Q.When 1 mol CrCl3⋅6H2O is treated with excess of AgNO3, 3 mol of AgCl are obtained. The formula of the complex is:
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🔒 Start your 14-day free trial to unlock the full solution →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 …
Concept: Coordination Compound Nomenclature — the number of AgCl moles precipitated equals the number of chloride ions present outside the coordination sphere (i.e., free or ionizable Cl−).
Reasoning:
- Excess AgNO3 precipitates only the chloride ions that are not coordinated to the metal centre.
- 3 mol of AgCl means 3 mol of free Cl− per mole of complex. …
The key is that only chloride ions outside the coordination sphere (counter ions) precipitate with AgNO3. Since 3 mol of AgCl form per mol of complex, all three chlorides are counter ions, so the formula must be [Cr(H2O)6]Cl3 — option (iv).
This problem tests your understanding of coordination compound nomenclature and the difference between coordination sphere and counter ions. When you treat a complex with AgNO3, the silver ions only precipitate chloride ions that are free — those outside the square brackets. Chloride ions inside the coordination sphere (bonded directly to the metal) do not dissociate and will not react with Ag+.
The question gives you a critical experimental fact: 1 mol of the complex yields 3 mol of AgCl. That means all three chloride ions present in the formula unit are outside the coordination sphere. None are bonded to chromium.
Let’s check each option systematically.
-
Option (i): [CrCl3(H2O)3]⋅3H2O
Here, all three chlorides are inside the coordination sphere. The three water molecules outside are just water of crystallization. So zero free Cl− ions — this would give 0 mol of AgCl. Eliminated.
-
Option (ii): [CrCl2(H2O)4]Cl⋅2H2O
Two chlorides are inside the sphere, one is outside as a counter ion. So only 1 mol of AgCl would precipitate. Not matching the given 3 mol. Eliminated.
-
Option (iii): [CrCl(H2O)5]Cl2⋅H2O
One chloride inside, two outside. That gives 2 mol of AgCl. Still not enough. Eliminated.
-
Option (iv): [Cr(H2O)6]Cl3 …
Method: Conductometric / Precipitation Analysis of Ionizable Chloride
This method uses the fact that only free (ionizable) chloride ions outside the coordination sphere react with AgNO3 to give AgCl precipitate. Chloride ions inside the coordination sphere (ligands) do not precipitate.
Steps
Step 1: Identify the given data
- 1 mol of complex → 3 mol of AgCl
- This means 3 mol of free Cl− are present per mol of complex.
Step 2: Count total chloride in each option
All options have total 3 Cl atoms per formula unit. But only those outside the square bracket are free.
Step 3: Check each option for number of free Cl−
-
(i) [CrCl3(H2O)3]⋅3H2O
→ All 3 Cl are inside coordination sphere → 0 free Cl → gives 0 mol AgCl ✗
-
(ii) [CrCl2(H2O)4]Cl⋅2H2O …
Common Mistakes & How to Avoid Them
Mistake 1: Confusing total chloride with ionizable chloride
The error: Students see 3 mol of AgCl formed and assume the complex contains 3 chloride ions in the coordination sphere — picking option (i) or (iv) without thinking.
Why it's wrong: AgNO₃ only precipitates free chloride ions (outside the coordination sphere). Chloride ions inside the coordination sphere (bonded to Cr) do not react with Ag⁺.
How to avoid: Always ask: "Which Cl⁻ are free to precipitate?" Only the counter ions (outside square brackets) react with AgNO₃.
Mistake 2: Forgetting water can be inside or outside the coordination sphere
The error: Students count all 6 water molecules as hydrate water (outside brackets) or all as coordinated water (inside brackets), ignoring that water can be both.
Why it's wrong: The formula CrCl3⋅6H2O tells total composition — it doesn't show how water is distributed. Some water may be coordinated to Cr, some may be outside as water of crystallization.
How to avoid: Remember: water molecules can be inside the coordination sphere (ligands) or outside (lattice water). The dot (⋅) separates the complex from crystallization water.
Mistake 3: Not balancing charge and coordination number
The error: Picking an option without checking if Cr's coordination number (usually 6) and oxidation state are consistent.
Why it's wrong: Cr³⁺ typically has coordination number 6. Each option must have exactly 6 ligands (Cl⁻ + H₂O) in the coordination sphere.
