Building MO Diagrams from Symmetry

Symmetry-based MO construction is the working method behind interpreting UV-visible spectra of transition metal complexes in analytical and materials labs: the position and intensity of d-d and charge-transfer bands are assigned by first building the symmetry-correct MO diagram, then applying the selection rules that come from the same group-theoretical framework (Laporte-forbidden $d$–$d$ transitions gain intensity through vibronic coupling, itself analyzed by exactly the same reduction-formula machinery used here, foreshadowing how Vibrational Spectroscopy and IR/Raman Selection Rules extend this toolkit to nuclear motion instead of electronic orbitals). In organometallic and catalytic chemistry, the $\pi$-acceptor/$\pi$-donor logic derived above is the everyday justification for ligand design: phosphines and CO analogues are tuned by adjusting how strongly their $\pi^*$ orbitals can symmetry-match and energy-match a metal's $t_{2g}$-type set, directly controlling catalytic activity in hydroformylation, cross-coupling, and metathesis catalysts. In bioinorganic chemistry, the same symmetry analysis (adapted to the lower symmetry of heme and non-heme iron sites) explains why oxygen binding to hemoglobin's iron is reversible rather than irreversible oxidative — the relevant $\pi$-backbonding is symmetry-limited by the porphyrin's approximate $C_4$ environment. In solid-state and materials chemistry, the identical SALC-matching procedure, generalized to extended lattices, underlies tight-binding and crystal-orbital band-structure pictures used to explain conductivity and color in transition-metal oxide materials. Computational chemistry software (Gaussian, ORCA, ADF) still labels computed molecular orbitals by irreducible representation for exactly this reason: it lets a chemist read off, at a glance, which orbitals are symmetry-forced to be pure metal, pure ligand, or genuinely mixed, without re-deriving the character table by hand every time.

Ask the tutor: Can you walk through building the full σ+π MO diagram for square-planar $\text{[PtCl}_4\text{]}^{2-}$ in $D_{4h}$ symmetry?

Every molecular orbital diagram answers the same question at heart: given a set of atomic orbitals, which linear combinations lower the energy (bonding), which raise it (antibonding), and which are untouched (nonbonding)? For a diatomic molecule this is trivial — two orbitals of the same symmetry (both $\sigma$, or both $\pi$) on the two atoms combine, in-phase and out-of-phase, into two new orbitals. Once you go to $\text{ML}_6$, $\text{ML}_4$, or any polyatomic fragment, the ligands no longer offer single orbitals — they offer six $\sigma$ orbitals, or twelve $\pi$ orbitals, and the metal's five d-orbitals plus one s and three p orbitals sit in the middle. Direct overlap of, say, the metal $d_{z^2}$ with each of the six individual ligand $\sigma$ orbitals is not how nature works; what actually mixes with $d_{z^2}$ is a specific symmetry-adapted combination of all six ligand orbitals that itself carries $d_{z^2}$'s symmetry label. This is where group theory earns its keep. The six ligand $\sigma$-orbitals in an octahedron transform as a 6-dimensional reducible representation, $\Gamma_\sigma$. Because $O_h$ has irreducible representations only up to dimension 3, $\Gamma_\sigma$ must decompose into a sum of smaller irreducible pieces — and the reduction formula tells you exactly which ones and how many times each appears. The result, worked below, is

Separately, you classify the metal's nine valence orbitals ($s$, three $p$, five $d$) by symmetry using the $O_h$ character table: $s$ transforms as $a_{1g}$, the $(p_x,p_y,p_z)$ set as $t_{1u}$, $(d_{z^2}, d_{x^2-y^2})$ as $e_g$, and $(d_{xy}, d_{xz}, d_{yz})$ as $t_{2g}$. Now the matching is automatic: $a_{1g}$ metal $s$ meets $a_{1g}$ ligand SALC, $t_{1u}$ metal $p$ meets $t_{1u}$ ligand SALC, $e_g$ metal $d$ meets $e_g$ ligand SALC. The $t_{2g}$ metal orbitals have no partner in $\Gamma_\sigma$ — there is no $t_{2g}$ ligand $\sigma$-SALC — so in a $\sigma$-only picture they are strictly nonbonding, sitting at their unperturbed atomic energy.

Once $\pi$-bonding ligands (CO, CN$^-$, halides with lone-pair p orbitals perpendicular to the M–L axis) are included, a second reducible representation $\Gamma_\pi$ is built from the twelve $\pi$-vectors (two per ligand) and reduced. In $O_h$ this gives $\Gamma_\pi = t_{1g} + t_{2g} + t_{1u} + t_{2u}$. Now $t_{2g}$ does have a ligand-orbital partner. If the ligand $\pi$ orbitals are low-lying filled orbitals (halide lone pairs, a $\pi$-donor), they push the resulting antibonding $t_{2g}^*$ up in energy, shrinking $\Delta_o$. If the ligand $\pi^*$ orbitals are empty and higher in energy than the metal d (CO, CN$^-$, a $\pi$-acceptor), the bonding combination sinks and the metal-based combination — now the lower member, stabilized — becomes the effective $t_{2g}$, enlarging $\Delta_o$. This is the symmetry-based derivation of the spectrochemical series, and it is the mechanistic reason CO is such a strong-field ligand despite being a poor $\sigma$-donor relative to amines.

