Q.Write the IUPAC names of the following coordination compounds:
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Werner Coordination Theory
Werner Coordination Theory: The Idea That Changed Inorganic Chemistry
Imagine you're looking at a salt like cobalt(III) chloride. The formula is written as CoClX3, and when you dissolve it in water, you expect to find CoX3+ and ClX− ions. But something strange happens: when you add silver nitrate (which precipitates chloride ions), only some of the chlorine comes out as silver chloride. Not all of it. And the amount that precipitates depends on how you made the compound.
This was the puzzle that faced chemists in the late 1800s. Compounds like CoClX3⋅6NHX3 (orange-yellow) and CoClX3⋅5NHX3 (purple) had the same metal and the same ligands (ammonia), but different colours, different conductivities in solution, and different numbers of chloride ions that could be precipitated. The old ideas of fixed valency couldn't explain it.
Alfred Werner proposed a radical solution in 1893. He said: a metal ion has two kinds of valency.
The Core Intuition
Think of a metal ion like a king in a castle. The king has two types of relationships:
-
Primary valency (today: oxidation state) — this is the king's royal authority. It's fixed, non-directional, and satisfied by negative ions. For cobalt(III), this is +3. It's like the king's crown: it doesn't change.
-
Secondary valency (today: coordination number) — this is the king's personal bodyguard. The king can have a fixed number of guards (usually 4 or 6) who stand in specific positions around him. These guards can be neutral molecules (like ammonia) or negative ions (like chloride). The key: these guards are directly attached to the metal, forming a stable cluster called the coordination sphere.
The revolutionary idea: the chloride ions that act as bodyguards (inside the coordination sphere) do not behave like free ions. They don't precipitate with silver nitrate. They don't conduct electricity. They are "locked" to the metal.
The Precise Statement
Werner Coordination Theory (1893)
-
Every metal atom has two types of valency:
- Primary valency (ionisable): corresponds to the oxidation state. It is satisfied by negative ions. These ions are outside the coordination sphere and behave as free ions in solution.
- Secondary valency (non-ionisable): corresponds to the coordination number. It is satisfied by neutral molecules or negative ions directly bonded to the metal. These are inside the coordination sphere and do not dissociate.
-
The secondary valencies are directional — they point to fixed positions in space around the metal, giving the complex a definite geometry (e.g., octahedral for coordination number 6, square planar for 4).
-
The primary valency is non-directional — it is just a number, not a spatial arrangement.
How It Explains the Puzzle
Take the compound CoClX3⋅6NHX3 (orange-yellow). Werner said:
- Cobalt has primary valency +3 (needs three negative charges to satisfy it).
- Cobalt has secondary valency 6 (can hold six ligands around it).
- The six ammonia molecules satisfy all six secondary valencies. So the chloride ions cannot be inside the coordination sphere — they must be outside, as free ions.
- Structure: [Co(NHX3)X6]ClX3. All three chlorides precipitate with AgNOX3.
Now take CoClX3⋅5NHX3 (purple):
- Again, primary valency +3, secondary valency 6.
- Five ammonia molecules satisfy five secondary valencies. One chloride ion must fill the sixth spot — it becomes a ligand inside the sphere.
- The other two chlorides are outside as free ions.
- Structure: [Co(NHX3)X5Cl]ClX2. Only two chlorides precipitate.
The number of free ions in solution determines the conductivity and the number of precipitable chlorides. Werner's theory predicted exactly these numbers — and experiments confirmed them.
The Geometry Insight …
Why this formula?
Werner Coordination Theory: Why the Key Formulas Hold
Werner Coordination Theory (1893) revolutionized inorganic chemistry by explaining how metal ions bind ligands. Let's build the reasoning from first principles — not just memorize formulas.
1. The Core Observation: Primary vs. Secondary Valence
Werner noticed that metal compounds had two types of bonding capacity:
- Primary valence (now oxidation state): Satisfies the metal's charge — ionic in nature.
- Secondary valence (now coordination number): Determines how many ligands attach — directional, spatial in nature.
Why this distinction?
Consider CoClX3 ⋅6NHX3 (one of Werner's classic compounds).
- The compound is electrically neutral overall.
- Adding AgNOX3 precipitates all 3 Cl⁻ as AgCl — meaning all chlorides are free ions.
- Therefore, the NHX3 molecules must be directly bonded to Co, not the chlorides.
This forces the idea: Co has a fixed capacity for direct ligand attachment (secondary valence = 6 here), separate from its charge balance (primary valence = +3).
2. The Key Formula: Coordination Number = Number of Ligands Attached
Formula:
Coordination number=number of donor atoms directly bonded to the metal
Why this holds:
- Werner's experiments showed that only a fixed number of ligands could be replaced without breaking the compound's identity.
- For CoClX3 ⋅6NHX3, adding acid doesn't remove NHX3 easily — they are coordinated.
- The maximum number of such tightly bound ligands is the coordination number — a property of the metal ion, not the counterions.
Derivation from data:
If you have [Co(NHX3)X6]ClX3, conductivity measurements show 4 ions in solution ([Co(NHX3)X6]X3+ + 3 Cl⁻).