How to avoid: For each option:
- Count total ligands inside brackets = must be 6
- Check charge balance: Cr³⁺ + (ligand charges) + (counter ion charges) = 0
Mistake 4: Rushing to pick the first option that "looks right"
The error: Seeing 3 mol AgCl and immediately choosing option (iv) [Cr(H2O)6]Cl3 because it has 3 Cl⁻ outside.
Why it's wrong: Option (iv) gives 3 mol AgCl — but so does option (iii)! Both have 3 ionizable Cl⁻. You must check all conditions.
How to avoid: Test every option systematically: …
Showing the 12 most recent of 21 on this concept.
- TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQQ.Which one of the following complexes has least number of stereoisomers? (A) [Co(NH3)5Cl]Cl2 (B) [Co(en)(NH3)2Cl2]Cl (C) [Co(en)2Cl2]Cl (D) [Co(en)3]Cl3
›Reveal solutionSolution
An octahedral MA5B complex like [Co(NH3)5Cl]2+ has exactly one spatial arrangement — zero stereoisomerism — while every other listed complex has 2–4 stereoisomers.
Concept. Stereoisomer counting for octahedral complexes: MA5B has a single form; M(AA)2B2 has cis/trans with the cis form chiral; M(AA)3 is chiral (Δ/Λ); M(AA)A2′B2-type mixed complexes have several geometric forms, some chiral.
Step-by-step count for each option:
- (A) [Co(NH3)5Cl]Cl2 — cation type MA5B. All six octahedral vertices are equivalent for placing the lone Cl, so there is only 1 form: no stereoisomers.
- (B) [Co(en)(NH3)2Cl2]Cl — type M(AA)B2C2. With en spanning a cis pair of sites, the two NH₃ and two Cl can be arranged as: both Cl trans; both NH₃ trans; or all-cis. That is 3 geometric isomers, and the all-cis form is chiral — 4 stereoisomers in total. …
- TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQQ.Which of the following exhibit both geometrical and optical isomerism? I. [Pt(en)2Cl2]2+ II. [Co(NH3)3(NO2)3] III. [Cr(H2O)2(C2O4)2]+ IV. [Co(en)3]Cl3 (A) I, III, IV only (B) I, II, III only (C) II, III, IV only (D) I, III only
›Reveal solutionSolution
Geometrical isomerism requires a non‑tetrahedral coordination geometry with at least two different ligands (or chelate rings) in a non‑equivalent arrangement; optical isomerism requires the absence of a plane of symmetry. Only complexes I and III satisfy both conditions, so the answer is (D).
Concept & Intuition
Geometrical isomerism (cis/trans or fac/mer) arises when ligands can occupy distinct positions around a metal centre, typically in square‑planar or octahedral complexes. Optical isomerism (non‑superimposable mirror images) occurs when the complex lacks any improper rotation axis (i.e., no plane or centre of symmetry). For a complex to show both, it must have a geometry that allows distinct spatial arrangements and be chiral. Chelating ligands (like en or oxalate) often create chirality by locking the structure into a helical twist.
Step‑by‑step analysis
-
Complex I: [Pt(en)2Cl2]2+
- Pt(IV) is octahedral. Two bidentate en ligands and two Cl⁻ ligands.
- The two Cl⁻ can be cis (adjacent) or trans (opposite) → geometrical isomerism.
- The cis isomer has no plane of symmetry (the two en rings create a chiral twist) → it exists as a pair of enantiomers → optical isomerism.
- The trans isomer has a plane of symmetry → not optically active.
- So overall, this complex exhibits both types.
-
Complex II: [Co(NH3)3(NO2)3]
- Octahedral Co(III) with three NH₃ and three NO₂ ligands.
- Geometrical isomers: fac (three identical ligands on one face) and mer (three in a meridian).
- Both fac and mer have planes of symmetry (e.g., the fac isomer has a C₃ axis and three vertical planes).
- No chiral centre → no optical isomerism.
- Hence, only geometrical isomerism, not both.
-
Complex III: [Cr(H2O)2(C2O4)2]+
- Octahedral Cr(III) with two water ligands and two oxalate (bidentate) ligands.
- The two waters can be cis or trans → geometrical isomerism.
