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Ligands inside the brackets, counter ions outside: silver precipitation, conductivity and freezing points count the free ions, as Werner used them to.
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.
By the end of this lesson you will be able to read a coordination compound's formula as ligands and counter ions, predict the silver chloride it precipitates and the particles it gives in solution, and work back from those measurements to the formula.
You can convert between mass and moles, and you know that silver nitrate precipitates dissolved chloride as white silver chloride, $\mathrm{AgCl}$, one mole for each mole of chloride ion. You know that an ionic compound separates into its ions in water, so a dissolved salt raises a solution's conductivity and lowers its freezing point in proportion to the number of particles. From the last lessons you can find a complex ion's charge and the ligands inside its brackets.
| Term | What it means |
|---|---|
| Coordination sphere | The metal and its ligands, written inside the square brackets of a formula. |
| Counter ion | An ion outside the brackets, held only by electrostatic attraction and free in solution. |
| Primary valence | Werner's term for what is now the oxidation state: the charge the counter ions balance. |
| Secondary valence | Werner's term for what is now the coordination number: the ligands the metal holds directly. |
| Molar conductivity | How well one mole of a dissolved salt carries current; it rises with the number of ions per formula unit. |
| van 't Hoff factor | The number of particles one formula unit gives in solution, $i$. |
| Freezing-point depression | $\Delta T = i\,K_f\,m$, the lowering of a solvent's freezing point by dissolved particles. |
In the 1890s chemists knew a series of cobalt compounds with the empirical formulas $\mathrm{CoCl_3\cdot 6NH_3}$, $\mathrm{CoCl_3\cdot 5NH_3}$ and $\mathrm{CoCl_3\cdot 4NH_3}$, differently colored, and nobody could say how ammonia, a stable molecule, attached itself to a salt. Alfred Werner proposed that the cobalt holds six groups directly, in a coordination sphere, and that anything beyond those six sits outside as a separate ion. In modern notation:
$$\mathrm{CoCl_3\cdot 6NH_3} = \mathrm{[Co(NH_3)_6]Cl_3}, \quad \mathrm{CoCl_3\cdot 5NH_3} = \mathrm{[Co(NH_3)_5Cl]Cl_2}, \quad \mathrm{CoCl_3\cdot 4NH_3} = \mathrm{[Co(NH_3)_4Cl_2]Cl}.$$
As ammonia is removed, a chloride moves inside the brackets to keep six groups on the cobalt, and each chloride that moves inside is one fewer free chloride ion.
Werner tested this with two measurements. Silver nitrate precipitates only the chloride that is free in solution, because a chloride bonded to cobalt is not available to silver. A mole of the first compound gives three moles of AgCl, the second two, the third one. Conductivity counts the ions: the first compound gives four ions per formula unit, the complex ion and three chlorides, the second three and the third two. Both measurements agree with the bracket formulas and with nothing else.
The rule that comes out of it is the one used every time a formula is read. Ligands are inside the brackets and bonded to the metal; counter ions are outside, held only by charge, and free in solution. The number of free counter ions is fixed by the complex ion's charge, and the number of particles one formula unit gives is
$$i = (\text{counter ions}) + 1,$$
the $1$ being the complex ion, which stays in one piece. A neutral complex such as $\mathrm{[Co(NH_3)_3Cl_3]}$ gives one particle, conducts no better than sugar, and gives no precipitate with silver at all.
Another way: picture
Picture the cobalt as a hub with six sockets, each plugged with an ammonia or a chloride. Chlorides plugged into a socket are fixed to the hub and travel with it. Chlorides not plugged in drift free. Silver can only catch the free ones, and a conductivity meter counts the hub as one charged particle and each free chloride as another.
