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Periodic Trends

Periodic Trends. Atomic Properties. Depend on: Number of protons – attractive nuclear force Number of electrons – shielding effect of electron shells Distance of electrons from nucleus – distance decreases nuclear force. Atomic Properties. Atomic radius. Metallic and covalent radii.

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Periodic Trends

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  1. Periodic Trends Barbara A. Gage PGCC CHM 1010

  2. Atomic Properties • Depend on: • Number of protons – attractive nuclear force • Number of electrons – shielding effect of electron shells • Distance of electrons from nucleus – distance decreases nuclear force Barbara A. Gage PGCC CHM 1010

  3. Atomic Properties • Atomic radius Metallic and covalent radii. Barbara A. Gage PGCC CHM 1010

  4. Atomic Properties Click on image above to go to the Periodic Trends excelet Barbara A. Gage PGCC CHM 1010

  5. Barbara A. Gage PGCC CHM 1010

  6. PROBLEM: Using the periodic table, rank each set of main group elements in order of decreasing atomic size: Ranking Elements by Atomic Size (a) Ca, Mg, Sr (b) K, Ga, Ca (c) Br, Rb, Kr (d) Sr, Ca, Rb Elements in the same group increase in size and you go down; elements decrease in size as you go across a period. SOLUTION: (a) Sr > Ca > Mg These elements are in Group 2A(2). (b) K > Ca > Ga These elements are in Period 4. (c) Rb > Br > Kr Rb has a higher energy level and is far to the left. Br is to the left of Kr. (d) Rb > Sr > Ca Ca is one energy level smaller than Rb and Sr. Rb is to the left of Sr. Barbara A. Gage PGCC CHM 1010

  7. Atomic Properties • Ionization energy energy required to remove an electron from a neutral, gaseous atom Cl (g)  Cl1+ + e- K (g)  K1+ + e- Barbara A. Gage PGCC CHM 1010

  8. First ionization energies of the main-group elements. Barbara A. Gage PGCC CHM 1010

  9. PROBLEM: Using the periodic table only, rank the elements in each of the following sets in order of decreasing IE1: Ranking Elements by First Ionization Energy (a) Kr, He, Ar (b) Sb, Te, Sn (c) K, Ca, Rb (d) I, Xe, Cs IE decreases as you proceed down in a group; IE increases as you go across a period. SOLUTION: (a) He > Ar > Kr Group 8A(18) - IE decreases down a group. (b) Te > Sb > Sn Period 5 elements - IE increases across a period. (c) Ca > K > Rb Ca is to the right of K; Rb is below K. (d) Xe > I > Cs I is to the left of Xe; Cs is furtther to the left and down one period. Barbara A. Gage PGCC CHM 1010

  10. Barbara A. Gage PGCC CHM 1010

  11. Barbara A. Gage PGCC CHM 1010

  12. Atomic Properties • Electron affinity energy change when an electron is added to a neutral atom to form a negative ion Cl + e-  Cl- K + e-  K1- Barbara A. Gage PGCC CHM 1010

  13. Electron affinities of the main-group elements. Barbara A. Gage PGCC CHM 1010

  14. Figure 8.15 and 8.16 Trends in four atomic properties. Barbara A. Gage PGCC CHM 1010

  15. Ionic Properties • Ionic radius radius of the ionic form of an atom Barbara A. Gage PGCC CHM 1010

  16. Ionic vs. atomic radii Cations Barbara A. Gage PGCC CHM 1010

  17. Ionic vs. atomic radii Anions Barbara A. Gage PGCC CHM 1010

  18. PROBLEM: Rank each set of ions in order of decreasing size, and explain your ranking: Ranking Ions by Size (a) Ca2+, Sr2+, Mg2+ (b) K+, S2-, Cl - (c) Au+, Au3+ Compare positions in the periodic table, formation of positive and negative ions and changes in size due to gain or loss of electrons. SOLUTION: (a) Sr2+ > Ca2+ > Mg2+ These are members of the same Group (2A/2) and therefore decrease in size going up the group. The ions are isoelectronic; S2- has the smallest Zeff and therefore is the largest while K+ is a cation with a large Zeff and is the smallest. (b) S2- > Cl - > K+ (c) Au+ > Au3+ The higher the + charge, the smaller the ion. Barbara A. Gage PGCC CHM 1010

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