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Charge balance inside a complex: counter ions fix the complex ion's charge, the ligands' charges fix the metal's oxidation state, and the group number gives the $d$ count.
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 find the oxidation state of the metal in any complex or salt of a complex by charge balance, and turn it into the metal's $d$ count.
From the last lesson you can turn a metal's charge into a $d$ count: a first-row ion of charge $+q$ is $d^{g - q}$. From general chemistry you can assign oxidation states in simple compounds, knowing that oxygen is usually $-2$, hydrogen $+1$ and the halogens $-1$, and that the oxidation states in an ion add up to its charge. This lesson applies the same bookkeeping to complexes, where the metal is bonded to whole molecules and ions at once.
| Term | What it means |
|---|---|
| Complex | A central metal atom or ion bonded to a set of surrounding molecules or ions. |
| Ligand | A molecule or ion that donates an electron pair to the metal in a complex. |
| Complex ion | A complex that carries an overall charge, written in square brackets with the charge outside. |
| Counter ion | An ion outside the brackets that balances the complex ion's charge in a salt. |
| Oxidation state | The charge the metal would carry if every ligand were removed with its donated electron pair. |
| Neutral ligand | A ligand with no charge, such as $\mathrm{H_2O}$, $\mathrm{NH_3}$, CO or en. |
| Anionic ligand | A ligand with a negative charge, such as $\mathrm{Cl^-}$, $\mathrm{CN^-}$, $\mathrm{OH^-}$ or oxalate, $\mathrm{C_2O_4^{2-}}$. |
A complex is a metal surrounded by ligands: molecules or ions that each donate a pair of electrons to it. In $\mathrm{[Fe(CN)_6]^{4-}}$ an iron is bonded to six cyanide ions; in $\mathrm{[Co(NH_3)_6]^{3+}}$ a cobalt is bonded to six ammonia molecules. The square brackets enclose the complex, and the charge outside them belongs to the whole unit, metal and ligands together.
The metal's oxidation state is the charge it would carry if every ligand were taken away with the electron pair it donated. Each ligand leaves with its own ordinary charge: a water or ammonia molecule leaves neutral, a chloride leaves as $\mathrm{Cl^-}$, a cyanide as $\mathrm{CN^-}$, an oxide as $\mathrm{O^{2-}}$. What is left on the metal is its oxidation state $x$, and the charges must add up:
$$x + (\text{sum of the ligands' charges}) = \text{charge of the complex ion}.$$
In $\mathrm{[Fe(CN)_6]^{4-}}$ the six cyanides carry $-6$, so $x + (-6) = -4$ and $x = +2$. In $\mathrm{[Co(NH_3)_6]^{3+}}$ the ammonias carry nothing, so $x = +3$. Only when every ligand is neutral does the metal's oxidation state equal the charge on the brackets; as soon as one ligand is charged, the two differ.
A complex ion found in a salt has its charge fixed by the counter ions outside the brackets. In $\mathrm{K_4[Fe(CN)_6]}$ four potassium ions carry $+4$, so the complex ion must be $-4$ for the salt to be neutral. In $\mathrm{[Co(NH_3)_5Cl]Cl_2}$ the two chlorides outside the brackets carry $-2$, so the complex ion is $+2$, and inside it the one chloride that is a ligand makes the cobalt $+3$. Working from the outside in, counter ions, then ligands, then metal, gives the oxidation state of any metal in any complex.
The oxidation state then gives the $d$ count by the rule from lesson 1: a metal in group $g$ with oxidation state $x$ has $g - x$ electrons in its d orbitals. Iron(II) in ferrocyanide is $d^6$; cobalt(III) in the ammine complexes is $d^6$ as well.
Another way: picture
Think of the complex ion as a sealed box with a price label on the outside, the overall charge. Inside are the metal and its ligands, each with its own price tag: water and ammonia cost nothing, chloride and cyanide are minus one, oxide and oxalate are minus two. The metal's tag is whatever makes the tags inside add up to the label on the box.