Another way: steps
| Empirical formula | Formula | Free $\mathrm{Cl^-}$ | Ions per formula unit |
|---|---|---|---|
| $\mathrm{CoCl_3\cdot 6NH_3}$ | $\mathrm{[Co(NH_3)_6]Cl_3}$ | $3$ | $4$ |
| $\mathrm{CoCl_3\cdot 5NH_3}$ | $\mathrm{[Co(NH_3)_5Cl]Cl_2}$ | $2$ | $3$ |
| $\mathrm{CoCl_3\cdot 4NH_3}$ | $\mathrm{[Co(NH_3)_4Cl_2]Cl}$ | $1$ | $2$ |
| $\mathrm{CoCl_3\cdot 3NH_3}$ | $\mathrm{[Co(NH_3)_3Cl_3]}$ | $0$ | $1$ |
| $\mathrm{PtCl_4\cdot 6NH_3}$ | $\mathrm{[Pt(NH_3)_6]Cl_4}$ | $4$ | $5$ |
| $\mathrm{PtCl_4\cdot 4NH_3}$ | $\mathrm{[Pt(NH_3)_4Cl_2]Cl_2}$ | $2$ | $3$ |
| $\mathrm{PtCl_4\cdot 2NH_3}$ | $\mathrm{[Pt(NH_3)_2Cl_4]}$ | $0$ | $1$ |
The platinum(IV) series follows the same pattern with four chlorides in all. In every row the coordination number stays six and the metal's oxidation state stays the same, $+3$ for cobalt and $+4$ for platinum; what changes is how the chlorides are shared between the coordination sphere and the counter ions. Werner called the oxidation state the primary valence and the coordination number the secondary valence, and the insight that a metal has both at once earned him the 1913 Nobel Prize in Chemistry.
Every dissolved particle lowers a solvent's freezing point by the same amount, whatever it is: $\Delta T = i\,K_f\,m$, with $K_f = 1.86$ K kg/mol for water and $m$ the molality. A $0.0500$ mol/kg solution of a compound giving four particles freezes $4 \times 1.86 \times 0.0500 = 0.372$ K below pure water; one giving two particles freezes only $0.186$ K below. Measured carefully, the two are easy to tell apart.
Conductivity reads the same count more quickly. The molar conductivity of a dilute salt rises with the number and charge of its ions, and Werner compared his compounds with simple salts of known type: $\mathrm{[Co(NH_3)_6]Cl_3}$ conducts like $\mathrm{LaCl_3}$, a four-ion salt, and $\mathrm{[Co(NH_3)_4Cl_2]Cl}$ like $\mathrm{NaCl}$, a two-ion salt. Real solutions give somewhat less than the ideal count because ions of opposite charge pair up a little, but the steps between the compounds are unmistakable.
The brackets are not decoration. $\mathrm{[Co(NH_3)_5Cl]Cl_2}$ and $\mathrm{CoCl_3\cdot 5NH_3}$ describe the same atoms, but only the first says that one chloride is bonded to cobalt and two are not. The conventions are:
Given a formula written this way, everything in this lesson follows from counting: ligands inside, counter ions outside, and the complex ion as one particle.
A Werner formula makes predictions that can be tested at the bench, which is how Werner persuaded chemists he was right, and the same predictions check an answer on paper.
Silver nitrate precipitates only the chloride ions that are free in solution. For every formula unit, $\mathrm{[Co(NH_3)_6]Cl_3}$ gives three moles of silver chloride, $\mathrm{[Co(NH_3)_5Cl]Cl_2}$ two and $\mathrm{[Co(NH_3)_4Cl_2]Cl}$ one, although all three contain three chlorines. If your formula puts a chloride outside the brackets, it must be one that silver nitrate can reach.
The conductivity of the solution rises with the number of ions a formula unit releases. A salt that gives two ions conducts like sodium chloride, one that gives four conducts much more strongly, and a neutral complex such as $\mathrm{[Co(NH_3)_3Cl_3]}$ hardly conducts at all. Werner used exactly this series to show that the chlorides inside the brackets were bound to the metal.
Finally, count the atoms. A Werner formula only regroups the atoms of the empirical formula, so the metal, ligand and counter-ion atoms must add up to the same totals, and the molar mass must come out the same either way. The coordination number in the brackets should also be one the metal normally shows: six for cobalt(III) and chromium(III), four for platinum(II). A formula that fails any of these checks describes a compound that is not in the bottle.
The checks work together. A cobalt(III) complex written with five ligands in the brackets fails the coordination-number check at once; one written with three chlorides outside the brackets and a chloride inside fails the atom count. Only the formula that passes every check at once is the one to trust, and it is the one the precipitate, the conductivity and the molar mass all point to.
Cisplatin, $\mathrm{[Pt(NH_3)_2Cl_2]}$, is one of the most widely used cancer drugs, given for testicular, ovarian and bladder cancers among others. Its formula shows it is a neutral complex: platinum(II) with two ammonias and two chlorides all inside the brackets and no counter ions. A neutral molecule crosses a cell membrane far more easily than an ion, and in the blood, where chloride is about $100$ mmol/L, the chloride ligands stay on the platinum.
