Which of the following complexes are heteroleptic? (i) [Cr(NH₃)₆]³⁺ (ii) [Fe(NH₃)₄Cl₂]⁺ (iii) [Mn(CN)₆]⁴⁻ (iv)

NCERT Exemplar · Q37

Q. Which of the following complexes are heteroleptic?

(i)

[Cr(NH_3)_6]^{3+}

(ii)

[Fe(NH_3)_4Cl_2]^{+}

(iii)

[Mn(CN)_6]^{4-}

(iv)

[Co(NH_3)_4Cl_2]

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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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A heteroleptic complex has two or more different types of ligands attached to the central metal ion. Among the given options, only (ii) $[Fe(NH_3)_4Cl_2]^{+}$ and (iv) $[Co(NH_3)_4Cl_2]$ contain more than one kind of ligand, so they are heteroleptic.

Why this classification matters

In coordination chemistry, we classify complexes based on how many different kinds of ligands are bonded to the metal centre. This isn't just a naming exercise — it affects symmetry, isomerism, and often the chemical properties of the complex.

Homoleptic complexes: all ligands are identical.

Heteroleptic complexes: at least two different types of ligands are present.

The trick is to look past the numbers and focus on the identity of the ligands. A complex with six ammonia molecules is homoleptic; one with four ammonias and two chlorides is heteroleptic — even if the total number of ligands is the same.

Step-by-step analysis

Option (i): $[Cr(NH_3)_6]^{3+}$

The only ligand present is ammonia ( $NH_3$ ). All six ligands are identical.

→ Homoleptic.
Option (ii): $[Fe(NH_3)_4Cl_2]^{+}$

Here we have two different ligands: ammonia ( $NH_3$ ) and chloride ( $Cl^-$ ). Four of one kind, two of another — but the presence of even one different ligand makes it heteroleptic.

→ Heteroleptic.
Option (iii): $[Mn(CN)_6]^{4-}$

Only cyanide ions ( $CN^-$ ) are present. All six ligands are the same.

→ Homoleptic.
Option (iv): $[Co(NH_3)_4Cl_2]$ …

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Common Mistakes in Identifying Heteroleptic Complexes

Mistake 1: Confusing "Heteroleptic" with "Mixed Oxidation State"

The Error: Students think heteroleptic means the metal has different oxidation states or that the complex has different charges.

The Fix: Heteroleptic refers only to the number of different types of ligands attached to the central metal atom/ion. It has nothing to do with charge or oxidation state.

Heteroleptic = Two or more different ligand types
Homoleptic = Only one type of ligand

Mistake 2: Counting Only the First Ligand Type

The Error: Students see [Co(NH₃)₄Cl₂] and think "NH₃ is one type, Cl is another — so it's heteroleptic." That part is correct, but they often miss that [Fe(NH₃)₄Cl₂]⁺ is also heteroleptic.

The Fix: Check every complex systematically:

(i) [Cr(NH₃)₆]³⁺ → Only NH₃ → Homoleptic
(ii) [Fe(NH₃)₄Cl₂]⁺ → NH₃ and Cl → Heteroleptic ✓
(iii) [Mn(CN)₆]⁴⁻ → Only CN⁻ → Homoleptic
(iv) [Co(NH₃)₄Cl₂] → NH₃ and Cl → Heteroleptic ✓

Correct Answer: (ii) and (iv)

Mistake 3: Thinking Charge Affects Ligand Type

The Error: Students assume that because [Fe(NH₃)₄Cl₂]⁺ has a +1 charge and [Co(NH₃)₄Cl₂] is neutral, they must be different in some fundamental way.

The Fix: Charge is irrelevant to the heteroleptic/homoleptic classification. Only the identity of ligands matters. Both have NH₃ and Cl — both are heteroleptic.

Mistake 4: Confusing "Heteroleptic" with "Ambidentate" …

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

KEAM 2026Set eng-2026-04184 marksMCQ
$[Ag(NH_3)_2][Ag(CN)_2]$
›Reveal solutionSolution

Both silver centres are +1; use current IUPAC ligand names (ammine, cyanido) and the -ate suffix for the anion.

