Name: Date: Blk: NOTES: PERIODIC TRENDS Examine your periodic table to answer these questions and fill-in-the-blanks. Use drawings to support your answers where needed: I. ATOMIC RADIUS (Size) Going from left to right a given period, the atomic number (# of protons) increases/decreases (circle one). Thus, the positive charge on the nucleus increases. This trend also brings an increase in the number of electrons surrounding the nucleus. 1. What would happen to the atomic radius as the number of electron shells increases? Why? Then, what should happen to the atomic radius going DOWN a group? The atomic radius increases down a group. The more electrons around the nucleus, the greater the volume needed to contain them (since electrons repel each other and can t easily be compacted ). Therefore, going down a group the atomic radius of the elements in the group should increase. 2. (a) What happens to the size of the positive nuclear charge felt by the electrons surrounding the nucleus in an element, going from left to right across a row of the table? Going across a row, the positive charge on the nucleus increases, and EACH electron experiences a greater attraction to the nucleus. (b) What effect would you expect this change in charge to have on the average distance between the nucleus and a given electron going across the table? This results in a smaller distance between the individual electrons and the nucleus. (c) What should happen to the radii of the elements, going from left to right ACROSS a row of the periodic table? This causes the atomic radius to decrease going across each period.
II. IONIZATION ENERGY Ionization Energy: energy required to REMOVE an electron from a neutral atom *The e - removed is the outermost and therefore the most easily removed 1. (a) What happens to the distance between the nucleus and outermost electrons going DOWN a chemical family? The distance between the nucleus and the outermost electrons increases. The greater the number of electrons, the larger the atomic radius. (b) What happens to the electrostatic attraction of the nucleus to an electron in the outermost shell going DOWN a family? The nucleus has a decreasing hold on the outer electrons because of the increasing nucleus-electron distance. (c) Thus, what happens to the ionization energy going DOWN a family? The ionization energy will decrease going down a family. The less hold the nucleus has on the outer electron, the easier it is to remove one of them. 2. (a) What happens to the distance between the nucleus and outermost electrons going ACROSS a period? (refer to above section on atomic radius!) The distance decreases as the increasing nuclear charge pulls the electrons closer. (b) What happens to the nuclear charge going ACROSS the period? The nuclear charge increases. (c) What happens to the electrostatic attraction of the nucleus to an electron in the outermost shell going ACROSS a period? The nucleus can hold more strongly to a given outer electron because of the greater charge and the smaller electron-nucleus distance. (d) Thus, what happens to the ionization energy going ACROSS a period? The ionization energy (energy required to remove and outer electron) will therefore increase going across a period. 3. What member of each of the following pairs should have a greater ionization energy? (a) Br or Cl (b) Al or Cl (c) Ne or Xe (d) Mg or Ba (e) F or Ne (f) Rb or I
Extension Questions: Answer the following questions based on the graph you ve made of ionization energy versus atomic number. 1. Why are the ionization energies for He, Ne and Ar so high? Electrons are being removed from full shells, which is very difficult to do. 2. Why do the ionization energies decrease going from He to Ne to Ar? The outermost electrons are farther from the nucleus, so that the attraction between the nucleus and outer electrons is decreased and less energy is required for electron removal. 3. Why is there a general increase in ionization energy going from Li to Ne? The outer electrons are drawn closer to the nucleus, causing an increased attraction between the larger nuclear charge and the outer most electrons. The increased attraction requires a greater energy to be applied before and electron can be removed. 4. Filled subshells and half-filled subshells have special stability which requires extra energy to be applied before electron removal can occur. This general statement is supported by the existence of the electron configuration exceptions found for Cu and CR. What experimental evidence exists in the graph to support this general statement? Be and Mg have filled s-subshells, so that their ionization energies are higher than those of the elements immediately before and after them. Similarly, N and P have half-filled p-subshells and their ionization energies are higher than those of the elements immediately before and after them. The filling of the p-subshells (Ne and Ar) is a special case of increased stability leading to increased ionization energy.
III. ELECTRONEGATIVITY Electronegativity: the tendency of an atom to attract electrons from a neighbouring atom. *The investigation of electronegativity focuses on the atoms Li, F and I. 1. This exercise compares Li and F: (a) Which atom is larger: Li or F? (refer to above section on atomic radius!) Li (b) Which atom has the stronger attraction to the outer electrons on a neighbouring atom based ONLY on atomic radius? WHY? F (its atomic radius is smaller, which means there is a smaller distance between the positive nucleus and the neighbouring electrons) (c) Which atom has the greater nuclear charge? F (greater number of protons) (d) Which atom can attract electrons from an adjacent atom most strongly based on BOTH size and nuclear charge? WHY? F (smaller atomic radius AND greater nuclear charge) (e) Summarize the above by filling in the blank below. IN GENERAL, when going from left to right across the periodic table the electronegativity of the atoms will INCREASE. 2. This exercise compares F and I: (a) Which atom is larger: F or I? I (more electron shells, larger radius) (b) Which atom has a stronger attraction to the outer electrons of another atom? WHY? F (less nuclear charge, but smaller radius still results in a stronger attraction) (e) Summarize the above by filling in the blank below. IN GENERAL, when going from going down a family of the periodic table the electronegativity of the atoms will DECREASE.
If an atom has HIGH electronegativity, it has a high/low (circle one) ionization energy. WHY? - If an atom has a high electronegativity, it strongly attracts electrons from a neighbouring atom and may completely remove an electron from the neighbouring atom. - This high attraction to a neighbouring atom s electrons also means that atoms with high electronegativity strongly attract their own valence electrons. - As a result, these valence electrons are difficult to remove and the atom has a high ionization energy. If an atom has LOW electronegativity, it has a high/low (circle one) ionization energy. WHY? - If an atom has a low electronegativity, it has little attraction to the electrons of a neighbouring atom and little tendency to remove electrons from a neighbour. - Such an atom also has a relatively small attraction to its own valence electrons. - Therefore, these valence electrons are relatively easy to remove and the atom has a low ionization energy.