Another way: steps
| Ligand | Formula | Charge | Name in a complex |
|---|---|---|---|
| water | $\mathrm{H_2O}$ | $0$ | aqua |
| ammonia | $\mathrm{NH_3}$ | $0$ | ammine |
| carbon monoxide | CO | $0$ | carbonyl |
| ethylenediamine | en, $\mathrm{H_2NCH_2CH_2NH_2}$ | $0$ | ethylenediamine |
| chloride | $\mathrm{Cl^-}$ | $-1$ | chlorido |
| cyanide | $\mathrm{CN^-}$ | $-1$ | cyanido |
| hydroxide | $\mathrm{OH^-}$ | $-1$ | hydroxido |
| nitrite | $\mathrm{NO_2^-}$ | $-1$ | nitrito |
| oxide | $\mathrm{O^{2-}}$ | $-2$ | oxido |
| oxalate | $\mathrm{C_2O_4^{2-}}$ | $-2$ | oxalato |
The charge of a ligand is the charge of the free molecule or ion it came from, because the oxidation-state convention hands each ligand back with its donated pair. That is why the table is short to learn: neutral molecules count zero, and the anions count whatever they carry on their own. The names in the last column are the ones the naming lesson uses. Ethylenediamine, en, bonds through both of its nitrogen atoms at once, so three of them fill the six positions around a metal in $\mathrm{[Ni(en)_3]^{2+}}$, but it is still a neutral molecule and still counts zero.
Oxide is the ligand that lets a metal reach its highest oxidation states, because each one takes two electrons' worth of charge. In permanganate, $\mathrm{[MnO_4]^-}$, the four oxides carry $-8$, so manganese is $+7$, which for a group-7 metal means $d^0$: every valence electron gone. Chromate, $\mathrm{[CrO_4]^{2-}}$, is chromium $+6$, also $d^0$.
These oxidation states are real in the bookkeeping sense and useful for balancing redox equations, but they are not charges sitting on the metal. The Mn–O bonds in permanganate are strongly covalent, and the actual charge on the manganese is far less than $+7$. The oxidation state is a count of electrons the metal has given up to more electronegative partners, which is exactly what makes $d^0$ metals such strong oxidants: permanganate and dichromate both pull electrons back readily. Their intense colors are also not $d$–$d$ colors, since they have no $d$ electrons; the color lesson returns to them.
Check the sign. A transition metal in an ordinary complex is in a positive oxidation state, usually $+1$ to $+3$ for the first row, higher only with oxide or fluoride ligands. If the balance gives a negative number, a ligand's charge has been added instead of subtracted, or a counter ion has been counted as a ligand.
Check against the metal. Each metal takes only certain oxidation states. Iron is found as $+2$ and $+3$; cobalt as $+2$ and $+3$; nickel and copper almost always as $+2$; chromium as $+3$ and $+6$; manganese as $+2$, $+4$ and $+7$. An answer of $\mathrm{Fe}(+5)$ or $\mathrm{Cu}(+4)$ is a sign to recount the ligands. The two checks together catch the commonest slip of all, reading the overall charge as the oxidation state, whenever a complex has anionic ligands.
Three checks catch almost every slip. First, add the charges back up: the oxidation state plus the charges of all the ligands must give the charge written outside the brackets, or zero for a neutral complex. If it does not, a ligand's charge was misread, and the usual culprit is a neutral ligand, water, ammonia, carbon monoxide or ethylenediamine, counted as if it were charged.
Second, the oxidation state cannot exceed the group number. A metal in group $n$ has only $n$ valence electrons to give, so the highest state it reaches is $+n$: scandium $+3$, titanium $+4$, vanadium $+5$, chromium $+6$, manganese $+7$. A result above that, or a negative $d$ count, means the arithmetic has gone wrong somewhere.
Third, compare with the states the metal is known to take. Iron is almost always $+2$ or $+3$, cobalt $+2$ or $+3$, copper $+1$ or $+2$, and platinum $+2$ or $+4$. An answer such as iron $+5$ in an ordinary salt is not impossible, but it is rare enough that the working deserves a second look before it is believed. None of these checks proves an answer right, but each one that fails proves it wrong.
Prussian blue, the first synthetic blue pigment (1706) and the blue of old architectural blueprints, has the formula $\mathrm{Fe_4[Fe(CN)_6]_3}$. The charge balance shows at once that it contains iron in two oxidation states. Each complex ion $\mathrm{[Fe(CN)_6]^{4-}}$ holds iron(II), $d^6$, since six cyanides carry $-6$. Three of those complex ions carry $-12$ in all, so the four iron ions outside the brackets must carry $+12$: each is iron(III), $d^5$.
The deep blue comes from an electron hopping between the two kinds of iron when light is absorbed, a mixed-valence charge transfer that neither iron(II) nor iron(III) cyanide shows alone. The same compound is on the World Health Organization's list of essential medicines: taken by mouth, it binds cesium and thallium ions in the gut, and it was used after the 1987 Goiânia cesium-137 accident to speed the removal of radioactive cesium from the body.
