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Structural isomers of complexes

Ionization, hydrate, linkage and coordination isomers: what moves in each, the measurement that tells them apart, and how to count linkage isomers.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to classify structural isomers of complexes, identify an isomer from its precipitation and heating data, and count the linkage isomers of a complex.

2. What you already have

You know from organic chemistry that isomers are compounds with the same formula and different structures, and that some differ in which atoms are bonded (structural isomers) while others differ only in arrangement in space (stereoisomers). From lesson 4 you can tell a complex's ligands from its counter ions and count what silver nitrate precipitates. This lesson uses exactly those counts to tell structural isomers of complexes apart.

3. Words for this lesson

4. One formula, several compounds

The empirical formula $\mathrm{CrCl_3\cdot 6H_2O}$ describes at least three different compounds, a violet one and two greens, and chemists in the 1890s could not explain that until Werner's coordination theory showed what differed. Each is chromium(III) with six ligands, but the six are shared differently between water and chloride, and whatever does not fit in the coordination sphere sits outside it. These are structural isomers: same atoms, different bonding. Complexes show four kinds.

Ionization isomers swap an anion between ligand and counter ion. $\mathrm{[Co(NH_3)_5Br]SO_4}$ has bromide bonded to cobalt and sulfate outside; $\mathrm{[Co(NH_3)_5SO_4]Br}$ has them the other way round. In water the first releases sulfate, which barium chloride precipitates, and the second releases bromide, which silver nitrate precipitates.

Hydrate isomers do the same with water. In the violet $\mathrm{[Cr(H_2O)_6]Cl_3}$ all six waters are ligands and all three chlorides are free. In the dark green $\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot 2H_2O}$ two chlorides are ligands, so only one is free, and two waters are pushed out into the crystal lattice. The silver chloride per formula unit, $3$, $2$ or $1$, tells the three isomers apart, and so does the water lost on gentle heating, which is only the lattice water.

Linkage isomers use an ambidentate ligand's other donor atom. Nitrite can bond to cobalt through nitrogen, giving the yellow-brown nitrito-N complex, or through an oxygen, giving a red nitrito-O complex with the same formula, $\mathrm{[Co(NH_3)_5(NO_2)]^{2+}}$. Thiocyanate can bond through sulfur or nitrogen. With $n$ identical ambidentate ligands, the linkage isomers differ in how many bond through each atom, from none to all, so there are $n + 1$ of them.

Coordination isomers exist only in salts where both ions are complexes: $\mathrm{[Co(NH_3)_6][Cr(CN)_6]}$ and $\mathrm{[Cr(NH_3)_6][Co(CN)_6]}$ trade their ligands between the two metals.

What makes these isomers separable at all is that chromium(III) and cobalt(III) complexes are inert: their ligands exchange slowly, over hours or days, so each isomer keeps its identity long enough to be crystallized and measured.

5. The three chromium(III) chloride hydrates

The rows follow one rule: each chloride that moves into the coordination sphere pushes one water out into the lattice and removes one free chloride and one particle. The color changes because the ligands around chromium change, and the crystal-field lessons explain why chloride ligands shift the color from violet toward green. Three measurements confirm which isomer a sample is: silver chloride, water lost on gentle heating, and conductivity or freezing-point depression.

6. Telling ionization isomers apart with two reagents

Silver nitrate precipitates chloride, bromide and iodide that are free in solution; barium chloride precipitates free sulfate as white barium sulfate. A ligand is invisible to both. So for the pair $\mathrm{[Co(NH_3)_5Br]SO_4}$ and $\mathrm{[Co(NH_3)_5SO_4]Br}$:

• the first gives a precipitate with barium chloride and none with silver nitrate;
• the second gives a cream silver bromide precipitate and none with barium chloride.

The two are also different colors, red-violet and red, because a bromide ligand and a sulfate ligand affect cobalt's d orbitals differently. Every structural isomer pair is like this: the difference in bonding shows up in some measurable property, and choosing the measurement is choosing which bond to look at.