If you had [Co(NHX3)X5Cl]ClX2, conductivity shows 3 ions.
The number of chlorides inside the coordination sphere (non-precipitable) plus those outside must sum to the total chlorides. This gives the coordination number directly.
3. The Geometry Formula: Coordination Number Determines Shape
Werner proposed that secondary valences are directed in space — leading to specific geometries.
| Coordination Number | Geometry | Why? |
|---|---|---|
| 2 | Linear | Minimizes repulsion between 2 ligands |
| 4 | Tetrahedral or Square planar | 4 points in space — two arrangements possible |
| 6 | Octahedral | 6 ligands at 90° angles — most symmetric |
Why octahedral for 6?
- 6 ligands around a central atom must be placed to maximize separation.
- The octahedron (6 vertices, all equidistant from center, 90° between adjacent bonds) is the only regular polyhedron with 6 vertices.
- This explains why [Co(NHX3)X6]X3+ is octahedral — no other arrangement gives equal bond angles and distances.
4. The Isomer Counting Formula: Why 2n or n! Appears
Werner used isomer counts to confirm geometry. For an octahedral complex [MaX2bX2cX2]:
Number of geometrical isomers = 5 (not 6, not 4)
Why this formula?
- Place the two 'a' ligands: they can be cis (90°) or trans (180°).
- For each, place 'b' and 'c' in remaining positions — but symmetry reduces duplicates. …
Concept: Werner Coordination Theory — the central metal ion has a fixed primary valence (oxidation state) and a variable secondary valence (coordination number), which determines the complex’s geometry and naming.
Reasoning steps:
- Identify the complex ion (inside square brackets) and the counter ions outside.
- For the complex ion: name ligands in alphabetical order (ignoring prefixes like di-, tri-), then the metal with its oxidation state in Roman numerals in parentheses.
- For anionic complexes, add the suffix -ate to the metal name. Cationic or neutral complexes use the metal name as is.
IUPAC names:
- Hexaamminecobalt(III) chloride
- Pentaamminechloridocobalt(III) chloride
- Potassium hexacyanidoferrate(III)
- Potassium trioxalatoferrate(III)
- Potassium tetrachloridopalladate(II) …
Werner’s coordination theory tells us to name the complex ion first (ligands alphabetically, then metal with its oxidation state in Roman numerals), followed by the counter ion. The IUPAC names are: (i) Hexaamminecobalt(III) chloride,
(ii) Pentaamminechloridocobalt(III) chloride,
(iii) Potassium hexacyanidoferrate(III),
(iv) Potassium trioxalatoferrate(III),
(v) Potassium tetrachloridopalladate(II),
(vi) Diamminechlorido(methanamine)platinum(II) chloride.
Werner’s coordination theory is the foundation here. He showed that metal ions have two types of valency: primary (ionic, satisfied by counter ions) and secondary (coordinate, satisfied by ligands). In naming, we treat the coordination sphere (the metal + its attached ligands) as a single entity. The name of the complex ion is built by listing ligands in alphabetical order (ignoring prefixes like di-, tri-), then the metal, then its oxidation state in Roman numerals in parentheses. If the complex ion is an anion, we add “-ate” to the metal’s name. Counter ions are named last.
Let’s go through each compound step by step.
-
[Co(NH3)6]Cl3
The coordination sphere is [Co(NH3)6]3+ — a cation. Ligands: six ammonia molecules, which are neutral. The ligand name is “ammine” (note the double m). So “hexaammine” (hexa- for six). Metal: cobalt. Oxidation state: each NH3 is neutral, so Co must be +3 to balance the 3 Cl− ions outside. Hence cobalt(III). The counter ion is chloride (three of them).
Name: Hexaamminecobalt(III) chloride.
-
[Co(NH3)5Cl]Cl2
Complex ion: [Co(NH3)5Cl]2+. Ligands: five ammines and one chloride. Alphabetical order: “ammine” before “chlorido” (IUPAC uses “chlorido” for Cl− as a ligand, not “chloro”). So “pentaamminechlorido”. Metal: cobalt. Oxidation state: five neutral NH3 + one Cl− ligand (charge -1) gives total ligand charge = -1. Complex charge is +2, so Co must be +3 (since +3 – 1 = +2). Hence cobalt(III). Counter ions: two chlorides.
Name: Pentaamminechloridocobalt(III) chloride.
-
K3[Fe(CN)6]
Complex ion: [Fe(CN)6]3− — an anion. Ligands: six cyanide ions, named “cyanido” (IUPAC prefers “cyanido” over “cyano”). So “hexacyanido”. Metal: iron. Since it’s an anionic complex, we use “ferrate” (Latin root). Oxidation state: each CN− is -1, total ligand charge = -6. Complex charge is -3, so Fe must be +3 (since +3 – 6 = -3). Hence ferrate(III). Counter ion: potassium (three).
Name: Potassium hexacyanidoferrate(III).
