Ligand field theory · printable packet

The molecular orbitals of an octahedral complex, bonding, nonbonding and antibonding, the eighteen-electron rule from the orbitals, net bonds and lability, and how $\pi$ donors and acceptors set the spectrochemical series.

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1. What you will learn

By the end of this lesson you will be able to place a complex's electrons in its octahedral molecular orbital diagram, count its net bonds, and explain the spectrochemical series from $\pi$ bonding.

2. What you already have

Lessons 10 and 11 split the d orbitals into $t_{2g}$ and $e_g$ by treating the ligands as points of negative charge, and lesson 15 counted eighteen electrons as a stable number without saying where it came from. Lessons 24 and 25 built molecular orbitals from overlapping atomic orbitals and showed carbon monoxide donating and accepting. This lesson puts the two together: the molecular orbitals of a whole octahedral complex.

3. Words for this lesson

4. The molecular orbitals of an octahedron

Term	What it means
Ligand field theory	The molecular orbital description of a complex, which keeps the crystal field's splitting and explains its size.
Ligand group orbital	A combination of the six ligand lone pairs with the right symmetry to overlap one metal orbital.
Nonbonding orbital	A metal orbital with no ligand orbital to overlap; in a $\sigma$-only octahedron, the $t_{2g}$ set.
$e_g^*$	The antibonding combination of the metal's $d_{z^2}$ and $d_{x^2-y^2}$ with ligand orbitals.
$\pi$ donor	A ligand with filled orbitals at right angles to the bond that interact with $t_{2g}$ and raise it.
$\pi$ acceptor	A ligand with empty $\pi^*$ orbitals that take density from $t_{2g}$ and lower it.
Inert complex	One that exchanges its ligands slowly, typically with an empty $e_g^*$ set.

In an octahedral complex the six ligands each point one lone pair at the metal. Those six lone pairs combine into ligand group orbitals of particular symmetries, and each overlaps only the metal orbital of matching symmetry: one combination matches the metal's 4s orbital, three match the three 4p orbitals, and two match $d_{z^2}$ and $d_{x^2-y^2}$, the $e_g$ pair that points at the ligands. Each overlap makes a bonding orbital, low in energy, and an antibonding one, high.

The three orbitals $d_{xy}$, $d_{xz}$ and $d_{yz}$, the $t_{2g}$ set, point between the ligands and have no ligand $\sigma$ orbital to overlap. They stay nonbonding, at about their original energy. So the diagram, from the bottom, has six bonding orbitals, then the three nonbonding $t_{2g}$, then the two antibonding $e_g^$, then higher antibonding orbitals from 4s and 4p. The gap between $t_{2g}$ and $e_g^$ is exactly the $\Delta_o$ of crystal field theory.

Now count electrons. The six ligand lone pairs bring twelve electrons, which fill the six bonding orbitals. The metal's $d$ electrons go into $t_{2g}$ and $e_g^*$, high spin or low spin as lesson 11 decided. So a complex has $12 + d$ valence electrons in the diagram, and when that total is eighteen, every bonding and nonbonding orbital is full and every antibonding one empty: the eighteen-electron rule of lesson 15, now explained.

Because $e_g^$ is antibonding, every electron in it cancels half of a metal-ligand bond. Twelve bonding electrons make six $\sigma$ bonds; with $e_g^$ electrons present, the net is

A low-spin $d^6$ complex such as $\mathrm{[Co(NH_3)_6]^{3+}}$ has an empty $e_g^$ and six full bonds; high-spin $\mathrm{[Co(H_2O)_6]^{2+}}$, with two $e_g^$ electrons, has five.

Ligands with $\pi$ orbitals change the picture at $t_{2g}$. A $\pi$ donor such as chloride has filled p orbitals at right angles to the bond; they overlap $t_{2g}$, lie below it, and push it up, narrowing $\Delta_o$. A $\pi$ acceptor such as carbon monoxide has empty $\pi^*$ orbitals above $t_{2g}$; they pull it down, widening $\Delta_o$. That is the spectrochemical series explained: $\pi$ donors at the weak end, $\sigma$-only ligands in the middle, $\pi$ acceptors at the strong end.

Another way: picture

Picture three floors in a building. On the ground floor are the six bonding orbitals, always filled by the ligands' twelve electrons. On the first floor are the three $t_{2g}$ orbitals, where the metal's electrons go first. On the second floor are the two $e_g^*$ orbitals, and any electron that has to live there pulls against the bonds holding the building together. $\pi$ ligands move the first floor: donors raise it toward the second, acceptors lower it away.

Another way: steps

Twelve ligand electrons fill the six bonding orbitals.
Place the metal's d electrons in $t_{2g}$ (nonbonding) and $e_g^*$ (antibonding) by the spin state.
Total valence electrons $= 12 + d$; eighteen fills everything below $e_g^*$.
Net $\sigma$ bonds $= \frac{1}{2}(12 - n_{e_g^*})$.
$\pi$ donors raise $t_{2g}$ (smaller $\Delta_o$); $\pi$ acceptors lower it (larger $\Delta_o$).

