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B1 ILTS — 08-30-2026 · Chemistry · Error notes

Coordination Compounds

Chemistry · 11 questions · scored 126 / 180

Every question flagged on this paper, worked through in full: what was given, what was asked, the concept, the formula, the steps written out, an animated diagram, and the shortcut that gets there faster. The weak areas for the chapter are set out at the end.

11Questions flagged
5Attempted, wrong
6Left blank
126Marks scored

Contents

Q46Q47Q52Q54Q61Q64Q69Q73Q80Q81Q90

Dark = attempted and missed · amber = left blank

Q 46Left blankWriting the formula from the name

The formula of the compound tris(ethane-1,2-diamine)cobalt(III) sulphate is

Correct answer (left blank)[Co(en)₃]₂(SO₄)₃
1  Given
  • tris(ethane-1,2-diamine): three en ligands, each neutral and bidentate.
  • cobalt(III): oxidation state +3.
  • sulphate: SO₄²⁻, the counter ion.
2  Asked

The complete formula.

3  Concept

Build the cation first, find its charge, then balance against the anion. Since en is neutral, the whole complex ion carries the metal's charge: [Co(en)₃]³⁺. Balancing 3+ against 2− needs the least common multiple, giving 2 cations to 3 anions.

4  Formula or rule
charge on complex = metal oxidation state + sum of ligand charges
5  Baby steps
  1. en is neutral, so the complex ion is [Co(en)₃]³⁺.
  2. The counter ion is SO₄²⁻.
  3. Balance: 2 × (3+) = 6+ and 3 × (2−) = 6−.
  4. [Co(en)₃]₂(SO₄)₃.
6  Diagram
Inside the square bracket, and outside itMLLLLLLSECONDARY valency — fixed, directional, non-ionisableX⁻X⁻X⁻PRIMARY valencyionisable, free in solutiononly what is OUTSIDE the bracket ionises — that sets the conductance,the precipitate with AgNO₃, and the primary/secondary count
7  Shortcuts
Cross-multiply the charges, exactly as for any ionic formula: 3+ and 2− give subscripts 2 and 3. Writing the charge on the complex ion explicitly before balancing is what prevents the common slip of [Co(en)₃](SO₄)₃​.

Ligand charges worth memorising: neutral — en, NH₃, H₂O, CO. Negative — Cl⁻, CN⁻, NO₂⁻, OH⁻ (all 1−), C₂O₄²⁻ (2−), EDTA (4−).
Q 47Left blankPrimary and secondary valency

Select the correct statements about the complex salt [Pt(NH₃)₄Cl₂]Cl₂.
I. All the chlorides are bonded by primary valency only
II. All the chlorides are bonded by secondary valency
III. All the NH₃ are bonded by secondary valency
IV. Half of the chlorides are bonded with secondary and primary valency

Correct answer (left blank)III and IV
1  Given
  • [Pt(NH₃)₄Cl₂]Cl₂.
  • Inside the bracket: 4 NH₃ and 2 Cl.
  • Outside the bracket: 2 Cl.
  • Total chlorides = 4.
2  Asked

Which statements are correct.

3  Concept

Werner's two valencies map exactly onto the square bracket. Secondary valency = inside the bracket, directional, non-ionisable. Primary valency = outside the bracket, ionisable.

Here the chlorides are split evenly: two inside (secondary) and two outside (primary). All four ammonias are inside.

4  Formula or rule
inside the bracket = SECONDARY  ·  outside = PRIMARY (ionisable)
5  Baby steps
  1. I. Two of the four chlorides sit inside the bracket, so “primary only” is false.
  2. II. Two chlorides are outside, so “all secondary” is also false.
  3. III. All four NH₃ are inside the bracket — all secondary. Correct.
  4. IV. Two of four chlorides are secondary and two are primary — exactly half each. Correct.
  5. Answer: III and IV.
6  Diagram
Inside the square bracket, and outside itMLLLLLLSECONDARY valency — fixed, directional, non-ionisableX⁻X⁻X⁻PRIMARY valencyionisable, free in solutiononly what is OUTSIDE the bracket ionises — that sets the conductance,the precipitate with AgNO₃, and the primary/secondary count
7  Shortcuts
Count what is inside the bracket and what is outside, and write both numbers down. Here it is 4 NH₃ + 2 Cl inside, 2 Cl outside. Every statement can then be checked in a second.

The experimental consequence: only the outside chlorides precipitate with AgNO₃. This compound would give 2 mol of AgCl per mole, not 4 — a very common follow-up question.
Q 52Attempted · wrongConductance and the number of ions

Out of the following complex compounds, which of the compound will be having the minimum conductance in solution?

