🧪 Contents · Interactive Lab Chapter 19 · Transition Metals and Coordination Chemistry
Chemistry · Chapter 19

Transition Metals and Coordination Chemistry

Why is ruby red and emerald green? Why does blood carry oxygen while rust does nothing of the kind? Both answers live in the middle of the periodic table. The metals there hold electrons in d orbitals, and those orbitals are close enough in energy to shift under a nudge. A ring of molecules around the metal gives that nudge. The result is color, magnetism and a huge family of compounds called complexes. This chapter shows you how to build one, name one, count its shapes and predict its color from the ligands alone.

By the end of this chapter you can

  • Write the electron configuration of a d-block atom and of its ions, taking the 4s electrons away first.
  • Pick out the metal, the ligands and the counter ions in a coordination compound, and find the metal's oxidation number.
  • Count donor atoms to get a coordination number, and sort ligands by denticity.
  • Name a coordination compound from its formula, and write the formula back from a name.
  • Spot structural, geometric and optical isomers, and say which ones are chiral.
  • Split the d orbitals with a crystal field, fill them high spin or low spin, and read the splitting off the spectrochemical series.
  • Turn a splitting energy into an absorbed wavelength, then name the color you would see.
  • Count unpaired electrons to get a magnetic moment, and explain why chelating ligands bind so tightly.

Section 1Transition metal properties and configurations

Where the d-block sits

Walk across the middle of the periodic table and you cross the d-block. It holds groups 3 through 12. Scandium through zinc make up its first row. A transition metal is a metal whose atoms or common ions carry a partly filled d subshell. That one trait drives everything else in this chapter.

These metals act alike in handy ways. They are hard and shiny. They melt at high temperatures. Tungsten melts at 3422 °C, the highest of any metal. They carry current well. Most of them show more than one charge. Iron gives you both Fe2+ and Fe3+. Manganese runs from +2 up to +7. Their compounds are often brightly colored, and many are tugged by a magnet.

Zinc, cadmium and mercury sit in the block but hold a full d10 set in every normal ion. Strict rules leave them out of the transition metal club. Most people still group them with the block for convenience.

Filling up: 4s goes in first

Energy order sets the filling order. In a neutral potassium or calcium atom, the 4s orbital lies a little below 3d. So 4s takes electrons first. Scandium is the first element with a 3d electron. Its configuration is [Ar]3d14s2.

Add protons one at a time and the pattern rolls on. Titanium is [Ar]3d24s2. Vanadium is [Ar]3d34s2. Notice that we write 3d before 4s, because 3 comes before 4. We fill 4s first, yet we list it second.

Two metals in the first row break the tidy pattern. Chromium is [Ar]3d54s1, not 3d44s2. Copper is [Ar]3d104s1, not 3d94s2. The two levels sit so close that a half filled or a filled d set is worth the shift. Molybdenum and silver do the same thing one row down.

Making ions: 4s leaves first

Here is the fact that trips people up most. When a transition metal forms a cation, the 4s electrons leave first. Every time.

Iron is [Ar]3d64s2. Pull two electrons off and you have Fe2+. The answer is [Ar]3d6, not [Ar]3d44s2. Pull a third and Fe3+ is [Ar]3d5.

Why the flip? The gap between 4s and 3d is tiny, and its size shifts with the number of d electrons. Once the 3d set holds electrons, 3d drops below 4s. The 4s orbital is also fatter and reaches further out, so its electrons are easier to strip. Both facts point the same way.

The 4s and 3d levels almost tie. In a neutral atom 4s lies lower and fills first. In a cation 3d lies lower, so the 4s electrons leave first.1 · The 4s and 3d near tie, and which way it tipsenergyNeutral Fe atom3d (five orbitals)4s (one orbital)tiny gap4s lies a little lower, so 4s fills firstenergyFe2+​ ion3d (five orbitals)4s (one orbital)tiny gap3d has dropped lower, so 4s empties firstThe same idea in boxes: strip the 4s pair, keep all six 3d electronsFe3d4s[Ar]3d6​4s2​Fe2+​3d4s[Ar]3d6​Fe3+​3d4s[Ar]3d5​
Figure 19.1 The 4s and 3d levels almost tie. In the neutral atom 4s wins by a hair, so it fills first. In the ion 3d has slipped below 4s, so the 4s pair is the first to go. Iron keeps all six of its 3d electrons through both steps.

Counting d electrons is then easy. Take the group number and subtract the charge. Iron sits in group 8, so Fe3+ has 8 − 3 = 5 d electrons. Copper sits in group 11, so Cu2+ has 9. You will lean on this count in every later section.

Key points
  • The d-block spans groups 3 to 12. A transition metal has a partly filled d subshell.
  • In neutral atoms 4s fills before 3d, but we always write 3d first.
  • Chromium and copper shift one electron to get a half filled or filled d set.
  • In cations the 4s electrons leave first, so Fe2+ is [Ar]3d6.
  • d electrons in an ion = group number − charge.

Worked example · configurations for cobalt and its ions

Write the electron configuration of Co, Co2+ and Co3+. Then count the unpaired electrons in each one.

