The complete reaction of 2.60 g of chromium metal with 50.00 mL of 1.200 M H2SO4 in the absence of air gave a blue solution and a colorless gas that was collected at 25°C and a pressure of 735 mm Hg. (e) When an excess of KCN is added to the solution, the color changes, and the paramagnetism of the solution
decreases. Explain.

Question

The complete reaction of 2.60 g of chromium metal with 50.00 mL of 1.200 M H2SO4 in the absence of air gave a blue solution and a colorless gas that was collected at 25°C and a pressure of 735 mm Hg. (e) When an excess of KCN is added to the solution, the color changes, and the paramagnetism of the solution
decreases. Explain.

decreases. Explain.

Accepted Answer

Welcome back everyone. Our next problem says the complete reaction of molybdenum metal with 1.0 molar H two. So four in the absence of air produced a blue solution in a colorless gas. Upon the addition of excess can 02, the solution changed color and its paramagnetism decreased. What is the reason for this observation? A this observation is due to the conversion of the complex mo h2o 6 with the two plus charge from a low spin complex with two unpaired electrons to a high spin complex with four unpaired electrons. By the addition of Xskno two B, this observation is due to the conversion of that same complex molybdenum with six waters from a high spin complex with two unpaired electrons to a low spin complex with four unpaired electrons. By the addition of Xskno two. See this observation is due to the conversion of minum with the six water ligands from a low spin complex with four unpaired electrons to a high spin complex with two unpaired electrons. By the addition of excess ko two or d. This observation is due to the conversion of molybdenum with six water ligands from a high spin complex with four unpaired electrons to a low spin complex with two unpaired electrons by the addition of xskno two. Well, we have a lot of words there, but we can use our answer choices to give us a few little shortcuts. We don't need to worry about what our reaction is. We're told what our initial product is of that reaction of molybdenum metal with H two. So four H two. So four being a strong acid, completely dissociates to form H 30 plus, which reacts with our molybdenum metal. It produces hydrogen gas, that would be the colorless gas. And this complex of molybdenum with six water ligands. So let's look at how many D electrons we have to know whether we're going to have two unpaired or four unpaired electrons. So molybdenum with are six waters, we're told we have an overall charge of positive two, water is a neutral ligand. And so therefore, molybdenum must be the source of the positive two charge. Lets look at our electron configuration, elemental and elliptic. It has an electron configuration of its nearest noble glasses krypton. So Kr and brackets five S two 44. So mo two plus has lost those two S electrons. It is krypton and brackets four D four. So we know we have four D electrons. Let's think about how they film those D orbitals. Well, the key is we'll need to think about do we have weak field or strong field splitting because we know our D orbitals, we'll fill in with three lower energy levels and to higher energy levels. Well, for that, we need to look at our ligand and water is a weak field ligand. So we have a small delta between these two energy levels. So we have a high spin complex. So as we fill in our d electrons, we fill in 123 on the bottom. And then we put our fourth electron on the top because again, it's just a small difference between those energy levels. So we started out with four unpaired electrons. So we can go ahead and eliminate some answer. Choices here. Choice A says that we will have a low spin complex which is incorrect. We'll have a high spin complex with two unpaired electrons, but we know we have four. So choice A is incorrect. B says we start out with high spin complex with two unpaired electrons. Well, we do have a high spin complex but we have four unpaired electrons. So choice B is incorrect. Choice C says low spin with four unpaired electrons, four is correct but not the low spin we have a high spin complex. So choice C is incorrect. So we've actually already eliminated three of our answer. Choices. Choice D says correctly, high spin complex with four impaired electrons. So on a test, I'm choosing B and moving on, but we'll be thorough and keep going here. So when we add our excess cano two, we would expect that the water ligands, I'll put a can or two would be replaced by nitrate by an 02 minus because that can or two will dissociate. And we know that no two minus is a strong field ligand. So in that case, we look at our de orbitals, we have a large delta when we have a strong field spudding there. So now my four D electrons from the molybdenum fill in 123 on the bottom. But the fourth one gets paired up down on the bottom with a larger energy level difference, it will not go into the top two energy levels. So now we just have two unpaired electrons. So that fits with their observation of paramagnetism decreasing because we now have fewer unpaired electrons. But what about the color chains? Well, we know that color is due to electrons being excited when they absorb wavelengths of light. And of course, here, when the electrons get excited, they go up a different amount. So with the water ligands, they absorb, they need to absorb less light to jump up to the higher energy level here. And with nitrate ligands, that field splitting is now larger, they absorb a different wavelength of light as they jump up to that elevated energy level. So that goes along with our answer choice here and choice d again, this observation is due to the conversion of molybdenum with those six water ligands from a high spin complex with four unpaired electrons to a low spin complex with two unpaired electrons by the addition of the excess kno two. So again, drop in unpaired electrons change its paramagnetism. And the difference in the crystal field splitting explains a difference in color as they absorb a different wavelength of light. See you in the next video.

Verified Solution

Key Concepts

Redox Reactions

Paramagnetism

Complex Ion Formation