Write the formulas for the following coordination compounds: (i) Tetraamminediaquacobalt(III) chloride (ii) Po

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Q. Write the formulas for the following coordination compounds:

(i) Tetraamminediaquacobalt(III) chloride

(ii) Potassium tetracyanidonickelate(II)

(iii) Tris(ethane-1,2-diamine) chromium(III) chloride

(iv) Amminebromidochloridonitrito-N-platinate(II)

(v) Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate

(vi) Iron(III) hexacyanidoferrate(II)

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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 $\ce{CoCl3}$ , and when you dissolve it in water, you expect to find $\ce{Co^{3+}}$ and $\ce{Cl^-}$ 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 $\ce{CoCl3·6NH3}$ (orange-yellow) and $\ce{CoCl3·5NH3}$ (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 $\ce{CoCl3·6NH3}$ (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: $[\ce{Co(NH3)6}]\ce{Cl3}$ . All three chlorides precipitate with $\ce{AgNO3}$ .

Now take $\ce{CoCl3·5NH3}$ (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: $[\ce{Co(NH3)5Cl}]\ce{Cl2}$ . Only two chlorides precipitate.

Note

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

Werner didn't just count ligands — he placed them in space. For coordination number 6, he proposed an octahedral arrangement (ligands at the six corners of an octahedron). This explained why $[\ce{Co(NH3)4Cl2}]^+$ exists as two different compounds (isomers): one where the two chlorides are next to each other (cis) and one where they are opposite (trans). No other geometry could produce exactly two isomers.

Important

Werner's theory was the first to show that complexes have definite three-dimensional structures. This was decades before X-ray crystallography could confirm it directly.

What It Replaced

Before Werner, chemists thought bonding was simple: each atom had a fixed valency (like carbon always forms four bonds). They tried to write chain structures for coordination compounds (like organic molecules), but it failed — you couldn't explain why $\ce{CoCl3·6NH3}$ and $\ce{CoCl3·5NH3}$ were different compounds with the same metal and ligands.

Werner's key break: the metal can bond to more species than its oxidation state would suggest, and those bonds are not all the same type.

The Legacy

Werner won the Nobel Prize in 1913. His theory:

Introduced the concept of coordination number and coordination sphere.
Explained isomerism in complexes (geometric, optical).
Laid the foundation for modern coordination chemistry, crystal field theory, and ligand field theory.
Showed that inorganic compounds could have complex, predictable geometries — not just simple salts.

Tip

When you see a formula like $\ce{[Co(NH3)6]Cl3}$ , read the square brackets as "the castle walls". Everything inside is tightly bound to the metal; everything outside is free. That's Werner's idea in a nutshell.

Werner's coordination theory is the historical and conceptual foundation of the NCERT/CBSE Class 12 Chemistry chapter on Coordination Compounds, and ‘Werner's theory of coordination compounds’ is one of the most frequently asked important questions in board exams, JEE Main and NEET. Understanding primary and secondary valency as Werner defined them is essential groundwork for every other topic in this chapter.

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 $\ce{CoCl3 \cdot 6 NH3}$ (one of Werner's classic compounds).

The compound is electrically neutral overall.
Adding $\ce{AgNO3}$ precipitates all 3 Cl⁻ as $\ce{AgCl}$ — meaning all chlorides are free ions.
Therefore, the $\ce{NH3}$ molecules must be directly bonded to $\ce{Co}$ , 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:

\text{Coordination number} = \text{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 $\ce{CoCl3 \cdot 6 NH3}$ , adding acid doesn't remove $\ce{NH3}$ 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 $\ce{[Co(NH3)6]Cl3}$ , conductivity measurements show 4 ions in solution ( $\ce{[Co(NH3)6]^{3+}}$ + 3 Cl⁻).

If you had $\ce{[Co(NH3)5Cl]Cl2}$ , 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 $\ce{[Co(NH3)6]^{3+}}$ is octahedral — no other arrangement gives equal bond angles and distances.

4. The Isomer Counting Formula: Why $2^n$ or $n!$ Appears

Werner used isomer counts to confirm geometry. For an octahedral complex $\ce{[Ma2b2c2]}$ :

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.
The count comes from combinatorial reasoning on a fixed octahedral framework, not arbitrary permutation.

General principle:

\text{Number of isomers} = \frac{\text{total arrangements}}{\text{symmetry factor}}

This is not a simple $n!$ — it depends on the point group symmetry of the complex.

5. The Valence Sum Rule: Primary + Secondary = Constant?

Not a fixed sum!

Werner's theory does not say primary + secondary valence = constant.

Example:

$\ce{Co^{3+}}$ has primary valence = 3, secondary = 6 → sum = 9
$\ce{Pt^{4+}}$ has primary = 4, secondary = 6 → sum = 10

Why no fixed sum?

Primary valence depends on the metal's oxidation state (variable).
Secondary valence depends on the metal's size and electronic configuration (also variable).
They are independent properties — the only link is that both must be satisfied for a stable complex.

Summary: The Core Insight

Werner's formulas hold because:

Coordination number is an experimentally determined maximum — not a theoretical guess.
Geometry follows from minimizing ligand-ligand repulsion on a sphere.
Isomer counts follow from symmetry constraints on a fixed polyhedron.
Primary and secondary valences are independent — no single formula links them.

The real power of Werner's theory: it turned coordination chemistry from a list of random compounds into a predictive, spatial science — long before X-ray crystallography confirmed the geometries.