-
K3[Fe(C2O4)3]
Complex ion: [Fe(C2O4)3]3−. Ligands: three oxalate ions (C2O42−), which are bidentate. The ligand name is “oxalato”. So “trioxalato”. Metal: iron → ferrate. Oxidation state: each oxalate is -2, total ligand charge = -6. Complex charge = -3, so Fe must be +3 (since +3 – 6 = -3). Hence ferrate(III). Counter ion: potassium.
Name: Potassium trioxalatoferrate(III).
-
K2[PdCl4]
Complex ion: [PdCl4]2−. Ligands: four chloride ions → “tetrachlorido”. Metal: palladium → “palladate” (anionic form). Oxidation state: each Cl− is -1, total = -4. Complex charge = -2, so Pd must be +2 (since +2 – 4 = -2). Hence palladate(II). Counter ion: potassium (two). …
Method: IUPAC Nomenclature of Coordination Compounds
This follows the Additive Nomenclature method under IUPAC rules (Red Book). The core principle: name the cation first, then the anion — and within the coordination sphere, name ligands alphabetically before the central metal (with its oxidation state in Roman numerals in parentheses).
Steps
- Identify the cation and anion of the compound.
- For the coordination sphere (inside square brackets):
- Name ligands in alphabetical order (ignoring prefixes like di-, tri-).
- Anionic ligands end in -o, in the modern -ido style (e.g., chlorido, cyanido, oxalato).
- Neutral ligands keep their name (e.g., ammine, water).
- Write the central metal name last.
- Add the oxidation state of the metal in Roman numerals in parentheses.
- If the complex is an anion, the metal name ends in -ate.
- Outside the sphere: name counter-ions normally.
(i) [Co(NH3)6]Cl3
- Cation: [Co(NH3)6]3+ — hexaamminecobalt(III)
- Anion: three chloride ions → chloride
- Full name: Hexaamminecobalt(III) chloride
(ii) [Co(NH3)5Cl]Cl2
- Cation: [Co(NH3)5Cl]2+ — ligands: ammine (×5) + chlorido Alphabetical order: ammine before chlorido → Pentaamminechloridocobalt(III) (chlorido is IUPAC for Cl⁻ ligand)
- Anion: two chloride ions → chloride
- Full name: Pentaamminechloridocobalt(III) chloride
(iii) K3[Fe(CN)6]
- Cation: three potassium ions → potassium
- Anion: [Fe(CN)6]3− — six cyanido ligands → hexacyanido Metal is iron, but as anion → ferrate Oxidation state: Fe is +3 → (III)
- Full name: Potassium hexacyanidoferrate(III)
(iv) K3[Fe(C2O4)3]
- Cation: potassium
- Anion: [Fe(C2O4)3]3− — oxalate ligand (C2O42−) is oxalato Three → trioxalato Metal: iron → ferrate, Fe is +3 → (III)
- Full name: Potassium trioxalatoferrate(III)
(v) K2[PdCl4]
- Cation: potassium
- Anion: [PdCl4]2− — four chlorido → tetrachlorido Metal: palladium → palladate, Pd is +2 → (II) …
Here are the common mistakes students make when naming coordination compounds under Werner’s theory, using your six examples as case studies. Each mistake is followed by a clear how-to-avoid tip.
1. Wrong order: Naming the anion before the cation
Mistake: Writing the name of the complex ion first, then the counter ion — especially when the complex is anionic (e.g., K3[Fe(CN)6]).
Example of error:
“Potassium hexacyanidoferrate(III)” is correct, but some write “hexacyanidoferrate(III) potassium” or “tripotassium hexacyanidoferrate(III)”.
How to avoid:
- Always name the cation first, then the anion — just like in simple salts (NaCl = sodium chloride).
- For K3[Fe(CN)6]: cation = K+ → “potassium”, anion = [Fe(CN)6]3− → “hexacyanidoferrate(III)”.
- For [Co(NH3)6]Cl3: cation = [Co(NH3)6]3+ → name first, then “chloride”.
2. Forgetting to change the suffix for anionic complexes
Mistake: Using the metal’s English name (e.g., “iron”, “cobalt”) even when the complex is an anion.
Example of error:
K3[Fe(CN)6] named as “potassium hexacyanoiron(III)” instead of “potassium hexacyanidoferrate(III)”.
How to avoid:
- If the complex ion is negative, replace the metal name’s ending with -ate.
- Iron → ferrate
- Cobalt → cobaltate
- Copper → cuprate
- Silver → argentate
- Tin → stannate
- Lead → plumbate
- Exception: Some metals use Latin roots (e.g., Fe → ferrate, Cu → cuprate). Memorise the common ones.
3. Omitting the oxidation state of the metal
Mistake: Leaving out the Roman numeral for the metal’s oxidation state, or writing it in the wrong place.
Example of error:
[Co(NH3)5Cl]Cl2 named as “pentaamminechloridocobalt chloride” — missing the (III).
How to avoid:
- Always calculate the oxidation state of the central metal before naming.
- For [Co(NH3)5Cl]Cl2:
- NH3 = neutral, Cl− inside complex = −1, two outer Cl− = −2 total.
- Let Co = x: x+0+(−1)+(−2)=0⇒x=+3.
- Write it as cobalt(III).