5. What the crystal field could not explain

Crystal field theory predicts that the more negative a ligand, the larger the splitting it causes, since a point charge's push should grow with its charge. The spectrochemical series says otherwise: neutral water and ammonia split the d orbitals more than hydroxide and fluoride, and neutral carbon monoxide more than any anion. A point charge cannot tell the difference between a lone pair that is spare and one that is not, or between a ligand with empty orbitals and one without.

Ligand field theory can. Fluoride and hydroxide carry extra filled p orbitals that interact with $t_{2g}$ and raise it, which a point charge has no way to represent. Ammonia has no spare lone pair and leaves $t_{2g}$ alone. Carbon monoxide has empty $\pi^*$ orbitals that lower $t_{2g}$. The order of the series, $\pi$ donors below $\sigma$ donors below $\pi$ acceptors, follows directly, and so does the observation that the covalent, back-bonding ligands are the ones that split most.

6. Bond strength and lability

The count of net bonds explains one of the most practical facts in coordination chemistry: some complexes swap their ligands in a fraction of a second and others keep them for days. Henry Taube, who won the Nobel Prize in Chemistry in 1983 for work on how complexes react, found that the slow ones, which he called inert, are usually those with no electrons in the antibonding $e_g^*$ orbitals, such as low-spin $d^6$ cobalt(III), and $d^3$ chromium(III), whose three electrons all sit in $t_{2g}$.

Complexes with electrons in $e_g^$ are labile: the antibonding electrons weaken the bonds, and a ligand leaves easily. High-spin $d^4$ chromium(II), with one $e_g^$ electron, exchanges its water ligands trillions of times faster than chromium(III). Chemists exploit the difference constantly: they build a complex quickly on a labile metal, then oxidize it to an inert one to lock the ligands in place, which is how many cobalt(III) complexes, including those Werner studied, are made.

7. The eighteen-electron rule, from the orbitals

Lesson 15 stated the eighteen-electron rule as a count; the diagram shows why it works. Six bonding orbitals and three nonbonding $t_{2g}$ make nine low-lying orbitals, and eighteen electrons fill them exactly. A nineteenth would go into $e_g^*$ and weaken the bonds; a seventeenth leaves a $t_{2g}$ orbital half empty.

The rule works best when the $t_{2g}$ orbitals are lowered by $\pi$ acceptors, which makes the gap to $e_g^$ large and the nine low orbitals clearly separate from the rest: the metal carbonyls. With $\pi$ donors such as water or halides, $t_{2g}$ is high, the gap is small, and electrons in $e_g^$ cost little, which is why aqua complexes are stable with anywhere from fifteen to twenty electrons. The rule and its exceptions come from the same diagram.

Other shapes give other diagrams, built the same way. In a square-planar complex, four ligand lone pairs fill four bonding orbitals with eight electrons, and the $d_{x^2-y^2}$ orbital that points at them is strongly antibonding. Eight d electrons fill the four d orbitals below it, for a total of sixteen, which is why square-planar $d^8$ complexes of platinum, palladium and rhodium are stable at sixteen electrons rather than eighteen. In a tetrahedral complex no d orbital points straight at a ligand, the splitting is small, and the counting rules are much looser, which fits the observation of lesson 12 that tetrahedral complexes are almost always high spin.

8. Checking a ligand field answer

Three checks catch most slips. First, the ligand electrons are always twelve in an octahedral $\sigma$ picture, one pair per ligand, whatever the ligand's charge; the charge matters to the oxidation state and so to $d$, not to the twelve.

Second, the d electrons must add up to $d$ across $t_{2g}$ and $e_g^$, and $e_g^$ can hold at most four. Third, net bonds lie between four and six: six when $e_g^*$ is empty, four when it is full, as in a $d^{10}$ octahedral ion. A value outside that range has miscounted the antibonding electrons.

Finally, the direction of a $\pi$ effect must be right. Donors narrow the gap and acceptors widen it; an answer that puts chloride above cyanide in field strength has the $\pi$ interactions backward.

9. In the world: why a cancer drug is platinum and a contrast agent is gadolinium

Designers of metal drugs choose metals partly by how fast they swap ligands. A drug must survive long enough in the blood to reach its target, but it must also react once it arrives. Cisplatin, $\mathrm{[PtCl_2(NH_3)_2]}$, is square-planar platinum(II), whose antibonding orbital is empty and whose ligand exchange takes hours: slow enough to reach the cell nucleus intact, fast enough to lose its chlorides there and bind DNA.

Chemists developing new drugs often use cobalt(III) and platinum(IV) complexes, low-spin $d^6$ with no $e_g^$ electrons, as prodrugs that stay completely intact in the bloodstream until the lower oxygen levels inside a tumor reduce them to a labile form that releases the active piece. The inertness that ligand field theory predicts for an empty $e_g^$ set is the whole design principle.

