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Review Sheet on Determining Term Symbols

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Review Sheet on Determining Term Symbols Review Sheet on Determining Term Symbols Term symbols for electronic configurations are useful not only to the spectroscopist but also to the inorganic chemist interested in understanding electronic and magnetic properties of molecules. We will concentrate on the method of Douglas and McDaniel (p. 26ff); however, you may find other treatments equal or superior to the D & M method. Term symbols are a shorthand method used to describe the energy, angular momentum, and spin multiplicity of an atom in any particular state. The general form is a given as Tj where T is a capital letter corresponding to the value of L (the angular momentum quantum number) and may be assigned as S, P, D, F, G, … for |L| = 0, 1, 2, 3, 4, … respectively. The superscript “a” is called the spin multiplicity and can be evaluated as a = 2S +1 where S is the spin quantum number. The subscript “j” is the numerical value of J, a new quantum number defined as: J = l +S, which corresponds to the total orbital and spin angular momentum of the system. The term symbol 3P is read as triplet – Pee state and indicates that there are two unpaired electrons in a state with maximum orbital angular momentum, L=1. The number of microstates (N) of a system corresponds to the total number of distinct arrangements for “e” number of electrons to be placed in “n” number of possible orbital positions. N = # of microstates = n!/(e!(n-e)!) For a set of p orbitals n = 6 since there are 2 positions in each orbital. Therefore, p2 (e = 2 and n = 6) so, N = 6!/(2!(6-2)!) = (6*5*4*3*2*1)/((2*1) * (4*3*2*1)) = 15 We can introduce the idea of a hole formalism which states that for many electronic properties one may consider systems with e or (n-e), the number of unoccupied sites or “holes”, to be equivalent. Thus, P4 (e = 4 and n = 6) N = 6!/(4!(6-4)!) = 15 so P4 (2 holes) and P2 (4 holes) give equivalent values of N. The same is true for all p, d, f, … systems such as d1/d9 or f2/f12. If one should consider an atom with multiple unoccupied orbitals, the total number of microstates will equal the product of the microstates for the individual orbitals. (N = π N*a; a = 1 to i where a = orbital index). For an s1p2 configuration N = 2 (from S1) * 15 (from p2) = 30. We can define component ML in two ways. First, ML = m1 + m2 + m3 + … where ml corresponds to the l angular momentum quantum number for each electron in a given orientation. Second, ML = L, L-1, …, 0, …, -L where L is the atomic orbital angular momentum. Thus, ML has 2L + 1 values. Similar definitions for MS, representing the spin of the system may be made with allowable values of MS = 2S+1 The total angular momentum quantum number, J, is given as J = L+S, L+S – 1, …, |L-S|. It is the absolute magnitude of J that determines the total angular momentum. Determination of Spectroscopic Terms 1. Determine total number of microstates 2. Determine the possible values of ML and MS 3. Determine the Pauli – allowed microstates 4. Set up a chart of microstates 5. Resolve this group of microstates into atomic states The following steps will be illustrated for a d2 configuration. STEP 1 (Number of Microstates) N = n!/(e!(n-e)! = 10!/(2!*8!) = 45 STEP 2 (Values of ML and MS) For a d orbital, ml may range from +2 through –2 so there are 5 acceptable ml values. The maximum value of ML for 2 electrons is when m1 = +2 and when m2 = +2 or ML = 4 (this can only occur if MS = 0. Therefore, ML will range +4, +3, …, -4. The allowed values for MS are 1, 0, -1 since two electrons may be spin aligned (up or down) or paired. STEP 3 The Pauli allowed microstates can be drawn as shown on the following pages. Remember that certain combinations of quantum numbers are not allowed (e.g., ML = -4 and MS = +1 since this would require two parallel spins in the same orbital). ms= ms= +1/2 -1/2 ML MS ml +2 +1 0 -1 -2 Ms ml +2 +1 0 -1 -2 = = 3+1 ↑↑ -1 ↓↓ 2+1 ↑↑ -1 ↓↓ 1+1 ↑↑-1 ↓↓ 0+1 ↑↑-1 ↓↓ 1+1 ↑↑ -1 ↓↓ 0+1 ↑↑-1 ↓↓ -1 +1 ↑↑-1 ↓↓ -1 +1 ↑↑ -1 ↓↓ -2 +1 ↑↑-1 ↓↓ -3 +1 ↑↑ -1 ↓↓ By symmetry an equivalent pattern must be observed for ms = -1/2 and MS = -1. 