- For [Co(NH3)5Cl]Cl2:
- Place the Roman numeral in parentheses right after the metal name (or metal-ate name), with no space before the parenthesis.
4. Incorrect ligand order (alphabetical vs. charge-based)
Mistake: Listing ligands by charge (anionic first, then neutral) instead of alphabetical order (ignoring prefixes like di-, tri-).
Example of error:
[Pt(NH3)2Cl(NH2CH3)]Cl named with the ligands out of alphabetical order (e.g. “chlorido” placed before “ammine”) — the correct order is ammine, then chlorido, then methanamine.
How to avoid:
- Alphabetise ligands by their name (not by symbol or charge).
- “ammine” (A), “chlorido” (C), “methanamine” (M).
- Correct order: diamminechlorido(methanamine) — ammine (A) comes before chlorido (C), which comes before methanamine (M).
- Prefixes like di-, tri-, tetra- are ignored when alphabetising. Only the ligand root counts.
- Exception: If two ligands start with the same letter, then use the full name.
5. Wrong use of numerical prefixes (di, tri, tetra, etc.)
Mistake: Using “di”, “tri” for complex ligands like ethylenediamine (en) or oxalate (C2O42−) — or forgetting them entirely.
Example of error:
K3[Fe(C2O4)3] named as “potassium trioxalatoferrate(III)” — correct. But some write “potassium oxalatoferrate(III)” (missing “tri”) or “potassium tris(oxalato)ferrate(III)” (using “tris” incorrectly for a simple ligand).
How to avoid:
- For simple ligands (e.g., Cl−, NH3, CN−): use di, tri, tetra, penta, hexa.
- For ligands whose own name already contains a multiplying prefix (e.g., ethylenediamine, which contains “di”): use bis, tris, tetrakis, pentakis and put the ligand name in parentheses, to avoid ambiguity — e.g. [Co(en)3]3+ → tris(ethylenediamine)cobalt(III).
- Oxalato does not need bis/tris — NCERT’s own convention simply uses tri (e.g. potassium trioxalatoaluminate(III), potassium trioxalatochromate(III)). So [Fe(C2O4)3]3− → trioxalatoferrate(III), matching the correct answer above. (The 2004 IUPAC recommendations also treat tris(oxalato)ferrate(III) as an equally unambiguous style, but NCERT — and the answer to this question — uses trioxalato; in the board exam, write the NCERT form.)
- Check the ligand: only use bis/tris/tetrakis when the ligand’s own name already contains di/tri/tetra, or when a simple prefix would be genuinely ambiguous.
6. Forgetting to enclose complex ligands in parentheses
Mistake: Writing “methanamine” without parentheses when it’s a complex ligand, or using parentheses for simple ligands.
Example of error:
[Pt(NH3)2Cl(NH2CH3)]Cl named as “diamminechloridomethanamineplatinum(II) chloride” — no parentheses around methanamine.
How to avoid:
- Always put parentheses around the name of a ligand that is itself a molecule with a systematic name (e.g., methanamine, ethylenediamine, oxalate).
- Simple ligands like NH3 (ammine), Cl− (chlorido), CN− (cyanido) do not need parentheses. …
Showing the 12 most recent of 30 on this concept.
- TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQQ.Which of the following will give maximum number of isomers? (A) [Co(NH3)4Cl2]+ (B) [Ni(en)(NH3)4]2+ (C) [Ni(C2O4)(en)2] (D) [Cr(SCN)2(NH3)4]+
›Reveal solutionSolution
The key is to count all possible stereoisomers (geometrical and optical) plus linkage isomers for each complex. The complex with the most isomers is [Cr(SCN)2(NH3)4]+, which gives 6 isomers (three S/N binding combinations, each with cis and trans forms), so the answer is (D).
Concept & Intuition
Isomers in coordination chemistry arise from different spatial arrangements of ligands (geometrical isomers) and, for chiral complexes, non-superimposable mirror images (optical isomers). Additionally, ambidentate ligands like SCN⁻ can bind through different atoms (S or N), creating linkage isomers. To find which complex gives the maximum number, we systematically count all distinct isomers for each option.
-
Option (A): [Co(NH3)4Cl2]+
- This is an octahedral complex with two identical monodentate ligands (Cl) and four identical NH₃ ligands.
- Geometrical isomers: Only cis and trans arrangements of the two Cl ligands.
- Optical isomers: Neither cis nor trans is chiral (cis has a plane of symmetry, trans has a center of symmetry).
- Total isomers = 2.
-
Option (B): [Ni(en)(NH3)4]2+
- en (ethylenediamine) is a bidentate ligand. The complex is octahedral with one en and four NH₃.
- The en ligand must occupy two adjacent positions (cis), so there is no trans isomer.
- The complex is not chiral (it has a plane of symmetry through the en ring and opposite NH₃).
- Total isomers = 1.
-
Option (C): [Ni(C2O4)(en)2]
- Oxalate (C₂O₄²⁻) is a bidentate ligand, and en is also bidentate. The complex is octahedral with three bidentate ligands.
- Geometrical isomers: For three bidentate ligands, only the cis arrangement is possible (all three bidentates must occupy adjacent positions).