10. In the world: building complexes that last

Industrial and research chemists who need a complex that will not fall apart use the same principle in reverse. Chromium(III), $d^3$ with all three electrons in the nonbonding $t_{2g}$ set, is inert, which is why chromium(III) complexes are used in leather tanning: the chromium binds to the collagen of the hide and stays bound, making leather that resists rot and heat.

Rhodium(III) and iridium(III), low-spin $d^6$ with large splittings, are so inert that their complexes serve as light-absorbing dyes in solar cells and as the emitters in phosphorescent organic light-emitting diodes, the OLED screens of many phones and televisions. An iridium complex in an OLED must survive years of electrical excitation without losing a ligand, and it can because it has no electrons in its antibonding $e_g^*$ orbitals.

11. The upper set is antibonding, not just higher

Crystal field theory describes the $e_g$ orbitals as raised by repulsion, which makes an electron there sound merely uncomfortable. In the molecular orbital picture $e_g^$ is antibonding: an electron in it actively weakens the metal-ligand bonds. That is why high-spin complexes have longer metal-ligand bonds than low-spin ones, and why complexes with $e_g^$ electrons swap their ligands so much faster.

The related error is treating $t_{2g}$ as bonding. In a $\sigma$-only complex it is nonbonding; electrons there neither help nor hurt the bonds. It becomes slightly antibonding with $\pi$ donors and slightly bonding with $\pi$ acceptors, which is why the eighteen-electron rule holds best for carbonyls.

12. Hexaamminecobalt(III)

13. Hexaaquacobalt(II)

14. Fluoride against cyanide on iron(III)

15. Your turn: how many net $\sigma$ bonds does high-spin $\mathrm{[Fe(H_2O)_6]^{2+}}$, $d^6$, have?

16. Guided practice

17. Guided practice

Complete the worked solution: $\mathrm{[Co(H_2O)_6]^{2+}}$ has $\Delta_o = 9300$ and pairing energy $22500$ cm$^{-1}$. Place its d electrons and count its valence electrons.

18. Guided practice

The same metal ion forms octahedral complexes with $\mathrm{CO}$, $\mathrm{CH_3^-}$ and $\mathrm{I^-}$. Which ligand gives the largest $\Delta_o$?

19. Practice

$\mathrm{[Fe(H_2O)_6]^{3+}}$ is $d^{5}$ with $\Delta_o = 13700$ and $P = 30000$ cm$^{-1}$; $\mathrm{[Fe(CN)_6]^{3-}}$ is $d^{5}$ with $\Delta_o = 35000$ and $P = 30000$ cm$^{-1}$. For each, in that order, fill in the electrons in $t_{2g}$, the electrons in $e_g^*$ and the total valence electrons in its molecular orbital diagram.

20. Practice

How many valence electrons are there in the octahedral molecular orbital diagram of $\mathrm{[Fe(CN)_6]^{4-}}$, a $d^{6}$ complex?

21. Practice

In $\mathrm{[Mn(H_2O)_6]^{3+}}$, a $d^{4}$ complex with $\Delta_o = 21000$ cm$^{-1}$ and pairing energy $28000$ cm$^{-1}$, how many net metal-ligand $\sigma$ bonds does the $\sigma$-only molecular orbital picture give?

22. Somewhere new

A medicinal chemist wants a metal complex that will not lose its ligands in the bloodstream, and compares candidates by their antibonding electrons. For $\mathrm{[Mn(H_2O)_6]^{3+}}$ ($d^{4}$, $\Delta_o = 21000$, $P = 28000$), $\mathrm{[Fe(H_2O)_6]^{3+}}$ ($d^{5}$, $13700$, $30000$) and $\mathrm{[Fe(CN)_6]^{4-}}$ ($d^{6}$, $32850$, $17600$), all in cm$^{-1}$, fill in the $e_g^*$ electrons and the net $\sigma$ bonds, in that order.

23. Lesson test

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

24. Test question

$\mathrm{[Mn(H_2O)_6]^{2+}}$ is $d^{5}$ with $\Delta_o = 7800$ and $P = 25500$ cm$^{-1}$; $\mathrm{[Fe(CN)_6]^{3-}}$ is $d^{5}$ with $\Delta_o = 35000$ and $P = 30000$ cm$^{-1}$. For each, in that order, fill in the electrons in $t_{2g}$, the electrons in $e_g^*$ and the total valence electrons in its molecular orbital diagram.

25. What you can do now

You can use ligand field theory. Explain why cyanide splits the d orbitals more than fluoride, and why hexaamminecobalt(III) holds its ligands so tightly.

Working for the steps left to you

15. Your turn: how many net $\sigma$ bonds does high-spin $\mathrm{[Fe(H_2O)_6]^{2+}}$, $d^6$, have?, step 3

	π donor	σ donor only	π acceptor
$\mathrm{OH^-}$
$\mathrm{NH_3}$
$\mathrm{PF_3}$

	t2g electrons	eg* electrons	valence electrons in the diagram
the first complex
the second complex

	eg* electrons	net σ bonds
the first complex
the second complex
the third complex

	t2g electrons	eg* electrons	valence electrons in the diagram
the first complex
the second complex