- The cis isomer: the two oxalate rings create a chiral arrangement (like a propeller) with no plane of symmetry → optical isomerism.
- The trans isomer has a plane of symmetry → not chiral.
- So again, both types are possible. …
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- TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQQ.Identify the correct orders of covalent character of given molecules I. KI > KBr > KCl > KF II. LiCl > NaCl > KCl > RbCl III. CsCl > CaCl2 > MgCl2 > AlCl3 The correct answer is (only —) (A) I, II, III (B) I, II only (C) II, III only (D) I, III only
›Reveal solutionSolution
Covalent character in ionic compounds increases with smaller cation size, larger anion size, and higher cation charge (Fajan’s rules). The correct orders are I and II only, making option (B) the answer.
Concept & Intuition
Covalent character in predominantly ionic compounds arises when the cation strongly polarizes the anion’s electron cloud. Fajan’s rules tell us:
- Smaller cation → greater polarizing power → more covalent character.
- Larger anion → more easily distorted → more covalent character.
- Higher cation charge → stronger pull on electrons → more covalent character. We apply these rules to each series to check if the given order matches.
Step-by-step reasoning
-
Series I: KI > KBr > KCl > KF
- All have the same cation (K⁺), so only the anion changes.
- Anion size order: I⁻ > Br⁻ > Cl⁻ > F⁻.
- Larger anion is more polarizable → more covalent character.
- So covalent character should decrease as anion gets smaller: KI > KBr > KCl > KF.
- This matches the given order → I is correct.
-
Series II: LiCl > NaCl > KCl > RbCl
- All have the same anion (Cl⁻), so only the cation changes.
- Cation size order: Li⁺ < Na⁺ < K⁺ < Rb⁺ (increasing down group 1).
- Smaller cation has greater polarizing power → more covalent character.
- So covalent character should decrease as cation gets larger: LiCl > NaCl > KCl > RbCl.
- This matches the given order → II is correct.
-
Series III: CsCl > CaCl₂ > MgCl₂ > AlCl₃
- Here both cation size and charge vary.
- Cs⁺ is large (+1), Ca²⁺ is smaller (+2), Mg²⁺ is even smaller (+2), Al³⁺ is smallest (+3). …
- TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQQ.In the metallurgy of silver, silver is leached with a dilute solution of KCN in the presence of air to form a complex ion X. This in the presence of zinc converts into another complex ion Y. The ratio of CNX− ligands in X and Y is (A) 1:1 (B) 1:2 (C) 2:1 (D) 2:3
›Reveal solutionSolution
The key is to recall the cyanide complexes formed during silver extraction: X is [Ag(CN)X2]− (2 CN⁻ ligands) and Y is [Zn(CN)X4]2− (4 CN⁻ ligands), so the ratio of CN⁻ ligands in X to Y is 2:4 = 1:2, making the correct option (B).
The problem describes the classic cyanide process for silver extraction. Understanding the coordination chemistry of silver and zinc with cyanide is essential.
Concept and Intuition:
Silver forms a very stable linear dicyanoargentate(I) ion when leached from its ore. Zinc, being more reactive, displaces silver from this complex and itself forms a stable tetracyanozincate(II) ion. The number of cyanide ligands bound to each metal is fixed by their oxidation states and coordination preferences: Ag⁺ typically coordinates two CN⁻, while Zn²⁺ coordinates four CN⁻. Thus, the ratio of CN⁻ ligands in the silver complex (X) to the zinc complex (Y) is simply 2:4, which simplifies to 1:2.
Step-by-step reasoning:
- Identify complex X (silver leaching): Silver is leached with dilute KCN in the presence of air (oxygen). The reaction is:
4Ag+8KCN+OX2+2HX2O→4K[Ag(CN)X2]+4KOH
Here, the soluble complex formed is potassium dicyanoargentate(I), [Ag(CN)X2]−. So X contains 2 CN⁻ ligands per silver ion.
- Identify complex Y (zinc displacement): Zinc metal is added to the solution of X to precipitate silver. The reaction is:
2K[Ag(CN)X2]+Zn→KX2[Zn(CN)X4]+2Ag
The zinc complex formed is potassium tetracyanozincate(II), [Zn(CN)X4]2−. So Y contains 4 CN⁻ ligands per zinc ion.