- Optical isomers: The cis arrangement is chiral (no plane of symmetry), so it exists as a pair of enantiomers.
- Total isomers = 2 (a pair of optical isomers).
-
Option (D): [Cr(SCN)2(NH3)4]+
- SCN⁻ is ambidentate: it can bind through sulfur (thiocyanato) or nitrogen (isothiocyanato). …
-
- TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQQ.S2O32−(aq)+OH−(aq)→SO42−(aq)+H2O(l)+e− After the above half reaction is balanced, which of the following are the coefficients of OH− and SO42− respectively? (A) 8, 3 (B) 6, 2 (C) 10, 2 (D) 5, 2
›Reveal solutionSolution
Balancing the half‑reaction in basic solution gives coefficients 10 for OH⁻ and 2 for SO₄²⁻, so the correct choice is (C).
We are balancing a half‑reaction in basic solution. The key idea: first balance atoms other than H and O, then balance O by adding H₂O, balance H by adding H⁺ (as if in acid), and finally neutralize H⁺ by adding the same number of OH⁻ to both sides. This method works because every H⁺ added in the acidic step is converted to H₂O when we add OH⁻.
- Identify the atoms to balance The skeleton is:
S2O32−+OH−→SO42−+H2O+e−
Sulfur (S) is unbalanced: left has 2 S, right has 1 S. So put a coefficient 2 in front of SO₄²⁻:
S2O32−+OH−→2SO42−+H2O+e−
-
Balance oxygen atoms
Left: 3 O (from S₂O₃²⁻) + 1 O (from OH⁻) = 4 O total (coefficient of OH⁻ unknown yet).
Right: 2 × 4 = 8 O (from 2 SO₄²⁻) + 1 O (from H₂O) = 9 O.
To balance O, we need more O on the left. Add H₂O? No — we add H₂O to the side that needs oxygen. Actually, we balance O by adding H₂O to the side that is short of oxygen.
Left has 4 O, right has 9 O → left needs 5 more O. So add 5 H₂O to the left? That would add 5 O but also 10 H — messy. Better to use the standard acidic-then-basic method.
-
Use the acidic method first
Ignore OH⁻ for a moment. Write the half‑reaction as if in acid:
S2O32−→2SO42−
Balance O: left 3 O, right 8 O → add 5 H₂O to left:
S2O32−+5H2O→2SO42−
Balance H: left now has 10 H, right has 0 → add 10 H⁺ to right:
S2O32−+5H2O→2SO42−+10H+
Balance charge: left –2, right 2×(−2) + 10 = +6. Difference = 8 electrons needed on right:
S2O32−+5H2O→2SO42−+10H++8e−… - TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQQ.Which of the following aqueous solution has highest freezing point? (A) 0.1m Al2(SO4)3 (B) 0.1m BaCl2 (C) 0.1m NH4Cl (D) 0.1m AlCl3
›Reveal solutionSolution
The freezing point is highest for the solution with the smallest total particle concentration after dissociation. Since all are 0.1 m, the one that dissociates into the fewest ions gives the highest freezing point. That is 0.1 m NH₄Cl, which yields 0.2 m particles.
Concept & Intuition
Freezing point depression is a colligative property — it depends only on the number of solute particles in solution, not on their identity. The formula is
ΔTf=i⋅Kf⋅m
where i is the van’t Hoff factor (number of particles per formula unit after dissociation), Kf is the cryoscopic constant (same for water here), and m is the molality.
Since all solutions have the same molality (0.1 m), the one with the smallest i will have the smallest ΔTf, and therefore the highest freezing point. So we just need to count ions produced per formula unit in water.
Step‑by‑step reasoning
-
Determine the van’t Hoff factor for each salt (assuming complete dissociation in dilute aqueous solution).
- (A) Al2(SO4)3→2Al3++3SO42− → total ions = 2+3=5 → i=5
- (B) BaCl2→Ba2++2Cl− → total ions = 1+2=3 → i=3
- (C) NH4Cl→NH4++Cl− → total ions = 1+1=2 → i=2
- (D) AlCl3→Al3++3Cl− → total ions = 1+3=4 → i=4
-
Compare the effective particle concentrations
Since m=0.1 for all, the total particle molality is i×0.1:
- (A) 5×0.1=0.5m
- (B) 3×0.1=0.3m
- (C) 2×0.1=0.2m …
-
- TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQQ.Identify the complex ion which does not exist (A) [SiF6]2− (B) [GeCl6]2− (C) [Sn(OH)6]2− (D) [SiCl6]2−
›Reveal solutionSolution
The key idea is that the central atom must have available d-orbitals to accommodate six ligands in an octahedral geometry. Silicon lacks accessible d-orbitals for chlorine ligands, so [SiCl6]2− does not exist. The answer is (D).
The question tests your understanding of coordination chemistry and the availability of d-orbitals in elements of the carbon family (Group 14). For a complex ion with six ligands (octahedral geometry), the central atom must use sp3d2 hybridization. This requires empty d-orbitals of suitable energy.