Many towns that draw water from wells add potassium permanganate, $\mathrm{KMnO_4}$, before filtration. Dissolved iron(II) and manganese(II) in groundwater are colorless but stain laundry and fixtures once air oxidizes them; permanganate oxidizes them deliberately so the insoluble products can be filtered out.
The oxidation states show why it works. In $\mathrm{[MnO_4]^-}$ manganese is $+7$, $d^0$, and in neutral water it is reduced to manganese dioxide, $\mathrm{MnO_2}$, in which it is $+4$: each permanganate takes three electrons. Each iron(II) gives up one electron to become iron(III), so the dose is set by that ratio, three iron(II) ions for every permanganate: by mass, $0.94$ mg of permanganate for each milligram of iron. A plant treating water with $1.5$ mg/L of dissolved iron therefore needs about $1.4$ mg/L of permanganate, and the operators watch for a faint pink tint after the filters, the color of unreacted permanganate, as the sign they have added too much.
Because so many common complexes have neutral ligands, such as $\mathrm{[Cu(H_2O)_6]^{2+}}$ and $\mathrm{[Co(NH_3)_6]^{3+}}$, it is easy to learn the shortcut "the oxidation state is the charge on the complex" and then apply it where it fails. In $\mathrm{[Fe(CN)_6]^{4-}}$ it would give iron $-4$, a state iron never takes; in $\mathrm{[CoCl_4]^{2-}}$ it would give cobalt $-2$. The ligands own part of the charge, and only the balance equation separates the metal's share from theirs.
The second slip is to count the counter ions of a salt as ligands. In $\mathrm{[Co(NH_3)_5Cl]Cl_2}$ only one chloride is inside the brackets and bonded to the cobalt. The other two are separate ions in the crystal and in solution, and their job is only to fix the charge of the complex ion. Counting all three as ligands would give cobalt $+5$, which the check against known oxidation states catches at once.
Identify the ligands and their charges.
$6 \times \mathrm{H_2O},\ \text{charge } 0\ \text{each}$
Water is a neutral molecule.
Add up the ligands' charges.
$6 \times 0 = 0$
Six neutral ligands carry no charge in total.
Write the charge balance.
$x + 0 = +3$
The complex ion is $\mathrm{[Cr(H_2O)_6]^{3+}}$.
Solve for the oxidation state.
$x = +3$
With neutral ligands the metal carries the whole charge.
Find the $d$ count.
$d = 6 - 3 = 3$
Chromium is in group $6$, so chromium(III) is $d^3$.
Find the charge of the complex ion in $\mathrm{[Co(NH_3)_5Cl]Cl_2}$.
$2 \times (-1) = -2 \Rightarrow \text{complex ion } {+2}$
The two chlorides outside the brackets are counter ions, and the salt is neutral.
List the ligands.
$5 \times \mathrm{NH_3}\ (0), \quad 1 \times \mathrm{Cl^-}\ (-1)$
Only the chloride inside the brackets is a ligand.
Add up the ligands' charges.
$5 \times 0 + 1 \times (-1) = -1$
Ammonia counts zero; chloride counts minus one.
Write and solve the charge balance.
$x + (-1) = +2 \Rightarrow x = +3$
Subtract the ligand charge from the complex ion's charge.
Find the $d$ count.
$d = 9 - 3 = 6$
Cobalt is in group $9$.
Compare with the tempting wrong answer.
$\text{complex charge} = +2 \ne x = +3$
Reading the bracket charge as the oxidation state would make this cobalt(II) and $d^7$.
Find the charge of the complex ion in $\mathrm{K_3[Fe(C_2O_4)_3]}$.
$3 \times (+1) = +3 \Rightarrow \text{complex ion } {-3}$
Three potassium counter ions must be balanced.
Add up the ligands' charges.
$3 \times (-2) = -6$
Oxalate, $\mathrm{C_2O_4^{2-}}$, carries $-2$, and there are three of them.
Write the charge balance.
$x + (-6) = -3$
Metal plus ligands equals the complex ion's charge.
Solve for the oxidation state.
$x = -3 + 6 = +3$
Adding six to both sides.
Find the $d$ count.
$d = 8 - 3 = 5$
Iron is in group $8$, so iron(III) is $d^5$.
Count the unpaired electrons, as in lesson 1.
