Q.(a) Write the IUPAC names of the following : (2 × 1)
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🔒 Start your 14-day free trial to unlock the full solution →Part (a)Concept understanding — Coordination Compound Nomenclature
Coordination Compound Nomenclature: From Intuition to Precision
Imagine you're naming a person. You'd say "Ravi Sharma" — family name first, then given name. Coordination compounds have a similar logic, but the "family name" is the metal, and the "given names" are the groups attached to it. The rules are just a systematic way of writing that name so any chemist anywhere can draw the exact structure from it.
The Core Idea
A coordination compound has a central metal ion surrounded by molecules or ions called ligands. Think of the metal as the nucleus and ligands as planets orbiting it. The entire assembly (metal + ligands) is called the coordination sphere, and it's written inside square brackets: [Co(NH₃)₆]Cl₃.
The nomenclature rules tell you:
- What order to list things
- How to name each ligand
- How to indicate the metal's oxidation state
- How to handle the counter-ions outside the brackets
The Rules, Step by Step
1. Cation before anion (just like NaCl is sodium chloride)
If the complex ion is positive, it's named first. If it's negative, it's named last. Simple.
2. Within the coordination sphere: ligands first, then metal
This is the big rule. Ligands are named before the metal, in alphabetical order (ignoring prefixes like di-, tri-).
Alphabetical order is based on the ligand's name, not its formula. So NH₃ (ammine) comes before H₂O (aqua), even though N comes after H in the alphabet.
3. Naming ligands
| Ligand type | Name | Example |
|---|---|---|
| Neutral molecule (NH₃) | ammine | [Co(NH₃)₆]³⁺ → hexaamminecobalt(III) |
| Neutral molecule (H₂O) | aqua | [Cu(H₂O)₄]²⁺ → tetraaquacopper(II) |
| Neutral molecule (CO) | carbonyl | [Ni(CO)₄] → tetracarbonylnickel(0) |
| Negative ion (Cl⁻) | chloro | [PtCl₆]²⁻ → hexachloroplatinate(IV) |
| Negative ion (CN⁻) | cyano | [Fe(CN)₆]⁴⁻ → hexacyanoferrate(II) |
| Negative ion (OH⁻) | hydroxo | [Al(OH)₄]⁻ → tetrahydroxoaluminate(III) |
ammine (with two m's) is for NH₃ as a ligand. amine (one m) is for organic compounds like ethylamine. Don't mix them up — exam setters love this trap.
4. Prefixes for multiple ligands
Use Greek prefixes: di-, tri-, tetra-, penta-, hexa-, hepta-, octa-.
If the ligand name already contains a number (like ethylenediamine), use bis-, tris-, tetrakis- instead.
[Co(en)₃]³⁺ is tris(ethylenediamine)cobalt(III), not triethylenediaminecobalt(III). The parentheses around the ligand name are mandatory when using bis/tris/tetrakis.
5. Oxidation state of the metal
Write it in Roman numerals in parentheses right after the metal name. No space.
[Fe(CN)₆]³⁻ → hexacyanoferrate(III) (iron is in +3 state)
6. If the complex is an anion, change the metal's ending
| Metal | Anionic form |
|---|---|
| Cobalt | cobaltate |
| Copper | cuprate |
| Iron | ferrate |
| Nickel | nickelate |
| Platinum | platinate |
| Zinc | zincate |
General pattern:
[M(L)ₙ]Xₘ → cation name = [prefix-ligands]metal(oxidation state)
anion name = [prefix-ligands]metalate(oxidation state)
Worked Examples
Example 1: K₃[Fe(CN)₆]
- Cation: potassium (K⁺)
- Complex anion:
[Fe(CN)₆]³⁻ - Ligands: 6 cyano → hexacyano
- Metal: iron → ferrate (because it's an anion)
- Oxidation state: Fe is +3 (since 6 CN⁻ = -6, total charge -3, so Fe must be +3)
- Answer: Potassium hexacyanoferrate(III)
Example 2: [Co(NH₃)₅Cl]Cl₂
- Cation:
[Co(NH₃)₅Cl]²⁺ - Ligands: 5 ammine + 1 chloro → alphabetical: ammine before chloro → pentaamminechloro
- Metal: cobalt
- Oxidation state: Co is +3 (5 NH₃ neutral, 1 Cl⁻ = -1, total +2, so Co = +3) …
Why this formula?
