Q.Write IUPAC name of the complex [Pt(en)2Cl2]. Draw structures of geometrical isomers for this complex.
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🔒 Start your 14-day free trial to unlock the full solution →Part (a)Concept understanding — Geometrical Isomerism Conditions
Geometrical Isomerism: The Intuition
Imagine you have two friends standing on opposite sides of a door. If the door is open, they can walk around and swap places easily — there's no real difference between who is on the left and who is on the right. But if the door is locked shut, they are stuck. One is permanently on the left side, the other on the right. That locked door creates two distinct arrangements: Friend A on the left, Friend B on the right versus Friend A on the right, Friend B on the left.
That locked door is the key idea behind geometrical isomerism.
In chemistry, molecules are three-dimensional. Atoms connected by a single bond can rotate freely — like an open door. But a double bond (or a ring structure) locks the atoms in place. If you have two different groups attached to each carbon of a double bond, you get two distinct spatial arrangements that cannot interconvert without breaking the bond. These are geometrical isomers (also called cis-trans or E-Z isomers).
The Precise Conditions
For a molecule to show geometrical isomerism, it must satisfy two conditions simultaneously:
Condition 1: There must be a restricted rotation around a bond — typically a carbon-carbon double bond (C=C) or a ring structure.
Condition 2: Each of the two atoms (or groups) involved in that restricted rotation must have two different substituents attached to it.
Let's unpack each.
Condition 1: Restricted Rotation
A single bond (C−C) allows free rotation — the atoms spin around the bond axis like a wheel. So no geometrical isomers exist there. A double bond (C=C) has a pi (π) bond that locks the molecule flat. Rotation would break the π bond, which requires a lot of energy (about 250–270 kJ/mol). At room temperature, this rotation simply does not happen.
Rings (like cyclopropane, cyclobutane, etc.) also restrict rotation because the ring is a closed loop — atoms cannot rotate past each other without breaking the ring.
Condition 2: Two Different Substituents on Each End
This is the "different groups" rule. Look at each carbon of the double bond (or each ring carbon involved). If both carbons have two different groups attached, geometrical isomers exist. If even one carbon has two identical groups, there is only one possible arrangement.
A common mistake: students check only one carbon. Both carbons must have two different substituents. If one carbon has two identical groups (like two hydrogens), the molecule is identical in both arrangements — no isomerism.
How to Check: A Step-by-Step Method
Take any molecule with a double bond. Follow these steps:
- Identify the double bond (or ring). Mark the two carbon atoms involved.
- List the two groups attached to the first carbon. Are they different from each other? If yes, proceed. If no → no geometrical isomerism.
- List the two groups attached to the second carbon. Are they different from each other? If yes → geometrical isomerism exists. If no → no geometrical isomerism.
If the two groups on a carbon are identical, the molecule is symmetric about that carbon. Flipping the other side gives the same molecule — no isomers.
Examples to Cement the Idea
Example 1: But-2-ene (CH3CH=CHCH3)
- Carbon 1 of the double bond: attached to CH3 and H → different ✓
- Carbon 2 of the double bond: attached to CH3 and H → different ✓
Result: Two geometrical isomers exist — cis (both methyl groups on the same side) and trans (methyl groups on opposite sides).
Example 2: 1,2-Dichloroethene (ClCH=CHCl)
- Carbon 1: attached to Cl and H → different ✓
- Carbon 2: attached to Cl and H → different ✓
Result: cis and trans isomers exist.
Example 3: 1,1-Dichloroethene (Cl2C=CH2)
- Carbon 1: attached to Cl and Cl → identical ✗ …
Why this formula?
Geometrical Isomerism: Why the Conditions Hold
Geometrical isomerism (also called cis-trans or E-Z isomerism) arises when atoms or groups are arranged differently in space around a rigid part of a molecule — typically a double bond or a ring. The key is that rotation is restricted, so the spatial positions become fixed and distinct.
Let’s break down why the conditions are what they are.
1. The Core Requirement: Restricted Rotation
For two molecules to be geometrical isomers, they must have the same connectivity but different spatial arrangement due to a barrier to rotation.
- Double bonds (C=C): The π-bond locks the two carbons in place — rotation requires breaking the π-bond (energy ~250 kJ/mol), so it doesn’t happen at room temperature.
- Rings (e.g., cycloalkanes): The ring structure physically prevents free rotation about C–C single bonds within the ring.
Why this matters: Without restricted rotation, the molecule would freely interconvert between arrangements — no distinct isomers exist.
2. Condition 1: Two Different Groups on Each Carbon (for C=C)
Consider a general alkene:
C=C
Each carbon must have two different substituents (not counting the other carbon of the double bond).
Why?
- If one carbon has two identical groups (e.g., both H), then swapping the groups on that carbon produces the same molecule — no isomerism.
Example:
- 1,2-dichloroethene (ClHC=CHCl): Each carbon has H and Cl (different) → geometrical isomers exist.