Concept: Werner Coordination Theory — the central metal ion has a fixed primary valence (oxidation state) and a variable secondary valence (coordination number), satisfied by ligands in a specific geometry.

Reasoning steps:

Identify the central metal ion and its oxidation state from the name.
Determine the coordination number from the ligands listed (prefixes like tetra-, hexa-).
Write the complex ion in square brackets: metal first, then the ligands in alphabetical order — a ligand's position does not depend on its charge (NCERT §5.3.1).
Balance the overall charge with counter ions outside the brackets.

Tetraamminediaquacobalt(III) chloride Cobalt(III) with 4 NH $_3$ (neutral) and 2 H $_2$ O (neutral) → complex charge = +3. Three Cl $^-$ counter ions. $[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})_2]\text{Cl}_3$
Potassium tetracyanidonickelate(II) Nickel(II) with 4 CN $^-$ → complex charge = 2 $-$ (since 4 × (−1) = −4, plus Ni $^{2+}$ gives −2). Two K $^+$ counter ions. $\text{K}_2[\text{Ni}(\text{CN})_4]$
Tris(ethane-1,2-diamine) chromium(III) chloride Chromium(III) with 3 en (neutral bidentate) → complex charge = +3. Three Cl $^-$ counter ions. $[\text{Cr}(\text{en})_3]\text{Cl}_3$
Amminebromidochloridonitrito-N-platinate(II) Platinate(II) indicates anionic complex. Ligands: NH $_3$ (neutral), Br $^-$ , Cl $^-$ , NO $_2^-$ (nitrito-N, N-bonded). Total charge from ligands = −3, plus Pt $^{2+}$ gives −1. Counter ion not named, so formula is the anion. $[\text{Pt}(\text{NH}_3)\text{Br}\text{Cl}(\text{NO}_2)]^-$ (as the anion)
Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate Platinum(IV) with 2 Cl $^-$ and 2 en (neutral) → complex charge = 2+ (since 2 × (−1) = −2, plus Pt $^{4+}$ gives +2). Two NO $_3^-$ counter ions. $[\text{Pt}(\text{Cl})_2(\text{en})_2](\text{NO}_3)_2$
Iron(III) hexacyanidoferrate(II) First iron is Fe $^{3+}$ (cation), second is Fe $^{2+}$ in the complex anion with 6 CN $^-$ → complex charge = 4 $-$ (6 × (−1) = −6, plus Fe $^{2+}$ gives −4). Formula: Fe $_4$ [Fe(CN) $_6$ ] $_3$ (Prussian blue). $\text{Fe}_4[\text{Fe}(\text{CN})_6]_3$
✓Final answer
$[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})_2]\text{Cl}_3$

Werner’s coordination theory tells us that the primary valence (oxidation state) is satisfied by anions, while the secondary valence (coordination number) is satisfied by ligands. The key is to identify the central metal, its oxidation state, the ligands (with their charges and names), and then assemble the formula — the complex ion is written inside square brackets, and counter ions are placed outside. The final formulas are given below.

Werner’s coordination theory is the foundation here. He proposed that metals have two types of valences: the primary valence (ionisable, corresponds to oxidation state) and the secondary valence (non-ionisable, corresponds to coordination number). In writing formulas, we place the central metal first, then the ligands in alphabetical order — a ligand's position in the list does not depend on its charge (NCERT §5.3.1) — and enclose the coordination sphere in square brackets. Counter ions (ions outside the bracket) balance the overall charge.

Let’s go through each compound step by step.

(i) Tetraamminediaquacobalt(III) chloride

Identify the central metal and its oxidation state.

The name ends with “cobalt(III)”, so cobalt is in the +3 oxidation state. The ligands are:
- “tetraammine” = four $\text{NH}_3$ (neutral)
- “diaqua” = two $\text{H}_2\text{O}$ (neutral) All ligands are neutral, so the charge on the complex ion comes only from the metal: $+3$ .
Write the complex cation.

The complex is a cation: $[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{3+}$ .
Add the counter ion.

“Chloride” means $\text{Cl}^-$ ions. To balance the $+3$ charge, we need three chloride ions.
Final formula:

[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})_2]\text{Cl}_3

Watch out

A common mistake is to write water as $\text{H}_2\text{O}$ inside the bracket but forget it’s neutral — it does not affect the charge. Also note the IUPAC formula order: metal first, then the ligands in alphabetical order, irrespective of their charge (NCERT §5.3.1) — “ammine” (a) before “aqua” (a-m before a-q), which is exactly the order the printed formula uses.

(ii) Potassium tetracyanidonickelate(II)

Identify the central metal and its oxidation state.

“nickelate(II)” means nickel is in the +2 oxidation state. The suffix “-ate” indicates the complex is an anion.
Identify the ligands.

“tetracyanido” = four $\text{CN}^-$ ions. Each cyanide ligand carries a $-1$ charge.
Calculate the charge on the complex anion.

Nickel: $+2$ ; four $\text{CN}^-$ : $4 \times (-1) = -4$ ; total charge = $+2 - 4 = -2$ .
Write the complex anion.

[\text{Ni}(\text{CN})_4]^{2-}

Add the counter ion.

“Potassium” is $\text{K}^+$ . To balance the $-2$ charge, we need two potassium ions.
Final formula:

\text{K}_2[\text{Ni}(\text{CN})_4]

Tip

Notice that the ligand name “cyanido” (from IUPAC) is used instead of the older “cyano”. The formula uses $\text{CN}^-$ as the ligand.

