1. { Chapter 8 Sections 6 and 7 By: Nader Alkhabbaz

2. Section 8.6: Periodic Trends in the size of the atoms and Effective nuclear charge  Main Ideas of Section  Trends in Atomic Radius  Effective Nuclear Charge  Atomic Radii and Transition Elements

3. Trends in Size of Atoms/Atomic Radius  Ways to define the size of an atom  Nonbonding Atomic Radius (also known as the van Der Waals radius)- represents the radius of an atom when it is not bonded to another atom  Bonding atomic radius (covalent radius)-defined as one half the distance between two of the atoms bonded together for nonmetals and one half the distance between two of the atoms next to each other in a crystal of the metal for metals  Atomic radius refers to a set of average bonding radii determined or based on measuring large numbers of elements and compounds  As you move down a column on the periodic table atomic radius increases  As you move right across a row on the periodic table atomic radius decreases (pg 335 for chart)

4. Trend in Atomic Radius Cont.  Atomic Radius increases as you go down a group  As you move down a column in the periodic table, the highest principal quantum number of the valence electrons increases  Atomic Radius decreases as you go from left to right across a period  The valence shell is held closer making it more difficult to remove the making the ion smaller as a result  The coulombic attraction b/w a nucleus and electron with increasing nuclear charge

5. Effective Nuclear Charge  Any one electron in a multielectron atom is experiencing both the positive attractive force of the nucleus and the negative repulsive force of the other electrons  Electrons in the outer shell are shielded from the charge/strength of the nucleus (screening effect)  The effective nuclear charge is defined as the average or net charge experienced by an electron  Effective nuclear charge=actual nuclear charge (Z) minus the charge shielded by other electrons  As you move across a row in the periodic table Z eff experienced by the outermost principal energy level increases resulting in stronger nucleus and valence electron attraction MEANING smaller radii  Z(eff)=Z-S *tip: Z is the number of protons in the nucleus*

6. Quick Example  Find the Effective nuclear charge for Lithium.  How to solve:  Li 1s2 2s1  3+ charge of nucleus  2- charge of the core electrons in 1s2  3-2=1+ (Effective nuclear charge)  Main Idea to understand: Core electrons efficiently shield electrons in the outermost principal energy level but electrons in the same energy level don’t contribute a lot to screening

7. Transition Elements and trends in their atomic radius  The size of the atomic radius increases as you go down a group  However the atomic radii stays roughly the same size as you go across each row  Due to the effective nuclear charge on the outermost principal energy level being nearly constant  Basically the number of outermost electrons stays constant and they experience a constant Z eff keeping the radius approx. same

8. Quick Examples  Choose the larger atom from the pairs.  C or Ge  Si or As  O or F

9. Section 8.7: Ions and their electron config., magnetic properties, radii and ionization energy  Electron Configuration of Cations in Ground State  **Cations form when the atom loses electrons from the valence shell**  Al atom= 1s2 2s2 2p6 3s2 3p1  Al (3+) ion= 1s2 2s2 2p6 since plus 3 charge gets rid of 3 electrons in the outermost shell  Magnetic Properties of Transition Metal Atoms and Ions  If an atom or ion contains unpaired electrons it is attracted by an extrenal magnetic field and this is called paramagnetism  Fe 3+ 1s2 2s2 2p6 3s2 3p6 4s0 3d5 (removed electrons from 4s orbital before 3d orbital since transition metal)  If an atom or ion in which all electrons are paired there will be no magnetic field and this is called diamagnetism  Al 3+ 1s2 2s2 2p6 Draw ORBITAL DIAGRAM  Since there are no unpaired electrons it is diamagnetic

10. Trends in Ionic Radii  Ion size increases as you go down a group because of higher valence shell  Cations are much smaller than their corresponding atoms  Less electrons means less repulsions among valence electrons  Anions are much larger than their corresponding atoms  Extra electrons increase repulsions among valence electrons  Cations are smaller than anions generally  A larger positive charge means a smaller cation  Isoelectronic series of ions-ions with the same number of electrons

11. Ionization Energy  The energy required to remove an electron from the atom or ion in the gaseous state  Similar to [positive (delta) H] endothermic process  Valence electrons are easiest to remove  Example: Just understand the concept  Na --> Na^(+) +1e^(-1) IE=496 kJ/mol  Na^(+)--> Na^(2+) +1e^(-1) IE=4560 kJ/mol  First ionization energy=energy required to remove electron from neutral atom  2 nd ionization energy= energy required to remove the second electron

12. Trends in First Ionization Energy  If the effective nuclear charge on the electron is larger then more energy is required to remove it  The farther the electron is from the nucleus means the less energy is required to remove it  For Main Group Elements:  *1 st Ionization energy decreases as you go down the group –due to farther valence electrons from the nucleus  *1 st Ionization energy increases as you go across a row due to increasing effective nuclear charge  Ex: Which has the higher first ionization energy? N or Si?

13. Exceptions to Trends in 1 st IE  Exceptions are b/w boron and beryllium  When you ionize boron you get a full sublevel, which requires less energy  When you ionize Be you must break up a full sublevel requiring extra energy  (Relationships can be seen by drawing orbital diagrams and realizing both of these principles)  Exceptions are b/w nitrogen and oxygen  When you ionize Nitrogen you must break up a half sublevel requiring more energy  When you ionize Oxygen you get a half full sublevel requiring less energy  (Draw orbital diagrams and see how these two points apply)

14. Trends in Second and Successive IE  Removing each successive electron will cost more energy  Valence electrons are closer to the nucleus making it harder to remove them  Causes shrinkage in size due to more protons than electrons present  *There is an increase in ionization energy for each successive valence electron (When you start to remove core electrons you get a larger IE)  [Pg 347 top right chart shows clear indication of this]

15. More Practice More practice in this section lies within questions 55-76 on page 359 and 360 With practice and understanding the underlying concepts this will become second nature