7. Linkage isomers and how to count them

Jørgensen made the red nitrito-O cobalt complex in 1894 and found that it turned slowly into the yellow-brown nitrito-N form on standing, the first linkage isomerism observed. The N-bonded form is the more stable, because nitrogen is the better donor to cobalt(III), and the O-bonded form is a kinetic product that forms first and rearranges.

When a complex has several identical ambidentate ligands, count linkage isomers by how many bond through the first atom. $\mathrm{[Pt(NH_3)_2(SCN)_2]}$ can have both thiocyanates S-bonded, one S and one N, or both N: three possibilities, $2 + 1$. Choosing an atom for each ligand separately would give $2 \times 2 = 4$, but "first S, second N" and "first N, second S" are the same compound until the arrangement in space is considered, which is the next lesson's job.

8. Checking a pair of structural isomers

Two compounds are structural isomers only if they have exactly the same formula, so the first check is to count every atom in both. $\mathrm{[Co(NH_3)_5Br]SO_4}$ and $\mathrm{[Co(NH_3)_5SO_4]Br}$ each hold one cobalt, five ammonia molecules, one bromine and one sulfate: isomers. A pair that differs by a water molecule is not a pair of isomers at all but two different compounds.

The second check is a test that tells the pair apart, because isomers that no experiment could distinguish would not be two compounds. Ionization isomers release different ions into water: barium chloride precipitates barium sulfate from the first, whose sulfate is free, and silver nitrate precipitates silver bromide from the second, whose bromide is free. Hydrate isomers differ in the chloride silver nitrate can reach and in color: violet $\mathrm{[Cr(H_2O)_6]Cl_3}$, blue-green $\mathrm{[Cr(H_2O)_5Cl]Cl_2\cdot H_2O}$ and dark green $\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot 2H_2O}$. Linkage isomers differ in which atom binds, which infrared spectra show: a nitrogen-bound nitro ligand absorbs at different frequencies from an oxygen-bound nitrito one.

The third check is that the metal's oxidation state and coordination number are the same in both. Structural isomers move a ligand in or out of the sphere or turn it round; they do not change the metal.

Last, remember what structural isomers are not. Two complexes with the same ligands bound in the same way, arranged differently in space, are stereoisomers, the subject of lessons 8 and 9, and no test with silver nitrate or barium chloride can tell them apart.

9. In the world: a blood-pressure drug that switches with light

Sodium nitroprusside, $\mathrm{Na_2[Fe(CN)_5(NO)]}$, is given by intravenous drip in emergencies to bring dangerously high blood pressure down within minutes: the nitric oxide ligand is released in the body and relaxes the walls of blood vessels. The same compound is a textbook case of linkage isomerism. In 1977 physicists found that blue-green light turns crystals cooled below about $-75$ °C into long-lived new states, and in 1997 X-ray crystallography showed what the main one is: the Fe–NO bond has become an Fe–ON bond, a metastable linkage isomer that survives in the dark until the crystal is warmed.

Because the two isomers absorb light differently and switch with light of different colors, nitroprusside crystals were studied as a medium for holographic data storage. In the hospital, the drug is protected from light for a more ordinary reason: light also releases cyanide from it, and drip bags and tubing are wrapped in opaque covers to keep a patient from receiving it.

10. In the world: why a chromium solution changes color on the shelf

Dissolve dark green chromium(III) chloride in water and the solution is green. Leave it for a few days and it turns steadily toward the gray-violet of $\mathrm{[Cr(H_2O)_6]^{3+}}$, as water slowly replaces the chloride ligands. Chromium(III) is so inert that each chloride takes hours to days to leave, which is why the green isomer survives dissolving at all and why the change is slow enough to watch.

Laboratories that titrate chloride or use chromium solutions as standards have to allow for it. A freshly made solution of the green salt gives one third of the silver chloride a week-old one does, because only the free chloride reacts; a method that does not specify how old the solution is will give different answers on different days. The fix is either to use the solution at once and account for the bound chloride, or to heat it so the exchange reaches equilibrium before use, and the isomer arithmetic of this lesson is what tells the analyst which correction to apply.