(iii) Tris(ethane-1,2-diamine) chromium(III) chloride

Identify the central metal and its oxidation state.

“chromium(III)” means Cr is in the +3 state.
Identify the ligands.

“tris(ethane-1,2-diamine)” = three molecules of $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$ , abbreviated as “en”. This is a neutral bidentate ligand.
Charge on the complex.

All ligands are neutral, so the complex cation has charge $+3$ .
Write the complex cation.

[\text{Cr}(\text{en})_3]^{3+}

Add the counter ion.

“Chloride” = $\text{Cl}^-$ . Three chloride ions balance the charge.
Final formula:

[\text{Cr}(\text{en})_3]\text{Cl}_3

Note

Ethane-1,2-diamine is often written as “en” in formulas for brevity, but in the full formula, it’s acceptable to write $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$ or simply “en”. The IUPAC name uses the full ligand name.

(iv) Amminebromidochloridonitrito-N-platinate(II)

Identify the central metal and its oxidation state.

“platinate(II)” means platinum is in the +2 state, and the complex is an anion (suffix “-ate”).
Identify the ligands.

The name lists four ligands:
- “ammine” = $\text{NH}_3$ (neutral)
- “bromido” = $\text{Br}^-$ (charge $-1$ )
- “chlorido” = $\text{Cl}^-$ (charge $-1$ )
- “nitrito-N” = $\text{NO}_2^-$ bonded through nitrogen (charge $-1$ ). The “-N” indicates the bonding atom is nitrogen — written as $\text{NO}_2^-$ , not $\text{ONO}^-$ (the O-bonded form is written $\text{ONO}^-$ and named “nitrito-O”).
Calculate the charge on the complex anion.

Pt: $+2$ ; $\text{NH}_3$ : $0$ ; $\text{Br}^-$ : $-1$ ; $\text{Cl}^-$ : $-1$ ; $\text{NO}_2^-$ : $-1$ ; total = $+2 - 3 = -1$ .
Write the complex anion.

The ligands are listed alphabetically by name, irrespective of charge: ammine (a), bromido (b), chlorido (c), nitrito-N (n). So:

[\text{Pt}(\text{NH}_3)(\text{Br})(\text{Cl})(\text{NO}_2)]^-

Counter ion. The name given is only for the complex anion itself (it ends in “-ate”, with no cation named) — no counter ion is specified, so the answer is the complex anion on its own, exactly as named.

Final formula (complex anion):

[\text{Pt}(\text{NH}_3)(\text{Br})(\text{Cl})(\text{NO}_2)]^-

Watch out

A common mistake is to write the N-bonded nitrite ligand as $\text{ONO}^-$ . Writing oxygen first denotes O-bonding (“nitrito-O”); since this name specifies “-N” bonding, the ligand formula is written $\text{NO}_2^-$ , nitrogen first.

(v) Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate

Identify the central metal and its oxidation state.

“platinum(IV)” means Pt is in the +4 state.
Identify the ligands.
- “dichlorido” = two $\text{Cl}^-$ (each $-1$ )
- “bis(ethane-1,2-diamine)” = two “en” molecules (neutral)
Calculate the charge on the complex cation.

Pt: $+4$ ; two $\text{Cl}^-$ : $-2$ ; two en: $0$ ; total = $+2$ .
Write the complex cation.

[\text{Pt}(\text{Cl})_2(\text{en})_2]^{2+}

Add the counter ion.

“nitrate” = $\text{NO}_3^-$ . To balance $+2$ , we need two nitrate ions.
Final formula:

[\text{Pt}(\text{Cl})_2(\text{en})_2](\text{NO}_3)_2

(vi) Iron(III) hexacyanidoferrate(II)

Identify the two metals.

This is a double salt or a coordination compound with two different metal ions. The name “Iron(III) hexacyanidoferrate(II)” means:
- The cation is iron(III), i.e., $\text{Fe}^{3+}$ .
- The anion is “hexacyanidoferrate(II)”, which is $[\text{Fe}(\text{CN})_6]^{4-}$ (since iron in the anion is in the +2 state, and six $\text{CN}^-$ ligands give a total charge of $+2 - 6 = -4$ ).
Balance the charges.

Cation: $\text{Fe}^{3+}$ (charge $+3$ ). Anion: $[\text{Fe}(\text{CN})_6]^{4-}$ (charge $-4$ ). To get a neutral compound, find the least common multiple of 3 and 4, which is 12. So we need 4 $\text{Fe}^{3+}$ (total $+12$ ) and 3 $[\text{Fe}(\text{CN})_6]^{4-}$ (total $-12$ ). That gives the formula $\text{Fe}_4[\text{Fe}(\text{CN})_6]_3$ .
Final formula:

\text{Fe}_4[\text{Fe}(\text{CN})_6]_3

Watch out

This is the classic “Prussian blue” formula. A common mistake is to think the iron in the anion is also Fe(III), but the name says “ferrate(II)”, so it’s Fe(II). Also, the ratio is 4:3, not 1:1.

✓Final answer

[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})_2]\text{Cl}_3

Werner Coordination Theory — Naming & Formula Writing

Method: Reverse Translation Method

Translate the IUPAC name back into its constituent parts: cation first, then anion; use oxidation states to determine charges and subscripts.