Inside the cell the chloride concentration falls to about $4$ to $20$ mmol/L, and the chlorides are slowly replaced by water, turning the neutral drug into a charged aqua complex. That charged species is the one that binds to the nitrogen atoms of guanine in DNA, cross-linking the strands and stopping the cell from dividing. The distinction this lesson is about, ligand chloride against free chloride, is the switch that turns cisplatin on only after it is inside the cell.
Chemical suppliers still sell chromium(III) chloride hexahydrate under the label $\mathrm{CrCl_3\cdot 6H_2O}$, a formula that hides three different compounds. The common dark green solid is $\mathrm{[Cr(H_2O)_4Cl_2]Cl\cdot 2H_2O}$, with two chlorides on the chromium; a pale blue-green form is $\mathrm{[Cr(H_2O)_5Cl]Cl_2\cdot H_2O}$; and a violet form is $\mathrm{[Cr(H_2O)_6]Cl_3}$, with every chloride free.
The difference matters in the laboratory. A chemist titrating chloride with silver nitrate in a freshly made solution of the green salt finds only one third of the chloride the label suggests, because the rest is bonded to chromium, and the ligand chlorides are released only slowly as water replaces them over hours. A procedure that assumes three free chlorides per chromium, as the empirical formula invites, would be off by a factor of three. The bracket formula, and the silver-chloride arithmetic of this lesson, is how the analyst knows what to expect.
The empirical formula $\mathrm{CoCl_3\cdot 5NH_3}$ has three chlorines, and the habit from simple salts says that $\mathrm{CoCl_3}$ dissolves to give three chloride ions. It does not: one of the chlorides is bonded to the cobalt, and silver ions cannot reach it. The compound gives two moles of AgCl, not three, and three ions in solution, not four. Every chloride inside the brackets is part of the complex ion.
The reverse error treats the whole salt as one molecule because it is a "complex". The counter ions outside the brackets are ordinary ions, held only by charge; they leave in water as surely as the chloride of table salt. The brackets are the dividing line, and the two measurements Werner used, precipitation and conductivity, are the evidence that they are drawn in the right place.
Find the counter ions outside the brackets.
$3\ \mathrm{Cl^-}$
Everything after the closing bracket is a counter ion.
Find the charge of the complex ion.
$3 \times (-1) = -3 \Rightarrow \mathrm{[Co(NH_3)_6]^{3+}}$
The salt is neutral, so the complex ion is $+3$.
Count the free chlorides.
$3$
No chloride is inside the brackets.
Count the particles.
$3 + 1 = 4$
Three chlorides and one complex ion.
Predict the silver chloride from one mole.
$1 \times 3 = 3\ \text{mol AgCl}$
Each free chloride gives one AgCl.
Separate ligands from counter ions.
$\text{inside: } 5\,\mathrm{NH_3},\ 1\,\mathrm{Cl^-}; \quad \text{outside: } 2\,\mathrm{Cl^-}$
One chloride is bonded to cobalt.
Take $0.0100$ mol of the compound and find the free chloride.
$0.0100 \times 2 = 0.0200\ \text{mol}$
Two free chlorides per formula unit.
Find the moles of AgCl.
$n_{\mathrm{AgCl}} = 0.0200\ \text{mol}$
One AgCl per free chloride.
Convert to a mass.
$0.0200 \times 143.32 = 2.866\ \text{g}$
Moles times molar mass.
Compare with the wrong count.
$0.0300 \times 143.32 = 4.300\ \text{g}$
Counting all three chlorides as free would predict half as much again.
A $0.4670$ g sample of $\mathrm{CoCl_3\cdot 4NH_3}$ ($233.42$ g/mol) is dissolved. Find its moles.
$n = \dfrac{0.4670}{233.42} = 0.002001\ \text{mol}$
Mass divided by molar mass.
Silver nitrate gives $0.2868$ g of AgCl. Find its moles.
$n_{\mathrm{AgCl}} = \dfrac{0.2868}{143.32} = 0.002001\ \text{mol}$
The same conversion for the precipitate.
Find the ratio.
$\dfrac{0.002001}{0.002001} = 1.00$
One free chloride per formula unit.
Place the chlorides.
$3 - 1 = 2\ \text{chlorides inside}$
Three chlorides in all, one free, so two are ligands.
Write the formula.
$\mathrm{[Co(NH_3)_4Cl_2]Cl}$
Four ammonias and two chlorides fill the six positions.
Check with the charge.
$x + 4 \times 0 + 2 \times (-1) = +1 \Rightarrow x = +3$
Cobalt stays $+3$, as in every member of the series.