Coordination Compound Nomenclature: Why the Rules Work
Coordination compound nomenclature isn't about a single formula — it's a system of rules built on a few core principles. Let's understand the why behind each major rule, so you never have to memorise blindly.
1. The Central Idea: Ligands as "Guests" Around a Metal "Host"
A coordination compound has a central metal atom/ion surrounded by ligands (molecules or ions that donate electron pairs). The naming reflects this relationship:
- Cation first, then anion (like normal ionic compounds)
- Ligands named before the metal (because they modify the metal's identity)
Why?
In chemistry, we name the more electropositive part first (cation). The metal-ligand complex is treated as a single unit — the ligands are "attached" to the metal, so they come first in the complex name.
2. Key Rule: Ligand Order — Alphabetical, Not by Charge
Rule: Ligands are named in alphabetical order (ignoring prefixes like di-, tri-).
Why?
- If we ordered by charge or size, the name would change every time a ligand is replaced.
- Alphabetical order is universal and unambiguous — it doesn't depend on the metal or oxidation state.
- Example:
[Co(NH₃)₄Cl₂]⁺is tetraamminedichlorocobalt(III) — "ammine" (a) before "chloro" (c).
3. Oxidation State: Why Roman Numerals?
Rule: The metal's oxidation state is written in Roman numerals in parentheses after the metal name.
Why?
- The oxidation state tells you the charge on the metal after accounting for ligand charges.
- Roman numerals avoid confusion with Arabic numbers (which are used for ligand counts).
- Example:
[Fe(CN)₆]³⁻→ hexacyanoferrate(III) — the iron is Fe³⁺, not Fe²⁺.
Derivation of oxidation state:
Let the complex charge = Q, ligand charges = sum of ligand charges L, number of ligands = n.
Then:
Metal oxidation state=Q−L
For [Fe(CN)₆]³⁻: CN⁻ has charge -1, so L=6×(−1)=−6, Q=−3.
Fe oxidation state=−3−(−6)=+3
4. Anionic Ligands: The "-o" Ending
Rule: Anionic ligands (negative ions) end in -o (e.g., Cl⁻ → chloro, CN⁻ → cyano, OH⁻ → hydroxo).
Why?
- This distinguishes them from neutral ligands (e.g., NH₃ → ammine, H₂O → aqua).
- The suffix -o signals "this ligand came from an anion" — crucial for charge balance.
Common examples:
| Anion | Ligand name |
|---|---|
| Cl⁻ | chloro |
| CN⁻ | cyano |
| OH⁻ | hydroxo |
| SO₄²⁻ | sulfato |
5. Neutral Ligands: Special Names
Rule: Neutral ligands keep their molecular name, except for a few with special names:
- NH₃ → ammine (not "ammonia")
- H₂O → aqua
- CO → carbonyl
- NO → nitrosyl
Why?
- "Ammine" avoids confusion with ammonia (NH₃) as a free molecule.
- These special names are historical but standardised — you must memorise them for exams.
6. Prefixes: di-, tri-, tetra-, etc.
Rule: Use Greek prefixes to indicate the number of each ligand:
- 2 → di, 3 → tri, 4 → tetra, 5 → penta, 6 → hexa
Why?
- Without prefixes,
[Co(NH₃)₆]³⁺would be "hexaamminecobalt(III)" — the "hexa" tells you there are six ammines. - For ligands with complex names (e.g., ethylenediamine), use bis-, tris-, tetrakis- to avoid confusion.
Example:
[Co(en)₃]³⁺ → tris(ethylenediamine)cobalt(III) — "tris" because "triethylenediamine" would sound like three ethylenediamine molecules (which is correct, but "tris" is clearer).