Silicon, germanium, and tin all belong to Group 14. As you go down the group, the size increases and the d-orbitals become more accessible. But here’s the catch: the nature of the ligand also matters. Small, highly electronegative ligands like fluorine can stabilize the high oxidation state and pull electron density, making d-orbital participation feasible even for silicon. Larger, less electronegative ligands like chlorine cannot do this effectively for silicon.
Let’s examine each option step by step.
-
Option (A): [SiF6]2−
Silicon is in the +4 oxidation state here. Fluorine is tiny and extremely electronegative. It strongly withdraws electron density, allowing silicon to use its 3d orbitals for sp3d2 hybridization. This complex is well-known and stable — it exists as salts like Na2[SiF6].
-
Option (B): [GeCl6]2−
Germanium is larger than silicon and has accessible 4d orbitals. Chlorine, though larger than fluorine, can still coordinate because germanium’s d-orbitals are energetically available. This complex exists, for example as (NH4)2[GeCl6].
-
Option (C): [Sn(OH)6]2−
Tin is even larger, with accessible 5d orbitals. The hydroxide ligand is a reasonable ligand for tin(IV). This complex is known — for instance, K2[Sn(OH)6] is a stable compound.
-
Option (D): [SiCl6]2− …
-
- TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQQ.Consider the following HF, H2O, BeCl2, CO2, BF3, NF3, CCl4, CHCl3 The number of polar molecules is (A) 2 (B) 3 (C) 4 (D) 5
›Reveal solutionSolution
The key idea is that a molecule is polar if it has polar bonds and an asymmetric shape (net dipole moment). After checking each molecule, exactly 4 are polar, so the answer is (C).
Concept & Intuition
Polarity depends on two things: (1) the presence of bonds between atoms with different electronegativities (polar bonds), and (2) the molecular geometry not cancelling those bond dipoles. Symmetric shapes like linear, trigonal planar, or tetrahedral can be nonpolar if all bonds are identical and arranged symmetrically. Asymmetric shapes or different substituents leave a net dipole.
Step-by-step reasoning
-
HF – Hydrogen and fluorine have a large electronegativity difference. The molecule is diatomic and linear, so the bond dipole is not cancelled. Polar.
-
H₂O – Bent shape (V-shaped) due to two lone pairs on oxygen. The two O–H bond dipoles do not cancel; they add to a net dipole pointing toward oxygen. Polar.
-
BeCl₂ – Linear molecule (Be is central, no lone pairs). The two Be–Cl bond dipoles are equal and opposite, cancelling exactly. Nonpolar.
-
CO₂ – Linear molecule (O=C=O). The two C=O bond dipoles are equal and opposite, cancelling. Nonpolar.
-
BF₃ – Trigonal planar (B is central, no lone pairs). The three B–F bond dipoles are symmetrically arranged at 120°, so their vector sum is zero. Nonpolar.
-
NF₃ – Trigonal pyramidal (N has one lone pair). The three N–F bond dipoles do not cancel because the lone pair pushes them downward, creating a net dipole. Polar. …
-
- TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQQ.For the alkyne with formula C6H10, the number of alkynes with acidic hydrogens is x and number of alkynes with no acidic hydrogens is y. x and y are respectively (A) 2,5 (B) 3,4 (C) 4,3 (D) 5,2
›Reveal solutionSolution
C6H10 has 4 alkynes with acidic (terminal) hydrogens and 3 without, so x,y=4,3 — option (C).
An alkyne has an acidic hydrogen only when the triple bond is terminal (≡C–H). Enumerating the acyclic hexyne isomers:
Terminal alkynes (acidic H), x:
- Hex-1-yne, HC≡C-CH2CH2CH2CH3
- 3-Methylpent-1-yne, HC≡C-CH(CH3)CH2CH3
- 4-Methylpent-1-yne, HC≡C-CH2CH(CH3)2
- 3,3-Dimethylbut-1-yne, HC≡C-C(CH3)3
⇒x=4.
Internal alkynes (no acidic H), y: …
- TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQQ.IUPAC name of [Co(NH3)4(H2O)Cl]Cl2 is (A) Tetraammineaquachlorocobalt (III) dichloride (B) Tetraamminechloroaquacobalt (III) chloride (C) Tetraammineaquachlorocobalt (III) chloride (D) Aquachlorotetraamminecobalt (III) chloride
›Reveal solutionSolution
The complex is a cationic coordination entity with cobalt in the +3 oxidation state; the correct IUPAC name lists ligands alphabetically (ignoring prefixes) and ends with the counterion. The answer is Tetraammineaquachlorocobalt(III) chloride.
The first thing to recognise is that this is a coordination compound with a complex cation and two chloride counterions. The square brackets enclose the coordination sphere, and the Cl2 outside the brackets tells us there are two chloride ions balancing the charge. So the name must end with "chloride" — not "dichloride" — because the counterion is simply chloride, and the number is implied by the formula or stated separately if needed.
Now, the central metal is cobalt. To name it correctly, we need its oxidation state. The complex inside the brackets is [Co(NH3)4(H2O)Cl]2+ (since the two external chlorides contribute −2 charge, the cation must be +2). Ammonia (NH3) and water (H2O) are neutral ligands; chloride inside the coordination sphere is anionic (Cl−). Let the oxidation state of Co be x. Then:
x+4(0)+0+(−1)=+2⇒x−1=+2⇒x=+3.