Step-by-Step Process

Identify the coordination sphere (the complex ion in brackets)
Count ligands using prefixes (di-, tri-, tetra-, etc.)
Determine the metal's oxidation state from the name suffix (-ate for anionic complexes) and overall charge balance
Write the formula — cation first, then anion; use square brackets for the coordination sphere

Solved Examples

(i) Tetraamminediaquacobalt(III) chloride

Cation: $\ce{[Co(NH3)4(H2O)2]^{3+}}$ $[Co (NH X_{3}) X_{4} (H X_{2} O) X_{2}] X^{3 +}$
- 4 ammine ( $\ce{NH3}$ ) + 2 aqua ( $\ce{H2O}$ ) ligands
- Cobalt(III) = +3 charge
Anion: 3 chloride ions ( $\ce{Cl-}$ )
Formula: $\boxed{\ce{[Co(NH3)4(H2O)2]Cl3}}$

(ii) Potassium tetracyanidonickelate(II)

Cation: $\ce{K+}$ (potassium)
Anion: $\ce{[Ni(CN)4]^{2-}}$ $[Ni (CN) X_{4}] X^{2 -}$
- 4 cyanido ( $\ce{CN-}$ ) ligands
- Nickel(II) = +2; each $\ce{CN-}$ = -1 → net charge = $+2 - 4 = -2$
Formula: $\boxed{\ce{K2[Ni(CN)4]}}$

(iii) Tris(ethane-1,2-diamine) chromium(III) chloride

Cation: $\ce{[Cr(en)3]^{3+}}$ $[Cr (en) X_{3}] X^{3 +}$ (en = $\ce{H2N-CH2-CH2-NH2}$ $H X_{2} N - CH X_{2} - CH X_{2} - NH X_{2}$ )
- 3 bidentate ethane-1,2-diamine ligands
- Chromium(III) = +3
Anion: 3 chloride ions
Formula: $\boxed{\ce{[Cr(en)3]Cl3}}$

(iv) Amminebromidochloridonitrito-N-platinate(II)

Cation: None named in the given name — it describes the complex anion only.
Anion: $\ce{[Pt(NH3)(Br)(Cl)(NO2)]^{-}}$ $[Pt (NH X_{3}) (Br) (Cl) (NO X_{2})] X^{-}$
- Ligands: ammine ( $\ce{NH3}$ ), bromido ( $\ce{Br-}$ ), chlorido ( $\ce{Cl-}$ ), nitrito-N ( $\ce{NO2-}$ )
- Platinum(II) = +2; total ligand charge = $0 + (-1) + (-1) + (-1) = -3$
- Net charge = $+2 - 3 = -1$
Counterion: None is named in the given compound name, so no cation should be assumed or added — the correct answer is the bare complex anion, exactly as named.
Formula: $\boxed{\ce{[Pt(NH3)(Br)(Cl)(NO2)]^{-}}}$

(v) Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate

Cation: $\ce{[PtCl2(en)2]^{2+}}$ $[PtCl X_{2} (en) X_{2}] X^{2 +}$
- 2 chlorido ( $\ce{Cl-}$ ) + 2 en ligands
- Platinum(IV) = +4; ligand charge = $2 \times (-1) = -2$ → net = $+4 - 2 = +2$
Anion: 2 nitrate ions ( $\ce{NO3-}$ )
Formula: $\boxed{\ce{[PtCl2(en)2](NO3)2}}$

(vi) Iron(III) hexacyanidoferrate(II)

Cation: $\ce{Fe^{3+}}$ (iron(III))
Anion: $\ce{[Fe(CN)6]^{4-}}$ $[Fe (CN) X_{6}] X^{4 -}$
- 6 cyanido ( $\ce{CN-}$ ) ligands
- Iron(II) = +2; ligand charge = $6 \times (-1) = -6$ → net = $+2 - 6 = -4$
Charge balance: $3+$ and $4-$ → need 4 $\ce{Fe^{3+}}$ and 3 $\ce{[Fe(CN)6]^{4-}}$ for neutrality
Formula: $\boxed{\ce{Fe4[Fe(CN)6]3}}$ (Prussian blue)

Key Exam Tips

Suffix	Meaning
-ate	Complex is anionic (e.g., ferrate, nickelate)
No suffix	Complex is cationic or neutral
Oxidation state	Always given in Roman numerals in parentheses

Remember: Werner's theory explains that primary valency (ionic bond) equals oxidation state, while secondary valency (coordination number) is fixed for a given metal.

Here is a breakdown of the common mistakes for each compound, with the correct formula and the "why" behind the rule.

General Mistake #1: Forgetting the Charge Balance (The Golden Rule)

The Mistake: Writing a formula that is electrically impossible. For example, writing a complex cation with a total charge of +3 and only one chloride ion ( $Cl^-$ ) outside the coordination sphere.

How to Avoid: Always, always calculate the net charge on the complex ion first. Then, add the correct number of counter-ions to make the total compound charge zero.

(i) Tetraamminediaquacobalt(III) chloride

Common Mistake: Writing the formula as $[Co(NH_3)_4(H_2O)_2]Cl$ or forgetting the oxidation state.

Why it's wrong:

Ligands: 4 neutral $NH_3$ + 2 neutral $H_2O$ = total ligand charge = 0.
Metal: Cobalt is in the +3 oxidation state ( $Co^{3+}$ ).
Complex Ion Charge: $+3 + 0 = +3$ . The complex is $[Co(NH_3)_4(H_2O)_2]^{3+}$ .
Counter-ion: "Chloride" means $Cl^-$ . To balance a +3 charge, you need three chloride ions.