- Determine the ratio of CN⁻ ligands in X and Y: …
- TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQQ.The correct formula of hexaammine chromium (III) hexafluoronickelate (II) is (A) [Cr(NH3)6][NiF6] (B) [Cr(NH3)6]4[NiF6]3 (C) [Cr(NH3)6]3[NiF6]2 (D) [Cr(NH3)6]2[NiF6]
›Reveal solutionSolution
The name "hexaammine chromium(III) hexafluoronickelate(II)" describes a coordination compound with a cationic complex and an anionic complex. The key is to balance the charges: [Cr(NH3)6]3+ and [NiF6]4− combine in a 4:3 ratio to give a neutral formula, so the correct option is (B).
The name of this compound tells you it contains two distinct coordination entities — one positive and one negative. The trick is to decode each part, find its charge, and then figure out how many of each you need to make a neutral salt.
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Identify the cationic complex. The first part is "hexaammine chromium(III)". "Hexaammine" means six ammonia (NH3) ligands attached to the metal. "Chromium(III)" tells you the oxidation state of chromium is +3. Ammonia is a neutral ligand, so the charge on the complex comes entirely from the metal. Therefore, the cationic complex is [Cr(NH3)6]3+.
-
Identify the anionic complex. The second part is "hexafluoronickelate(II)". "Hexafluoro" means six fluoride (F−) ligands. The suffix "-ate" tells you this is an anionic complex. "Nickelate(II)" means nickel is in the +2 oxidation state. Each fluoride ligand carries a −1 charge. So the total charge on the complex is: +2 (from Ni) plus 6×(−1) (from six F−) = +2−6=−4. Therefore, the anionic complex is [NiF6]4−.
-
Balance the charges to get the neutral formula. You have a cation with charge +3 and an anion with charge −4. To form a neutral compound, the total positive charge must equal the total negative charge. The smallest whole numbers that satisfy this are found by taking the least common multiple of 3 and 4, which is 12.
- To get +12 from +3 ions, you need 12/3=4 cations. …
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- TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQQ.When 100 mL of 0.2 M solution of CoCl3.x NH3 is treated with excess of AgNO3 solution, 3.6×1022 ions are precipitated. The value of x is (N = 6×1023 mol−1) (A) 5 (B) 6 (C) 4 (D) 3
›Reveal solutionSolution
The key idea is that only the chloride ions outside the coordination sphere (free Cl⁻) precipitate with Ag⁺; the number of free Cl⁻ per formula unit is found from the moles of AgCl precipitated, giving x = 6.
Concept & Intuition
In coordination compounds like CoCl₃·xNH₃, the ammonia molecules and some chloride ions are bonded directly to the cobalt ion inside the coordination sphere. Only chloride ions that are outside the sphere (counterions) are free in solution and can react with AgNO₃ to form AgCl precipitate. The problem gives the number of Ag⁺ ions precipitated, which equals the number of free Cl⁻ ions. From that, we find how many free Cl⁻ are released per formula unit, and then deduce x.
Step-by-step solution
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Find moles of Ag⁺ (and thus free Cl⁻) precipitated
Number of Ag⁺ ions precipitated = 3.6×1022.
Avogadro’s number NA=6×1023 mol−1.
Moles of Ag⁺ = 6×10233.6×1022=0.06 mol.
Since each Ag⁺ reacts with one Cl⁻, moles of free Cl⁻ = 0.06 mol.
-
Moles of CoCl₃·xNH₃ in the sample
Volume = 100 mL = 0.100 L, concentration = 0.2 M.
Moles of complex = 0.100×0.2=0.020 mol.
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Free Cl⁻ per formula unit
From 0.020 mol of complex, we get 0.060 mol of free Cl⁻.
So per mole of complex, free Cl⁻ = 0.0200.060=3 moles.
That means each formula unit releases 3 free chloride ions.
-
Determine the coordination sphere
The compound is CoCl₃·xNH₃. Total Cl atoms = 3.
If 3 Cl⁻ are free, then 0 Cl⁻ are inside the coordination sphere. …
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- TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQQ.Total number of geometrical isomers possible for the complexes [NiCl4]2−, [CoCl2(NH3)4]+, [Co(NH3)3(NO2)3] and [Co(NH3)5Cl]2+ is (A) 2 (B) 3 (C) 4 (D) 5
›Reveal solutionSolution
The total number of geometrical isomers across all four complexes is 4, so the correct option is (C).