So it's cobalt(III).
The ligands are: four ammines (NH3), one aqua (H2O), and one chloro (Cl−). In IUPAC nomenclature, ligands are named in alphabetical order regardless of charge or number. The prefixes (tetra-, etc.) are ignored when alphabetising. So we compare the ligand names: "ammine", "aqua", "chloro". Alphabetically: ammine (a), aqua (a), chloro (c). But "ammine" and "aqua" both start with 'a' — we go to the second letter: 'm' vs 'q'. 'm' comes before 'q', so ammine comes before aqua. Then chloro comes last.
Thus the order is: tetraammineaquachloro.
Watch outA common mistake is to write "tetraamminechloroaqua" because chlorido ligands are sometimes placed before aqua in older conventions, or because students list by negative charge first. IUPAC rules are strictly alphabetical by ligand name (ignoring prefixes), so aqua comes before chloro.
Now assemble the name: Tetraammineaquachlorocobalt(III) chloride. Notice that the metal name is attached directly after the ligands (no space), and the oxidation state is in Roman numerals in parentheses. The counterion is named as a separate word: "chloride".
Let's check the options: …
- TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQQ.Choose the correct statements from the following I) In vapour phase BeCl2 exists as chlorobridged dimer II) BeSO4 is readily soluble in water III) BeO is completely basic in nature IV) BeCO3, being unstable, is kept in the atmosphere of CO2 V) BeCO3 is less soluble among all the carbonates of group 2 elements (A) II, III, IV (B) I, II, IV (C) I, IV, V (D) II, III, V
›Reveal solutionSolution
Statements I, II and IV are correct; III and V are false. The correct option is (B).
Beryllium is anomalous in group 2 because of its very small size and high polarising (charge/size) power, giving covalent character and a diagonal relationship with aluminium.
I) BeCl2 exists as a chloro-bridged dimer in the vapour phase - TRUE.
On heating the solid (polymeric) chloride, the vapour first forms a chloro-bridged dimer Be2Cl4, which dissociates to the linear Cl-Be-Cl monomer only at high temperature (∼1200 K). So over the ordinary vapour range the bridged dimer exists.
II) BeSO4 is readily soluble in water - TRUE.
The tiny Be2+ ion has a very high hydration enthalpy that exceeds the lattice enthalpy, so BeSO4 (and MgSO4) are highly soluble, unlike the heavier, insoluble BaSO4.
III) BeO is completely basic - FALSE.
BeO is amphoteric: it dissolves in acids to give Be2+ salts and in strong alkali to give beryllate, e.g. Na2[Be(OH)4]. Only the heavier group-2 oxides are purely basic. …
- TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQQ.Solubility of A3X4 in pure water is ‘S’ mol L−1. Its solubility product is (A) S7 (B) 108S5 (C) 5184S7 (D) 6912S7
›Reveal solutionSolution
The solubility product (Ksp) for a sparingly soluble salt A3X4 is derived by considering its dissociation into 3A4+ and 4X3− ions. If its solubility is S mol L−1, then [A4+]=3S and [X3−]=4S. Substituting these into the Ksp expression, Ksp=[A4+]3[X3−]4, yields Ksp=6912S7.
When a sparingly soluble ionic compound dissolves in water, it establishes an equilibrium between the undissolved solid and its constituent ions in solution. The solubility product constant, Ksp, quantifies this equilibrium. It is a measure of how much of the solid dissolves to form a saturated solution.
The key idea is to understand how the stoichiometry of the ionic compound dictates the relative concentrations of its ions in solution. If 'S' represents the molar solubility of the compound (i.e., the number of moles of the compound that dissolve per liter of solution), then the concentration of each ion in the saturated solution will be a multiple of 'S', determined by the coefficients in the balanced dissociation equation.
For a general sparingly soluble salt AmXn, its dissociation equilibrium is:
AmXn(s)⇌mAn+(aq)+nXm−(aq)
If the molar solubility of AmXn is S mol L−1, then at equilibrium:
[An+]=mS
[Xm−]=nS
The solubility product expression is then given by:
Ksp=[An+]m[Xm−]n
Substituting the concentrations in terms of S:
Ksp=(mS)m(nS)n=mmnnSm+n
This general formula is what we will apply to the specific compound A3X4.
-
Write the dissociation equilibrium for A3X4:
The compound A3X4 dissociates into its constituent ions. To maintain charge neutrality, if there are 3 A ions and 4 X ions, and the formula is A3X4, then the charge on A must be +4 and the charge on X must be −3.
So, the dissociation equilibrium is:
A3X4(s)⇌3A4+(aq)+4X3−(aq)
-
Relate ion concentrations to solubility (S):
If the molar solubility of A3X4 is S mol L−1, it means that S moles of A3X4 dissolve in one liter of water. According to the stoichiometry of the dissociation reaction:
For every 1 mole of A3X4 that dissolves, 3 moles of A4+ ions and 4 moles of X3− ions are produced.
