Oxidation state and coordination number

Look at $[Cr(en)_2Cl_2]^+$ and two questions arise immediately: what is chromium's actual charge buried inside that formula, and how many bonds is it actually making? Neither answer is obvious by inspection — you cannot just read off the metal's oxidation state or count ligand molecules and expect the right number, because ligands can carry their own charge and some of them latch onto the metal through more than one point. Oxidation state and coordination number are the two bookkeeping numbers coordination chemists compute for every complex before doing anything else with it — naming it, predicting its geometry, or explaining its magnetism. This article builds a systematic charge-balance method for oxidation state, a systematic donor-atom count for coordination number, and works both through three complexes of increasing subtlety: a simple bidentate case, a neutral square-planar drug, and a single hexadentate ligand that alone fills an octahedral coordination sphere. Get these two counts right and every later topic in Coordination Compounds — nomenclature, geometry, magnetism — becomes a matter of reading off a known number rather than guessing.

The core insight

Oxidation state comes from solving $x_M + \sum_i n_i l_i = z$ (metal charge plus ligand charges equals overall charge), while coordination number comes from summing $n_i \times \text{denticity}_i$ (ligand count times donor atoms per ligand) — two independent counts answering different questions from the same formula.

The core idea

Oxidation state of the metal is the charge the metal would carry if every ligand were stripped away together with the electron pair it donated — i.e., if the bonding were treated as fully ionic rather than covalent/dative. For a coordination entity of overall charge $z$ , containing metal $M$ (unknown oxidation state $x_M$ ) and ligands $L_i$ of charge $l_i$ each, with $n_i$ ligands of that type present:

x_M + \sum_i n_i\, l_i = z

where: $x_M$ is the metal's oxidation state (the unknown to solve for), $n_i$ is how many ligands of type $i$ are in the formula, $l_i$ is the charge on one ligand of that type, and $z$ is the net charge written on the whole coordination entity (0 for a neutral complex, the ionic charge otherwise). Neutral ligands — $NH_3$ , $H_2O$ , $CO$ , ethylenediamine ("en") — contribute $l_i = 0$ ; anionic ligands — $Cl^-$ , $CN^-$ , oxalate $C_2O_4^{2-}$ — contribute their own negative charge.

Coordination number (CN) counts donor atoms, not ligand molecules. It is the number of coordinate (dative) bonds running from ligand donor atoms to the central metal:

\text{CN} = \sum_i n_i \times (\text{denticity of ligand } i)

where: denticity is how many donor atoms one molecule/ion of that ligand offers to the metal simultaneously. A monodentate ligand ( $NH_3$ , $Cl^-$ , $H_2O$ ) has denticity 1. A bidentate ligand (ethylenediamine "en", oxalate) has denticity 2 — it wraps around and bonds through two atoms at once, forming a chelate. Higher-denticity chelating ligands exist too; EDTA is the standard hexadentate example, donating through six atoms from a single molecule. This is exactly why counting ligand formula units instead of donor atoms is the single most common error in this topic. Common coordination numbers are 4 (tetrahedral or square planar) and 6 (octahedral), with 2 seen for some $d^{10}$ ions such as $Ag^+$ .

Figure

A metal centre with six coordinate bonds shown as two chloride donors and two ethylenediamine ligands each donating through two nitrogen atoms totaling six bonds for an octahedral coordination number of six

Derivation ASolve oxidation state and coordination number together for a real complex ion

The goal is to take a single complex-ion formula and extract both numbers by a method that generalises to any mononuclear complex a student will meet in this unit, so the reasoning below is written once and can be replayed mechanically: identify overall charge, identify each ligand's charge and denticity, then apply the two formulas from the core idea in turn. We use $[Cr(en)_2Cl_2]^+$ as the running example, since its two ligand types (one neutral bidentate, one anionic monodentate) exercise both formulas fully.

Step 1. Read off the overall charge and the ligand inventory directly from the formula. $[Cr(en)_2Cl_2]^+$ has overall charge $z = +1$ . Inside the brackets sit two "en" ligands (ethylenediamine, neutral, so $l = 0$ ) and two $Cl^-$ ligands (each $l = -1$ ). Chromium's oxidation state, $x$ , is the unknown.

Step 2. Apply the charge-balance equation with these values. $x_{Cr} + 2(0) + 2(-1) = +1$ , which simplifies to $x_{Cr} - 2 = 1$ , so $x_{Cr} = +3$ . Chromium is present as chromium(III), and this number will later feed directly into the complex's name, "chloridobis(ethylenediamine)chromium(III) ion."

Step 3. Switch to the coordination-number question, which uses denticity rather than charge. Ethylenediamine is bidentate — one molecule of en donates through two nitrogen atoms at once — so two en ligands contribute $2 \times 2 = 4$ to the count. Each $Cl^-$ is monodentate, contributing $1$ each, so two chlorides contribute $2 \times 1 = 2$ .