Step 1 · place the neutral atomCobalt is element 27, in group 9. Argon covers 18 electrons. That leaves 9 to place. Fill 4s with 2, then put 7 in 3d. So Co is [Ar]3d74s2.

Step 2 · check the count18 + 7 + 2 = 27. The electrons are all accounted for.

Step 3 · take the 4s pair off firstCo2+ loses both 4s electrons. That gives [Ar]3d7. Co3+ loses the 4s pair plus one 3d electron, giving [Ar]3d6.

Step 4 · spread the d electrons by Hund's ruleFive d orbitals, filled singly first. With 7 electrons: 5 single, then 2 of them get a partner. That leaves 3 unpaired. With 6 electrons: 5 single, then 1 gets a partner, leaving 4 unpaired.

AnswerCo is [Ar]3d74s2. Co2+ is [Ar]3d7 with 3 unpaired. Co3+ is [Ar]3d6 with 4 unpaired. Those unpaired counts hold for a free ion. Section 5 shows that ligands can change them.

Worth knowing. The 4s and 3d levels are so close that a tiny change in charge reorders them. This is why d-block chemistry is rich and why main group chemistry is tidy.
Interactive Fill d orbitals and handle the ion exceptions. Open full ↑

Section 2Coordination compounds and ligands

A metal with a ring of partners

Drop a crystal of CuSO4 into water and the blue you see is not bare Cu2+. Six water molecules have latched onto the ion. The real species is [Cu(H2O)6]2+. Any species like that, a central metal ion with a ring of attached partners, is a coordination compound. If the whole unit carries a charge, we call it a complex ion.

Each attached partner is a ligand. A ligand always has at least one lone pair to share. It hands that pair to the metal, so the ligand is a Lewis base and the metal ion is a Lewis acid. The bond that forms is a coordinate covalent bond. Both electrons in it came from the ligand.

The atom on the ligand that actually holds the lone pair is the donor atom. In water it is the oxygen. In ammonia it is the nitrogen. In the cyanide ion it is the carbon. The metal plus its ligands make up the coordination sphere, and we wrap that in square brackets. Ions that just balance the charge from outside the brackets are counter ions.

Denticity: how many teeth does a ligand have?

Some ligands grab once. Some grab several times. The number of donor atoms one ligand uses is its denticity. The word comes from teeth, and the picture is a bite.

A monodentate ligand bites once. Water, ammonia, chloride, cyanide and hydroxide all work this way. A bidentate ligand bites twice with one body. Ethylenediamine, written en, has a nitrogen at each end and a short carbon chain between. The oxalate ion, C2O42-, bites with two oxygens.

A ligand with more teeth is polydentate. The champion in everyday use is EDTA, which bites six times with two nitrogens and four oxygens. One EDTA can take over a metal ion all by itself.

An octahedral complex with six donor atoms around one cobalt, and three ligands that bite with one, two or six donor atoms.2 · Six bonds to the metal, however many ligands that takesNH3​ClNH3​NH3​NH3​NH3​Cothe coordination sphere, written inside [ ][Co(NH3​)5​Cl]2+​coordination number 6: five ammonia plus one chlorideone donor atom (monodentate)Mammoniagrabs the metal oncetakes 1 of the six slotstwo donor atoms (bidentate)Methylenediaminetwo grabs make one ringtakes 2 of the six slotssix donor atoms (hexadentate)MEDTAwraps the ion completelytakes 6 of the six slotsThe dots are lone pairs. Every lone pair that lands on the metal counts as one bond.
Figure 19.2 Left: six donor atoms around one cobalt, boxed off as the coordination sphere. Right: the same six slots filled by six monodentate ligands, by three bidentate ligands, or by a single hexadentate EDTA.

Coordination number counts bonds, not ligands

The coordination number is the number of donor atoms bonded to the metal. It is a bond count, not a ligand count. In [Co(NH3)6]3+ both counts are 6, since each ammonia bites once. In [Co(en)3]3+ there are only 3 ligands, yet the coordination number is still 6. Three bidentate ligands supply six donor atoms.

Coordination number sets the shape. A count of 2 gives a line, as in [Ag(NH3)2]+. A count of 4 gives either a tetrahedron, as in [Zn(NH3)4]2+, or a flat square, as in [Ni(CN)4]2-. A count of 6 gives an octahedral shape, which is by far the most common. Think of the metal at the center of a cube with one ligand poking out through each face.

Coordination numberShapeExampleBond angles
2linear[Ag(NH3)2]+180°
4tetrahedral[CoCl4]2-109.5°
4square planar[Pt(NH3)2Cl2]90°
6octahedral[Fe(CN)6]3-90° and 180°
Key points
  • A coordination compound is a metal ion plus ligands. Charged, it is a complex ion.
  • Ligands donate a lone pair, so they are Lewis bases. The metal is the Lewis acid.
  • Denticity is the number of donor atoms on one ligand: mono, bi, or poly.
  • Coordination number counts bonds to the metal, not ligands.
  • Count 6 means octahedral, count 4 means tetrahedral or square planar, count 2 means linear.