The essential mechanism in both panels is identical: mix two orbitals of matching symmetry, the lower starting orbital drops further, the higher one rises further, and the metal-dominant combination is whichever one is nearer the metal's original energy. Whether $t_{2g}$ ends up net-stabilized or net-destabilized depends only on whether the ligand $\pi$ partner started below or above the metal d in energy — filled, low-lying halide lone pairs versus empty, higher CO $\pi^*$.

The reduction formula extracts the number of times irreducible representation $i$ appears in a reducible representation $\Gamma$:

where: $h$ is the order of the group, the sum runs over classes $R$, $g_R$ is the number of operations in class $R$, $\chi(R)$ is the character of the reducible representation, and $\chi_i(R)$ is the character of irreducible representation $i$, both evaluated at that class. Reading: this is just an orthogonality projection — you are asking "how much of the pure basis vector $i$ is hiding inside the messy vector $\Gamma$," and the character table rows are an orthonormal (after normalization by $h$) basis for that projection. The great orthogonality theorem underlying it guarantees

where: $\delta_{ij}$ is 1 if $i=j$ and 0 otherwise, and the sum is again over classes with multiplicity $g_R$. Reading: irreducible representations behave like an orthonormal basis of "vectors" (their character strings), which is exactly why the reduction formula works as a dot product divided by the norm $h$. The symmetry-matching selection rule for mixing two orbitals $\phi_1$ (metal) and $\phi_2$ (ligand SALC) is

where: $\Gamma_{\phi_1}, \Gamma_{\phi_2}$ are the irreducible representations spanned by the two orbitals, and the integral is the Hamiltonian matrix element (interaction energy) between them. Reading: this is the master rule of the whole article — the integrand transforms as $\Gamma_{\phi_1}\otimes\Gamma_{\phi_2}$, and only if this product contains the totally symmetric representation can the integral survive; for real orbitals this reduces to requiring $\Gamma_{\phi_1}=\Gamma_{\phi_2}$. The two-orbital mixing energetics, once symmetry allows an interaction, follow the standard two-level (near-degenerate perturbation theory) result:

where: $E_1, E_2$ are the unperturbed orbital energies, $H_{12}$ is the (symmetry-allowed, nonzero) interaction matrix element, and $E_\pm$ are the resulting bonding/antibonding MO energies. Reading: bigger $|H_{12}|$ (better overlap, matched symmetry) or smaller energy gap $|E_1-E_2|$ both push the split $E_+-E_-$ wider apart — this is why good $\sigma$-donors split $a_{1g}, e_g, t_{1u}$ strongly, while poorly-matched-energy $\pi$ sets shift $t_{2g}$ only modestly. The SALC wavefunctions themselves are generated by the projection operator

where: $\ell_i$ is the dimension of irreducible representation $i$, $\phi$ is a chosen "generating" ligand orbital, and $\hat{R}\phi$ is the orbital produced by applying symmetry operation $R$ to $\phi$. Reading: you throw a single ligand orbital at every operation of the group, weight each result by how strongly that operation belongs to representation $i$ (via its character), and sum — the survivors are exactly the SALC combination of that symmetry.

Example 1 — Foundational: σ-only bonding count in $\text{[Co(NH}_3\text{)}_6\text{]}^{3+}$. $\text{Co}^{3+}$ is low-spin d$^6$ in this strong-field amine complex. Using the $\Gamma_\sigma = a_{1g}+e_g+t_{1u}$ result derived above, the metal contributes one $a_{1g}$ ($4s$), three $t_{1u}$ ($4p_x,4p_y,4p_z$), and two $e_g$ ($3d_{z^2}, 3d_{x^2-y^2}$) orbitals that each find a same-symmetry ligand SALC partner; the three $t_{2g}$ d-orbitals ($3d_{xy},3d_{xz},3d_{yz}$) are left nonbonding. Six ligand lone pairs (12 electrons) fill the six bonding MOs ($1a_{1g}, 1t_{1u}\times3, 1e_g\times2$) completely. The remaining $6$ d-electrons of Co$^{3+}$ then fill the nonbonding $t_{2g}$ set (6 electrons, all paired, $t_{2g}^6$) with the antibonding $e_g^*$ left empty — reproducing the familiar low-spin d$^6$, $t_{2g}^6 e_g^0$ crystal-field picture, but now with the extra information that $e_g^*$ is not a "pure" metal orbital: it carries substantial ligand-SALC character mixed in by the bonding interaction. As a limiting-case check, if all six ligand $\sigma$ orbitals were degenerate with the metal d orbitals and had zero net overlap ($H_{12}\to 0$), the two-level formula $E_\pm = \frac{E_1+E_2}{2}\pm\sqrt{(\ldots)^2+H_{12}^2}$ would collapse to $E_+=E_-=E_1=E_2$ — no splitting at all, and $\Delta_o\to 0$, consistent with the physical statement that $\Delta_o$ vanishes only if metal and ligand orbitals do not actually interact.

