When 1 mol $CrCl_3 \cdot 6H_2O$ is treated with excess of $AgNO_3$, 3 mol of $AgCl$ are obtained. The formula of the complex is: (i) $[CrCl_3(H_2O)_3] \cdot 3H_2O$ (ii) $[CrCl_2(H_2O)_4]Cl \cdot 2H_2O$ (iii) $[CrCl(H_2O)_5]Cl_2 \cdot H_2O$ (iv) $[Cr(H_2O)_6]Cl_3$

When 1 mol CrCl₃ · 6H₂O is treated with excess of – Class 12 Chemistry

NCERT Exemplar · Q4

CrCl_3 \cdot 6H_2O

(i)

[CrCl_3(H_2O)_3] \cdot 3H_2O

(ii)

[CrCl_2(H_2O)_4]Cl \cdot 2H_2O

(iii)

[CrCl(H_2O)_5]Cl_2 \cdot H_2O

(iv)

[Cr(H_2O)_6]Cl_3

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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-).

Note

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)

Watch out

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.

Tip

[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 \to cation name = [prefix-ligands]metal(oxidation state) anion name = [prefix-ligands]metal ate (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:

\text{Metal oxidation state} = Q - L

For [Fe(CN)₆]³⁻: CN⁻ has charge -1, so $L = 6 \times (-1) = -6$ , $Q = -3$ .

\text{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 …

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The key is that only chloride ions outside the coordination sphere (counter ions) precipitate with $AgNO_3$ . Since 3 mol of $AgCl$ form per mol of complex, all three chlorides are counter ions, so the formula must be $[Cr(H_2O)_6]Cl_3$ — 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 $AgNO_3$ , 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): $[CrCl_3(H_2O)_3] \cdot 3H_2O$

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): $[CrCl_2(H_2O)_4]Cl \cdot 2H_2O$

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): $[CrCl(H_2O)_5]Cl_2 \cdot H_2O$

One chloride inside, two outside. That gives 2 mol of $AgCl$ . Still not enough. Eliminated.
Option (iv): $[Cr(H_2O)_6]Cl_3$ …

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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 $CrCl_3 \cdot 6H_2O$ 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 ( $\cdot$ ) 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) $[Cr(H_2O)_6]Cl_3$ 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: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$[Co(en)_3][Fe(CN)_6]$
›Reveal solutionSolution
This tests naming-to-formula conversion for coordination compounds — figuring out ligand charges, metal oxidation states, and then balancing overall charge between cation and anion. The answer is $[Co(en)_3]_4[Fe(CN)_6]_3$ .

Concept and Intuition

When a coordination compound's name specifies both a cationic complex and an anionic complex (as here — cobalt complex is the cation, ferrate complex is the anion), the overall salt must be electrically neutral. You find each complex ion's own charge first, then combine cation and anion in whatever whole-number ratio makes the total charge zero — exactly like balancing a simple ionic salt such as $Al_2O_3$ (from $Al^{3+}$ and $O^{2-}$ ).

Step-by-Step Solution

Cation — tris(ethane-1,2-diamine)cobalt(III): "ethane-1,2-diamine" (en) is a neutral bidentate ligand ( $NH_2CH_2CH_2NH_2$ ), and "tris" means three of them are attached: $[Co(en)_3]$ . The oxidation state of cobalt is given as (III), i.e. $+3$ . Since en is neutral, the whole complex ion carries the metal's charge: $[Co(en)_3]^{3+}$ .

Anion — hexacyanidoferrate(II): "hexacyanido" means six $CN^-$ ligands, each carrying $-1$ charge, i.e. $-6$ total from ligands. "ferrate(II)" tells us iron is in the $+2$ state. Net charge of the complex ion $= +2 + (-6) = -4$ : $[Fe(CN)_6]^{4-}$ . …
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AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$[PtCl_2(H_2N{-}CH_2{-}CH_2{-}NH_2)_2](NO_3)_2$
›Reveal solutionSolution
This tests IUPAC coordination nomenclature: computing the metal's oxidation state and applying the alphabetical-ligand-order + no-prefix-on-a-Stock-named-counter-ion rules. Answer: Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate.

Concept and Intuition

A coordination compound name is built as [cationic complex or ligands][metal(oxidation state)] [anion]. Two separate rules govern it here: (1) ligands inside the complex are cited in alphabetical order by the ligand's own name, ignoring any multiplying prefix (mono/di/bis, tris...); (2) once the metal's oxidation state is stated in Roman numerals (Stock notation), the charge of the whole complex ion is fixed, so the number of counter-ions needed to balance it is implicit and is NOT restated with a multiplying prefix on the counter-ion's name.

Step-by-Step Solution

Ligands: $Cl^-$ (chlorido, charge $-1$ each, two of them) and $en$ = ethane-1,2-diamine (neutral, a bidentate ligand named with "bis" since its own name contains a numerical locant "1,2").

Overall salt is $[PtCl_2(en)_2](NO_3)_2$ , so the complex cation carries $+2$ charge (two nitrate anions balance it).