Step 4. Add the two contributions: $\text{CN} = 4 + 2 = 6$ . Note that this uses four ligand units (2 en + 2 Cl) but six donor bonds — the mismatch between "count the formula units" (wrong, gives 4) and "count the donor atoms" (right, gives 6) is exactly the trap this method is built to avoid. Six-coordinate chromium(III) complexes are characteristically octahedral, consistent with CN = 6. Result: $x_{Cr} = +3$ , CN $= 6$ , matching the octahedral geometry expected for this complex.

Worked example

Find the oxidation state and coordination number of the central metal in $[Pt(NH_3)_2Cl_2]$ , then in $[Ni(EDTA)]^{2-}$ , where EDTA is the hexadentate anion $EDTA^{4-}$ .

Step 1. For $[Pt(NH_3)_2Cl_2]$ : the complex is written with no charge outside the bracket, so $z = 0$ . Two $NH_3$ ligands are neutral ( $l = 0$ each); two $Cl^-$ ligands each carry $l = -1$ .

Step 2. Solve the charge balance: $x_{Pt} + 2(0) + 2(-1) = 0$ , so $x_{Pt} = +2$ . Platinum is in the $+2$ oxidation state.

Step 3. Coordination number for the platinum complex: $NH_3$ is monodentate (2 ligands contribute $2 \times 1 = 2$ ) and $Cl^-$ is monodentate (2 ligands contribute $2 \times 1 = 2$ ), giving $\text{CN} = 2 + 2 = 4$ . This four-coordinate $d^8$ platinum(II) complex — the core of the anticancer drug cisplatin — is square planar, the geometry typical for CN = 4 with $d^8$ metals.

Step 4. For $[Ni(EDTA)]^{2-}$ : overall charge $z = -2$ , and there is only one ligand, $EDTA^{4-}$ , so $n = 1$ and $l = -4$ . Charge balance: $x_{Ni} + 1(-4) = -2$ , giving $x_{Ni} = +2$ . Nickel is in the $+2$ oxidation state.

Step 5. Coordination number for the nickel complex: EDTA is hexadentate, so one EDTA molecule alone supplies $1 \times 6 = 6$ donor bonds. Final answer: Pt is $+2$ with CN $= 4$ (square planar); Ni is $+2$ with CN $= 6$ (octahedral), even though only one ligand formula unit is present.

Common traps

Counting the number of ligand species written in the formula instead of the number of donor atoms — one EDTA or one oxalate contributes far more than 1 to CN.
Assigning a nonzero charge to a neutral ligand ( $NH_3$ , $H_2O$ , $CO$ , en) in the charge-balance equation, which throws off the oxidation state.
Treating oxidation state and coordination number as the same kind of quantity — oxidation state is a signed charge-accounting number (it can be zero or even negative in some complexes) computed from charges, while CN is always a positive count of bonds computed from denticity; mixing up the two methods gives wrong answers for both.
Misreading the charge on a complex polyatomic ligand before balancing — e.g. treating oxalate $C_2O_4^{2-}$ as $-1$ or EDTA $^{4-}$ as $-2$ instead of its actual charge.

Ask the tutor: Give the AI tutor a complex-ion formula of your own choosing and ask it to check your oxidation-state and coordination-number working step by step.

Why does ethylenediamine, en, count as 2 toward coordination number when it is written as a single ligand in the formula?

Because coordination number counts donor atoms bonded to the metal, not ligand molecules. Ethylenediamine has two nitrogen atoms, each with a lone pair, and both nitrogens bond to the metal at the same time, forming a five-membered ring called a chelate. So one en molecule occupies two coordination positions even though it is one chemical species — that is exactly what "bidentate" means, and it is why you sum denticity rather than just counting ligands.

Can oxidation state ever come out negative or zero for a metal in a complex?

Yes. The charge-balance equation $x_M + \sum n_i l_i = z$ can solve to zero (as in some neutral carbonyl complexes) or even to a negative value in unusual cases, because oxidation state is purely an electron-accounting convention, not a physical bond count. It reflects what charge the metal would have if you assigned every bonding electron pair fully to the more electronegative ligand atom. Coordination number, by contrast, is always positive since it counts real coordinate bonds — so a negative CN is never possible, but a low or zero oxidation state is.

If EDTA fills all six coordination positions by itself, does that mean no other ligand can bind the metal at the same time?

Correct, in the simplest case: once EDTA's six donor atoms occupy all six positions of an octahedral coordination sphere, as in $[Ni(EDTA)]^{2-}$ , there is no room left for another ligand, because CN = 6 is already used up. This is exactly why EDTA is such an effective chelating agent in titrations and in treating heavy-metal poisoning — it wraps around a metal ion and locks out water or other competing ligands almost completely, forming a very stable 1:1 complex.

Oxidation state and coordination number

The core idea

Worked example

Common traps

Before this

Where this leads