Her answer[Co(NH₃)₅Cl]Cl₂
Correct answer[Co(NH₃)₃Cl₃]
1  Given
  • [Co(NH₃)₆]Cl₃ → 4 ions.
  • [Co(NH₃)₅Cl]Cl₂ → 3 ions.
  • [Co(NH₃)₃Cl₃] → no ions at all.
  • [Co(NH₃)₄Cl₂]Cl → 2 ions.
2  Asked

Which has the lowest conductance.

3  Concept

Conductance depends on the number of ions released in solution, which is entirely determined by what lies outside the square bracket.

A complex with nothing outside the bracket is a neutral molecule — a non-electrolyte. It produces no ions and its conductance is essentially zero, which is the minimum possible.

4  Formula or rule
conductance ∝ number of ions = 1 (complex ion) + number of counter ions
5  Baby steps
  1. Count the ions for each, by looking only outside the bracket:
  2. [Co(NH₃)₆]Cl₃ → one complex cation + 3 Cl⁻ = 4 ions.
  3. [Co(NH₃)₅Cl]Cl₂ → 1 + 2 = 3 ions.
  4. [Co(NH₃)₄Cl₂]Cl → 1 + 1 = 2 ions.
  5. [Co(NH₃)₃Cl₃] → nothing outside the bracket = 0 ions. Minimum conductance.
6  Diagram
Inside the square bracket, and outside itMLLLLLLSECONDARY valency — fixed, directional, non-ionisableX⁻X⁻X⁻PRIMARY valencyionisable, free in solutiononly what is OUTSIDE the bracket ionises — that sets the conductance,the precipitate with AgNO₃, and the primary/secondary count
7  Shortcuts
Scan for the option with nothing written outside the bracket. If one exists, it is a neutral non-electrolyte and it is automatically the answer to any “minimum conductance” question — no counting needed.

The same reasoning inverted answers “maximum conductance”: pick the one with the most counter ions outside.
Q 54Attempted · wrongEDTA donor sites

The donor sites of (EDTA)⁴⁻ are?

Her answerThree N atoms and three O atoms
Correct answerTwo N atoms and four O atoms
1  Given
  • EDTA⁴⁻ = ethylenediaminetetraacetate.
  • It is a well-known hexadentate ligand — six donor sites.
2  Asked

Which atoms act as the donors.

3  Concept

The name itself gives the structure. Ethylenediamine supplies two nitrogen atoms; tetraacetate supplies four carboxylate groups, each donating one oxygen — so four oxygens.

Two plus four is six, which matches EDTA being hexadentate.

4  Formula or rule
EDTA⁴⁻: hexadentate = 2 N + 4 O
5  Baby steps
  1. Break the name apart: ethylenediamine + tetraacetate.
  2. Diamine means two amine nitrogens → 2 N donors.
  3. Tetraacetate means four acetate arms, each contributing one carboxylate oxygen → 4 O donors.
  4. Total = 6, confirming hexadentate.
  5. Two N atoms and four O atoms.
6  Diagram
Inside the square bracket, and outside itMLLLLLLSECONDARY valency — fixed, directional, non-ionisableX⁻X⁻X⁻PRIMARY valencyionisable, free in solutiononly what is OUTSIDE the bracket ionises — that sets the conductance,the precipitate with AgNO₃, and the primary/secondary count
7  Shortcuts
Read the prefixes in the name and add them. diamine gives 2, tetraacetate gives 4, and 2 + 4 = 6 = hexadentate. The name is self-checking.

Denticity worth memorising: en = 2, oxalate = 2, dien = 3, EDTA = 6. And the 4− charge on EDTA comes from the four deprotonated carboxylic acids.
Q 61Attempted · wrongWhich complex shows no geometrical isomerism

Which of the following will not show geometrical isomerism?

Her answer[Pt(NH₃)₂Cl₂]
Correct answer[Co(NH₃)₅NO₂]Cl₂
1  Given
  • [Cr(NH₃)₄Cl₂]Cl — octahedral MA₄B₂.
  • [Co(en)₂Cl₂]Cl — octahedral M(AA)₂B₂.
  • [Co(NH₃)₅NO₂]Cl₂ — octahedral MA₅B.
  • [Pt(NH₃)₂Cl₂] — square planar MA₂B₂.
2  Asked

Which one has no geometrical isomers.

3  Concept

Geometrical isomerism needs at least two positions that are genuinely different for a given ligand. In an octahedron with five identical ligands and one different (MA₅B), every vertex is equivalent by symmetry — wherever you put B, you get the same molecule. So MA₅B has no isomers at all.

