Representing multi-electron systems Understanding the Absorption Electronic Spectra of Coordination Compounds at greater depth Representing multi-electron systems Chapter 20

Review of the Previous Lecture Discussed magnetism Used crystal field theory to explain electronic transitions

To understand d electron transitions in coordination compounds it is necessary to revisit the representation of multi-electron systems.

1. Quantum numbers Are a representation of an electronic situation. Now let’s consider a situation that we have previously ignored. Define the quantum numbers for an electron in the valence shell of the p-orbital of oxygen O: 1s22s22p4 n = 2 l = 1 ml = -1,0,+1 +1 -1 The fourth electron can go into any of the three p-orbitals but how do we distinguish the positioning of that e-?

1. Quantum numbers +1 -1 Need to introduce new quantum numbers. Are a representation of an electronic situation. Now let’s consider a situation that we have previously ignored. Define the quantum numbers for an electron in the valence shell of the p-orbital of oxygen O: 1s22s22p4 n = 2 l = 1 ml = -1,0,+1 +1 -1 Need to introduce new quantum numbers. Consider the interaction of electrons by coupling the magnetic fields generated by their spin or orbital motion.

1A. Quantum number L The maximum value of ML is equal to L. L : Total orbital angular momentum quantum number Related to the values of the quantum number l L parallels l ML : Resultant orbital magnetic quantum number ML = Σ ml = L, (L-1)…0…-(L-1), -L Possible values The maximum value of ML is equal to L.

1A. Quantum number L The value of L defines the atomic states.

1B. Quantum number S The maximum value of MS is equal to S. S : Total spin quantum number. The sum of the electron spin. MS : Resultant spin angular momentum magnetic quantum number MS = Σ ms = S, (S-1)…0…-(S-1), -S Possible Values The maximum value of MS is equal to S.

1C. Multiplicity or or Multiplicity = 2S + 1 Spin = 0 Multiplicity = 1 Singlet State Spin = 1/2 Multiplicity = 2 Doublet State Spin = 1 Multiplicity = 3 Triplet State Either no electrons or all electrons paired or or

1D. Hund’s 1st rule Hund’s rule of maximum multiplicity states that the ground state of an atom will be the one having the greatest multiplicity. Maximize the number of unpaired e-

2. Microstates and Russell-Saunders Term Symbols Now we will tackle how to more specifically define the electronic situation. Let’s focus on carbon. C: 1s22s22p2 Consider only the open shells. ml Degenerate orbitals +1 -1

2A. Microstates S = 1 2S + 1 = 3 Maximum multiplicity, most stable C: p2 There are different microstate arrangements of the p2 electrons. Let’s consider only 3 possibilities. S = 1 2S + 1 = 3 Maximum multiplicity, most stable microstate ml +1 -1 S = 0 2S + 1 = 1 Penalty: Coulombic repulsion (πc) ml +1 -1 Penality: Exchange Energy (πe) ml S = 0 2S + 1 = 1 +1 -1

2B. Russell-Saunders Term Symbols C: p2 S = 1 2S + 1 = 3 Maximum multiplicity, most stable microstate ml +1 -1 Let’s determine the Russell-Saunders Term Symbol to represent this most stable microstate, which would be the ground state. The ground state will always be identified by the occupancy of electrons in ml of highest values in a manner which respects Hund’s 1st rule. Calculate the value of L and then identify the atomic state L = │1(+1) + 1(0)│ = 1 P state Calculate the multiplicity Term symbol: 3P

2C. Hund’s 2nd rule If two states have the same multiplicity, the one with the higher value of L will lie lower in energy.

2D. Rigorous approach to determining all of the possible microstates of an electronic situation C: p2 The following is true for any p2 system. ml +1 -1 How many microstates? Basically how many way can you arrange the open shell valence electrons. Introduce terms: x = # of electrons = 2 l = quantum # for orbital = 1 for p orbital Nl = 2(2l + 1) = 6 Maximum # of electrons possible for the particular orbital set - In this case, you have 3 p orbitals

ml +1 -1 C: p2 The following is true for any p2 system. -1 How many microstates? # of microstates = Nl! = 6! = 6x5x4x3x2x1 x!(Nl – x)! 2!x4! 2x1 x 4x3x2x1

ml +1 -1 II. ML = Σ ml ml MS = Σ ms C: p2 The following is true for any p2 system. ml +1 -1 How many microstates? # of microstates = Nl! = 6! = 6x5x4x3x2x1 = 15 possible arrangements x!(Nl – x)! 2!x4! 2x1 x 4x3x2x1 II. ML = Σ ml ML 2 0 -2 1 1 0 0 1 1 0 0 -1 -1 -1 -1 +1   MS 0 0 0 1 0 1 0 0 -1 0 -1 1 0 0 ml MS = Σ ms