Correct Formula: $\mathbf{[Co(NH_3)_4(H_2O)_2]Cl_3}$

How to Avoid: Write the complex ion with its charge first. Then ask: "How many chlorides do I need to cancel this charge?"

(ii) Potassium tetracyanidonickelate(II)

Common Mistake: Writing the formula as $K[Ni(CN)_4]$ — adding only one potassium without checking the charge — or writing the wrong metal symbol.

Why it's wrong:

Ligands: 4 cyanide ions ( $CN^-$ ). Total ligand charge = $4 \times (-1) = -4$ .
Metal: Nickel is in the +2 oxidation state ( $Ni^{2+}$ ).
Complex Ion Charge: $+2 + (-4) = -2$ . The complex is $[Ni(CN)_4]^{2-}$ .
Counter-ion: "Potassium" means $K^+$ . To balance a -2 charge, you need two potassium ions.

Correct Formula: $\mathbf{K_2[Ni(CN)_4]}$

How to Avoid: Remember that cyanide is anionic ( $CN^-$ ). Don't confuse it with neutral ligands like $NH_3$ or $H_2O$ .

(iii) Tris(ethane-1,2-diamine) chromium(III) chloride

Common Mistake: Writing the formula as $[Cr(en)_3]Cl_3$ (which is actually correct in shorthand) but forgetting the parentheses around the ligand in the full formula, or writing the wrong ligand formula.

Why it's wrong (in full form):

Ligand: Ethane-1,2-diamine is $H_2NCH_2CH_2NH_2$ . It is a bidentate ligand (two donor atoms). The prefix "tris" means three.
Complex Ion Charge: 3 neutral ligands + $Cr^{3+}$ = +3 charge.
Counter-ion: Need 3 chlorides.

Correct Formula: $\mathbf{[Cr(H_2NCH_2CH_2NH_2)_3]Cl_3}$ or $\mathbf{[Cr(en)_3]Cl_3}$ (acceptable shorthand).

How to Avoid: When writing the full formula, always put the entire ligand formula inside parentheses if it is polyatomic. For shorthand, use the standard abbreviation (en).

(iv) Amminebromidochloridonitrito-N-platinate(II)

Common Mistake: This is the trickiest one. The most common mistake is misplacing the "N" in the name or writing the wrong isomer.

Why it's wrong:

"Nitrito-N" means the nitrite ligand ( $NO_2^-$ ) is bonded through the Nitrogen atom. This is the nitro isomer ( $-NO_2$ ). If it were bonded through Oxygen, it would be "nitrito-O" or simply "nitrito" ( $-ONO$ ).
Metal: Platinum is in the +2 oxidation state ( $Pt^{2+}$ ).
Ligands: 1 neutral $NH_3$ + 1 $Br^-$ + 1 $Cl^-$ + 1 $NO_2^-$ (bonded through N).
Complex Ion Charge: $+2 + 0 + (-1) + (-1) + (-1) = -1$ . The complex is $[Pt(NH_3)BrCl(NO_2)]^-$ .
Counter-ion: The name ends in "-ate", meaning it is an anion. The counter-ion is implied (often potassium or hydrogen), but the question asks for the formula of the complex ion itself. The name "platinate(II)" tells you it's an anion.

Correct Formula: $\mathbf{[Pt(NH_3)BrCl(NO_2)]^-}$ (or as a salt, e.g., $K[Pt(NH_3)BrCl(NO_2)]$ )

How to Avoid: Pay close attention to the bonding atom in the name (e.g., "-N", "-O", "-S"). This determines the isomer. Also, remember that "-ate" ending means the complex is an anion.

(v) Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate

Common Mistake: Writing the formula as $[PtCl_2(en)_2]NO_3$ (forgetting the charge on the complex).

Why it's wrong:

Ligands: 2 $Cl^-$ + 2 neutral (en) = total ligand charge = -2.
Metal: Platinum is in the +4 oxidation state ( $Pt^{4+}$ ).
Complex Ion Charge: $+4 + (-2) = +2$ . The complex is $[PtCl_2(en)_2]^{2+}$ .
Counter-ion: "Nitrate" means $NO_3^-$ . To balance a +2 charge, you need two nitrate ions.

Correct Formula: $\mathbf{[PtCl_2(en)_2](NO_3)_2}$

How to Avoid: Never assume the counter-ion count is 1. Always calculate the complex charge first.

(vi) Iron(III) hexacyanidoferrate(II)

Common Mistake: This is the most famous "trap" question. The common mistake is to write the formula as $Fe_3[Fe(CN)_6]_2$ (iron(II) hexacyanidoferrate(III) — nominally Turnbull's blue, a different compound; Prussian blue is the correct answer $Fe_4[Fe(CN)_6]_3$ itself) or to get the oxidation states reversed.

Why it's wrong:

Cation: "Iron(III)" means $Fe^{3+}$ .
Anion: "hexacyanidoferrate(II)" means the complex is $[Fe(CN)_6]^{4-}$ (because Fe is +2, and 6 CN- = -6, so +2 -6 = -4).
Charge Balance: $Fe^{3+}$ $F e^{3 +}$ and $[Fe(CN)_6]^{4-}$ $[F e (C N)_{6}]^{4 -}$ . The lowest common multiple of 3 and 4 is 12.
- You need 4 $Fe^{3+}$ ions to get +12.
- You need 3 $[Fe(CN)_6]^{4-}$ ions to get -12.