7. Anionic Complexes: The "-ate" Suffix …
Part (b)Concept understanding — Complex Formation
Complex Formation: The Intuition
Imagine you have a metal ion — say, a copper ion (Cu2+) — floating in water. It's positively charged, so it attracts anything negative or electron-rich nearby. Water molecules themselves have lone pairs of electrons on oxygen, so they crowd around the copper ion, each one donating a pair of electrons to form a coordinate bond. That cluster — the metal ion surrounded by water molecules — is already a complex ion: [Cu(H2O)6]2+.
Now, suppose you add ammonia (NH3) to the solution. Ammonia also has a lone pair on nitrogen, and it's a stronger electron donor than water. One by one, the ammonia molecules push the water molecules aside, replacing them. You end up with a deep blue complex: [Cu(NH3)4]2+.
That process — the stepwise replacement of one set of molecules (or ions) around a central metal atom by another set — is complex formation. The central metal is the Lewis acid (electron-pair acceptor), and the molecules or ions that attach to it are ligands (Lewis bases, electron-pair donors). The resulting species is a coordination compound or complex.
The word "complex" doesn't mean complicated. It just means a central atom (usually a metal) bonded to surrounding molecules or ions.
The Precise Statement
Complex formation is the reversible, stepwise reaction in which a central metal atom or ion (usually a transition metal) accepts electron pairs from one or more ligands to form a coordination entity. Each step has its own equilibrium constant, and the overall stability of the complex is measured by the formation constant (Kf) or stability constant.
For a general reaction:
M+nL⇌MLn
The overall formation constant is:
Kf=[M][L]n[MLn]
A large Kf means the complex is very stable — the ligands bind tightly and are hard to remove.
Kf=[metal][ligand]n[complex]
Key Features to Remember
-
Stepwise nature: Complexes don't form all at once. First one ligand binds, then another, and so on. Each step has its own constant (K1,K2,…). The product of all stepwise constants equals the overall Kf.
-
Coordination number: The number of ligand donor atoms directly bonded to the metal. Common values are 4 (tetrahedral or square planar) and 6 (octahedral). For [Cu(NH3)4]2+, the coordination number is 4.
-
Ligand denticity: A ligand can have one donor atom (monodentate, like NH3 or H2O) or multiple donor atoms (polydentate, like EDTA, which wraps around the metal with six donor atoms). Polydentate ligands form especially stable complexes — this is the chelate effect.
-
Reversibility: Complex formation is an equilibrium. Change the concentration of ligand, pH, or temperature, and the complex can break apart or form a different one. …
Why this formula?
Complex Formation: Understanding the Why Behind the Key Formulas
Complex formation is a fundamental concept in coordination chemistry and equilibrium. Let's build the reasoning step-by-step, starting from the simplest idea.
1. What is Complex Formation?
A complex forms when a central metal ion (Lewis acid) accepts electron pairs from surrounding molecules or ions called ligands (Lewis bases).
Example:
Cu2++4NH3⇌[Cu(NH3)4]2+
The key question: Why do we get a specific formula for the equilibrium constant?
2. The Stepwise Formation (The Core Reason)
Complex formation does not happen in one giant leap. It occurs in successive, reversible steps — each step adding one ligand.
For a metal M and ligand L:
Step 1:
M+L⇌ML
Equilibrium constant: K1=[M][L][ML]
Step 2:
ML+L⇌ML2
K2=[ML][L][ML2]
Step 3:
ML2+L⇌ML3
K3=[ML2][L][ML3]
... and so on up to MLn.
Each Ki is called a stepwise formation constant.
3. The Overall Formation Constant (The Key Formula)
Now, what if we want the equilibrium constant for the overall reaction:
M+nL⇌MLn
We can multiply the stepwise equilibria (because when you add reactions, you multiply their equilibrium constants):
Kf=K1×K2×K3×⋯×Kn
So:
Kf=[M][L]n[MLn]
Why this form?