Find the counter ions.
$2\ \mathrm{Cl^-}\ \text{outside}$
Two chlorides after the bracket.
Count the complex ion.
$1\ \mathrm{[Pt(NH_3)_4Cl_2]^{2+}}$
The two ligand chlorides stay on the platinum.
Add them up.
How many ions does one formula unit of $\mathrm{[Pt(NH_3)_4Cl_2]Cl_2}$ give when it dissolves in water?
Complete the worked solution: how many particles does one formula unit of $\mathrm{[Cr(H_2O)_6]Cl_3}$ give in water?
Count the chloride ions outside the brackets.
$\text{free chlorides} = \text{charge of the complex ion} =$ a
Each balances one unit of the complex ion's charge.
Add the complex ion itself.
$\text{particles} = (\text{free chlorides}) + 1 =$ i
The complex ion stays whole, with all its ligands.
Check against the formula's chlorides.
$\text{free chlorides} \le \text{all chlorides in the formula}$
Chlorides inside the brackets are ligands, and they stay on the metal in water.
Solutions of four cobalt(III) ammine chlorides each hold $9$ mmol of compound. Match each to the silver chloride that excess silver nitrate precipitates from it.
| $27$ mmol AgCl | $18$ mmol AgCl | $9$ mmol AgCl | no precipitate | |
|---|---|---|---|---|
| $\mathrm{[Co(NH_3)_6]Cl_3}$ | ||||
| $\mathrm{[Co(NH_3)_5Cl]Cl_2}$ | ||||
| $\mathrm{[Co(NH_3)_4Cl_2]Cl}$ | ||||
| $\mathrm{[Co(NH_3)_3Cl_3]}$ |
Fill in the table for $\mathrm{[Co(NH_3)_4Cl_2]Cl}$, $\mathrm{[Pt(NH_3)_4Cl_2]Cl_2}$ and $\mathrm{[Pt(NH_3)_5Cl]Cl_3}$, in that order.
| charge of the complex ion | chlorides precipitated per formula unit | particles per formula unit | |
|---|---|---|---|
| the first compound | |||
| the second compound | |||
| the third compound |
A solution contains $34$ mmol of $\mathrm{[Pt(NH_3)_3Cl_3]Cl}$. What mass of silver chloride ($143.32$ g/mol) does excess silver nitrate precipitate?
Answer: unit: g / kg / mg
A $8441/10000$ g sample of $\mathrm{PtCl_4\cdot 5NH_3}$ (molar mass $422.05$ g/mol) is dissolved and treated with excess silver nitrate. It gives $10749/12500$ g of silver chloride ($143.32$ g/mol). How many chloride ions per formula unit are outside the brackets?
Answer: chloride ions outside the brackets per formula unit
A stockroom finds three old bottles labeled $\mathrm{CoCl_3\cdot 6NH_3}$, $\mathrm{CoCl_3\cdot 5NH_3}$ and $\mathrm{CrCl_3\cdot 6H_2O}$. A technician suspects they are $\mathrm{[Co(NH_3)_6]Cl_3}$, $\mathrm{[Co(NH_3)_5Cl]Cl_2}$ and $\mathrm{[Cr(H_2O)_6]Cl_3}$ and tests each by the freezing point of a $0.0500$ mol/kg solution in water ($K_f = 1.86$ K kg/mol). For each, in that order, fill in the particles per formula unit and the freezing-point depression it predicts.
| particles per formula unit | freezing-point depression (K) | |
|---|---|---|
| the first bottle | ||
| the second bottle | ||
| the third bottle |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Fill in the table for $\mathrm{[Pt(NH_3)_3Cl_3]Cl}$, $\mathrm{[Co(NH_3)_5Cl]Cl_2}$ and $\mathrm{[Cr(H_2O)_6]Cl_3}$, in that order.
| charge of the complex ion | chlorides precipitated per formula unit | particles per formula unit | |
|---|---|---|---|
| the first compound | |||
| the second compound | |||
| the third compound |
You can tell ligands from counter ions in a formula. Explain how silver nitrate shows that $\mathrm{CoCl_3\cdot 5NH_3}$ is $\mathrm{[Co(NH_3)_5Cl]Cl_2}$.
15. Your turn: how many ions does one formula unit of $\mathrm{[Pt(NH_3)_4Cl_2]Cl_2}$ give?, step 3
$2 + 1 = 3$
Three particles per formula unit.