Example of the use of Tanabe-Sugano Diagrams
## Example of the use of Tanabe-Sugano Diagrams

For the use of Tanabe-Sugano diagrams we will be using Tables 17.1 and 17.2 (see the resources for Test 3). 10 Dq_{o} = *f* x *g*.
Let us consider the complex Co(NH_{3})_{6}^{2+}. The oxidation state of the cobalt is +2, so the the metal isconsidered a
d^{7}. To figure out 10 Dq_{o} (also known as delta octahedral), from Table 17.1 we multiply *f* from the ligand
column by *g* from the metal ion column. This gives 1.25 x 9000 = 11,250 cm^{-1} which is the size of 10 Dq_{o}.
The next step is to determine the reduced Racah parameter for the complex. The reduced Racah parameter is called beta.

beta=(*B*_{complex})/(B_{ free ion}) = 1 - *h*^{.}k

The quantities *h* and *k* can also be found in Table 17.1 for many ligands and metal centers. For the current example

beta=(*B*_{complex})/(B_{ free ion}) = 1 - *h*^{.}k = 1 - (1.4)(0.09) = 0.874

From this it easy to rearrange things to get *B*_{complex} and use the value of *B*_{free ion} for Co^{2+}
from Table 17.2

(beta)(*B*_{free ion}) = *B*_{complex} = (0.874)(971 cm^{-1}) = 849 cm^{-1}

To use a Tanabe-Sugano diagram, you mustdivide the value of 10 Dq_{o} by *B*_{complex}.

(10 Dq_{o})/*B*_{complex} = 11,250 cm^{-1}/849 cm^{-1>} = 13.25

This is the value that will be read on the x-axis of the Tanabe-Sugano diagram. Using the correct Tanabe-Sugano diagram
(d^{7} in this case) is critical. Looking at the Tanabe- Sugano diagram quickly reveals that
the term symbol for a free Co^{2+} ion is ^{4}F. Also looking at the Tanabe Sugano diagram, we notice that the value
of 13.25 is to the left of the point of inflection. This means that the complex Co(NH_{3})_{6}^{2+} is a
high spin complex (if the value was to the left of the inflection point, it would be a low spin complex). Spin allowed ttransitions from the ground state
will therefore all be from quadruplet to quadruplet. The allowed transitions are:

^{4}T_{1g} -----> ^{4}T_{2g}

^{4}T_{1g} -----> ^{4}T_{1g}

^{4}T_{1g} -----> ^{4}A_{2g}

Reading straight up from 13.25 on the x-axis until it crosses the line corresponding to the other quadruplet states will give us E/*B*_{complex} on the
y-axis.

^{4}T_{1g} -----> ^{4}T_{2g} E/*B*_{complex} =12.4

^{4}T_{1g} -----> ^{4}T_{1g} E/*B*_{complex} = 25.6

^{4}T_{1g} -----> ^{4}A_{2g} E/*B*_{complex} = 25.6

To get the energy of the transitions in cm^{-1}, each of these must be multiplied by *B*_{complex}

^{4}T_{1g} -----> ^{4}T_{2g} E/*B*_{complex} =12.4 ; 12.4 x 849 cm^{-1} = 10,528 cm^{-1}

^{4}T_{1g} -----> ^{4}T_{1g} E/*B*_{complex} = 25.6; 25.6 x 849 cm^{-1} = 21,734 cm^{-1}

^{4}T_{1g} -----> ^{4}A_{2g} E/*B*_{complex} = 25.6, 25.6 x 849 cm^{-1} = 21,734 cm^{-1}

The last step is to convert the wave number (reciprocal centimeters, cm^{-1}) to namometers

^{4}T_{1g} -----> ^{4}T_{2g} 10,528 cm^{-1}; 1/(10,528 cm^{-1}) = 9.50 x 10^{-5} cm; (9.50 x 10^{-5} cm)(10^{7} nm/cm) = 950 nm

^{4}T_{1g} -----> ^{4}T_{1g} 21,734 cm^{-1}; 1/(21,734 cm^{-1}) = 4.60 x 10^{-5} cm; (4.60 x 10^{-5} cm)(10^{7} nm/cm) = 460 nm

^{4}T_{1g} -----> ^{4}A_{2g} 21,734 cm^{-1}; 1/(21,734 cm^{-1}) = 4.60 x 10^{-5} cm; (4.60 x 10^{-5} cm)(10^{7} nm/cm) = 460 nm

All of these transitions are d-d transitions. The first transition at 950 nm is in the near IR just above the red portion of the visible spectrum.
The two transitions at 460 nm correspond to an absorbance of blue (very slightly shaded to green) light in the visible spectrum.