III. Group into microstates Arrange into columns of MS Arrange into rows of ML   MS +1 -1 +2 x xx ML xxx -2

One atomic state IV. Resolve the chart into appropriate atomic states An atomic state forms an array of microstates consisting of 2L+1 rows and 2S+1 columns   MS +1 -1 +2 x xx ML xxx -2 Look at the first row of ML. There is only one entry at Ms = 0. Recall highest ML values equals L Recall highest MS values equals S 2L + 1 rows by 2S + 1 columns 2(2)+1 rows by 2(0)+1 columns 5 by 1 One atomic state

One atomic state IV. Resolve the chart into appropriate atomic states An atomic state forms an array of microstates consisting of 2L+1 rows and 2S+1 columns   MS +1 -1 +2 x x x ML x x x -2 Look at the first row of ML. There is only one entry at Ms = 0. Recall highest ML values equals L Recall highest MS values equals S 2L + 1 rows by 2S + 1 columns 2(2)+1 rows by 2(0)+1 columns 5 by 1 One atomic state

One atomic state IV. Resolve the chart into appropriate atomic states An atomic state forms an array of microstates consisting of 2L+1 rows and 2S+1 columns   MS +1 -1 +2 x ML x x -2 Look at the next row of ML. All three Ms columns are occupied. Recall highest ML values equals L Recall highest MS values equals S 2L + 1 rows by 2S + 1 columns 2(1)+1 rows by 2(1)+1 columns 3 by 3 One atomic state

One atomic state IV. Resolve the chart into appropriate atomic states An atomic state forms an array of microstates consisting of 2L+1 rows and 2S+1 columns   MS +1 -1 +2 x ML x x -2 Look at the next row of ML. All three Ms columns are occupied. Recall highest ML values equals L Recall highest MS values equals S 2L + 1 rows by 2S + 1 columns 2(1)+1 rows by 2(1)+1 columns 3 by 3 One atomic state

One atomic state IV. Resolve the chart into appropriate atomic states An atomic state forms an array of microstates consisting of 2L+1 rows and 2S+1 columns   MS +1 -1 +2 ML x -2 Look at the next row of ML. Only the Ms = 0 columns is occupied. Recall highest ML values equals L Recall highest MS values equals S 2L + 1 rows by 2S + 1 columns 2(0)+1 rows by 2(0)+1 columns 1 by 1 One atomic state

One atomic state IV. Resolve the chart into appropriate atomic states An atomic state forms an array of microstates consisting of 2L+1 rows and 2S+1 columns   MS +1 -1 +2 ML x -2 Look at the next row of ML. Only the Ms = 0 columns is occupied. Recall highest ML values equals L Recall highest MS values equals S 2L + 1 rows by 2S + 1 columns 2(0)+1 rows by 2(0)+1 columns 1 by 1 One atomic state

MS = 0; S = 0; Multiplicity 2S + 1 = 1 1D V. Determine the atomic state Atomic State ML = +2; L = 2 MS = 0; S = 0; Multiplicity 2S + 1 = 1 1D ML = +1; L = 1 MS = +1; S = 1; Multiplicity 2S + 1 = 3 3P ML = 0; L = 0 MS = 0; S = 0; Multiplicity 2S + 1 = 1 1S

MS = 0; S = 0; Multiplicity 2S + 1 = 1 1D V. Determine the atomic state Atomic State ML = +2; L = 2 MS = 0; S = 0; Multiplicity 2S + 1 = 1 1D How to arrange these by energy? Recall Hund’s 1st and 2nd rules ML = +1; L = 1 MS = +1; S = 1; Multiplicity 2S + 1 = 3 3P ML = 0; L = 0 MS = 0; S = 0; Multiplicity 2S + 1 = 1 1S

VI. Rank the atomic states by energy 1D 3P