Because each step contributes one [L] in the denominator and one [MLi] in the numerator, but intermediate species cancel out when multiplied.
4. Why the Denominator Has [L]n (Not n[L])
This is a common confusion. Let's derive it explicitly:
From step 1: [ML]=K1[M][L]
From step 2: [ML2]=K2[ML][L]=K1K2[M][L]2
From step 3: [ML3]=K3[ML2][L]=K1K2K3[M][L]3
Continuing:
[MLn]=(K1K2…Kn)[M][L]n
Therefore:
[M][L]n[MLn]=K1K2…Kn=Kf
The exponent n comes from repeated multiplication, not addition. Each ligand adds one factor of [L] in the denominator.
5. The Stability Connection
A larger Kf means:
- The complex is more stable
- Equilibrium lies far to the right (products favoured) …
Part (a)
(i) [Co(NH3)5(ONO)]2+
Ligands: five ammine + one nitrito-κO (O-bonded ONO−). Oxidation state of Co: x+5(0)+(−1)=+2⇒x=+3.
Name: pentaamminenitrito-κO-cobalt(III) ion.
(ii) K2[NiCl4]
Anionic complex [NiCl4]2−; Ni: x+4(−1)=−2⇒x=+2; anionic ⇒ "-ate". …
- [Co(NH3)5(ONO)]2+ = pentaamminenitrito-κO-cobalt(III) ion; K2[NiCl4] = potassium tetrachloridonickelate(II).
- A chelate has one polydentate ligand forming a ring ([Cu(en)2]2+); a heteroleptic complex has more than one type of ligand ([Co(NH3)4Cl2]+).
Part (a)
Rules: name ligands alphabetically (ignore the multiplying prefixes), anionic ligands end in "-o/-ido", give the metal oxidation state in Roman numerals; an anionic complex takes the suffix "-ate".
(i) [Co(NH3)5(ONO)]2+
- Ligands: 5 × NH3 = pentaammine; 1 × ONO− bonded through oxygen = nitrito-κO.
- Oxidation state: x+5(0)+(−1)=+2⇒x=+3.
- Alphabetical order: ammine before nitrito. Name: pentaamminenitrito-κO-cobalt(III) ion.
ONO− is an ambidentate ligand: O-bonded = nitrito-κO, N-bonded = nitrito-κN (these are linkage isomers).
(ii) K2[NiCl4]
- Complex ion [NiCl4]2−; ligands: 4 × Cl− = tetrachlorido.
- Oxidation state: x+4(−1)=−2⇒x=+2.
- Anionic complex ⇒ nickel → nickelate. …
Showing the 12 most recent of 79 on this concept.
- CBSE 2026Set 56/3/11 markMCQQ.The oxidation number of Co in [Co(en)3]2(SO4)3 is : (A) +3 (B) +2 (C) +4 (D) +6
›Reveal solutionSolution
The complex cation [Co(en)3]n+ must balance three sulfate anions (SO42−); since ethylenediamine is neutral, cobalt carries +3.
Why oxidation numbers matter in coordination compounds
Oxidation state tells us the formal charge on the central metal after we've assigned all bonding electrons to the more electronegative atom. In coordination chemistry, we treat ligands as intact units: neutral ligands contribute zero, anionic ligands contribute their charge. The sum of oxidation states in the entire complex must equal its net charge.
The formula [Co(en)3]2(SO4)3 shows us a salt: two complex cations paired with three sulfate anions. Our job is to figure out what charge the cobalt must carry to make the arithmetic work.
Step-by-step determination
1. Identify the ionic components
The compound dissociates into:
- Two [Co(en)3]n+ cations (where n is unknown)
- Three SO42− anions
2. Apply charge neutrality
The entire salt is neutral, so total positive charge equals total negative charge:
2×(charge on one cation)=3×2
2n=6
n=+3
Each complex cation carries a +3 charge: [Co(en)3]3+.
