and a single valence electron.
  • Which element would have the highest second ionization energy?
  • A) Mg
  • B) Ca
  • C) Na
  • Answer: (C), Na. Sodium is an alkali metal, and removing a second electron from Na involves removing a core '2p' electron. The second electron removed from Ca and Mg, both alkali earth metals, would still be removing a second 's' valence electron, which would require a lower ionization energy than removing core electrons.



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