Concept & Intuition
Geometrical isomerism in coordination compounds arises when ligands can occupy different positions around a central metal ion, leading to distinct spatial arrangements that are not superimposable. The key is to identify the coordination number and geometry of each complex, then systematically count distinct arrangements (cis/trans, fac/mer, etc.). For square planar and octahedral complexes, we must be careful: square planar complexes with identical ligands may have no isomers, while octahedral complexes with mixed ligands can have multiple.
Step-by-step reasoning
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Complex 1: [NiCl4]2−
- Nickel(II) has a d8 configuration. For d8, tetrahedral geometry is common, but with strong-field ligands like chloride, square planar is also possible. However, Ni2+ with four chlorides is known to be tetrahedral (paramagnetic, no crystal field stabilization favoring square planar).
- In a tetrahedral geometry, all positions are equivalent; swapping any two ligands gives the same arrangement. Thus, no geometrical isomers exist.
- Count so far: 0.
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Complex 2: [CoCl2(NH3)4]+
- Cobalt(III) is d6 and almost always octahedral. Here, we have two chlorides and four ammines.
- In an octahedron, two identical ligands (Cl) can be adjacent (cis) or opposite (trans). These are non-superimposable.
- Cis and trans are the only possibilities. So 2 geometrical isomers.
- Count so far: 0 + 2 = 2.
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Complex 3: [Co(NH3)3(NO2)3]
- Again octahedral Co(III). Three ammines and three nitrites.
- For a MA3B3 complex, two distinct arrangements exist: facial (fac) where three identical ligands occupy one face of the octahedron, and meridional (mer) where they lie in a plane including the metal.
- These are geometrical isomers. So 2 isomers.
- Count so far: 2 + 2 = 4. …
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- TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQQ.Total number of geometrical isomers possible for the complexes [NiCl4]2−, [CoCl2(NH3)4]+, [Co(NH3)3(NO2)3] and [Co(NH3)5Cl]2+ is (A) 2 (B) 3 (C) 5 (D) 4
›Reveal solutionSolution
The total number of geometrical isomers across the four complexes is 4, so the correct option is (D).
Concept & Intuition
Geometrical isomerism in coordination compounds arises when ligands can occupy different positions around the central metal ion, leading to distinct spatial arrangements that are not mirror images. The key is to identify the geometry (tetrahedral, square planar, octahedral) and then count distinct arrangements of identical vs. different ligands. For tetrahedral complexes, all positions are equivalent, so no geometrical isomers exist. For square planar and octahedral complexes, we look for cis/trans or fac/mer possibilities.
Step-by-step reasoning
-
Complex 1: [NiCl4]2−
- Nickel(II) is d8, and with four chloride ligands, the geometry is tetrahedral (common for d8 with weak-field ligands like Cl⁻).
- In a tetrahedron, all vertices are equivalent; swapping any two ligands gives the same arrangement.
- Result: 0 geometrical isomers.
-
Complex 2: [CoCl2(NH3)4]+
- Cobalt(III) is d6 (low-spin with strong-field NH₃), and with six ligands, the geometry is octahedral.
- The formula shows two Cl⁻ and four NH₃ ligands. The two Cl⁻ can be:
- Adjacent (90° apart) → cis isomer.
- Opposite (180° apart) → trans isomer.
- No other distinct arrangements exist because all NH₃ are identical.
- Result: 2 geometrical isomers.
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Complex 3: [Co(NH3)3(NO2)3]
- Again, Co(III) is octahedral. Here we have three NH₃ and three NO₂ ligands.
- The two possible arrangements are:
- fac (facial): each set of three identical ligands occupies one face of the octahedron (all three NH₃ mutually cis to each other, same for NO₂).
- mer (meridional): the three identical ligands lie in a plane that goes through the metal (two trans to each other, one cis to both).
- These are non-superimposable and not mirror images (they are diastereomers).
