MHT-CET Chemistry · Coordination Compounds

Oxidation State, Coordination Number, Werner's Counter Ions and IUPAC Names

The oxidation state of the central metal is what is left after the ligand charges are subtracted from the sphere charge, its coordination number is the number of donor atoms bonded to it, only the ions outside the square bracket ionise in solution, and an IUPAC name lists the ligands alphabetically before the metal, gives the metal's oxidation state in Roman numerals and ends in -ate when the sphere is an anion.

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Why this matters

23 PYQs, none HARD. Of the fifteen set from 2021 to 2025, eight are oxidation state and coordination number — Pt in [PtCl₆]²⁻, Fe in hexacyanoferrate, Cr in Cr(CO)₆, the coordination number of an octahedral complex, of cisplatin and of trioxalatoferrate, and one on Stock notation; two count the chloride that precipitates with silver nitrate; five turn a name into a formula or a formula into a name. Three cards.

Concept 1 of 3: Oxidation State and Coordination Number

Two numbers, two different counts. The OXIDATION STATE comes from charge balance: sphere charge = metal oxidation state + the sum of ligand charges, so in [PtCl₆]²⁻, x + 6(−1) = −2 and x = +4; in [Fe(CN)₆]⁴⁻, x − 6 = −4 and x = +2; in Cr(CO)₆ every ligand is neutral, so Cr is 0 and has lost no electrons. The COORDINATION NUMBER counts donor atoms, not ligands: six monodentate ligands give 6, but three bidentate oxalates also give 6, and cisplatin's two NH₃ and two Cl⁻ give 4. Geometry follows the coordination number: 6 is octahedral, 4 is tetrahedral or square planar, 2 is linear.

Definition

Oxidation state: $x + \sum(\text{ligand charges}) = \text{sphere charge}$ . $[\text{PtCl}_6]^{2-}$ → +4; $[\text{Fe(CN)}_6]^{4-}$ → +2; $\text{Cr(CO)}_6$ , $\text{Ni(CO)}_4$ , $\text{Fe(CO)}_5$ → 0 (no electrons lost).
Coordination number = number of donor atoms bonded to the metal. $[\text{Fe(C}_2\text{O}_4)_3]^{3-}$ → 3 × 2 = 6; cisplatin $[\text{Pt(NH}_3)_2\text{Cl}_2]$ → 4; $[\text{Ag(CN)}_2]^-$ → 2.
Geometry by coordination number: 6 → octahedral ( $d^2sp^3$ or $sp^3d^2$ ); 4 → tetrahedral ( $sp^3$ ) or square planar ( $dsp^2$ ); 2 → linear.
Stock notation writes the oxidation state in Roman numerals after the element: aurous chloride is Au(I)Cl, auric chloride Au(III)Cl₃; mercurous chloride $\text{Hg}_2\text{(I)Cl}_2$ is the reduced form of mercuric Hg(II)Cl₂. The notation shows at a glance whether a species is in its oxidised or reduced form.

Oxidation state and coordination number

x + \sum q_{\text{ligand}} = q_{\text{sphere}};\qquad \text{CN} = \sum (\text{ligands} \times \text{denticity})

Worked example

Give the oxidation state and the coordination number of the metal in K₃[Cr(C₂O₄)₃] and in [Co(en)₂Cl₂]Cl.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 1 · Q59Easy

Example 1 · Coordination Compounds · Oxidation State, Coordination Number and IUPAC Nomenclature

What is coordination number of central metal ion in

[\text{Fe}(\text{C}_2\text{O}_4)_3]^{3-}

Counting ligands instead of donor atoms

[Fe(C₂O₄)₃]³⁻ has three ligands but six donor atoms, so its coordination number is 6, not 3. The coordination number counts bonds to the metal.

Reading -ous and -ic backwards

-ous is the LOWER oxidation state (aurous Au⁺, cuprous Cu⁺, ferrous Fe²⁺, mercurous Hg₂²⁺), -ic the higher (auric Au³⁺, cupric Cu²⁺, ferric Fe³⁺, mercuric Hg²⁺).

Drill 7 more on oxidation state and coordination number

Concept 2 of 3: Counter Ions: What Precipitates with Silver Nitrate

Werner's test. Only the ions OUTSIDE the square bracket are free in solution; a chloride inside is bonded to the metal and does not react with Ag⁺. So excess AgNO₃ precipitates one AgCl for every chloride written after the bracket. [Co(NH₃)₅CO₃]Cl has one outer chloride — 1 mol AgCl per mol. Run it backwards: a complex that gives 2 mol AgCl has 2 ionisable chlorides. The same outer ions set the number of ions the complex gives in water, and so its molar conductivity.

Definition

Moles of AgCl per mole of complex = number of Cl⁻ outside the coordination sphere.
Werner's cobalt ammines: $\text{CoCl}_3\cdot6\text{NH}_3 = [\text{Co(NH}_3)_6]\text{Cl}_3$ → 3 AgCl, 4 ions; $\text{CoCl}_3\cdot5\text{NH}_3 = [\text{Co(NH}_3)_5\text{Cl}]\text{Cl}_2$ → 2 AgCl, 3 ions; $\text{CoCl}_3\cdot4\text{NH}_3 = [\text{Co(NH}_3)_4\text{Cl}_2]\text{Cl}$ → 1 AgCl, 2 ions.
$[\text{Co(NH}_3)_5\text{CO}_3]\text{Cl}$ (pentaamminecarbonatocobalt(III) chloride) → 1 mol AgCl: carbonato is inside, one chloride outside.
The metal's valencies in Werner's terms: the primary valency is the oxidation state (satisfied by the counter ions and anionic ligands), the secondary valency is the coordination number.