11. One formula does not mean one compound

A reagent bottle labeled $\mathrm{CrCl_3\cdot 6H_2O}$ tells you the atoms, not how they are bonded, and it is easy to assume the formula settles what is inside. For coordination compounds it does not: the dot formula hides whether a chloride is a ligand or a free ion and whether a water is bonded to the metal or held in the lattice, and those differences change the color, the ions in solution and every calculation based on them.

The related slip is to count linkage isomers by giving each ambidentate ligand its own choice of atom, $2^n$. That treats identical ligands as if they were labeled. Two thiocyanates bonded one through sulfur and one through nitrogen are one arrangement of bonds, whichever thiocyanate is which; only where they sit in space, the next lesson's question, can make two compounds of it.

12. Which ionization isomer is in the bottle?

1. Write the two candidates.

   $\mathrm{[Co(NH_3)_5Br]SO_4} \quad \text{or} \quad \mathrm{[Co(NH_3)_5SO_4]Br}$

   Same formula, $\mathrm{CoBrSO_4\cdot 5NH_3}$.

2. Predict the silver nitrate test.

   $\text{first: no AgBr}; \quad \text{second: AgBr precipitates}$

   Only free bromide meets the silver.

3. Predict the barium chloride test.

   $\text{first: } \mathrm{BaSO_4} \text{ precipitates}; \quad \text{second: none}$

   Only free sulfate meets the barium.

4. Run the tests on the sample.

   $\mathrm{AgNO_3}: \text{cream precipitate}; \quad \mathrm{BaCl_2}: \text{none}$

   Suppose these are the observations.

5. Identify the isomer.

   $\mathrm{[Co(NH_3)_5SO_4]Br}$

   Bromide is free, so it is the counter ion; sulfate is a ligand.

13. Water lost from the dark green chromium chloride

1. Write the isomer's formula.

   $\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot 2H_2O}$

   Two chloride ligands, four water ligands.

2. Count the lattice water.

   $6 - 4 = 2$

   Six waters in the formula, four of them ligands.

3. Take $10.0$ mmol and find the water lost on gentle heating.

   $10.0 \times 2 = 20.0\ \text{mmol} = 0.0200\ \text{mol}$

   Only lattice water leaves at low temperature.

4. Convert to a mass.

   $0.0200 \times 18.02 = 0.360\ \text{g}$

   Moles times the molar mass of water.

5. Compare with the violet isomer.

   $\mathrm{[Cr(H_2O)_6]Cl_3}: 0\ \text{g}$

   All its water is bonded to chromium, so gentle heating removes none.

14. Counting linkage isomers of a trinitrito complex

1. Identify the ambidentate ligands in $\mathrm{[Co(NH_3)_3(NO_2)_3]}$.

   $3 \times \mathrm{NO_2^-}, \ \text{each through N or O}$

   Ammonia has only one donor atom.

2. Say what distinguishes the linkage isomers.

   $\text{number of N-bonded nitrites}$

   The three nitrites are identical, so only the count matters.

3. List the possibilities.

   $3,\ 2,\ 1,\ 0\ \text{N-bonded}$

   From all three through nitrogen to none.

4. Count the isomers.

   $3 + 1 = 4$

   One more than the number of ambidentate ligands.

5. Reject the overcount.

   $2^3 = 8 \ne 4$

   Choosing an atom for each nitrite separately counts rearrangements of identical ligands as different.

6. Note what remains.

   $\text{each of the four has fac and mer forms}$

   Three identical ligands in an octahedron can sit on a face or a meridian, lesson 8.

15. Your turn: which isomer of $\mathrm{CrCl_3\cdot 6H_2O}$ gives $2$ mol of AgCl per mole?

1. Read the free chlorides.

   $2\ \mathrm{Cl^-}\ \text{free}$

   Silver found two per formula unit.