- Result: 2 geometrical isomers. …
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- TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQQ.The number of non-ionizable valences of Co3+ and Pt4+ ions in the complexes CoCl3.5NH3 and PtCl4.5NH3 is respectively (A) 4, 4 (B) 3, 5 (C) 5, 6 (D) 6, 6
›Reveal solutionSolution
The key idea is to determine the number of chloride ions that are outside the coordination sphere (i.e., ionizable) by finding the coordination number and charge balance. For CoCl₃·5NH₃, the number of non-ionizable valences is 3; for PtCl₄·5NH₃, it is 5. Thus the correct pair is (3, 5).
Concept & Intuition
In coordination chemistry, “non-ionizable valences” refer to the number of chloride ions directly bonded to the metal ion (i.e., inside the coordination sphere) that do not dissociate in solution. The formula written as CoCl₃·5NH₃ means there are three chlorides and five ammonia molecules associated with the cobalt. However, not all chlorides are necessarily ligands — some may be outside the coordination sphere as counterions. To find how many are inside, we use the metal’s oxidation state, the charge of the complex, and the fact that NH₃ is neutral. The total charge of the complex ion plus any counterions must be zero.
Step-by-step reasoning
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Analyze CoCl₃·5NH₃
- Cobalt is in the +3 oxidation state (Co³⁺).
- NH₃ is neutral, so it contributes no charge.
- Let x be the number of Cl⁻ ions inside the coordination sphere (as ligands). Then the complex ion is [Co(NH3)5Clx](3−x)+.
- The remaining (3−x) Cl⁻ ions are outside, balancing the charge.
- The total compound is neutral, so the charge on the complex ion must equal the total negative charge from the outside chlorides: (3−x)=(3−x) — this is automatically satisfied.
- However, the coordination number of Co³⁺ is typically 6. With 5 NH₃ already, only 1 chloride can fit inside to reach coordination number 6. So x=1.
- Therefore, the number of non-ionizable (inside) chlorides = 1. But wait — the question asks for “non-ionizable valences” of the metal ion. This means the number of bonds from the metal to ligands that are not ionizable. Since each chloride inside uses one valence, and NH₃ also uses one valence, the total non-ionizable valences = number of ligands × 1. Here, 5 NH₃ + 1 Cl = 6 valences? No — careful: “non-ionizable valences” refers specifically to the chloride ions that are not free to ionize. So it’s just the number of Cl⁻ inside the coordination sphere. That is 1? But the options are larger numbers.
- Let’s re-read: “non-ionizable valences of Co³⁺ and Pt⁴⁺ ions” — this means the number of coordination sites occupied by non-ionizable groups (i.e., ligands that are not easily replaced by water or that do not dissociate). In older terminology, “valences” here means the number of bonds from the metal to ligands that are not ionizable. For Co³⁺, all six coordination sites are occupied by NH₃ and Cl⁻, but only the Cl⁻ inside are non-ionizable? Actually, NH₃ is also non-ionizable. So the total non-ionizable valences = total number of ligands that are not free ions. That would be 5 (NH₃) + 1 (Cl) = 6. But then the answer would be 6 for Co.
- However, the options include (3,5) and (5,6). Let’s check the second complex to decide.
-
Analyze PtCl₄·5NH₃
- Pt is in the +4 oxidation state.
- NH₃ is neutral.
- Let y be the number of Cl⁻ inside the coordination sphere. Then the complex ion is [Pt(NH3)5Cly](4−y)+.
- The remaining (4−y) Cl⁻ are outside.
- Coordination number of Pt⁴⁺ is typically 6. With 5 NH₃, only 1 chloride can fit inside to make coordination number 6. So y=1.
- Then non-ionizable valences = 5 (NH₃) + 1 (Cl) = 6 again. That would give (6,6) — option (D).
- But wait — the formula is PtCl₄·5NH₃. If only one Cl is inside, then there are 3 outside Cl⁻. The complex ion would be [Pt(NH3)5Cl]3+, and three Cl⁻ outside. That is plausible. So both would be 6? That seems too trivial.
-
Re-examine the meaning of “non-ionizable valences”
- In older coordination theory (Werner), “primary valences” are ionizable (oxidation state), and “secondary valences” are non-ionizable (coordination number). The question likely means: how many of the chloride ions are directly bonded to the metal (i.e., non-ionizable)? Because NH₃ is always non-ionizable.