3. Determine cobalt's oxidation state within the cation
Now look inside [Co(en)3]3+. Ethylenediamine (en=H2NCH2CH2NH2) is a neutral bidentate ligand—it donates two lone pairs but carries no charge. Three en ligands contribute:
3×0=0
The oxidation state of cobalt plus the ligand contributions must equal the cation charge:
xCo+0=+3
xCo=+3 …
- CBSE 2026Set 56/2/11 markMCQQ.The correct IUPAC name of the complex [Pt(NH3)2Cl2] is : (A) diamminedichloridoplatinum (IV) (B) diamminedichloridoplatinum (II) (C) dichloridodiammineplatinum (IV) (D) dichloridodiammineplatinum (II)
›Reveal solutionSolution
The complex [Pt(NH3)2Cl2] is neutral, so the oxidation state of Pt must be +2. Ligands are named alphabetically (ammine before chlorido), and the metal is named without a suffix. The correct IUPAC name is diamminedichloridoplatinum(II) — option (B).
The key to naming coordination compounds is to follow the IUPAC rules in order: identify the oxidation state of the metal, list ligands alphabetically (ignoring prefixes like di-, tri-), and then name the metal with its oxidation state in parentheses.
Let’s break this down step by step.
-
Determine the oxidation state of platinum.
The complex [Pt(NH3)2Cl2] is neutral — no overall charge.
- NH3 is a neutral ligand (charge 0).
- Cl is a negatively charged ligand (chlorido, charge –1). Let the oxidation state of Pt be x. Then: x+2(0)+2(−1)=0⟹x−2=0⟹x=+2. So platinum is in the +2 oxidation state.
-
Name the ligands in alphabetical order.
IUPAC rules: ligands are named alphabetically by their name (not by prefix).
- NH3 is called ammine (note the double 'm').
- Cl is called chlorido (the anionic ligand name for chloride). Alphabetically, "ammine" comes before "chlorido". So the ligand order is: diammine then dichlorido.
-
Name the metal.
Since the complex is anionic? No — it’s neutral. For neutral complexes, the metal is called by its usual name (platinum), followed by the oxidation state in Roman numerals in parentheses: platinum(II).
-
Assemble the full name.
Ligands first (with prefixes di- for two identical ligands), then metal + oxidation state:
diamminedichloridoplatinum(II). …
-
- CBSE 2026Set 56/2/11 markMCQQ.Which of the following is heteroleptic complex ? (A) [Co(NH3)6]3+ (B) [Cr(NH3)6]3+ (C) [Ni(H2O)6]2+ (D) [Co(NH3)4Cl2]+
›Reveal solutionSolution
A heteroleptic complex contains more than one type of ligand. Among the given options, only [Co(NH3)4Cl2]+ has two different ligands (NH3 and Cl−), making (D) the answer.
The distinction between homoleptic and heteroleptic complexes is fundamental to coordination chemistry and comes down to ligand diversity.
A homoleptic complex (from Greek homo = same, leptos = taking) contains only one kind of ligand attached to the central metal ion. Think of it as a "uniform" coordination sphere where every ligand is identical.
A heteroleptic complex (from Greek hetero = different) contains two or more different types of ligands. The coordination sphere is "mixed."
This classification matters because heteroleptic complexes exhibit richer isomerism (geometrical, optical) and more varied chemical behavior than their homoleptic counterparts.
Now let's examine each option systematically:
-
Option (A): [Co(NH3)6]3+
The cobalt(III) ion is surrounded by six ammonia molecules. Every ligand is NH3—no variation whatsoever. This is a textbook homoleptic complex.
-
Option (B): [Cr(NH3)6]3+
Chromium(III) coordinated to six identical ammonia ligands. Again, uniform ligand environment. Homoleptic.
-
Option (C): [Ni(H2O)6]2+
Nickel(II) surrounded by six water molecules. All ligands are the same. Homoleptic.
-
Option (D): [Co(NH3)4Cl2]+ …
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- CBSE 2026Set ANNUAL1 markMCQQ.The oxidation number of nickel in [Ni(CO)4] will be:(a) 1(b) 0(c) 2(d) 3
›Reveal solutionSolution
CO is a neutral ligand, so the oxidation number of Ni in [Ni(CO)₄] is 0.