Werner's precipitation count

n(\text{AgCl}) = n(\text{Cl}^- \text{ outside } [\ ])\ \text{per mole of complex}

Worked example

One mole of a complex CrCl₃·5H₂O gives 2 mol of AgCl with excess silver nitrate. Write its formula.

Practise this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift II · Q86Moderate

Example 2 · Coordination Compounds · Oxidation State, Coordination Number and IUPAC Nomenclature

What is the number of moles of silver chloride precipitated when excess silver nitrate is treated with one mole of pentaamine carbonato cobalt (III) chloride?

Counting every chlorine in the formula

[Co(NH₃)₅Cl]Cl₂ has three chlorines but gives only 2 AgCl — the chloro ligand inside the bracket is held by the metal and never meets the Ag⁺.

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Concept 3 of 3: From Name to Formula and Back

A name is built in a fixed order: cation first, then anion; inside the complex, ligands in ALPHABETICAL order (ignoring di-, tri-, tetra-) before the metal; the metal's oxidation state in Roman numerals; and -ate on the metal when the sphere is an anion (cobaltate, ferrate, aluminate, platinate, argentate for Ag, cuprate for Cu). To go from a name to a formula, read off the ligands and the oxidation state, add the charges to get the sphere charge, and put enough counter ions outside to make the salt neutral. Isothiocyanato means the N end binds, so it is written NCS; nitro is N-bound NO₂, nitrito the O-bound ONO.

Definition

Order in the name: cation before anion. Ligands alphabetically (ammine before aqua before chloro), multiplying prefixes ignored; metal last; oxidation state as (II), (III)… Examples: $[\text{Co(H}_2\text{O)}_2(\text{NH}_3)_4]\text{Cl}_3$ → tetraamminediaquacobalt(III) chloride.
-ate for an anionic sphere: $\text{Na}_3[\text{Co(NO}_2)_6]$ sodium hexanitrocobaltate(III); $\text{Na}_3[\text{AlF}_6]$ sodium hexafluoroaluminate(III); $\text{K}_3[\text{Al(C}_2\text{O}_4)_3]$ potassium trioxalatoaluminate(III); $\text{K}_4[\text{Fe(CN)}_6]$ potassium hexacyanoferrate(II).
Name → formula: sphere charge = oxidation state + ligand charges, then counter ions to neutralise. Pentaaquaisothiocyanatoiron(III): +3 + 5(0) + (−1) = +2 → $[\text{Fe(H}_2\text{O)}_5(\text{NCS})]^{2+}$ .
A neutral salt carries no charge on its formula: $\text{K}_3[\text{Al(C}_2\text{O}_4)_3]$ is right; $\text{K}_3[\ldots]^{2-}$ is impossible, and potassium never goes inside the bracket.

Name to formula

q_{\text{sphere}} = \text{(Roman numeral)} + \sum q_{\text{ligand}};\quad \text{counter ions} \times \text{their charge} = -q_{\text{sphere}}

Worked example

Write the formula of potassium hexacyanoferrate(III) and name [Pt(NH₃)₄Cl₂]Cl₂.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q61Moderate

Example 3 · Coordination Compounds · Oxidation State, Coordination Number and IUPAC Nomenclature

Identify the formula of pentaaquaisothiocyanatoiron(III) ion from following.

Ordering ligands by charge or by prefix

Ligands go alphabetically by the ligand name alone: tetraammine before diaqua because 'ammine' precedes 'aqua' — the tetra- and di- do not count. Neutral-before-anionic is the formula-writing habit, not the naming rule.

Dropping the -ate or the oxidation state

An anionic sphere always ends in -ate (hexacyanoferrate, not hexacyanoiron), and the oxidation state is the metal's, not the sphere's charge: [Fe(CN)₆]⁴⁻ is ferrate(II), not ferrate(IV).

Drill 4 more on from name to formula and back

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Oxidation State and Coordination Number
Oxidation state and coordination number
$x + \sum q_{\text{ligand}} = q_{\text{sphere}};\qquad \text{CN} = \sum (\text{ligands} \times \text{denticity})$
Counter Ions: What Precipitates with Silver Nitrate
Werner's precipitation count
$n(\text{AgCl}) = n(\text{Cl}^- \text{ outside } [\ ])\ \text{per mole of complex}$
From Name to Formula and Back
Name to formula
$q_{\text{sphere}} = \text{(Roman numeral)} + \sum q_{\text{ligand}};\quad \text{counter ions} \times \text{their charge} = -q_{\text{sphere}}$

Watch out for (5)

Counting ligands instead of donor atoms
→ Oxidation State and Coordination Number
Reading -ous and -ic backwards
→ Oxidation State and Coordination Number
Counting every chlorine in the formula
→ Counter Ions: What Precipitates with Silver Nitrate
Ordering ligands by charge or by prefix
→ From Name to Formula and Back
Dropping the -ate or the oxidation state
→ From Name to Formula and Back

Test yourself on Coordination Compounds

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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Related notes

Ligands — names, charges and denticity →