4  Formula or rule
MA₅B → no isomers  ·  MA₄B₂, MA₃B₃, M(AA)₂B₂ → cis / trans
5  Baby steps
  1. MA₄B₂ (Cr complex): the two B ligands can be adjacent or opposite → cis and trans. Shows isomerism.
  2. M(AA)₂B₂ (Co–en complex): likewise cis and trans, and the cis form is even optically active. Shows isomerism.
  3. Square planar MA₂B₂ (Pt complex): the classic cisplatin / transplatin pair. Shows isomerism.
  4. MA₅B (Co(NH₃)₅NO₂): all six octahedral positions are equivalent, so there is only one structure. No geometrical isomerism.
6  Diagram
Which complexes can show geometrical isomerismMA₄B₂cis and transYESM(AA)₂B₂cis and transYESMA₂B₂ square planarcis and transYESMA₅Bonly ONE arrangementNOwith FIVE identical ligands and one different, every position isequivalent — so MA₅B has no isomers at all
7  Shortcuts
Reduce each option to its MA₃B₃-style formula and ignore the chemistry. Then apply one short list of octahedral cases that DO show geometrical isomerism: MA₄B₂, MA₃B₃, MA₂B₂C₂, M(AA)₂B₂.

Anything of the form MA₆ or MA₅B has none. That single rule answers this question without thinking about cobalt or platinum at all.
Q 64Left blankIsomer count and IUPAC naming

Statement-I: Number of possible isomers for the complex with the formula [CoCl₂(en)₂]⁺ will be two.
Statement-II: IUPAC name of [CoCl₂(en)₂]Cl is dichlorido bis(ethane-1,2-diamine)cobalt(III) chloride

Correct answer (left blank)Statement-I is false but Statement-II is true
1  Given
  • [CoCl₂(en)₂]⁺ is octahedral, of type M(AA)₂B₂.
  • en is a bidentate neutral ligand.
2  Asked

The truth of each statement.

3  Concept

Statement-I undercounts. The complex has cis and trans geometrical isomers — but the cis form is chiral and exists as a pair of optical isomers. So the total is three, not two.

Statement-II is a correct IUPAC name: ligands alphabetically (chlorido before ethane-1,2-diamine), bis because the ligand name already contains a numerical prefix, and cobalt(III) from the charge balance.

4  Formula or rule
[CoCl₂(en)₂]⁺: trans (achiral) + cis (a chiral pair) = 3 isomers
5  Baby steps
  1. Geometrical isomers: the two Cl can be cis or trans — that is 2.
  2. The trans form has a plane of symmetry, so it is optically inactive.
  3. The cis form has no plane of symmetry — it is chiral and gives two enantiomers.
  4. Total = 1 (trans) + 2 (cis pair) = 3. So Statement-I is false.
  5. Statement-II follows the naming rules correctly and is true.
6  Diagram
Which complexes can show geometrical isomerismMA₄B₂cis and transYESM(AA)₂B₂cis and transYESMA₂B₂ square planarcis and transYESMA₅Bonly ONE arrangementNOwith FIVE identical ligands and one different, every position isequivalent — so MA₅B has no isomers at all
7  Shortcuts
Always ask whether the cis form is chiral before quoting an isomer count. For M(AA)₂B₂ the answer is yes, and the count jumps from 2 to 3.

Naming reminders: use bis, tris, tetrakis when the ligand name already contains di/tri (as in ethane-1,2-diamine); order ligands alphabetically ignoring those prefixes; and anionic ligands end in -ido (chlorido, not chloro, in current IUPAC usage).
Q 69Left blankWerner's theory

Consider the following statements related to Werner's theory:
(a) Ligands are linked to the central metal ion through ionic bonds.
(b) Secondary valencies possess directional properties.
(c) Secondary valencies are non-ionisable.
Which of the above statements are correct according to Werner's theory?

Correct answer (left blank)b and c are correct
1  Given
  • Three statements about Werner's coordination theory.
2  Asked

Which are correct.

3  Concept

Werner's central insight is that secondary valencies are directional — they point to fixed positions in space, which is what gives complexes their geometry and makes isomerism possible. They are also non-ionisable, since the ligands stay attached in solution.

What is wrong in (a) is the word ionic: ligands are attached by coordinate (dative) bonds, in which the ligand donates a lone pair.

4  Formula or rule
primary valency: ionisable, non-directional  ·  secondary valency: non-ionisable, DIRECTIONAL
5  Baby steps
  1. (a) Ligands bond through coordinate bonds, not ionic ones. Incorrect.
  2. (b) Secondary valencies are directed towards fixed positions — that is why octahedral and square planar geometries exist. Correct.
  3. (c) Secondary valencies do not ionise; only primary valencies do. Correct.
  4. Answer: b and c.
6  Diagram
Inside the square bracket, and outside itMLLLLLLSECONDARY valency — fixed, directional, non-ionisableX⁻X⁻X⁻PRIMARY valencyionisable, free in solutiononly what is OUTSIDE the bracket ionises — that sets the conductance,the precipitate with AgNO₃, and the primary/secondary count
7  Shortcuts
Directional is the word to attach to secondary valency — it is Werner's single most important claim, because it explains geometry and isomerism.