Correct Formula: $\mathbf{Fe_4[Fe(CN)_6]_3}$

How to Avoid: This is a mixed-valence compound. The two iron atoms have different oxidation states. Do not cancel them out. Write the cation first, then the complex anion, and cross-multiply the charges.

Summary Table of Mistakes & Fixes

Compound	Common Mistake	Correct Formula	Key Fix
(i)	$[Co(NH_3)_4(H_2O)_2]Cl$	$[Co(NH_3)_4(H_2O)_2]Cl_3$	Balance the +3 charge with 3 $Cl^-$
(ii)	$K[Ni(CN)_4]$	$K_2[Ni(CN)_4]$	4 $CN^-$ give -4 charge; need 2 $K^+$
(iii)	$[Cr(en)_3]Cl$	$[Cr(en)_3]Cl_3$	3 neutral ligands; $Cr^{3+}$ needs 3 $Cl^-$
(iv)	$[Pt(NH_3)BrCl(ONO)]^-$	$[Pt(NH_3)BrCl(NO_2)]^-$	"Nitrito-N" means N-bonded ( $-NO_2$ )
(v)	$[PtCl_2(en)_2]NO_3$	$[PtCl_2(en)_2](NO_3)_2$	$Pt^{4+}$ + 2 $Cl^-$ = +2 charge; need 2 $NO_3^-$
(vi)	$Fe_3[Fe(CN)_6]_2$	$Fe_4[Fe(CN)_6]_3$	Cross-multiply charges: $Fe^{3+}$ and $[Fe(CN)_6]^{4-}$

Final Pro-Tip: For every coordination compound, write down the charge of the metal, the charge of each ligand, and the total charge of the complex. Then, add the counter-ions. This three-step process will eliminate 90% of your errors.

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

Showing the 12 most recent of 66 on this concept.

CBSE 2026Set 56/3/11 markMCQ
$Q. According to Werner's theory, the primary valencies of the central metal atom : (A) are satisfied by neutral molecules or negative ions. (B) are equal to its coordination number. (C) are satisfied by negative ions. (D) are non-ionisable.$
›Reveal solutionSolution
Werner’s theory distinguishes primary valencies (ionisable, satisfied only by negative ions, equal to oxidation state) from secondary valencies (non-ionisable, satisfied by neutral molecules or negative ions, equal to coordination number). The correct answer is (C).

Werner’s Coordination Theory was a breakthrough because it explained why compounds like $\ce{CoCl3·6NH3}$ (which we now write as $\ce{[Co(NH3)6]Cl3}$ ) behave so differently from simple salts. The key insight: a metal ion has two kinds of bonding capacity — not just one.

Primary valency (now called oxidation state) is the metal’s ionic bonding capacity. It is satisfied only by negative ions (anions), because it arises from the metal’s need to neutralise its positive charge. These bonds are ionisable — they break apart in solution, giving the familiar conductivity and precipitation tests.

Secondary valency (now called coordination number) is the metal’s ability to bind ligands directly. It is satisfied by neutral molecules (like $\ce{NH3}$ ) or negative ions (like $\ce{Cl-}$ ). These bonds are non-ionisable — they stay attached to the metal even in solution.

Now let’s apply this to the options.

Option (A): “Primary valencies are satisfied by neutral molecules or negative ions.”

This is false. Primary valencies are satisfied only by negative ions. Neutral molecules satisfy secondary valencies. For example, in $\ce{[Co(NH3)6]Cl3}$ , the three $\ce{Cl-}$ ions satisfy the primary valency (Co³⁺ needs three negative charges), while six $\ce{NH3}$ molecules satisfy the secondary valency.

Option (B): “Primary valencies are equal to its coordination number.”

This is false. Primary valency equals the oxidation state of the metal. Coordination number equals the secondary valency. They are often different. In $\ce{[Co(NH3)6]Cl3}$ , primary valency = 3, coordination number = 6.

Option (C): “Primary valencies are satisfied by negative ions.”

This is true. Werner explicitly stated that primary valencies are satisfied only by anions. This is the defining property.

Option (D): “Primary valencies are non-ionisable.”

This is false. Primary valencies are ionisable — they dissociate in solution. Secondary valencies are non-ionisable. In $\ce{[Co(NH3)6]Cl3}$ , the three $\ce{Cl-}$ ions (primary) are free in solution and can be precipitated by $\ce{Ag+}$ , while the six $\ce{NH3}$ (secondary) remain bound to cobalt.

Watch out

A common mistake is to confuse primary valency with coordination number. Remember: primary = oxidation state (charge neutralisation), secondary = coordination number (ligand binding). They are not the same.

Tip

A quick way to identify primary valency in a complex: count the number of ionisable anions outside the coordination sphere. That number equals the primary valency (oxidation state) of the metal.

✓Final answer
$The correct option is (C) — primary valencies are satisfied by negative ions.$
CBSE 2026Set 56/1/11 markMCQ
$[Co(NH_3)_5(NO_2)]^{2+}$
›Reveal solutionSolution
In Werner’s coordination theory, secondary valency equals the coordination number — the number of ligand donor atoms directly bonded to the metal. For $[Co(NH_3)_5(NO_2)]^{2+}$ , there are 5 NH₃ molecules (each donating one N) and one NO₂⁻ ligand (also donating one N), giving a total of 6 donor atoms. So the secondary valency is 6.