Worked example · pull apart [Cr(en)2Cl2]Cl

Find the coordination number, the charge on the complex ion, and the oxidation number of chromium.

Step 1 · split inside from outsideInside the brackets: [Cr(en)2Cl2]. Outside: one Cl-. That lone chloride is a counter ion.

Step 2 · count donor atomsEach en is bidentate, so 2 en supply 2 × 2 = 4 donor atoms. The two chloride ligands add 2 more. Total = 4 + 2 = 6, so the shape is octahedral.

Step 3 · balance the outsideOne Cl- sits outside. The whole compound is neutral, so the complex ion must be 1+.

Step 4 · solve for the metalLet the oxidation number of Cr be x. en is neutral, so it adds 0. Each chloride ligand adds −1.

x + 2(0) + 2(−1) = +1, so x − 2 = +1, so x = +3.

AnswerCoordination number 6, complex ion charge 1+, chromium is +3. Chromium has 6 electrons past argon, so Cr3+ is [Ar]3d3.

Interactive Attach ligands and count coordination number. Open full ↑

Section 3Naming coordination compounds

Five moves, in order

A name for a coordination compound looks long. It is built from small pieces, always in the same order. Learn the five moves and you can name any of them.

First, name the cation, then the anion. That is the rule for every ionic compound, and it does not change here. Second, inside the brackets, list the ligands in alphabetical order, then the metal. Third, show how many of each ligand there are. Fourth, add an ending to the metal if the complex is negative. Fifth, give the metal's oxidation number in Roman numerals inside parentheses.

The whole coordination sphere becomes one single word. No spaces inside it.

Ligand names and ligand counts

A ligand that carries a negative charge gets an -o ending. Chloride becomes chlorido, cyanide becomes cyanido, hydroxide becomes hydroxido, oxalate becomes oxalato. Older books drop the id and write chloro and cyano. Both forms are still in print, so you should know each one.

Neutral ligands usually keep their own name, but three of them have special names you simply have to learn. Water becomes aqua. Ammonia becomes ammine, with two letters m. Carbon monoxide becomes carbonyl.

For counts, use the Greek prefixes di, tri, tetra, penta and hexa. If the ligand name already contains a prefix, switch to bis, tris or tetrakis and wrap the ligand in parentheses. Two ethylenediamines become bis(ethylenediamine), never diethylenediamine. The prefix never affects alphabetical order. Sort by the ligand name itself, so tetraammine files under a, not under t.

The formula for potassium hexacyanidoferrate three, split into its parts, with five naming moves listed beside the name piece each one adds.3 · One formula, five naming movesK3​[Fe(CN)6​]3−​outside the bracketsthree K+​ ionsso the complex carries a 3− chargeinside the bracketsone Fe and six CN−​so it is a complex anion1Cation firstPotassium is the counter ion, so it leads.potassium2Ligands, alphabeticalSix cyanide ligands give hexacyanido.hexacyanido3Metal nextIron follows, inside the same word....ferr...4An anion ends in -ateThe complex is negative, so use ferrate....ferrate5Oxidation number lastx + 6(−1) = −3, so x = +3. Write (III).ferrate(III)Finished name: potassium hexacyanidoferrate(III)
Figure 19.3 Every piece of the name traces back to a piece of the formula. The counter ion leads, the ligands come next in alphabetical order, and the metal closes the word with its oxidation number.

When the complex is an anion

If the coordination sphere carries a negative charge, the metal name ends in -ate. Several metals switch to a Latin stem at the same time. Iron becomes ferrate. Copper becomes cuprate. Silver becomes argentate. Gold becomes aurate. Lead becomes plumbate and tin becomes stannate. Most other metals just take the ending, as in nickelate, cobaltate and chromate.

Three worked names show the pattern. [Co(NH3)6]Cl3 is hexaamminecobalt(III) chloride. [Cu(NH3)4]SO4 is tetraamminecopper(II) sulfate. Na2[Zn(OH)4] is sodium tetrahydroxidozincate(II). Only the third one is an anion, so only the third one picks up -ate.

Key points
  • Cation first, then anion. The coordination sphere is one word.
  • Ligands come before the metal, sorted alphabetically by ligand name.
  • Negative ligands end in -o. Water is aqua, ammonia is ammine, CO is carbonyl.
  • Use bis, tris and tetrakis with parentheses for ligands that hold their own prefix.
  • A negative complex gives the metal an -ate ending, often on a Latin stem.
  • The oxidation number goes last, in Roman numerals, in parentheses.

Worked example · name [Cr(H2O)4Cl2]Br

Step 1 · find the counter ionOne Br- sits outside the brackets. It is the anion, so it is named last as bromide. The complex is the cation, so it leads.

Step 2 · charge on the complexThe compound is neutral and the bromide is −1. So the complex ion is 1+.

Step 3 · oxidation number of chromiumx + 4(0) + 2(−1) = +1. So x = +3, and we write (III).

Step 4 · name and sort the ligandsFour waters become tetraaqua. Two chlorides become dichlorido. Sort by ligand name: aqua before chlorido, since a beats c.