2. Place the third chloride.

   $3 - 2 = 1\ \text{chloride ligand}$

   It is bonded to chromium.

3. Your turn: work this step out. Its working is at the end of the packet.

   Write the formula.

16. Guided practice

Each pair of compounds below is a pair of isomers. Match each pair to the kind of structural isomerism it shows. (A sample of $5$ mmol of each would give the same mass.)

17. Guided practice

Complete the worked solution: an isomer of $\mathrm{CrCl_3\cdot 6H_2O}$ has $1$ chloride ligands on its six-coordinate chromium. Where are its waters and its other chlorides?

1. Fill the rest of the six positions with water.

   $\text{water ligands} = 6 - (\text{chloride ligands}) =$ _w

   Chromium(III) is six-coordinate.

2. Put the remaining water in the lattice.

   $\text{lattice water} = 6 - (\text{water ligands}) =$ _l

   The formula has six waters in all.

3. Count the chlorides outside the brackets.

   $\text{free chlorides} = 3 - (\text{chloride ligands}) =$ _o

   The formula has three chlorides in all.

18. Guided practice

A freshly made solution of $2$ mmol of one isomer of $\mathrm{CrCl_3\cdot 6H_2O}$ gives $4$ mmol of silver chloride with excess silver nitrate. Which isomer is it?

• The form with 1 coordinated chlorides and 5 coordinated waters.
• The form with 2 coordinated chlorides and 4 coordinated waters.
• The form with 0 coordinated chlorides and 6 coordinated waters.

19. Practice

Samples of $3$ mmol of each isomer of $\mathrm{CrCl_3\cdot 6H_2O}$: the violet $\mathrm{[Cr(H_2O)_6]Cl_3}$, the pale green $\mathrm{[Cr(H_2O)_5Cl]Cl_2\cdot H_2O}$ and the dark green $\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot 2H_2O}$. For each, in that order, fill in the silver chloride it gives, the water it loses on gentle heating, and the particles per formula unit in solution.

20. Practice

A sample of $32$ mmol of $\mathrm{[Cr(H_2O)_5Cl]Cl_2\cdot H_2O}$ is warmed gently until it stops losing mass, which drives off only the water outside the brackets. What mass of water ($18.02$ g/mol) does it lose?

Answer: unit: g / kg / mg

21. Practice

In $\mathrm{[Co(NH_3)_3(NO_2)_3]}$ each of the $3$ $\mathrm{NO_2^{-}}$ ligands can bond through $\mathrm{N}$ or through $\mathrm{O}$. Leaving aside where the ligands sit in space, how many linkage isomers can the complex have?

Answer: linkage isomers

22. Somewhere new

A quality-control lab receives three lots, each labeled $\mathrm{CrCl_3\cdot 6H_2O}$. Freshly dissolved and treated with silver nitrate, $3$ mmol of lot A gives $9$ mmol of AgCl, $3$ mmol of lot B gives $3$ mmol, and $2$ mmol of lot C gives $4$ mmol. For each lot, in that order, fill in the chlorides outside the brackets per formula unit and the water ligands on the chromium.

23. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

24. Test question

Samples of $6$ mmol of each isomer of $\mathrm{CrCl_3\cdot 6H_2O}$: the violet $\mathrm{[Cr(H_2O)_6]Cl_3}$, the pale green $\mathrm{[Cr(H_2O)_5Cl]Cl_2\cdot H_2O}$ and the dark green $\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot 2H_2O}$. For each, in that order, fill in the silver chloride it gives, the water it loses on gentle heating, and the particles per formula unit in solution.

25. What you can do now

You can tell structural isomers apart by measurement. Explain how silver nitrate and gentle heating together identify which isomer of $\mathrm{CrCl_3\cdot 6H_2O}$ a sample is.

Working for the steps left to you

15. Your turn: which isomer of $\mathrm{CrCl_3\cdot 6H_2O}$ gives $2$ mol of AgCl per mole?, step 3