Let Pt oxidation state be $x$ : $x + 2(-1) + 2(0) = +2 \Rightarrow x = +4$ . So it's platinum(IV).

Ligand naming: "chlorido" for $Cl^-$ ×2 → "dichlorido"; "ethane-1,2-diamine" ×2 → "bis(ethane-1,2-diamine)" (bis is used, not "di", because the ligand name itself already has numbers in it — using "di" would be ambiguous).

Alphabetical citation order compares "chlorido" (c) vs "ethane-1,2-diamine" (e): c precedes e, so chlorido-containing part is written first: "Dichloridobis(ethane-1,2-diamine)platinum(IV)". …
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AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$K_3[\mathrm{Co(ox)_3}]$
›Reveal solutionSolution
IUPAC naming of coordination compounds: ligand 'oxalato' × 3 = 'trioxalato', anionic complex metal gets the '-ate' suffix ('cobaltate'), oxidation state by charge balance is +3, and the counter-cation 'potassium' is stated without a multiplying prefix — giving Potassium trioxalatocobaltate(III).

Concept and Intuition

Coordination-compound IUPAC names follow a fixed template: cation first, then ligands alphabetically with multiplying prefixes (bis/tris for complex ligand names, di/tri for simple ones like oxalato), then the metal name — with the special rule that if the complex ion is an anion, the metal name is modified to end in '-ate' (e.g. cobalt → cobaltate, iron → ferrate), followed by the oxidation state in Roman numerals in brackets.

Step-by-Step Solution

Formula: $K_3[\mathrm{Co(ox)_3}]$ , where 'ox' = oxalate ion ( $\mathrm{C_2O_4^{2-}}$ ), a bidentate ligand named 'oxalato' in coordination nomenclature.

Charge balance: complex ion is $[\mathrm{Co(ox)_3}]^{3-}$ (balanced by 3 $K^+$ ). Let Co oxidation state be $x$ : $x + 3(-2) = -3 \Rightarrow x = +3$ .

Ligand prefix: three oxalato ligands → 'trioxalato' (using tri- since oxalato is a simple, unsubstituted ligand name).

Since the complex ion is an anion, the metal name takes the anionic suffix: cobalt → 'cobaltate'.

Combine: 'trioxalatocobaltate', with oxidation state (III) in brackets: 'trioxalatocobaltate(III)'. …
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AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
$\mathrm{[Co(H_2NCH_2CH_2NH_2)_3]SO_4}$
›Reveal solutionSolution
The complex cation [Co(en)₃]³⁺ needs 2 cations for every 3 sulphate ions to balance charge, giving [Co(en)₃]₂(SO₄)₃.

Concept and Intuition

"Ethane-1,2-diamine" (en) is the neutral bidentate ligand $H_2NCH_2CH_2NH_2$ . Cobalt(III) with three such neutral ligands gives a complex cation charge equal to the metal's oxidation state alone: $[Co(en)_3]^{3+}$ . To form a neutral salt with sulphate ( $SO_4^{2-}$ ), the overall charges must balance using the smallest whole-number ratio — LCM of 3 and 2 is 6, so 2 cations (total +6) balance 3 sulphates (total −6).

Step-by-Step Solution

Write the complex cation: $[Co(H_2NCH_2CH_2NH_2)_3]^{3+}$ (Co³⁺ + 3 neutral en ligands).

Sulphate anion charge: $SO_4^{2-}$ .

Balance charges: need ratio such that $2\times3=3\times2=6$ → 2 cations : 3 sulphates. …
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AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
$[\mathrm{Pt(NH_3)_2Cl(NH_2CH_3)}]\mathrm{Cl}$
›Reveal solutionSolution
Naming ligands alphabetically (ammine, chloro, methanamine) and computing Pt's oxidation state as +2 gives Diamminechloro(methanamine)platinum(II) chloride.

Concept and Intuition

IUPAC coordination nomenclature: (1) name ligands alphabetically using IUPAC ligand names (ignoring di/tri prefixes for alphabetization); (2) anionic ligands get '-o' suffix (chloro for Cl-); (3) metal oxidation state deduced from charge balance; (4) counter-ion named last.

Step-by-Step Solution

Ligands in $[\mathrm{Pt(NH_3)_2Cl(NH_2CH_3)}]\mathrm{Cl}$ : two $\mathrm{NH_3}$ (ammine, neutral), one $\mathrm{Cl^-}$ (chloro, -1), one $\mathrm{NH_2CH_3}$ = methanamine (neutral, the IUPAC ligand name for methylamine).

Alphabetize ignoring 'di': ammine (a) < chloro (c) < methanamine (m) -> "diamminechloro(methanamine)". …
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AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
$[Co(NH_3)_4Br_2]^{\oplus}$
›Reveal solutionSolution
A homoleptic complex has all identical ligands; only $[\mathrm{Co(NH_3)_6}]^{3+}$ qualifies among the given options.