Cation $[Ag(NH_3)_2]^+$ : two $NH_3$ = diammine, Ag oxidation state $+1$ $\Rightarrow$ diamminesilver(I).

Anion $[Ag(CN)_2]^-$ : two $CN^-$ = dicyanido (modern IUPAC), Ag $+1$ , anionic complex takes -ate on the Latin stem argent $\Rightarrow$ dicyanidoargentate(I). …
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KEAM 2026Set eng-2026-04204 marksMCQ
$[CoCl_2(en)_2]Cl$
›Reveal solutionSolution
Correct IUPAC name = Dichloridobis(ethane-1,2-diamine)cobalt(III)chloride.

Oxidation state of Co: complex $[CoCl_2(en)_2]^{+}$ (charge $+1$ balancing the outer $Cl^{-}$ ); with $2(-1)$ from chlorido and neutral en, Co $=+3$ .

Naming rules:

Ligands cited alphabetically: chlorido (c) before ethane-1,2-diamine (e), so "dichlorido" then "bis(ethane-1,2-diamine)". …
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KEAM 2026Set eng-2026-04214 marksMCQ
$[Co(NH_3)_5(CO)]Cl_2$
›Reveal solutionSolution

Charge balance gives $[Co(NH_3)_5(CO_3)]Cl$ .

Co is $+3$ , five neutral $NH_3$ , carbonato $CO_3^{2-}$ :

$+3 + 5(0) + (-2) = +1.$ …
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KEAM 2026Set pha-2026-0420F4 marksMCQ
$Q. The IUPAC name of the complex [Co(NH3)5ONO]Cl2 is (A) Pentaamminenitritocobalt (III) chloride (B) Pentaamminenitritocobalt (II) chloride (C) Pentaamminenitrocobalt (III) chloride (D) O-Nitritopentaamminecobalt (III) chloride (E) Pentaamminemononitritocobalt (III) chloride$
›Reveal solutionSolution

Co is +3; O-bonded ONO is 'nitrito'; alphabetical ligand order gives Pentaamminenitritocobalt(III) chloride.

Determine the metal oxidation state: NH $_3$ is neutral, the ONO ligand (bonded through oxygen) is nitrito with charge $-1$ , and the two chlorides are counterions ( $-1$ each). For overall neutrality the complex cation is $+2$ , so

$\text{Co} + 5(0) + (-1) = +2 \quad\Rightarrow\quad \text{Co} = +3.$ …
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KEAM 2025Set eng-2025-04254 marksMCQ
$[Co(en)_3]_2(SO_4)_3$
›Reveal solutionSolution
With three neutral $en$ ligands and a sulphate counter-ion, cobalt is in the $+3$ state; the IUPAC name is tris(ethane-1,2-diamine)cobalt(III) sulphate.

In $[Co(en)_3]_2(SO_4)_3$ :

$en$ (ethane-1,2-diamine) is a neutral bidentate ligand; there are three of them $\rightarrow$ prefix tris (as the ligand name contains 'di').

Charge: three $SO_4^{2-}$ give $-6$ shared over two complex cations, so each $[Co(en)_3]^{3+}$ , making cobalt $+3$ . …
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KEAM 2025Set pha-2025-0424F4 marksMCQ
$[Co(NH_3)_3(H_2O)_3]Cl_3$
›Reveal solutionSolution
Name ligands alphabetically (ammine < aqua) and assign Co the +3 oxidation state, so the complex is triamminetriaquacobalt(III) chloride.

For $[Co(NH_3)_3(H_2O)_3]Cl_3$ :

Ligands are named in alphabetical order using the ligand name (ignoring the multiplying prefix): ammine (a) comes before aqua (a-q) → "triammine" is cited before "triaqua".

Oxidation state of cobalt: overall charge zero; three $Cl^-$ counter-ions give the complex ion a +3 charge; $NH_3$ and $H_2O$ are neutral, so Co is +3.