In a coordination compound, the oxidation number of the central metal is found by assigning charges to the ligands and balancing against the overall charge of the complex. Carbonyl (CO) is a neutral ligand — it donates a lone pair from carbon without carrying any charge itself. Since [Ni(CO)₄] is a ne …
- CBSE 2026Set ANNUAL1 markQ.Write the formula of the coordination compound tetraamine aquachlorido cobalt (III) chloride.
›Reveal solutionSolution
Build the octahedral coordination sphere from the ligands named, find the complex ion's net charge from the metal's oxidation state, then add counter-ions to balance that charge.
Naming breakdown: 'tetraammine' → 4 NH3 ligands (neutral); 'aqua' → 1 H2O ligand (neutral); 'chlorido' → 1 Cl− ligand (anionic, −1); 'cobalt(III)' → central metal Co3+. Coordination number =4+1+1=6 (octahedral), consistent with typical cobalt(III) ammine complexes.
…
- CBSE 2026Set ANNUAL1 markMCQQ.Mohr's salt is-(a)(i) Fe₂(SO₄)₃.(NH₄)₂SO₄.6H₂O(b)(ii) FeSO₄.(NH₄)₂SO₄.6H₂O(c)(iii) MgSO₄.7H₂O(d)(iv) FeSO₄.7H₂O
›Reveal solutionSolution
Mohr's salt is ferrous ammonium sulphate hexahydrate, FeSO4⋅(NH4)2SO4⋅6H2O. Correct option: (ii).
Concept. Mohr's salt is a double salt — a stoichiometric combination of two simple salts, ferrous sulphate FeSO4 and ammonium sulphate (NH4)2SO4 — that dissolves in water to release all its constituent ions independently (Fe2+, NH4+, SO42−).
Why the other options are wrong.
- (i) Fe2(SO4)3⋅(NH4)2SO4⋅6H2O contains ferric iron (Fe3+) — that is ferric alum-type, not Mohr's salt.
- (iii) MgSO4⋅7H2O is Epsom salt. …
- CBSE 2026Set ANNUAL1 markQ.Write answer in one word/sentence: Write chemical formula of Iron (III) hexacyanidoferrate (II).
›Reveal solutionSolution
Iron(III) hexacyanidoferrate(II) is Fe4[Fe(CN)6]3.
The complex anion hexacyanidoferrate(II) is [Fe(CN)6]4- (Fe in +2, six CN- ligands). The counter-cation is iron(III), Fe3+.
…
- CBSE 2026Set ANNUAL1 markMCQQ.Which of the following is a chelating ligand ?(a) NH3(b) H2O(c) Cl-(d) C2O4^2-
›Reveal solutionSolution
A chelating ligand grips the metal at more than one point; oxalate binds through two O atoms, forming a five-membered ring. Answer: (d) C2O4^2-.
- NH3, H2O and Cl- are all monodentate — each donates through a single atom, so they cannot chelate. …
- CBSE 2026Set ANNUAL1 markQ.A complex has the composition Co(NH₃)₄BrCl₂. Conductance measurement shows that there are two ions per formula unit and on treatment with silver nitrate it forms a yellow precipitate. Write the IUPAC name of the complex compound.
›Reveal solutionSolution
A yellow AgBr precipitate shows free Br⁻; two ions per formula unit fix the structure as [Co(NH₃)₄Cl₂]Br → tetraamminedichloridocobalt(III) bromide.
Deducing the structure:
- The composition is Co(NH3)4BrCl2.
- Conductance shows two ions per formula unit, so the complex ionises into one cation and one anion.
- With AgNO3 it gives a yellow precipitate, which is AgBr (silver chloride is white). So it is bromide (Br−) that is the free, ionisable counter-ion outside the coordination sphere, while both chloride ions are coordinated (non-ionisable) inside.
Hence the formula is [Co(NH3)4Cl2]Br, giving the two ions [Co(NH3)4Cl2]+ and Br−.