Step 5 · build one wordtetraaqua + dichlorido + chromium(III). The complex is positive, so chromium keeps its plain name.

Answertetraaquadichloridochromium(III) bromide. In the older style it is tetraaquadichlorochromium(III) bromide.

Worked example · go the other way

Write the formula for potassium hexafluoridocobaltate(III).

Step 1 · read the piecesPotassium is the counter cation. Hexafluorido means six F- ligands. Cobaltate means the complex is an anion. The (III) sets cobalt at +3.

Step 2 · add up the charge inside+3 + 6(−1) = −3. So the complex ion is [CoF6]3-.

Step 3 · balance with potassiumEach K is +1, so you need three of them.

AnswerK3[CoF6]

Worth knowing. Ammine has two letters m and means NH3. Amine has one and means an organic nitrogen group. One letter changes the whole meaning, so write it with care.
Interactive Name a complex piece by piece. Open full ↑

Section 4Isomerism in complexes

Same formula, different compound

Isomers share a formula but differ in how the atoms are joined or placed. Complexes offer more ways to do this than almost any other family of compounds. The isomers split into two camps. Structural isomers differ in which atoms are bonded to what. Stereoisomers have the same bonds but a different arrangement in space.

Four kinds of structural isomer

In an ionization isomer, a ligand and a counter ion trade places. [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br hold the same atoms. Only one of them drops free sulfate into water, so a simple test tube check tells them apart.

A hydrate isomer moves a water molecule in or out of the sphere. [Cr(H2O)6]Cl3 is violet. [Cr(H2O)5Cl]Cl2·H2O is pale green. Same formula, different color.

A coordination isomer swaps ligands between two metals. [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6] are a matched pair.

A linkage isomer uses a different donor atom on the same ligand. Thiocyanate, SCN-, can bond through its sulfur or through its nitrogen. Nitrite can bond through nitrogen or through oxygen. A ligand with that choice is called ambidentate.

Cis and trans

Geometric isomers differ only in which positions the ligands take. The classic case is a flat square with two ligands of each kind. If the two chlorides touch at a corner, that is cis. If they sit straight across from each other, that is trans.

This difference is not a detail. cis-[Pt(NH3)2Cl2] is cisplatin, a drug that treats several cancers. The trans form has no such use. The two chlorides in the cis form are close enough to bind two neighboring sites on a strand of DNA. In the trans form they point the wrong way.

Octahedral complexes do the same trick. In [Co(NH3)4Cl2]+ the two chlorides are either 90° apart, which is cis, or 180° apart, which is trans. A tetrahedron cannot do this at all. Every corner of a tetrahedron touches every other corner, so there is no far side.

Cis and trans square planar isomers, cis and trans octahedral isomers, and a pair of mirror image complexes that cannot be stacked on each other.4 · Same formula, different arrangementcis, square planarClClNH3​NH3​Ptcisplatin: it fights tumourstrans, square planarClNH3​ClNH3​Pttransplatin: no use in therapycis, octahedralClNClNNNCothe two Cl sit 90° aparttrans, octahedralClClNNNNCothe two Cl sit 180° apartA mirror image you cannot stack back on the originalNNNNNNCoNNNNNNComirrorleft handed formright handed formthree bidentate rings, twisted one waythe same rings, twisted the other way
Figure 19.4 Top: cis and trans forms in a square plane and in an octahedron. Bottom: two complexes built from three bidentate ligands. They are mirror images, and no amount of turning makes one match the other.

Left handed and right handed

Some complexes come in two forms that are mirror images and cannot be laid on top of each other. Your hands work the same way. Such a molecule is chiral, and the pair are enantiomers.

[Co(en)3]3+ is the standard example. Three bidentate rings wind around the metal like the blades of a propeller, and a propeller twists one way or the other. The two forms are chemically alike in almost every test. They part ways when they meet polarized light. One form turns the plane of the light to the right, the other turns it left by the same amount. That is optical isomerism.

An equal mixture of the two shows no net turn, because the two effects cancel. Chemists call that a racemic mixture. The quick test for chirality is to hunt for a mirror plane in the molecule. If one exists, the molecule is not chiral.

Key points
  • Structural isomers differ in bonding. Stereoisomers differ only in placement.
  • Ionization, hydrate, coordination and linkage isomers are the four structural types.
  • cis means neighboring, trans means across. Tetrahedral complexes have neither.
  • Cisplatin works and transplatin does not, because of the angle alone.
  • Enantiomers are mirror image pairs. They rotate polarized light in opposite directions.
  • If a molecule has a mirror plane, it is not chiral.

Worked example · count the stereoisomers of [Co(en)2Cl2]+

Step 1 · get the shapeTwo en give 4 donor atoms. Two chlorides add 2. The coordination number is 6, so the complex is octahedral.

Step 2 · place the two chloridesThey can be 90° apart or 180° apart. That is one cis form and one trans form.

Step 3 · test the trans form for a mirror planeWith the chlorides opposite, the two en rings lie in one plane through the metal. That plane is a mirror. So trans is not chiral.