10 microstates for MS = +1 10 microstates for MS = -1 ms= ms= +1/2 -1/2 ML MS ml +2 +1 0 -1 -2 ml +2 +1 0 -1 -2 = = 40 ↑↓ 30 ↑↓ 20 ↑↓ 10 ↑↓ 00 ↑↓ 30 ↑↓ 20 ↑↓ 10 ↑↓ 00 ↑↓ -1 0 ↑↓ 20 ↑↓ 10 ↑↓ 00 ↑↓ -1 0 ↑↓ -2 0 ↑↓ 10 ↑↓ 00 ↑↓ -1 0 ↑↓ -2 0 ↑↓ -3 0 ↑↓ 00 ↑↓ -1 0 ↑↓ -2 0 ↑↓ -3 0 ↑↓ -4 0 ↑↓ These are the 25 microstates for MS = 0 Therefore, 10 (MS = +1) + 10 (MS = -1) + 25 (MS = 0) = 45 microstates as required by Step 1 STEP 4. Chart Microstates On left column put range of ML values On bottom row put range of MS values Then enter the total number of microstates that have equivalent ML and MS values Table 1. +4 1 +3 121 +2 131 +1 242 ML 0 252 -1 242 -2 131 -3 121 -4 1 +1 0 -1 MS The sum of the values in this chart must total 45, the total number of microstates. STEP 5. Resolve Microstates We can immediately see that there is a microstate which must correspond to ML = -4 and MS = 0. This must be a G state since L = 4 and it must be an orbital singlet since 1 1 MS = 0. Therefore, the first term is G (We will assign J values later.). A G representation corresponds to a 9 x 1 matrix (ML x MS) since L can range from +4 through –4 (9 elements) and S can only equal 0 (1 element). Removing 9 microstates from Table 1 yields Table 2. Table 2. +3 1 11 +2 121 +1 232 ML 0 242 -1 232 -2 121 -3 111 +1 0 -1 MS Elements Removed: 1G (9 microstates) 1 (Notice that the elements removed for G came from the MS = 0 column, with one from each ML row as required.) Now go to the upper left corner (italicized and red) element. The ML and MS value for this element yield the next microstate representation. ML=3 therefore F, MS =1 therefore triplet, so the element is 3F. 3F is a 7 x 3 matrix = 21 microstates that are removed to give Table 3. Table 3. +2 1 +1 121 ML 0 131 -1 121 -2 1 +1 0 -1 MS Elements Removed: 1G, 3F (30 microstates) 1 Now got to the single element in row +2 (red) which gives D (ML = 2, MS = 0). This is a 5 x 1 matrix yielding Table 4. Table 4. +1 1 11 ML 0 121 -1 111 +1 0 -1 MS Elements Removed: 1G, 3F, 1D (35 microstates) 3 Upper left gives ML = 1, MS = 1 or a P which is a 3 x 3 matrix. This leaves table 5. Table 5. ML 0 1 0 MS Elements Removed: 1G, 3F, 1D, 3P (44 microstates) This table must correspond to 1S so the d2 configuration gives: 1G, 3F, 1D, 3P, 1S (45 microstates) Assigning J Values J = L + S: number of microstates = 2J + 1 1 1 For G ! G4 (L = 4, S = 0) 9 microstates 1 1 S ! S0 (L = 0, S = 0) 1 microstate 1 1 D ! D2 (L = 2, S = 0) 5 microstates 3 3 F ! F4 (L = 3, S = 1) 9 microstates 3 3 P ! P2 (L = 1, S = 1) 5 microstates Notice that the total number of microstates does not equal 45 at this point! All of the microstates associated with 1G, 1S, and 1D are accounted for, while 12 3F microstates and 4 3P microstates are missing. This is because we have only calculated the state of maximum multiplicity! Both the 3F and 3P states can be factored. We do this by lowering the J value by one until all microstates have been accounted for. Thus, 3 F4 (9 microstates) 3 F3 (7 microstates) 21 microstates 3 F2 (5 microstates) 3 P2 (5 microstates) 3 P1 (3 microstates) 9 microstates 3 P0 (1 microstate) We now have accounted for all 45 microstates: 1 1 1 3 3 3 3 3 3 G4, S0, D2, F4, F3, F2, P2, P1, P0 We must now arrange these in a predicted orbital energy scheme. Rules for Assigning Energies (Hund’s Rules) 1. Ground state will have the largest spin multiplicity i.e. 3P is lower in energy than 1P 2. Two states with the same spin multiplicity can be distinguished by L values. The state with the largest L is of lowest energy, i.e. 1D < 1S 3. After checking 1 and 2 above: If the subshell is less than 1/2 full, the lowest J corresponds to the lowest energy. If the subshell is greater than 1/2 full, the highest J corresponds to the lowest energy. 1S 1 S0 1S 1 D 1 D2 1G 1 G 1 G4 3P s2p6d2 1D 3 3 P P2 3 P1 3 3 F P0 3 F 3 F4 3 F3 3 Actual F2 3 Predicted ground state is F2. CAUTION! The energy difference between levels can be less than or equal to the splitting of states.
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