There are three salts that contain complex ions of chromium...

There are three salts that contain complex ions of chromium and have the molecular formula $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$. Treating $0.27 \mathrm{~g}$ of the first salt with a strong dehydrating agent resulted in a mass loss of $0.036 \mathrm{~g}$. Treating $270 \mathrm{mg}$ of the second salt with the same dehydrating agent resulted in a mass loss of $18 \mathrm{mg}$. The third salt did not lose any mass when treated with the same dehydrating agent. Addition of excess aqueous silver nitrate to $100.0-\mathrm{mL}$ portions of $0.100 M$ solutions of each salt resulted in the formation of different masses of silver chloride; one solution yielded 1430 $\mathrm{mg} \mathrm{AgCl} ;$ another, $2870 \mathrm{mg} \mathrm{AgCl}$; the third, $4300 \mathrm{mg} \mathrm{AgCl}$. Two of the salts are green and one is violet. Suggest probable structural formulas for these salts, defending your answer on the basis of the preceding observations. State which salt is most likely to be violet. Would a study of the magnetic properties of the salts be helpful in determining the structural formulas? Explain.

There are three salts that contain complex ions of chromium and have the molecular formula $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$. Treating $0.27 \mathrm{~g}$ of the first salt with a strong dehydrating agent resulted in a mass loss of $0.036 \mathrm{~g}$. Treating $270 \mathrm{mg}$ of the second salt with the same dehydrating agent resulted in a mass loss of $18 \mathrm{mg}$. The third salt did not lose any mass when treated with the same dehydrating agent. Addition of excess aqueous silver nitrate to $100.0-\mathrm{mL}$ portions of $0.100 M$ solutions of each salt resulted in the formation of different masses of silver chloride; one solution yielded 1430 $\mathrm{mg} \mathrm{AgCl} ;$ another, $2870 \mathrm{mg} \mathrm{AgCl}$; the third, $4300 \mathrm{mg} \mathrm{AgCl}$. Two
of the salts are green and one is violet. Suggest probable structural formulas for these salts, defending your answer on the basis of the preceding observations. State which salt is most likely to be violet. Would a study of the magnetic properties of the salts be helpful in determining the structural formulas? Explain.

Key Concepts

Electronic Transitions and Color in Transition Metal Complexes

The different colors exhibited by coordination compounds, such as green or violet hues, arise from electronic transitions between d-orbitals influenced by the ligand field. Variations in ligand type, geometry, and charge distribution result in different energy gaps, which in turn affect the wavelengths of light absorbed and the color observed.

Magnetic Properties and Their Relation to Electronic Structure

The magnetic behavior of a coordination compound, determined by the number and arrangement of unpaired electrons in the d-orbitals, provides valuable insights into its electronic configuration and geometry. A study of magnetic properties can help confirm the oxidation state and coordination environment of the central metal ion, offering additional evidence for the proposed structural formulas.

Precipitation Analysis for Anion Determination

Using precipitation reactions, such as the reaction of chloride ions with aqueous silver nitrate to form insoluble silver chloride, is a quantitative method for determining the number of free (non-coordinated) anions in a coordination compound. This information is crucial for establishing the positions of ligands relative to the metal center.

Hydration and Dehydration in Complex Salts

Complex salts often incorporate water molecules that can either be directly coordinated to the metal center or exist as water of crystallization. The process of dehydration, where water is removed using a dehydrating agent, helps distinguish between coordinated and non?coordinated water molecules, providing clues about the coordination environment of the metal center.

Inner Sphere versus Outer Sphere Ligands

In many coordination complexes, some ligands are directly attached to the central metal ion (inner sphere), while other ions, typically counter ions needed to balance charge, are not bonded directly (outer sphere). Recognizing which chloride ions or water molecules are coordinated versus those that are free is essential for deducing the correct structural formulas of the salts.

Coordination Chemistry

This concept involves compounds where a central metal atom or ion is bonded to surrounding molecules or ions called ligands. In coordination compounds, the arrangement of these ligands and the oxidation state of the metal are key factors that determine the overall structure, reactivity, and properties of the complex, such as color and magnetic behavior.

Transcript

00:01 A marathon problem for the unit.

00:02 And we're considering three salts that contain complex ions of chromium and have a particular molecular formula that i've given here.

00:14 Hang on just a second.

00:19 And i'm not going to read all this because it'll just take more time, but i'll use this information as we're solving the problem and identify what i'm doing.

00:27 Okay.

00:29 So let's get going here.

00:33 First, our cr, cl2, or seal 3, 6h2o, has nine possible ligands, and six form octahedral complexes.

01:20 And now remember we've got this and this.

01:23 So different ways that we can put these together.

01:27 The three species that aren't in the complex our ion have to be counter ions or waters of hydration.

01:37 There's three species not in the complex.

01:40 In the complex complexes are either counter ions, that'll be the chloride or water of hydration.

01:59 It could be waters of hydration.

02:07 Okay, and we can figure out the counter ions.

02:19 We can calculate that with the silver nitrate information and the waters of hydration, we can calculate with dehydration data.

02:58 Okay.

03:03 Okay, so for compound one, let's take our moles of our crcl6, 6h2o.

03:25 That's going to be 270 .7 grams we were given times, excuse me, this will be equal 1 .0 times 10 to the minus third moles of the hydrate.

03:55 And that's going to be this right here, just to save a little writing.

04:02 And my moles of my water of hydration are going to be equal, i shouldn't say moles of that.

04:19 From my dehydration data, that's going to be equal to 0 .036 grams of h2o divided by 18 .02 grams per mole.

04:34 This is 2 .0 times 10 to the minus third moles of water of hydration.

04:45 So for my moles of my h2o of hydration, divided by my moles of hydrate or compound, i will get two times 10 to the minus three divided by one times, i should go 2 .0, and that will be 2 .0.

05:21 So that means that there's two waters of hydration.

05:38 So four h2os are in the complex.

05:47 Okay.

05:52 So this will be crh2o4.

06:07 I'm going to have dot 2h2o.

06:14 And that means that i will need to have, let's see, let's see here.

06:44 I'm going to have a cl.

06:52 That'll give me a two.

06:53 And then, oops, maybe not.

06:58 I have to have two more and then a cl here.

07:02 Okay, i made way too much of you.

07:06 Okay, and this is my counter ion.