Step 4 · test the cis formWith the chlorides at 90°, no mirror plane survives. The cis form is chiral, so it exists as two enantiomers.

AnswerThree stereoisomers in all: trans, plus the two mirror forms of cis. Only the two cis forms turn polarized light, and they turn it in opposite directions.

Interactive Flip between cis, trans and optical isomers. Open full ↑

Section 5Crystal field theory and color

Ligands push the d orbitals apart

The five d orbitals in a free ion all share one energy. Bring six ligands up along the x, y and z axes and that changes. Each ligand carries a lone pair, and lone pairs repel the electrons already on the metal. The whole d set is pushed up, but not evenly. Crystal field theory treats each ligand as a small point of negative charge and tracks that uneven push.

Two of the five d orbitals point straight down the axes, right at the incoming ligands. Those two are dz2 and dx2-y2, and they get pushed up hard. We call the pair the eg set. The other three point into the gaps between the axes, so they feel less. Those three are dxy, dxz and dyz, and they form the t2g set.

The energy gap between the two sets is the crystal field splitting energy, written Δ. The average energy does not move, so the three lower orbitals drop by 0.4Δ while the two upper ones rise by 0.6Δ.

A tetrahedral field flips the picture. Now no orbital points straight at a ligand, and the set of three ends up above the set of two. The gap is also smaller, roughly four ninths of the octahedral value.

High spin or low spin

Picture filling the split levels with 6 d electrons. You have two choices. Spread them four below and two above, leaving four unpaired. Or cram all six into the lower set, with none unpaired.

Each choice has a price. Climbing to the upper set costs Δ. Pairing two electrons in one orbital costs the pairing energy, P. Whichever is cheaper wins. If Δ is small, climbing is cheap, so electrons spread out. That is a high-spin complex. If Δ is large, pairing is cheaper, so electrons stack in the lower set. That is a low-spin complex.

The choice only exists for d4 through d7. With fewer or more electrons, both routes give the same answer. Tetrahedral complexes have a small gap, so they are high spin in practice.

The spectrochemical series

So what sets Δ? Mostly the ligands. Rank them by the size of the split and you get the spectrochemical series. Weak field ligands sit on the left, strong field ligands on the right.

I- < Br- < Cl- < F- < OH- < H2O < NH3 < en < NO2- < CN- < CO

Two other rules help. A higher charge on the metal pulls the ligands closer, so Δ grows. And Δ grows going down a group, so 4d and 5d metals are nearly always low spin.

Octahedral d orbital splitting drawn twice, once with a weak field and high spin filling and once with a strong field and low spin filling, above a chart pairing each absorbed colour with the colour you see.5 · The split, the two fillings, and the colour you seeWeak field, high spin: d6​averageeg​ points straight at the ligandst2g​ points between the ligandsΔSmall Δ. Climbing costs less than pairing. Four unpaired.Strong field, low spin: d6​averageeg​ points straight at the ligandst2g​ points between the ligandsΔLarge Δ. Pairing costs less than climbing. None unpaired.Absorb one band and you see the opposite colour→violet410 nmyellowyou see→blue470 nmorangeyou see→green530 nmpurpleyou see→yellow580 nmvioletyou see→orange620 nmblueyou see→red700 nmgreenyou see
Figure 19.5 The same d6 ion in a weak field and in a strong field. A small gap leaves four unpaired electrons. A large gap pairs them all. Below, each absorbed band and the color that gets through.

Why complexes are colored

A d electron in the lower set can jump to the upper set if a photon hands it exactly Δ of energy. That jump is a d-d transition. The photon is swallowed, so that part of the light never reaches your eye. The color you see is what is left over.

So the color you see is the complementary color of the color absorbed. Absorb green and the solution looks purple. Absorb orange and it looks blue. Absorb violet and it looks yellow. This is the step most people get backwards. A green solution is green because it is not absorbing green.

Link the gap and the wavelength with one equation. Here h is 6.626 × 10-34 J·s and c is 2.998 × 108 m/s.

Δ = hc / λ

A big gap means a short wavelength. Swap in a stronger ligand and the band slides toward the blue end, so the color you see slides too.

Key points
  • Six ligands split the d orbitals into a lower t2g set of three and an upper eg set of two.
  • The lower set drops 0.4Δ and the upper set rises 0.6Δ.
  • Δ smaller than the pairing energy gives high spin. Δ larger gives low spin.
  • The spectrochemical series ranks ligands from weak field to strong field.
  • Δ = hc/λ, so a bigger gap absorbs a shorter wavelength.
  • You see the complementary color of the light that gets absorbed.

Worked example · from wavelength to Δ

[Ti(H2O)6]3+ absorbs most strongly at 493 nm. Find Δ in kJ/mol and name the color of the solution.

Step 1 · put the wavelength in meters493 nm × (1 m / 109 nm) = 4.93 × 10-7 m

Step 2 · energy of one photonΔ = hc/λ = (6.626 × 10-34 J·s)(2.998 × 108 m/s) / (4.93 × 10-7 m)

The top is 1.987 × 10-25 J·m. Divide: 1.987 × 10-25 / 4.93 × 10-7 = 4.03 × 10-19 J per photon.