$d^{5} \Rightarrow 5\ \text{unpaired (in a weak field)}$
Oxalate is a weak-field ligand, which lesson 11 explains; the five electrons stay unpaired.
Say why the salt is light-sensitive.
$\mathrm{Fe^{3+}} + e^- \to \mathrm{Fe^{2+}}$
Light moves an electron from an oxalate to the iron, reducing it, which is the chemistry of the old cyanotype photograph.
Add up the ligands' charges.
$6 \times (-1) = -6$
Cyanide is $\mathrm{CN^-}$.
Solve the charge balance.
$x + (-6) = -4 \Rightarrow x = +2$
Subtract the ligand charges from the ion's charge.
Find the $d$ count.
What is the oxidation state of the metal in $\mathrm{[Ni(CN)_4]^{2-}}$? Its $4$ ligands carry a total charge of $-4$.
Complete the worked solution: find the oxidation state and the $d$ count of the metal in $\mathrm{[MnO_4]^{-}}$, whose ligands carry $-8$ in all. The metal is in group $7$.
Subtract the ligands' charges from the ion's charge.
$x = (\text{ion charge}) - (\text{ligand charges}) =$ x
The metal's oxidation state is what is left of the ion's charge.
Subtract the oxidation state from the group number.
$d = (\text{group number}) - x =$ d
The metal keeps the electrons the oxidation state has not taken, all in $d$.
Check the pair against the group.
$x + d = \text{group number}$
The electrons the oxidation state took and the $d$ electrons left account for every valence electron.
Match each complex ion to the $d$ count of its metal.
| $d^{0}$ | $d^{1}$ | $d^{5}$ | $d^{8}$ | |
|---|---|---|---|---|
| $\mathrm{[CrO_4]^{2-}}$ | ||||
| $\mathrm{[Ti(H_2O)_6]^{3+}}$ | ||||
| $\mathrm{[Fe(C_2O_4)_3]^{3-}}$ | ||||
| $\mathrm{[Ni(en)_3]^{2+}}$ |
Fill in the table for the complex ions $\mathrm{[Cr(NH_3)_4Cl_2]^{+}}$, $\mathrm{[Fe(H_2O)_6]^{3+}}$ and $\mathrm{[Co(en)_3]^{3+}}$, in that order. The metals are in groups $6$, $8$ and $9$.
| total ligand charge | metal oxidation state | d electrons | |
|---|---|---|---|
| the first complex | |||
| the second complex | |||
| the third complex |
In the complex ion $\mathrm{[CrO_4]^{2-}}$ the ligands carry a total charge of $-8$. What is the oxidation state of $\mathrm{Cr}$? Give it as a number.
Answer: oxidation state of the metal
The salt $\mathrm{KMnO_4}$ contains the counter ion $\mathrm{K^{+}}$. How many $d$ electrons does the transition metal in it have? The metal is in group $7$.
Answer: d electrons on the metal
Three compounds met outside the laboratory: potassium ferrocyanide, the anticaking agent in some road salt, $\mathrm{K_4[Fe(CN)_6]}$; potassium permanganate, used to take iron out of well water, $\mathrm{KMnO_4}$; and Schweizer's reagent, which dissolves cotton, $\mathrm{[Cu(NH_3)_4](OH)_2}$. For the transition metal in each, in that order, fill in the charge of its complex, its oxidation state and its $d$ count. The metals are in groups $8$, $7$ and $11$; the ligands carry $-6$, $-8$ and $0$.
| charge of the complex | metal oxidation state | d electrons | |
|---|---|---|---|
| the first compound | |||
| the second compound | |||
| the third compound |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Fill in the table for the complex ions $\mathrm{[Ti(H_2O)_6]^{3+}}$, $\mathrm{[Fe(CN)_6]^{3-}}$ and $\mathrm{[CoCl_4]^{2-}}$, in that order. The metals are in groups $4$, $8$ and $9$.
| total ligand charge | metal oxidation state | d electrons | |
|---|---|---|---|
| the first complex | |||
| the second complex | |||
| the third complex |
You can find a metal's oxidation state and $d$ count in a complex. Explain why iron in $\mathrm{[Fe(CN)_6]^{4-}}$ is $+2$ and not $-4$, and name the check that would have caught the wrong answer.
15. Your turn: find the oxidation state and $d$ count of iron in $\mathrm{[Fe(CN)_6]^{4-}}$., step 3
$d = 8 - 2 = 6$
Iron(II) is $d^6$.