Werner’s coordination theory is the foundation here. He proposed that metal ions have two kinds of valency: primary (ionisable, corresponding to oxidation state) and secondary (non-ionisable, corresponding to coordination number). The secondary valency is simply the number of ligand donor atoms directly attached to the central metal ion — in other words, the coordination number.

For the complex $[Co(NH_3)_5(NO_2)]^{2+}$ , we need to count how many atoms are actually bonded to cobalt.

Identify the ligands and their denticity.

NH₃ (ammine) is a monodentate ligand — it binds through the lone pair on nitrogen. Each NH₃ contributes one donor atom.

NO₂⁻ (nitrito or nitro, depending on binding mode) is also monodentate when it binds through nitrogen (the more common nitro form) or through oxygen. In either case, it uses one donor atom per ligand. The problem doesn’t specify the linkage isomer, but that doesn’t change the count — it’s still one donor atom.

Count the total number of donor atoms.

Five NH₃ ligands: $5 \times 1 = 5$ donor atoms.

One NO₂⁻ ligand: $1 \times 1 = 1$ donor atom.

Total = $5 + 1 = 6$ donor atoms.

Relate this to secondary valency.

Werner defined secondary valency as the number of groups directly coordinated to the metal — exactly the coordination number. So here, secondary valency = 6.

Watch out

A common mistake is to confuse secondary valency with the charge on the complex or the number of ligands (5 ligands, not 6). But remember: secondary valency counts donor atoms, not ligand molecules. Even though there are only 6 ligands (5 NH₃ + 1 NO₂⁻), each is monodentate, so the count matches the number of ligands here — but that’s a coincidence. If a ligand were bidentate (like ethylenediamine), the number of ligands would be smaller than the secondary valency.

Tip

A quick check: the coordination number of cobalt in most octahedral complexes is 6. Since NH₃ and NO₂⁻ are both strong-field ligands and Co(III) is d⁶, an octahedral geometry is expected. So even without counting, you could guess 6 — but always verify by counting donor atoms.

✓Final answer
$[Co(NH_3)_5(NO_2)]^{2+}$
CBSE 2026Set DZ1 markMCQ
$K_3[Co(C_2O_4)_3]$
(a) 3
(b) 4
(c) 5
(d) 6
›Reveal solutionSolution

Oxalate is a bidentate ligand; three of them occupy six coordination sites, so the coordination number of Co is 6 — option (d).

The coordination number is the number of ligand donor atoms directly bonded to the central metal ion (not the number of ligand molecules).

In $K_3[Co(C_2O_4)_3]$ the ligand is the oxalate ion $C_2O_4^{2-}$ . Each oxalate binds the metal through two oxygen atoms, so it is a bidentate (chelating) ligand.

$\text{Donor atoms} = 3 \text{ oxalates} \times 2 = 6$

Hence Co is surrounded by six O-donor atoms in an octahedral geometry.

✓Final answer
$(d) 6 .$
CBSE 2026Set ANNUAL1 markMCQ
$Q. What is the coordination number of Cobalt in [Co(en)3]Cl3 ?$
(a) 6
(b) 5
(c) 4
(d) 3
›Reveal solutionSolution

Each 'en' ligand is bidentate, so [Co(en)3]Cl3 has coordination number 6.

In $[Co(en)_3]Cl_3$ , ethylenediamine (en, $H_2NCH_2CH_2NH_2$ ) acts as a bidentate ligand — it has two nitrogen donor atoms, each with a lone pair, that can simultaneously coordinate to the central metal ion, forming a five-membered chelate ring with cobalt.

Since there are three 'en' ligands attached to the cobalt centre, and each contributes 2 donor atoms, the total number of coordinate bonds (and hence the coordination number) is:

$3 \times 2 = 6$

The three $Cl^-$ ions are ionizable counter-ions lying outside the coordination sphere (as confirmed by the complex giving 3 moles of $AgCl$ precipitate per mole with $AgNO_3$ ) and do not count towards the coordination number.

✓Final answer
$[Co(en)_3]Cl_3$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Ambidentate ligand is$
(a) H2O
(b) NH3
(c) NO2-
(d) Cl-
›Reveal solutionSolution
An ambidentate ligand has two different donor atoms but binds through only one of them at a time, depending on conditions.

H2O, NH3 and Cl- each have only one type of donor atom, so they are simple monodentate ligands, not ambidentate.

NO2- has two possible donor atoms - N (giving a nitro complex) or O (giving a nitrito complex) - so it is the classic ambidentate ligand, along with SCN-/NCS- and CN-/NC-.

✓Final answer
$(c) NO2-.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Coordination number of Pt in [Pt(NH3)2 Cl (NO2)] complex is$
(a) 3
(b) 4
(c) 5
(d) 6
›Reveal solutionSolution

Coordination number = the total number of ligand donor atoms directly bonded to the central metal atom/ion.

In [Pt(NH3)2Cl(NO2)], the ligands bonded to Pt are: 2 x NH3 + 1 x Cl- + 1 x NO2- = 4 donor atoms.

✓Final answer
$(b) 4.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Identify the homoleptic complex among the following compounds:$
(a) [Co(NH3)4Cl2]+
(b) [Ni(CN)4]2-
(c) [Cr(NH3)2Cl2(en)]+
(d) Both (A) and (C)
›Reveal solutionSolution
A homoleptic complex is one in which the metal is bonded to only a single kind of donor ligand; a heteroleptic complex has more than one kind.

Checking each option:

[Co(NH3)4Cl2]+: contains two different ligands, NH3 and Cl- (heteroleptic).