Anion: chloride. …
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KEAM 2025Set pha-2025-0424A4 marksMCQ
$[Cr(H_2O)_3(NH_3)_3]Cl_3$
›Reveal solutionSolution
The IUPAC name is triamminetriaquachromium(III) chloride.

Naming $[Cr(H_2O)_3(NH_3)_3]Cl_3$ :

Ligands are cited in alphabetical order of ligand name: ammine (a) comes before aqua (a-q), so 'triammine' precedes 'triaqua'.

Oxidation state of Cr: the complex cation carries $+3$ (balanced by three $Cl^-$ ), so chromium is $Cr(III)$ . …
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KEAM 2024Set eng-2024-06054 marksMCQ
$[Co(NH_{3})_{4}(H_{2}O)Cl]Cl_{2}$
›Reveal solutionSolution

Alphabetical ligand order ammine < aqua < chlorido, Co oxidation state $+3$ , giving tetraammineaquachloridocobalt(III) chloride.

In $[Co(NH_3)_4(H_2O)Cl]Cl_2$ the two chloride counter-ions make the complex ion $+2$ . Charge balance: $x + 4(0) + 0 + (-1) = +2 \Rightarrow x = +3$ . …
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KEAM 2024Set eng-2024-06064 marksMCQ
$[Cr(NH_{3})_{3}(H_{2}O)_{3}]Cl_{3}$
›Reveal solutionSolution
In IUPAC nomenclature ligands are cited alphabetically (ammine < aqua), and the metal oxidation state is set by charge balance. For $[Cr(NH_3)_3(H_2O)_3]Cl_3$ , three chlorides outside give Cr = $+3$ .

Working it out:

Ligands: 3 $NH_3$ (ammine) and 3 $H_2O$ (aqua). Alphabetical order → triammine then triaqua. …
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KEAM 2024Set eng-2024-06084 marksMCQ
$Hg[Co(SCN)_4]$
›Reveal solutionSolution

$Hg[Co(SCN)_4]$ : Hg is $+2$ , so $[Co(SCN)_4]^{2-}$ makes Co $+2$ ; thiocyanate binds via sulphur, giving Mercury(II) tetrathiocyanato-S-cobaltate(II).

Assign oxidation states. Mercury is $Hg^{2+}$ , so the complex anion carries a $-2$ charge:
$[Co(SCN)_4]^{2-}$
Each thiocyanate ligand is $SCN^{-}$ , so four contribute $-4$ . Letting the cobalt state be $x$ :

$x + 4(-1) = -2 \;\Rightarrow\; x = +2$ …
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Q.Which of the following complexes are heteroleptic?

Concept understanding — Coordination Compound Nomenclature

Coordination Compound Nomenclature: From Intuition to Precision

The Core Idea

The Rules, Step by Step

1. Cation before anion (just like NaCl is sodium chloride)

2. Within the coordination sphere: ligands first, then metal

3. Naming ligands

4. Prefixes for multiple ligands

5. Oxidation state of the metal

6. If the complex is an anion, change the metal's ending

Worked Examples

Why this formula?

Coordination Compound Nomenclature: Why the Rules Work

1. The Central Idea: Ligands as "Guests" Around a Metal "Host"

2. Key Rule: Ligand Order — Alphabetical, Not by Charge

3. Oxidation State: Why Roman Numerals?

4. Anionic Ligands: The "-o" Ending

5. Neutral Ligands: Special Names

6. Prefixes: di-, tri-, tetra-, etc.

7. Anionic Complexes: The "-ate" Suffix …

Why this classification matters

Step-by-step analysis

Method: Classification by Ligand Type (Homoleptic vs Heteroleptic)

Steps:

Applying to each option:

Common Mistakes in Identifying Heteroleptic Complexes

Mistake 1: Confusing "Heteroleptic" with "Mixed Oxidation State"

Mistake 2: Counting Only the First Ligand Type

Mistake 3: Thinking Charge Affects Ligand Type

Mistake 4: Confusing "Heteroleptic" with "Ambidentate" …

Q.Which of the following complexes are heteroleptic?

Why this classification matters

Step-by-step analysis

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