Square-Planar Complexes and d⁸ Ions

Introduction

Four-coordinate complexes need not be tetrahedral. Many d⁸ ions, especially Pd²⁺ and Pt²⁺, form square-planar complexes, and Ni²⁺ can do so with a suitable ligand environment. The difference is more than a sketch of positions: square-planar ligand approach produces a large separation of d-orbital energies, often allowing eight d electrons to pair in four lower orbitals and leave the highest one empty.

Core explanation

Place four ligands on the +x, −x, +y and −y directions around a metal. The d(x²−y²) orbital points directly at all four and typically becomes the highest-energy, strongly σ-antibonding d-like orbital. The other four d-like orbitals lie lower, though their detailed order can vary with metal, ligands and bonding model. A d⁸ ion can put two electrons into each of the four lower levels and leave d(x²−y²) empty. That distribution has no unpaired d electrons in the simple picture and is diamagnetic.

Compare a tetrahedral d⁸ field. Its usual high-spin configuration e⁴t₂⁴ has two unpaired electrons. Thus magnetic measurement provides useful evidence about a four-coordinate d⁸ geometry: a strong diamagnetic response supports a paired square-planar assignment, whereas an approximately two-unpaired moment supports tetrahedral filling. Other electronic or temperature effects can complicate an exact moment, so structure and spectra also matter.

For 4d and 5d d⁸ ions such as Pd²⁺ and Pt²⁺, more extended d orbitals interact strongly with ligands and square-planar arrangements are very common. Ni²⁺ is 3d⁸ and can show either geometry depending on ligand field and other energetic factors. For example, [NiCl₄]²⁻ is commonly tetrahedral and paramagnetic, while [Ni(CN)₄]²⁻ is square planar and diamagnetic. Both contain Ni²⁺ with the same formal d⁸ count, so oxidation state alone does not choose the shape.

It is tempting to say square planar is always favoured because it “has more CFSE,” but a complete comparison includes the cost of pairing, ligand–metal bond energies, steric interactions and geometry-specific splitting. Unlike the octahedral and tetrahedral two-level diagrams, square-planar d orbitals split into several distinct energies; one universal coefficient such as −1.2Δₒ is not sufficient. State the qualitative large gap to d(x²−y²) unless actual level energies are supplied.

A square-planar structure supports geometrical isomerism. With two ligands A and two B around Pt(II), cis-A₂B₂ and trans-A₂B₂ arrangements are distinct. Cisplatin, cis-[PtCl₂(NH₃)₂], and its trans form have the same formula and formal Pt²⁺ d⁸ count but different spatial arrangement and biological behaviour. Electron count predicts neither isomer count nor physiological effect by itself.

The “d⁸ square planar” association is strong but not a universal rule for all d⁸ compounds. Six-coordinate octahedral d⁸ ions also exist, and four-coordinate nickel(II) can be tetrahedral. Always establish coordination number and compare plausible geometries before using a memorised electron diagram.

Step-by-step reasoning

Determine the metal oxidation state and d count, then count ligands. If four-coordinate d⁸, consider both tetrahedral and square-planar options. Sketch the square-planar ligand axes and put d(x²−y²) highest; check whether eight electrons can pair in the four lower levels. Compare magnetic prediction with a tetrahedral e⁴t₂⁴ diagram and available structural evidence.

Visual explanation

Draw an X-shaped tetrahedral projection beside a cross-shaped square plane. Over the square, show d(x²−y²) lobes pointing at four ligand positions. In the square-planar energy ladder, place four filled lower boxes and one empty high box; in the tetrahedral ladder, show two unpaired arrows in t₂.

Real-world analogy

Four guests can sit around a square table or around a tetrahedral display stand. The same people and number of seats produce different relationships and lines of sight. Similarly, identical formula and d count can support different orbital energies and magnetism when ligand positions change.

Real-world example

The contrast between [NiCl₄]²⁻ and [Ni(CN)₄]²⁻ is a standard laboratory comparison. Both have Ni²⁺ d⁸, but the former is commonly tetrahedral with two unpaired electrons, while the latter is square planar with paired electrons. Ligand environment helps select the geometry.

Why?

Why does d(x²−y²) sit so high in a square plane? Its lobes point directly along x and y toward all four ligands. Strong σ interaction creates a high-energy antibonding combination compared with orbitals directed less directly at those ligand positions.

Common misconception

“Every four-coordinate d⁸ ion is square planar and diamagnetic.” Tetrahedral d⁸ complexes exist, especially for some Ni²⁺ ligand environments, and their two unpaired electrons make them paramagnetic.

Worked example

Consider [Ni(CN)₄]²⁻. Four CN⁻ ligands contribute −4; the complex has charge −2, so Ni is +2 and d⁸. In a square-planar splitting scheme, eight electrons pair in four lower d-like levels and leave the high d(x²−y²) level empty. The simple predicted unpaired count is zero and spin-only moment zero. For tetrahedral [NiCl₄]²⁻, the same d⁸ count gives e⁴t₂⁴ and two unpaired electrons.

Quick check

1. Which d-like orbital is usually highest in a square-planar σ-ligand diagram? Answer: d(x²−y²), because it points directly toward the four in-plane ligands. 2. Is [PtCl₂(NH₃)₂] necessarily one unique arrangement? Answer: No. Its square-planar cis and trans geometrical isomers are distinct.

Exam focus

Do not reuse octahedral CFSE coefficients for a square plane. Use oxidation state, coordination number, a geometry-specific level sketch and a magnetic prediction to support your answer.

Advanced insight

The square-planar field can be pictured as an extreme tetragonal elongation of an octahedron in which axial ligands are removed. This connects Jahn–Teller reasoning to the high d(x²−y²) level, although the actual square-planar bonding pattern deserves its own molecular-orbital treatment.

Summary

Square-planar d⁸ complexes commonly pair eight electrons below a high, empty d(x²−y²)-like level and are diamagnetic. Pd²⁺ and Pt²⁺ often adopt this geometry; Ni²⁺ can be square planar or tetrahedral depending on ligands.

Practice questions

1. Predict the simple unpaired count for tetrahedral and square-planar d⁸. Answer: Tetrahedral high-spin e⁴t₂⁴ has two unpaired electrons. A typical square-planar d⁸ filling has all eight electrons paired in four lower levels, giving zero. 2. Why can two Ni²⁺ complexes differ in magnetism without a change in oxidation state? Answer: Their ligands can favour different four-coordinate geometries and splitting patterns. The same d⁸ count then fills orbitals differently. 3. Why is cisplatin’s cis designation structural rather than an oxidation-state label? Answer: It specifies that the two chloride ligands occupy adjacent positions in a square plane. The trans isomer has opposite chloride positions but the same Pt²⁺ formal oxidation state.