Concept and Intuition

Coordination complexes are classified as homoleptic (all ligands of one single type bound to the metal) or heteroleptic (two or more different kinds of ligands). This is purely a matter of reading off the ligands present in each formula and checking whether they are all the same species.

Step-by-Step Solution

$[\mathrm{Co(NH_3)_4Br_2}]^{+}$ — ligands present: $\mathrm{NH_3}$ and $\mathrm{Br^-}$ (two types) → heteroleptic.

$[\mathrm{Co(C_2O_4)(NH_3)_4}]^{+}$ — ligands present: oxalate ( $\mathrm{C_2O_4^{2-}}$ ) and $\mathrm{NH_3}$ (two types) → heteroleptic.

$[\mathrm{Co(NH_3)_6}]^{3+}$ — ligand present: only $\mathrm{NH_3}$ , six of them → homoleptic. …
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AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
$\mathrm{[Co(NH_2CH_2CH_2NH_2)_2Br_2]Br}$
›Reveal solutionSolution
The oxidation state of Co is $+3$ (from a $+1$ complex cation with two neutral en and two $\mathrm{Br^-}$ ligands), and IUPAC alphabetical ligand ordering puts "bromido" before "(ethylenediamine)," giving dibromidobis(ethylenediamine)cobalt(III) bromide.

Concept and Intuition

Naming a coordination compound requires: (1) determining the metal's oxidation state from overall charge balance, (2) naming ligands with multiplying prefixes (di-, tri- for simple ligands; bis-, tris- for more complex/substituted ligand names like ethylenediamine, to avoid ambiguity), and (3) listing ligands in strict alphabetical order by ligand name (ignoring the multiplying prefix), followed by the metal name+oxidation state, then the counter-ion name.

Step-by-Step Solution

The compound is $[\mathrm{Co(en)_2Br_2}]\mathrm{Br}$ — one $\mathrm{Br^-}$ is outside the coordination sphere (counter-ion), so the complex cation is $[\mathrm{Co(en)_2Br_2}]^{+}$ .

Charge balance inside the bracket: en is neutral (×2 = 0), $\mathrm{Br^-}$ ligands contribute $-2$ (×2), so $\mathrm{Co} + 0 - 2 = +1 \Rightarrow \mathrm{Co} = +3$ .

Ligand names: "ethylenediamine" (a bidentate amine, needs "bis" not "di" since its name itself could be ambiguous with a simple prefix) and "bromido" for $\mathrm{Br^-}$ .

Alphabetical order of ligand names (ignoring multiplying prefixes): "bromido" (b) before "ethylenediamine" (e). …
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AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
$[Cr(NH_3)_3(H_2O)_2Cl]Cl_2$
›Reveal solutionSolution
Naming $[\mathrm{Cr(NH_3)_3(H_2O)_2Cl}]\mathrm{Cl_2}$ requires listing ligands alphabetically (ammine, aqua, chlorido) with multiplying prefixes, then the oxidation state of Cr (found by charge balance) in Roman numerals, followed by the anion name — giving "triamminediaquachloridochromium(III) chloride".

Concept and Intuition

IUPAC naming of coordination compounds lists ligands in alphabetical order (by the first letter of the ligand name itself, ignoring multiplying prefixes like di-/tri-), followed by the name of the central metal with its oxidation state in Roman numerals in parentheses, and finally the counter-ion name (as a separate word) if the complex is a cation. The oxidation state of the metal is found by balancing charges: neutral ligands (NH₃, H₂O) contribute 0, anionic ligands contribute their charge, and the sum plus the metal's oxidation state must equal the overall charge of the complex ion.

Step-by-Step Solution

Identify ligands inside the bracket: 3 NH₃ ("ammine" ×3 → "triammine"), 2 H₂O ("aqua" ×2 → "diaqua"), 1 Cl⁻ ("chlorido").

Alphabetize by ligand name ignoring prefixes: compare "ammine", "aqua", "chlorido" — a-m-m... vs a-q-u-a: 'm' precedes 'q', so "ammine" comes before "aqua"; "chlorido" (starting with 'c') comes after both 'a' words. Order: ammine, aqua, chlorido.

Combine with multiplying prefixes in that order: "triamminediaquachlorido".

Determine oxidation state of Cr: the compound is neutral with 2 Cl⁻ as counter ions outside the bracket, so the complex ion charge is +2. Inside the bracket: 3(NH₃, 0) + 2(H₂O, 0) + 1(Cl⁻, −1) = −1. So Cr + (−1) = +2 → Cr = +3.

Complete metal name: "chromium(III)". …
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Q.When 1 mol CrCl3⋅6H2OCrCl_3 \cdot 6H_2OCrCl3​⋅6H2​O is treated with excess of AgNO3AgNO_3AgNO3​, 3 mol of AgClAgClAgCl are obtained. The formula of the complex is:

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Q.When 1 mol $CrCl_3 \cdot 6H_2O$ is treated with excess of $AgNO_3$ , 3 mol of $AgCl$ are obtained. The formula of the complex is:

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