- For CoCl₃·5NH₃: total Cl = 3. If coordination number is 6, and 5 NH₃ occupy 5 spots, then only 1 Cl can be bonded. So non-ionizable Cl = 1. But that’s not among the options.
- Perhaps the coordination number is not 6? For Co³⁺, it is almost always 6. For Pt⁴⁺, it is 6. So something is off.
-
Consider the possibility of different coordination numbers
- Actually, Pt⁴⁺ can have coordination number 6, but the formula PtCl₄·5NH₃ suggests 4 chlorides and 5 ammonias. If all 4 chlorides were inside, coordination number would be 9 — impossible. So only some are inside.
- Let’s solve systematically: Let the complex be [Co(NH3)aClb]c+ with a+b=6 (coordination number). The total number of Cl in the formula is 3, so outside Cl⁻ = 3−b. The charge balance: c=3−(3−b)=b. But also c=3−b (from metal charge minus ligand charges). So b=3−b → 2b=3 → b=1.5 — impossible. So coordination number cannot be 6.
- This means the complex is not simply hexacoordinate. In fact, CoCl₃·5NH₃ is known as the classic Werner complex: it is [Co(NH3)5Cl]Cl₂. Here, coordination number is 6 (5 NH₃ + 1 Cl), and there are 2 ionizable Cl⁻ outside. So non-ionizable Cl = 1. But again, that’s not an option.
- Wait — the question says “non-ionizable valences of Co³⁺” — this might mean the number of coordination sites that are non-ionizable, i.e., the total number of ligands (both NH₃ and Cl) that are not free ions. That would be 6 for Co.
- For PtCl₄·5NH₃, the known compound is [Pt(NH3)5Cl]Cl₃? That would give 5 NH₃ + 1 Cl inside = 6 non-ionizable valences. But then both are 6 → option (D).
- However, there is also the possibility that PtCl₄·5NH₃ is actually [Pt(NH3)4Cl2]Cl₂·NH₃? That would be messy.
-
Look up the classic result (since this is a standard question):
- For CoCl₃·5NH₃, the complex is [Co(NH3)5Cl]Cl₂ → non-ionizable valences = 5 (from NH₃) + 1 (from Cl) = 6? No, the “valences” of Co³⁺ are 6 total, but the question likely means the number of chloride ions that are non-ionizable. That would be 1. Not matching. …
-
- TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQQ.In which of the following, ions are correctly arranged in the increasing order of oxidizing power? (A) Cr2O72−<MnO4−<VO2+ (B) VO2+<Cr2O72−<MnO4− (C) VO2+<MnO4−<Cr2O72− (D) MnO4−<Cr2O72−<VO2+
›Reveal solutionSolution
The key idea is that oxidizing power increases with the standard reduction potential of the half-reaction. Using standard potentials, the correct increasing order is VO2+<Cr2O72−<MnO4−, which corresponds to option (B).
Concept & Intuition
Oxidizing power measures how readily a species gains electrons (i.e., gets reduced). The stronger the oxidizer, the more positive its standard reduction potential (E∘). So to rank VO2+, Cr2O72−, and MnO4− in increasing order of oxidizing power, we need to list them from the least positive E∘ to the most positive E∘. These are all common oxoanions in acidic solution, and their reduction potentials are well-known.
-
Recall the standard reduction potentials (in acidic medium)
- For MnO4−+8H++5e−→Mn2++4H2O, E∘=+1.51 V
- For Cr2O72−+14H++6e−→2Cr3++7H2O, E∘=+1.33 V
- For VO2++2H++e−→VO2++H2O, E∘=+1.00 V
These values are standard textbook data. The higher the E∘, the stronger the oxidizer.
-
Arrange them by increasing E∘
- VO2+: +1.00 V (lowest)
- Cr2O72−: +1.33 V (middle)
- MnO4−: +1.51 V (highest)
So increasing order of oxidizing power: VO2+<Cr2O72−<MnO4−.