Therefore, in a saturated solution:
[A4+]=3S mol L−1
[X3−]=4S mol L−1
-
Write the expression for the solubility product (Ksp): …
-
- TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQQ.What is the correct order of freezing point of the following solutions? I 0.1m Ba3(PO4)2 II 0.1m Na2SO4 III 0.1m NaCl (A) III > II > I (B) I > II > III (C) II > III > I (D) III > I > II
›Reveal solutionSolution
Freezing point depression depends on the number of particles in solution. The more particles, the lower the freezing point. The correct order is III > II > I, so option (A).
The key concept here is colligative properties — properties that depend only on the number of solute particles, not on their identity. Freezing point depression is one such property: adding a solute lowers the freezing point of a solvent, and the drop is proportional to the total molality of particles in solution.
For ionic compounds, each formula unit dissociates into a certain number of ions. The van’t Hoff factor i tells us how many particles each mole of solute produces. The freezing point depression is given by ΔTf=i⋅Kf⋅m, where m is the molality. Since Kf and m are the same for all three solutions here (all are 0.1m in water), the freezing point is lowest for the solution with the largest i, and highest for the one with the smallest i.
Let’s work through each solution.
- Solution I: 0.1m Ba3(PO4)2 This salt dissociates completely in water:
Ba3(PO4)2→3Ba2++2PO43−
That’s 3+2=5 ions per formula unit. So i=5.
The effective particle molality is 0.1×5=0.5m.
This gives the largest depression, hence the lowest freezing point.
- Solution II: 0.1m Na2SO4 Dissociation:
Na2SO4→2Na++SO42−
That’s 2+1=3 ions. So i=3.
Effective particle molality = 0.1×3=0.3m. …
- TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQQ.Consider the following about the Tyndall effect I) It is used to distinguish between a true and colloidal solution II) It is possible only when the dispersed medium and dispersed phase differ much in their refractive indices III) It is observed only when the size of colloidal particles is much smaller than the wavelength of the light used The correct statements are (A) I & III only (B) II & III only (C) I, II & III (D) I & II only
›Reveal solutionSolution
The Tyndall effect is the scattering of light by colloidal particles; it distinguishes true from colloidal solutions, requires a refractive index difference, and occurs when particle size is comparable to (not much smaller than) the light’s wavelength. Only statements I and II are correct.
The Tyndall effect is a classic optical phenomenon used to tell apart a true solution (like salt water) from a colloidal dispersion (like milk or fog). The key idea is that light scatters when it hits particles of a certain size — but the exact conditions matter. Let’s break down each statement.
-
Statement I: “It is used to distinguish between a true and colloidal solution.”
This is correct. In a true solution, solute particles are of molecular size (less than 1 nm) and do not scatter visible light — the beam passes through invisibly. In a colloid, particles are larger (1–1000 nm) and scatter light, making the beam visible (the Tyndall effect). So shining a light through a sample and looking for a visible beam is a simple test.
-
Statement II: “It is possible only when the dispersed medium and dispersed phase differ much in their refractive indices.”
This is also correct. Scattering intensity depends on the difference in refractive index between the particles and the surrounding medium. If they have nearly the same refractive index, light passes through with little scattering — the effect is weak or absent. A large difference (e.g., water droplets in air) gives strong scattering.
-
Statement III: “It is observed only when the size of colloidal particles is much smaller than the wavelength of the light used.” …
-
- TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQQ.According to Werner’s theory, the number of groups bonded to the central metal atom / ion in a coordination complex represent (A) Oxidation state (B) Primary valency (C) Secondary Valency (D) Polyhedron
›Reveal solutionSolution
Werner’s theory distinguishes primary valency (ionic, non-directional) from secondary valency (directional, fixed number of ligands directly bonded to the metal). The number of groups bonded to the central metal atom represents the secondary valency.
Werner’s theory was the first successful model of coordination compounds. Before it, chemists struggled to explain why compounds like CoClX3 ⋅6NHX3 existed — why would ammonia stick to cobalt chloride in such a fixed ratio? Werner’s key insight was that a metal ion has two kinds of valency:
- Primary valency (now called oxidation state) — satisfied by negative ions, non-directional, and corresponds to the charge on the metal.
- Secondary valency (now called coordination number) — satisfied by neutral molecules or negative ions directly bonded to the metal, directional, and fixed in number for a given metal.
The question asks: “the number of groups bonded to the central metal atom / ion” — that is, the count of ligands directly attached to the metal. That is exactly what Werner called secondary valency.
-
Identify what “groups bonded” means
In a complex like [Co(NHX3)X6]X3+, six ammonia molecules are directly bonded to cobalt. The number “6” is the count of groups bonded. This is not the charge (oxidation state) nor the shape (polyhedron) itself — it’s the number that determines the shape.
-
Eliminate the distractors
- (A) Oxidation state — This is the charge left on the metal after satisfying primary valency. For CoX3+, the oxidation state is +3, not 6.
- (D) Polyhedron — The arrangement of ligands (e.g., octahedron) is a consequence of the number of groups, not the number itself. …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.