Step 3 · scale to a mole4.03 × 10-19 J × 6.022 × 1023 /mol = 2.43 × 105 J/mol = 243 kJ/mol

Step 4 · name the color493 nm is blue-green light. Take blue-green out of white light and red plus violet are left. Your eye blends those into purple.

AnswerΔ = 243 kJ/mol, and the solution looks purple. Titanium(III) has one d electron, so it makes one clean jump. Ions with more d electrons absorb at several wavelengths, and the strongest band sets the color.

Worked example · from Δ back to a color

For [Cr(H2O)6]3+, Δ = 208 kJ/mol. What wavelength does it absorb, and what color is the solution?

Step 1 · get down to one photon208 kJ/mol = 2.08 × 105 J/mol. Divide by Avogadro's number: 2.08 × 105 / 6.022 × 1023 = 3.454 × 10-19 J.

Step 2 · rearrange for wavelengthλ = hc/Δ = 1.987 × 10-25 J·m / 3.454 × 10-19 J = 5.75 × 10-7 m

Step 3 · convert to nanometers5.75 × 10-7 m × (109 nm / 1 m) = 575 nm

Step 4 · read the color wheel575 nm is yellow. The complement of yellow is violet.

AnswerIt absorbs near 575 nm and looks violet. Swap water for ammonia and Δ grows to 258 kJ/mol. The band moves to 464 nm, which is blue, so [Cr(NH3)6]3+ looks yellow instead.

Interactive Split d orbitals with ligands and predict the color absorbed. Open full ↑

Section 6Magnetism, chelation and uses

Counting unpaired electrons with a magnet

An electron spins, and a spinning charge acts like a tiny magnet. Pair two electrons in one orbital and their magnets cancel. Leave one alone and its magnet shows. So a sample with unpaired electrons gets pulled into a magnetic field. We call that paramagnetic. A sample with every electron paired is pushed out a little instead, and we call it diamagnetic.

The pull can be weighed. Hang a sample between the poles of a magnet and record the change in apparent mass. From that you get the magnetic moment, μ, measured in Bohr magnetons. For most first row complexes a simple formula gets you close. Here n is the number of unpaired electrons.

μ = √ n(n + 2) Bohr magnetons

Run the numbers and you get 1.73 for one unpaired electron, 2.83 for two, 3.87 for three, 4.90 for four and 5.92 for five. Those values are far enough apart that a measurement tells you n with no guessing. That is how chemists prove a complex is high spin or low spin.

A few metals do something stronger. In iron, cobalt and nickel metal, whole blocks of atoms line their spins up and stay lined up. That is ferromagnetism, and it is what makes a permanent magnet.

Why a claw holds tighter

A chelate is a complex in which one ligand bites more than once and closes a ring. The name comes from the Greek word for a crab's claw. Chelating ligands bind far more tightly than the same number of single bite ligands.

Compare two nickel complexes. Both give the metal six nitrogen donors. [Ni(NH3)6]2+ uses six ammonia molecules. [Ni(en)3]2+ uses three ethylenediamines. The en version is nearly ten billion times more stable. The bonds themselves are nearly the same, so something else is at work.

That something is particle count. Replacing six waters with six ammonias trades six free molecules for six. Replacing six waters with three en molecules frees six waters while using only three ligands. The number of loose particles goes up, and rising disorder favors the change. This bonus is the chelate effect.

Heme holds iron inside a flat ring of four nitrogen donors with an oxygen molecule bound above it, and EDTA wraps a lead ion with two nitrogen and four oxygen donors.6 · Two claws that matter: heme and EDTAHeme: iron inside a flat ring of four nitrogensNNNNFeO2​rides on the sixth slotNthe protein holds on belowFour ring nitrogens and one protein nitrogen fill five slots. The sixth takes O2​ or CO.EDTA: one ligand, six donor atoms, five ringsNNOOOOPbTwo nitrogens and four oxygens take all six slots at once.Six waters leave and one EDTA arrives, so the particle count rises.
Figure 19.6 Left: heme holds iron in a flat ring of four nitrogens, with one slot left open on top for O2. Right: EDTA closes five rings around one metal ion, using all six slots by itself.

Where this shows up

Your own blood runs on a chelate. Heme wraps Fe2+ in a flat ring of four nitrogens. A nitrogen from the protein holds the fifth slot, and O2 rides on the sixth. Carbon monoxide fits that same slot and binds about 200 times tighter, which is exactly why it is so dangerous. Chlorophyll uses the same ring with Mg2+ in the middle. Vitamin B12 uses a cousin of it with cobalt.

EDTA is the workhorse of the lab and the grocery store. It takes all six slots on a metal ion at once. Salad dressing lists it because trace iron and copper ions speed up spoiling, and EDTA locks them away. Shampoo uses it to tame hard water. Doctors use a calcium form of it to pull lead out of a poisoned patient, a treatment called chelation therapy.