…
- CBSE 2025Set 56/6/11 markMCQQ.Which of the following complex ion is not optically active ? (A) [Co(ox)3]3− (B) cis-[Co(en)2Cl2]+ (C) trans-[Co(en)2Cl2]+ (D) [Co(en)3]3+
›Reveal solutionSolution
Optical activity in coordination complexes requires the absence of a plane of symmetry. Among the given options, trans-[Co(en)2Cl2]+ has a centre of symmetry and a plane of symmetry, making it optically inactive. The correct answer is (C).
Why Optical Activity Matters in Coordination Chemistry
Optical activity is a property of chiral molecules — those that are non-superimposable on their mirror image. In coordination compounds, chirality arises from the spatial arrangement of ligands around the central metal ion. A complex is optically active if it lacks an improper axis of rotation (specifically, a plane of symmetry or a centre of symmetry). The classic test: if a complex and its mirror image cannot be superimposed, they are enantiomers, and the complex is optically active.
For octahedral complexes, chirality often appears when:
- Bidentate ligands (like oxalate, ox2−, or ethylenediamine, en) create a helical twist.
- The arrangement of different ligands breaks symmetry.
Let’s examine each option systematically.
1. [Co(ox)3]3− — The Tris(oxalato) Complex
Oxalate (ox2−) is a bidentate ligand that forms a five-membered chelate ring. Three oxalate ions around Co(III) give an octahedral geometry. The complex has a propeller-like shape: each oxalate spans one edge of the octahedron, and the three rings are arranged in a helical fashion.
Think of it like a three-bladed fan. The complex exists as a pair of enantiomers — left-handed and right-handed helices. There is no plane of symmetry because the chelate rings lock the structure into a chiral twist. Therefore, [Co(ox)3]3− is optically active.
TipAny octahedral complex with three identical bidentate ligands (like [M(AA)3]) is always chiral — it’s a classic example of helical chirality. The same applies to [Co(en)3]3+ in option (D).
2. cis-[Co(en)2Cl2]+ — The Cis Isomer
Here, two ethylenediamine (en) ligands and two chloride ligands surround Co(III). The “cis” prefix means the two chlorides are adjacent (90° apart). In this geometry, the two en ligands are not equivalent in space — they create a non-superimposable mirror image.
Draw the structure: the two en rings lie in roughly perpendicular planes. The cis arrangement of Cl atoms breaks any plane of symmetry. The complex is chiral, and indeed, cis-[Co(en)2Cl2]+ has been resolved into enantiomers. So it is optically active.
Watch outA common mistake is to think that any complex with two identical bidentate ligands is automatically chiral. That’s only true for the cis isomer — the trans isomer is different, as we’ll see next.
3. trans-[Co(en)2Cl2]+ — The Trans Isomer …
- CBSE 2025Set ANNUAL1 markQ.Write formula for co-ordination compound Potassium trioxalatochromate (III).
›Reveal solutionSolution
Three bidentate oxalate ligands (each -2) plus Cr3+ gives a -3 complex ion balanced by 3 K+.
'Trioxalato' means three oxalate (C2O42−) ligands (bidentate, each carrying charge −2); 'chromate(III)' means the central metal is chromium in the +3 oxidation state.
…
- CBSE 2025Set D1 markMCQQ.The IUPAC name of complex compound [Co(NH3)6]Cl3 is(a) Hexa-ammine cobalt (III) chloride(b) Hexa-ammine cobalt (II) chloride(c) Hexa-ammine trichloridocobalt (III)(d) None of these
›Reveal solutionSolution
[Co(NH3)6]Cl3 = hexaamminecobalt(III) chloride.
Rules of IUPAC nomenclature:
- Name the cation first, then the anion.
- Within the complex, ligands are named alphabetically before the metal.
- NH3 as a ligand is 'ammine' (six of them -> hexaammine).
- Oxidation state of Co: three Cl- give -3; overall neutral, so Co = +3, written as (III).
- The chloride outside is the counter-anion. …
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