As shown in Figure 23.26, the d-d transition of [Ti(H₂O)₆]³⁺ produces an absorption maximum at a wavelength of about 500 nm .

a. What is the magnitude of ∆ for [Ti(H₂O)₆]³⁺ in kJ/mol?

b. How would the magnitude of ∆ change if the H₂O ligands in [Ti(H₂O)₆]]³⁺ were placed with NH₃ ligands?

Video transcript

All right, hello everyone. So this question says that the hypothetical complex A H 206 2 positive has ad D transition that produces an absorption maximum at a wavelength of about 550 nanometers. Part one is asking what is the value for or rather what is the value of delta four A H2O 62 positive and kilo joules per mole. And part two is asking how would the magnitude of delta change if the h2o ligands in A H2O 62 positive were replaced with no two negative? All right. So here recall first and foremost that delta represents the crystal field splitting energy and is the difference in energy between the higher level and lower level orbitals. And so first, let's go ahead and calculate the value for delta. Now, here we are given information about a photon. And so we have to recall the formula to find the energy of that photo. Here, delta E is equal to HC divided by lambda H is planks constant C is the speed of light and lambda is the wavelength in meters. So let's go ahead and substitute these values H which is planks constant is equal to 6.626 multiplied by 10 to the negative 34th power dual seconds C is the speed of light. That's 3.00 multiplied by 10 to the eighth meters per second. And all of this is divided by LAMBDA which is the wavelength provided in meters. And so the photon has a wavelength of 550 multiplied by 10 to the negative ninth meters. thats for the photo. So now evaluating this expression for the energy of the photon that's going to be equal to 3.614182 multiplied by 10th, the negative 19th power joules per photon. However, here we don't have the correct units just yet because our answer is supposed to be reported in the units of kilo jules per bolt. Because of this, we're going to use Avogadro's number to convert from photons tools. And we also have to go from jewels to kilo joules. So let me go ahead and rewrite the answer here for delta E. That's 3.614182 multiplied by 10th of the negative 19th power joules per photon. I am going to use Avo Gajos number as a conversion factor to cancel out photons. This means that I have Ava Garos number thats six point 022 multiplied by 10 to the 23rd power vs in the numerator of my conversion factor and one mole of photons in the denominator, this cancels out photons and leaves me with units of jules per m. So now to go from jewels to kilo jewels, I'm also going to go ahead and divide this quantity by 1000 since 1000 joules equals one kilojoule. And so now after evaluating this expression and also rounding my answer to two significant figures, the value for delta of the hypothetical complex is equal to 2.2 multiplied by 10 to the second power ketel joules per mole. And that would be our answer for part one, which I'm going to rewrite on the top of the screen here. So now lets go ahead and talk about part two. Part two is asking how would the magnitude of delta change if the H2O ligands of the hypothetical complex were replaced with no two negative? And so for this part, recall that the strength of the ligands themselves and the ability to increase the energy gap decrease in the following order, just scroll down here to open up some space. But the order is as follows. The ligand that causes the largest value for delta is cyanide or C and negative. This is followed by I know two negative, followed by en and then NH three followed by H2O followed by fluoride and then chloride bromide. And lastly iodide is the ligand that results in the smallest value for delta. Because of this, we can see that an 02 negative is stronger than water as a ligand and therefore has a greater ability to increase the value of delta. This means that replacing H2O with no two negative is going to increase the magnitude of delta because no two negative creates a stronger field ligand and there you have it. So just to go ahead and recap our answers. For part one, the value for delta was 2.2 multiplied by 10th, the second power kilo joules per mole. And as for part two, it will increase, referring to the value of delta. And so with that being said, thank you so very much for watching. And I hope you found this helpful.