-
Match with the options
- (A) Cr2O72−<MnO4−<VO2+ — wrong order …
-
- TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQQ.The molecular formula of a coordinate complex is CoH12O6Cl3. When one mole of this aqueous solution of complex is reacted with excess of aqueous AgNO3 solution, three moles of AgCl was formed. What is the correct formula of the complex? (A) [Co(H2O)6]Cl3 (B) [Co(H2O)5Cl]Cl2H2O (C) [Co(H2O)4Cl2]Cl(H2O)2 (D) [Co(H2O)3Cl3](H2O)3
›Reveal solutionSolution
The key is that only chloride ions outside the coordination sphere (free ions) precipitate with AgNO₃. Since 3 moles of AgCl form, all three chlorides are free, so the complex must have no coordinated Cl⁻. The correct formula is (A).
The concept here is ionization isomerism in coordination chemistry. When a coordination compound dissolves in water, the ligands bound directly to the metal (inside the coordination sphere) do not dissociate, while counter-ions outside the sphere do. Adding AgNO₃ tests for free chloride ions — each free Cl⁻ gives one mole of AgCl precipitate. So the number of moles of AgCl tells us exactly how many chlorides are outside the coordination sphere.
-
Interpret the molecular formula
The formula CoH12O6Cl3 has 6 oxygen atoms and 12 hydrogens — that strongly suggests six water molecules (H2O) are present, because 6×H2O=H12O6. So the compound is Co(H2O)6Cl3 in terms of composition, but we need to decide how these are arranged.
-
Use the AgNO₃ test result
One mole of the complex gives three moles of AgCl precipitate. That means all three chloride ions are free in solution — none are coordinated to cobalt. If any chloride were a ligand, it would not precipitate.
-
Eliminate options
- (A) [Co(H2O)6]Cl3: All three Cl⁻ are outside the sphere → 3 free Cl⁻ → 3 AgCl. ✓
- (B) [Co(H2O)5Cl]Cl2⋅H2O: One Cl⁻ is coordinated, only two are free → would give 2 AgCl. ✗
- (C) [Co(H2O)4Cl2]Cl⋅(H2O)2: Two Cl⁻ coordinated, one free → 1 AgCl. ✗ …
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- TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQQ.The number of amphoteric, basic and acidic oxides among the following respectively are CrO3, MgO, K2O, B2O3, Al2O3, In2O3, PbO, As2O3 (A) 3, 2, 3 (B) 3, 3, 2 (C) 3, 4, 1 (D) 2, 3, 3
›Reveal solutionSolution
The key is to classify each oxide by its acid-base character using periodic trends: metallic oxides are basic, non‑metallic oxides are acidic, and oxides of elements near the metal–non‑metal boundary (especially amphoteric ones like Al, In, Pb, As) are amphoteric. Counting gives 3 amphoteric, 3 basic, 2 acidic → option (B).
Concept & Intuition
Oxides reflect the element’s position in the periodic table.
- Basic oxides come from metals (Groups 1, 2, and many transition/post‑transition metals). They react with acids to form salts and water.
- Acidic oxides come from non‑metals. They react with bases to form salts and water.
- Amphoteric oxides behave as both acids and bases; they are typically oxides of elements near the diagonal line from Be to At (e.g., Al, Zn, Sn, Pb, As, Sb).
We classify each given oxide by the element’s metallic/non‑metallic character and its position relative to that diagonal.
Step‑by‑Step Classification
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CrO₃ (chromium trioxide)
Chromium is a transition metal, but in its +6 oxidation state it behaves as a non‑metal oxide. CrO₃ is strongly acidic (it forms chromic acid, H₂CrO₄).
→ Acidic
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MgO (magnesium oxide)
Magnesium is an alkaline earth metal (Group 2). MgO is a classic basic oxide.
→ Basic
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K₂O (potassium oxide)
Potassium is an alkali metal (Group 1). K₂O reacts vigorously with water to give KOH, a strong base.
→ Basic
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B₂O₃ (boron trioxide)
Boron is a metalloid, but its oxide is purely acidic (forms boric acid, H₃BO₃).
→ Acidic
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Al₂O₃ (aluminium oxide)
Aluminium lies on the amphoteric diagonal. Al₂O₃ reacts with both acids and bases (e.g., with HCl gives AlCl₃; with NaOH gives NaAlO₂).
→ Amphoteric
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In₂O₃ (indium oxide)
Indium is below aluminium in Group 13. Its oxide is predominantly basic (In is more metallic than Al), though it shows very weak amphoteric character. In standard classification, In₂O₃ is considered basic.
→ Basic
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PbO (lead(II) oxide) …
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