Very often, chemical reactivity can be thought of in terms of ionization energies.  For example, the lower an atom's ionization energy, the higher its reactivity.  For an atom X, its ionization energy, IE, can be defined as

X(g) -->  X+(g) + 1e-         ΔE = IE

Of course, ionization energies will depend on which atomic orbital an electron is removed.   The lowest ionization energy would correspond to removing an electron from the highest occupied orbital.   So given its importance, it is often useful to calculate an atom's ionization energy to predict or explain its chemical reactivity.  In this activity, you will calculate the ionization energy directly of a series of atoms by calculating explicitly the energy of X, and X+ and take energy differences:

IE = E(X+) − E(X)          (1)

The details of how electronic energies are calculated are beyond the scope of this activity.  In short, you will use the Hartree-Fock method with a 'DZP' atomic orbital basis.  For more accurate results, you may choose to use any method available in the QuantumChemistry package (i.e. DensityFunctional, MP2, Parametric2RDM).

Here we initialize the Maple packages we wish to use as well as define the experimental values of the quantities we wish to calculate for comparison.

In this section, we calculate the energies of the neutral and cation species for Group 1A, Group 7A, and Period 2 elements and use equations (1) and (2) to determine IEs.  We then plot the results along with experimental values to compare.    We will use Hartree Fock theory with a DZP basis, which allows us to calculate energies of Group 1A elements up to cesium and Group 7A elements up to iodine.  (For more accurate methods, see for example DensityFunctional, MP2, or Parametric2RDM. If desired, change AOmethod and AObasis in the Initialize subsection. Make sure the AObasis exists for the elements of interest.)

The following Maple input is divided into steps.  The first 3 steps are used to calculated the IE of the H atom.  Step 4 requires you to repeat Steps 1 - 3 to calculate IEs for the remaining elements in Group 1A.  Once you have completed the series, you can plot the results and then repeat the process for Group7A and Period 2.  For logistical reasons, we consider the following variables to be specified before each calculation:

Group1A series has n = 6 elements.  The index m labels which atom:  

   H (m = 1),  

   Li (m = 2),

   Na (m = 3),

   K (m = 4),

   Rb (m = 5), and

   Cs (m = 6).  

   Num_neutral = 1, Num_cation = 0.

Group7A series has n = 4 elements.  The index m labels which atom:  

   F (m = 1),  

   Cl (m = 2),

   Br (m = 3), and

   I (m = 4).  

   Num_neutral = 1,

   Num_cation = 2.

Period 2 series has n = 8 elements.  The index m labels which atom:  

   Li (m = 1),

   Be (m = 2),

   B (m = 3),

   C (m = 4),

   N (m = 5),

   O (m = 6),

   F (m = 7), and

   Ne (m = 8).  

   Num_neutral = varies according to ground electron configuration and Hund's Rule,

   Num_cation = varies according to ground electron configuration and Hund's Rule.

Step 1: Specify parameters

Step 2: Calculate energies of neutral atom and cation. Determine highest occupied atomic orbital.

Step 3: Calculate IE (in kJ/mol) using Equations (1) and (2) 

Step 4: Repeat steps 1 - 3 for each element in group or period

Step 5: Plot Ionization Energies and Compare

How do ionization energies compare with experimental values using Equations (1)?

	Answer
	We see that for the Hartree Fock method and the DZP basis, the trend in IE is correct, but as atomic number increases, the agreement decreases.

Repeat Steps 0 - 5 for Group 7A and Period 2.

How do ionization energies and trends compare with experimental values?

	Answer
	For both Group 7A and Period 2, the trend in calculated ionization energies compares well with experiment!