Example 2 — Applied: π-back-bonding in $\text{Cr(CO)}_6$, an octahedral π-acceptor complex. $\text{Cr(CO)}_6$ is $O_h$, $d^6$, and each CO ligand offers a filled $5\sigma$ (HOMO, mostly carbon lone pair) and two empty, degenerate $2\pi^*$ orbitals perpendicular to the C–O axis. The $2\pi^*$ set on all six CO ligands (12 orbitals total) reduces, by the same style of reduction-formula calculation used for $\Gamma_\sigma$ above but now tracking vectors perpendicular to each M–L axis, to $\Gamma_\pi = t_{1g}+t_{2g}+t_{1u}+t_{2u}$. Among these, only $t_{2g}$ matches a metal valence orbital (the metal $d_{xy}, d_{xz}, d_{yz}$ set). Since the CO $\pi^*$ orbitals are empty and lie above the filled metal $t_{2g}$ in energy, the two-level mixing formula

places the metal-dominant combination ($E_-$, since metal $t_{2g}$ starts lower) further below its unperturbed energy — the electrons drain some density into the ligand $\pi^*$ orbital (this is "back-bonding" in symmetry language) and in return the whole $t_{2g}$ level is stabilized. Because $t_{2g}$ is now bonding rather than nonbonding, and the antibonding $e_g^*$ is essentially unaffected (no $\pi$-partner reaches $e_g$ in $O_h$), the gap between the lowered $t_{2g}$ and unchanged $e_g^*$ — i.e. $\Delta_o$ — widens substantially relative to a pure-$\sigma$ amine ligand. This is exactly why CO sits at the extreme strong-field end of the spectrochemical series, is diamagnetic as $d^6$, $t_{2g}^6$, and why the C–O stretch in $\text{Cr(CO)}_6$ (about $2000\ \text{cm}^{-1}$) is red-shifted from free CO ($2143\ \text{cm}^{-1}$): the electron density pushed into $\pi^*$ weakens the C–O bond order. As a check on scope: if the ligand instead offered a filled, low-energy $\pi$ donor orbital (e.g. F$^-$ lone pairs) rather than an empty acceptor orbital, the same symmetry match would apply to $t_{2g}$, but now the filled ligand orbital starts below the metal d, so $E_1 - E_2$ flips sign and the metal-dominant combination is pushed up instead of down — correctly recovering the opposite, $\Delta_o$-shrinking behavior of halide ligands described in the overview.

Example 3 — Exam/competition level: full reduction and MO matching for tetrahedral $\text{[FeCl}_4\text{]}^{2-}$. $\text{[FeCl}_4\text{]}^{2-}$ belongs to $T_d$ ($h=24$; classes $E, 8C_3, 3C_2, 6S_4, 6\sigma_d$). Build $\Gamma_\sigma$ from the four Cl$^-$ $\sigma$-donor orbitals: under $E$ all four are fixed ($\chi=4$); under $8C_3$ (through a vertex and the opposite face center) one ligand is fixed, three cycle ($\chi=1$); under $3C_2$ no ligand is fixed, they swap in two pairs ($\chi=0$); under $6S_4$ no ligand is fixed ($\chi=0$); under $6\sigma_d$ (each plane contains one edge, i.e. two ligands) two ligands are fixed ($\chi=2$). This gives

Apply the reduction formula against the $T_d$ irreducible representations. For $a_1$ (all characters $+1$):

For $t_2$ (characters $3,0,-1,-1,1$):

For $e$ (characters $2,-1,2,0,0$):

For $a_2$ and $t_1$, the same procedure gives zero (skip the arithmetic; the class strings have no overlap once $a_1$ and $t_2$ are removed, and dimension count $1+3=4$ already exhausts $\Gamma_\sigma$). So

On the metal side, in $T_d$ the $s$ orbital is $a_1$, the $(p_x,p_y,p_z)$ set is $t_2$, and the d-orbitals split as $e\,(d_{z^2},d_{x^2-y^2})$ and $t_2\,(d_{xy},d_{xz},d_{yz})$. Matching symmetry labels: metal $s\,(a_1)$ and metal $p\,(t_2)$ each find a ligand $\sigma$-SALC partner and form bonding/antibonding pairs; metal $d(e)$ has no ligand $\sigma$ partner (since $\Gamma_\sigma$ contains no $e$) and stays nonbonding at low energy; metal $d(t_2)$ does find a partner (since $\Gamma_\sigma$ contains $t_2$) and is pushed up as the ligand-mixed, antibonding combination. This correctly reproduces — now from first principles rather than crystal-field assertion — the inverted tetrahedral splitting pattern: $e$ below $t_2$, with $\Delta_t$ (roughly $\frac{4}{9}\Delta_o$ for comparable ligands and metal) smaller than $\Delta_o$ because only three, not four, ligand orbitals reach the upper d-set, and the geometry gives poorer directional overlap than octahedral $\sigma$-bonding. High-spin d$^6$ Fe$^{2+}$ in this tetrahedral field places electrons as $e^3 t_2^3$, four unpaired spins, matching the paramagnetism observed for $\text{[FeCl}_4\text{]}^{2-}$.