Complexes do heavy industrial work too. Gold is pulled from crushed rock as [Au(CN)2]-. Film photography fixed images by washing silver salts away as [Ag(S2O3)2]3-. Nickel is purified by turning it into Ni(CO)4, a gas, then heating it back to metal. Cisplatin treats cancer. And gadolinium chelates light up an MRI scan, because Gd3+ carries seven unpaired electrons.

Key points
  • Unpaired electrons make a complex paramagnetic. All paired makes it diamagnetic.
  • μ = √n(n + 2) in Bohr magnetons, so a measured μ gives n.
  • A chelate ligand bites twice or more and closes a ring.
  • The chelate effect is mostly about particle count, not stronger bonds.
  • Heme, chlorophyll and vitamin B12 are all chelates of a flat nitrogen ring.
  • EDTA, cisplatin, gold mining and MRI dyes all run on coordination chemistry.

Worked example · read the spin state off a magnet

Two iron(II) complexes are measured. [Fe(H2O)6]2+ gives μ = 5.3 Bohr magnetons. [Fe(CN)6]4- gives μ = 0. Explain both results.

Step 1 · count the d electronsIron is group 8, so Fe2+ has 8 − 2 = 6 d electrons. Both complexes are d6 and octahedral.

Step 2 · test μ against the formulaFor n = 4, μ = √(4 × 6) = √24 = 4.90. For n = 0, μ = 0. The measured 5.3 is a little above 4.90, which is normal for iron(II), so n = 4.

Step 3 · match each to a spin stateFour unpaired means t2g4eg2, which is high spin. Zero unpaired means t2g6, which is low spin.

Step 4 · check against the seriesWater sits low in the spectrochemical series, so Δ is small and the electrons spread out. Cyanide sits near the top, so Δ is large and they pair up in the lower set.

AnswerThe water complex is high spin and paramagnetic. The cyanide complex is low spin and diamagnetic. Same metal, same charge, same d count. Only the ligand changed.

Worth knowing. The spin only formula ignores any help from orbital motion. For first row metals the match is usually good. Measured values for cobalt(II) and iron(II) often run a bit high, which is why 5.3 still points to four unpaired electrons.
Interactive Count unpaired electrons and see chelation at work. Open full ↑

The chapter in one breath

  • The d-block runs from group 3 to group 12. Its metals hold a partly filled d subshell, and that is the source of their color, their many charges and their magnetism.
  • In a neutral atom 4s fills before 3d, but we write 3d first. Chromium and copper shift one electron to reach a half filled or filled d set.
  • When a cation forms, the 4s electrons leave first. So Fe2+ is [Ar]3d6, and d electrons = group number − charge.
  • A coordination compound is a metal ion plus ligands. Ligands donate lone pairs, so they are Lewis bases.
  • Coordination number counts donor atoms, not ligands. Six donors give an octahedron, four give a tetrahedron or a square, two give a line.
  • Names run cation first, then ligands in alphabetical order, then the metal with its oxidation number. A negative complex ends in -ate.
  • Complexes come as structural isomers, which differ in bonding, and stereoisomers, which differ only in placement. cis and trans change what a complex can do, as cisplatin shows.
  • Ligands split the d orbitals by Δ. A small Δ gives a high-spin complex, a large Δ gives a low-spin one, and the spectrochemical series ranks the ligands.
  • Δ = hc/λ, and you see the complementary color of the light absorbed.
  • Unpaired electrons set μ = √n(n + 2), and chelating ligands such as EDTA, heme and chlorophyll hold on by closing rings.

ReferenceWords worth knowing

These are the terms this chapter leans on most. Come back to the list whenever a formula or a name stops making sense.

bidentate
A ligand that uses two donor atoms and closes one ring, such as ethylenediamine or oxalate.
chelate
A complex in which one ligand bites twice or more, forming a ring with the metal. Named for a crab's claw.
complementary color
The color you see when one band is taken out of white light. Absorb green and you see purple.
coordination compound
A central metal ion surrounded by bonded ligands, with the sphere written inside square brackets.
coordination number
The number of donor atoms bonded to the metal. It counts bonds, not ligands.
crystal field splitting energy
Δ, the energy gap that ligands open between the t2g and eg sets of d orbitals.
d-block
Groups 3 through 12 of the periodic table, where added electrons go into d orbitals.
denticity
How many donor atoms one ligand uses. Monodentate is one, bidentate is two, hexadentate is six.
diamagnetic
Having every electron paired, so the sample is pushed slightly out of a magnetic field.
donor atom
The atom on a ligand that actually holds the lone pair given to the metal.
EDTA
A hexadentate ligand with two nitrogen and four oxygen donors. It wraps a metal ion on its own.
eg orbitals
The upper pair in an octahedral field. They point straight at the ligands, so they rise 0.6Δ.
enantiomers
A pair of complexes that are mirror images and cannot be stacked. They rotate polarized light in opposite directions.
high-spin complex
A complex with a small Δ, so electrons spread into the upper set rather than pair up.
ligand
A molecule or ion that donates at least one lone pair to a metal ion. Every ligand is a Lewis base.
low-spin complex
A complex with a large Δ, so electrons pair in the lower set rather than climb.
paramagnetic
Having one or more unpaired electrons, so the sample is pulled into a magnetic field.
spectrochemical series
Ligands ranked by the size of the split they cause, from weak field iodide up to strong field carbon monoxide.
t2g orbitals
The lower three in an octahedral field. They point between the ligands, so they drop 0.4Δ.
transition metal
A metal whose atoms or common ions carry a partly filled d subshell.

