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A chloride ligand is coordinated $\mathrm{C}{\mathrm{l}}^{-}$. It is an anionic two-electron donor in ionic electron counting, with occupied orbitals that can contribute pi donation.

Four chloride ligands give Ti(IV), $d^0$. Ionic electron counting givestetrahedral molecular geometry separates the four donors without placing them at $90^{\circ }$. A $d^0$ centre has no occupied d orbitals providing a square-planar stabilization; the 18-electron rule is not a requirement for this early-metal compound.

The metal is Cr(II), $d^4$. A chloride ligand produces a relatively weak ligand field, so the octahedral center is a high-spin complex with $t_{2g}^3{e}_g^1$, $S=2$. The singly occupied $e_g$ pair is orbitally degenerate: $d_{z^2}$ and $d_{x^2-y^2}$ cannot retain identical occupancies once the distortion selects one. This is a strong setting for the Jahn-Teller effect.

On tetragonal elongation, the two axial Cr–Cl distances increase. Repulsion is reduced most for $d_{z^2}$, so it falls relative to $d_{x^2-y^2}$; $d_{xz},d_{yz}$ also fall relative to $d_{xy}$. The single $e_g$ electron occupies the lower $d_{z^2}$-derived level. The electronic stabilization offsets an elastic cost until a distorted equilibrium geometry is reached. The high-spin $d^4$ occupancy drives an elongated, rather than regular, octahedron. The three lower-set electrons remain one per orbital in this simple diagram; it is primarily the unequal $e_g$ occupancy that produces the strong distortion.

The usual qualitative order of increasing ligand field isA chloride ligand is a sigma-donor ligand and pi-donor ligand; its occupied pi orbitals raise the metal $t_{2g}$-derived levels and reduce the splitting. A carbonyl ligand is a particularly effective pi-acceptor ligand, lowering those levels and producing a large splitting. Both phosphine ligands are sigma donors with acceptor orbitals of mainly P–C antibonding molecular orbital character. Replacing phenyl groups by electron-releasing tert-butyl groups increases sigma basicity but generally weakens pi acceptance, so the alkyl-rich phosphine is placed below triphenylphosphine in this qualitative pi-bonding comparison. It would be wrong to equate a stronger sigma donor automatically with a stronger-field ligand, or to invoke empty phosphorus $3d$ orbitals as the essential acceptors.^[ref]

This is a qualitative spectrochemical series prediction, not a universal numerical ranking for all metals and geometries. The opposing sigma and pi trends mean that precise comparisons require matched complexes and measured splittings; a Tolman electronic parameter measures net effects on CO backbonding and is not itself a ligand-field splitting.

For electron counting, use the ionic method: subtract the metal oxidation state from its group number, then add two electrons for each neutral phosphine ligand or carbonyl ligand and two for each chloride ligand.

$\mathrm{Ni}(\mathrm{PP}{\mathrm{h}}_3)(\mathrm{CO}{)}_3$ is Ni(0), $d^{10}$, 18-electron and tetrahedral. The count is $10+2+3(2)=18$. With only one distinct ligand in an $\mathrm{M}{\mathrm{L}}_3{\mathrm{L}}^{\prime }$ tetrahedral molecular geometry, there is only one geometrical arrangement.

$\mathrm{Rh}(\mathrm{PP}{\mathrm{h}}_3{)}_2(\mathrm{CO})\mathrm{Cl}$ is Rh(I), $d^8$, 16-electron and square planar, with the two phosphines trans. Its count is $8+2(2)+2+2=16$. A square planar molecular geometry is characteristic of this low-spin $4d^8$ configuration. Opposite triphenylphosphine groups avoid their largest mutual steric hindrance; CO and chloride occupy the other trans pair.

$\mathrm{Pd}[\mathrm{P}{(}^{\mathrm{t}}\mathrm{Bu}{)}_2\mathrm{Ph}{]}_2\mathrm{C}{\mathrm{l}}_2$ is Pd(II), $d^8$, 16-electron and square planar, with the bulky phosphines trans. Its count is likewise $8+4+4=16$. The two di-tert-butylphenylphosphine groups are particularly crowded if cis. These predictions concern the favored intact four-coordinate species; ligand dissociation can occur under other conditions.

In $A$, the tridentate ligand contains one anionic phenoxide ligand, a neutral secondary amine with its N–H bond intact, and a neutral phosphine ligand. The three chloride ligands contribute another three negative charges. Consequently the neutral complex has Ti(IV), $d^0$, and 12 valence electrons:The N–H label is essential: counting this nitrogen as an anionic amido ligand would give the wrong oxidation state. A six-coordinate early-metal complex need not obey the 18-electron rule.

In $B$, the substituted cyclopentadienyl ligand is ${\eta }^5$ and anionic, while the three-carbon bent ligand is an anionic ${\eta }^3$ allyl ligand, a four-electron donor in ionic counting. The pendant pyridine nitrogen is uncoordinated. Therefore Mo(II), $d^4$, 18 electrons follows fromThe curved delocalization mark across the three-carbon ligand must not be read as a separate alkene or alkyne.

$E=\mathrm{Mo}({\eta }^5:\kappa N\text{-}L)(\mathrm{CO}{)}_2\mathrm{Cl}$, with chloride replacing the remaining acetonitrile. The pendant pyridine remains attached. The substitution removes the positive electric charge on $D$ but does not change the Mo(II) oxidation state. In ionic electron counting, both neutral acetonitrile and anionic chloride ligand donate two electrons, giving $4+6+2_N+4_{\mathrm{CO}}+2_{\mathrm{Cl}}=18$ in $E$.

acetonitrile is the remaining weak, exchangeable monodentate donor. Its displacement forms a favorable Mo–Cl bond at a cationic center, retains the intramolecular Mo–N bond, and restores an 18-electron product after substitution. The following formulas specify the coordination changes directly:Tetramethylammonium is a spectator counterion; it is not a carbon donor to Mo.