- 1,1-dichloroethene (Cl2C=CH2): One carbon has two Cl (identical) → no geometrical isomers.
Formal condition:
For a C=C bond, geometrical isomerism is possible iff each doubly bonded carbon bears two different substituents.
3. Condition 2: For Rings — Similar Logic
In a ring (e.g., cyclopropane, cyclohexane), the ring itself restricts rotation. Here, geometrical isomerism occurs when two substituents on different ring carbons can be on the same side (cis) or opposite sides (trans).
Why?
- The ring is a closed loop — you cannot rotate one carbon relative to another without breaking bonds.
- If the two substituents are on different carbons, their relative orientation (same side / opposite sides) is fixed.
Condition:
- The ring must have at least two substituents (could be same or different) on different carbons.
- If both substituents are on the same carbon, swapping them doesn’t change the molecule (no isomerism).
Example:
- 1,2-dimethylcyclopropane: Two methyl groups on adjacent carbons → cis and trans isomers exist.
- 1,1-dimethylcyclopropane: Both methyls on same carbon → no geometrical isomerism.
4. The E-Z Notation (Why It’s Needed)
When the four substituents on a C=C are all different, cis-trans naming fails. The Cahn-Ingold-Prelog priority rules assign E (opposite sides) or Z (same side).
Why this works:
- Priority is based on atomic number (higher = higher priority). …
Part (b)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 (a)
IUPAC name of [Pt(en)2Cl2]: Pt oxidation state =+2 (en neutral, two Cl−, complex neutral). Ligands alphabetically (chlorido before ethylenediamine); use "bis" for en:
dichloridobis(ethylenediamine)platinum(II).
Geometrical isomers — an octahedral [M(en)2Cl2]-type complex shows cis (the two Cl at 90°, adjacent) and trans (the two Cl at 180°, opposite) forms; each en chelates two adjacent sites.
cis: two Cl adjacent (90 deg) trans: two Cl opposite (180 deg)
Cl Cl
| |
en___Pt___Cl en___Pt___en
| |
en Cl …
- [Pt(en)2Cl2] is dichloridobis(ethylenediamine)platinum(II) and shows cis–trans geometrical isomerism.
- Hexaamminecobalt(III) sulphate = [Co(NH3)6]2(SO4)3; potassium trioxalatochromate(III) = K3[Cr(C2O4)3].
Part (a)
IUPAC name of [Pt(en)2Cl2]
- Ligands: two Cl− (dichlorido) and two en (bis(ethylenediamine); "bis" because en already contains "di").
- Oxidation state: x+2(0)+2(−1)=0⇒x=+2.
- Alphabetical order of ligand names: chlorido before ethylenediamine.
Name: dichloridobis(ethylenediamine)platinum(II).
Geometrical isomers
For a [M(AA)2X2] complex (AA = bidentate en, X = Cl) the two chlorido ligands can be:
- cis — the two Cl occupy adjacent positions (90° apart);
- trans — the two Cl occupy opposite positions (180° apart).
Each en must span two adjacent sites (it cannot bridge trans positions). Both arrangements are possible, distinguished by the relative positions of the two Cl ligands.
cis trans
Cl Cl
| |
en___Pt___Cl en___Pt___en
| |
en Cl
(two Cl at 90 deg) (two Cl at 180 deg)
``` …
Showing the 12 most recent of 67 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]+ …
-
- CBSE 2026Set ANNUAL1 markQ.Draw the structure of geometrical isomers of [Co(NH3)4Cl2].
›Reveal solutionSolution
[Co(NH3)4Cl2]+ is an octahedral complex of the type [MA4B2], which shows cis-trans geometrical isomerism depending on the relative positions of the two identical Cl ligands.
The complex [Co(NH3)4Cl2]+ has an octahedral geometry with 4 NH3 and 2 Cl- ligands around the central Co(III) ion. For an [MA4B2] type octahedral complex, two arrangements of the two B (Cl) ligands are possible:
- cis-isomer: the two Cl- ligands occupy adjacent positions on the octahedron, with a Cl-Co-Cl bond angle of 90 degrees. (Structure: picture an octahedron with NH3 on four positions and the two Cl ligands on two adjacent corners.) …
- 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 markMCQQ.Which complexes do not show geometrical isomerism?(a) Square planar complexes(b) Tetrahedral complexes(c) Octahedral complexes(d) All of the above
›Reveal solutionSolution
Geometrical (cis/trans, fac/mer) isomerism requires ligand positions that are not all equivalent/adjacent; a tetrahedral geometry has no such distinction, so it alone among these never shows geometrical isomerism.
- (a) Square planar complexes (e.g. [Pt(NH3)2Cl2], type MA2B2) do show cis–trans geometrical isomerism, since two positions can be adjacent (cis, 90∘) or opposite (trans, 180∘).
- (c) Octahedral complexes (types MA4B2, MA3B3, etc.) do show both cis–trans and facial–meridional (fac/mer) geometrical isomerism, since some positions are adjacent and some are directly opposite. …
- 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 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 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.
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- 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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