Check yourself

1. Write the electron configuration of Mn and of Mn2+, then count the unpaired electrons in the ion.
Mn is [Ar]3d54s2, Mn2+ is [Ar]3d5, with 5 unpaired. Manganese is element 25 in group 7, so argon's 18 plus 7 more. Put 2 in 4s and 5 in 3d: 18 + 5 + 2 = 25. The ion loses the 4s pair first, leaving 5 d electrons. Five electrons in five orbitals sit one apiece, so all 5 are unpaired.
2. Find the oxidation number of platinum in [PtCl6]2-.
+4 Each chloride ligand is −1, and there are six: 6 × (−1) = −6. Set up the balance: x + (−6) = −2. So x = −2 + 6 = +4. Platinum(IV) with six donor atoms is octahedral.
3. For [Co(en)2(NO2)2]Cl, give the coordination number, the charge on the complex ion and the oxidation number of cobalt.
Coordination number 6, complex charge 1+, cobalt +3. Each en is bidentate: 2 × 2 = 4 donor atoms. The two nitrite ligands add 2, so 4 + 2 = 6. One Cl- sits outside, so the complex must be 1+. Then x + 2(0) + 2(−1) = +1, which gives x = +3.
4. Why is Fe2+ written [Ar]3d6 and not [Ar]3d44s2?
Because the 4s electrons leave first. Once 3d holds electrons, the 3d level drops below 4s, and the 4s orbital reaches further from the nucleus anyway. Both effects make the 4s pair the easiest to remove. Filling order and removal order are not the same thing.
5. Name [Cr(NH3)4Cl2]Cl.
tetraamminedichloridochromium(III) chloride One chloride sits outside, so the complex is 1+ and chromium is +3, since x − 2 = +1. Ammonia becomes ammine and chloride becomes chlorido. Sort by ligand name, so ammine comes before chlorido. The older spelling is tetraamminedichlorochromium(III) chloride.
6. A complex absorbs light at 600 nm. Find Δ in kJ/mol, and say what color the solution looks.
Δ = 199 kJ/mol, and the solution looks blue. Δ = hc/λ. The product hc is (6.626 × 10-34)(2.998 × 108) = 1.987 × 10-25 J·m. Divide by 6.00 × 10-7 m to get 3.31 × 10-19 J per photon. Times 6.022 × 1023 gives 1.99 × 105 J/mol, or 199 kJ/mol. Light at 600 nm is orange, and the complement of orange is blue.
7. A complex has Δ = 290 kJ/mol. What wavelength does it absorb, and what color do you see?
412 nm, and it looks yellow. One photon carries 2.90 × 105 J/mol ÷ 6.022 × 1023 /mol = 4.816 × 10-19 J. Then λ = hc/Δ = 1.987 × 10-25 ÷ 4.816 × 10-19 = 4.12 × 10-7 m = 412 nm. That is violet light, and the complement of violet is yellow.
8. Which is low spin, [CoF6]3- or [Co(CN)6]3-? How many unpaired electrons does each have?
[Co(CN)6]3- is low spin with 0 unpaired. [CoF6]3- is high spin with 4. Both are Co3+, which is d6. Cyanide sits near the top of the spectrochemical series, so Δ beats the pairing energy and all six electrons crowd into t2g. Fluoride sits low, so Δ loses and the filling is t2g4eg2.
9. A complex is measured at μ = 3.9 Bohr magnetons. How many unpaired electrons does it have?
Three. Test values with μ = √n(n + 2). For n = 2: √(2 × 4) = √8 = 2.83. For n = 3: √(3 × 5) = √15 = 3.87. For n = 4: √(4 × 6) = √24 = 4.90. The measured 3.9 matches n = 3.
10. How many stereoisomers does square planar [Pt(NH3)2Cl2] have, and is either one chiral?
Two, cis and trans, and neither is chiral. The two chlorides either share a corner, which is cis, or sit opposite, which is trans. A flat square always has the molecular plane itself as a mirror plane, so neither form is chiral. The cis form is the drug cisplatin.
11. Why does one EDTA ion hold a metal far more tightly than six water molecules do?
The chelate effect, which is mostly about particle count. EDTA takes all six slots at once, so six waters are set free while only one ligand is used up. Two particles become seven. Rising disorder favors that trade, and once one arm of EDTA is attached the others are already nearby.
12. [CoCl4]2- is tetrahedral. Is it high spin or low spin, and how many unpaired electrons does it carry?
High spin, with 3 unpaired electrons. Cobalt is group 9, so Co2+ is 9 − 2 = 7 d electrons. A tetrahedral gap is only about four ninths of the octahedral one, and chloride is a weak field ligand on top of that, so Δ never beats the pairing energy. The lower pair of orbitals takes 4 electrons and the upper three take 1 each, leaving 3 unpaired.