[Ni(CN)4]2-: contains only CN- ligands, all of the same type (homoleptic).

[Cr(NH3)2Cl2(en)]+: contains NH3, Cl-, and ethylenediamine (en) - three different ligand types (heteroleptic).

Only [Ni(CN)4]2- has a single ligand type throughout.

✓Final answer
$(b) [Ni(CN)4]2-.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following is a bidentate ligand?$
(a) CN-
(b) NH3
(c) C2O4^2-
(d) CO3^2-
›Reveal solutionSolution
A bidentate ligand has two donor atoms that simultaneously coordinate to the same central metal atom, forming a ring (chelate).

CN- has one donor atom (C), so it is monodentate.

NH3 has one donor atom (N), so it is monodentate.

C2O4^2- (oxalate ion) has two oxygen atoms, one from each carboxylate group, that can each donate a lone pair to the metal simultaneously - this makes it bidentate, forming a five-membered chelate ring.

CO3^2- typically binds through one or two oxygens but is generally classified as monodentate or, in rarer bridging modes, differently - it is not the standard bidentate example here.

✓Final answer
$(c) C2O4^2-.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The co-ordination number of Cr in K3[Cr(C2O4)3] is: (as printed; the paper's option values are +3/+4/+2/+6)$
(a) +3
(b) +4
(c) +2
(d) +6
›Reveal solutionSolution

As printed, the options ( $+3,+4,+2,+6$ ) are oxidation-state values, so this question is really testing the oxidation number of Cr, which is $+3$ . (For completeness: the true coordination number — the count of donor atoms bonded to Cr — is 6, not among the given options, since oxalate is bidentate and there are three oxalate ligands.)

Oxidation number of Cr: In $K_3[Cr(C_2O_4)_3]$ , potassium contributes $+1$ each (three K $^+$ ions), and each oxalate ion $C_2O_4^{2-}$ carries a charge of $-2$ . Let the oxidation number of Cr be $x$ :

$3(+1) + x + 3(-2) = 0$

$3 + x - 6 = 0 \implies x = +3$

Coordination number (for clarity): Oxalate ( $C_2O_4^{2-}$ ) is a bidentate ligand — it binds through two oxygen donor atoms. With three oxalate ligands, Cr is bonded to $3 \times 2 = 6$ donor atoms, so the coordination number is 6 (octahedral geometry), not any of the printed options.

✓Final answer
$+3$
CBSE 2026Set ANNUAL1 mark
$[Fe(CN)_6]^{4-}$
›Reveal solutionSolution

Balancing the ligand charges against the overall complex-ion charge gives the oxidation state of iron.

Working it out

Let the oxidation number of Fe be $x$ . Cyanide is an anionic ligand, $CN^-$ , contributing $-1$ each; there are 6 of them:

$x + 6(-1) = -4\ (\text{overall charge of the complex ion})$

$x - 6 = -4 \implies x = +2$

So in $[Fe(CN)_6]^{4-}$ (the ferrocyanide ion), iron is present as $Fe^{2+}$ .

✓Final answer
$= +2$
CBSE 2026Set ANNUAL1 mark
$Q. What is the coordination number of Fe in [Fe(EDTA)]⁻?$
›Reveal solutionSolution

EDTA is a hexadentate chelating ligand (2 N + 4 O donor atoms), so a single EDTA fills all six coordination sites — coordination number of Fe = 6.

Concept. The coordination number of a metal ion is the number of ligand donor atoms directly bonded to it (the number of coordinate bonds), not the number of ligand molecules.

Reasoning. EDTA $^{4-}$ (ethylenediaminetetraacetate) is a hexadentate ligand: it donates through its two amine nitrogen atoms and four carboxylate oxygen atoms. Even though only one EDTA molecule is present, it forms six coordinate bonds to the central iron, wrapping around it as a chelate.

$\text{donor atoms} = 2\,\text{N} + 4\,\text{O} = 6$

✓Final answer
$[\text{Fe(EDTA)}]^{-}$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion (A): Glycinate ion H₂NCH₂COO⁻ is a chelating ligand. Reason (R): It can link through either nitrogen or anionic oxygen.$
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is correct.
›Reveal solutionSolution

A (glycinate is chelating) is true, but the reason given describes ambidentate behaviour (binding through one OR the other), not chelation (binding through both at once), so R is false — option (C).

Assertion (A): The glycinate ion $H_2N-CH_2-COO^-$ is a chelating ligand. This is true: it is a bidentate ligand that binds a metal through the nitrogen of the $-NH_2$ group and the anionic oxygen of the $-COO^-$ group at the same time, forming a stable five-membered chelate ring.

Reason (R): "It can link through either nitrogen or anionic oxygen." This describes an ambidentate ligand — one that attaches through one donor atom or the other, but only one at a time. That is not what makes a ligand chelating. A chelating ligand must use two (or more) donor atoms simultaneously to grip the same metal. So R, as worded, is a false explanation of A.

✓Final answer
$(C) A is true but R is false — glycinate chelates by using both N and O together, not "either/or".$

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Q.Write the formulas for the following coordination compounds:

(i) Tetraamminediaquacobalt(III) chloride

(ii) Potassium tetracyanidonickelate(II)

(iii) Tris(ethane-1,2-diamine) chromium(III) chloride

(iv) Amminebromidochloridonitrito-N-platinate(II)

(v) Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate

(vi) Iron(III) hexacyanidoferrate(II)

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