And be alert to the word ionic anywhere in a coordination-compound statement: the metal–ligand bond is coordinate, and calling it ionic is the standard planted error.
Q 73Left blankFormula from the name, again

The formula of the complex tris(ethane-1,2-diammine)cobalt(III) sulphate is

Correct answer (left blank)[Co(en)₃]₂(SO₄)₃
1  Given
  • Three neutral bidentate en ligands.
  • Cobalt in the +3 oxidation state.
  • Sulphate, SO₄²⁻.
2  Asked

The complete formula.

3  Concept

Identical in structure to Q46 on this same paper. The complex ion is [Co(en)₃]³⁺ because en is neutral, and balancing 3+ against 2− requires two cations to three anions.

4  Formula or rule
[Co(en)₃]³⁺ + SO₄²⁻ → cross-multiply the charges
5  Baby steps
  1. en is neutral, so the cation carries cobalt's charge: [Co(en)₃]³⁺.
  2. Sulphate is 2−.
  3. Cross-multiply: 2 cations (total 6+) and 3 anions (total 6−).
  4. [Co(en)₃]₂(SO₄)₃.
6  Diagram
Inside the square bracket, and outside itMLLLLLLSECONDARY valency — fixed, directional, non-ionisableX⁻X⁻X⁻PRIMARY valencyionisable, free in solutiononly what is OUTSIDE the bracket ionises — that sets the conductance,the precipitate with AgNO₃, and the primary/secondary count
7  Shortcuts
This is the same question as Q46 on this paper, worth 8 marks between them and both left blank. One rule covers both: write the complex ion's charge, then cross-multiply with the counter ion exactly as for NaCl or Al₂(SO₄)₃.

When a paper repeats a question, the second appearance is free marks — provided the first one was worked out rather than skipped.
Q 80Attempted · wrongLinkage isomerism and ambidentate ligands

Assertion (A): Linkage isomerism arises in coordination compounds containing ambidentate ligand.
Reason (R): Ambidentate ligand contains more than one donor atom.

Her answerBoth true, but (R) is not the correct explanation
Correct answerBoth true, and (R) is the correct explanation
1  Given
  • Assertion about linkage isomerism.
  • Reason defining an ambidentate ligand.
2  Asked

The truth of each statement and whether R explains A.

3  Concept

Both statements are true, and the second is precisely why the first holds. An ambidentate ligand has more than one kind of donor atom but binds through only one at a time — NO₂⁻ can attach through N (nitro) or through O (nitrito), and SCN⁻ through S or N.

Having that choice of donor atom is what creates linkage isomers. Remove the choice and the isomerism disappears.

4  Formula or rule
ambidentate: two possible donor atoms, one used at a time → LINKAGE isomerism
5  Baby steps
  1. Assertion. Linkage isomerism is defined as arising from ambidentate ligands. True.
  2. Reason. An ambidentate ligand does have more than one donor atom available. True.
  3. Does R explain A? Apply the “because” test: linkage isomerism arises because the ligand offers more than one donor atom — the sentence reads correctly and is the actual mechanism.
  4. So both are true and R is the correct explanation.
6  Diagram
Which complexes can show geometrical isomerismMA₄B₂cis and transYESM(AA)₂B₂cis and transYESMA₂B₂ square planarcis and transYESMA₅Bonly ONE arrangementNOwith FIVE identical ligands and one different, every position isequivalent — so MA₅B has no isomers at all
7  Shortcuts
Use the “because” test on every assertion-reason question. Read the assertion, say “because”, read the reason. If the sentence states the real mechanism, R explains A.

Do not confuse ambidentate with polydentate. A polydentate ligand (en, EDTA) uses several donors at once and gives chelation. An ambidentate ligand has a choice but uses one, and gives linkage isomerism.
Q 81Attempted · wrongHybridisation and unpaired electrons

The type of hybridization and the magnetic property of [MnCl₆]³⁻ are,

Her answersp³d², paramagnetic with two unpaired electrons
Correct answersp³d², paramagnetic with four unpaired electrons
1  Given
  • [MnCl₆]³⁻, octahedral.
  • Charge balance: x + 6(−1) = −3, so Mn is +3.
  • Mn = [Ar]3d⁵4s², so Mn³⁺ = 3d⁴.
  • Cl⁻ is a weak field ligand.
2  Asked

The hybridisation and the number of unpaired electrons.

3  Concept

Two decisions, and the ligand settles both. Cl⁻ is a weak field ligand, so it cannot force the d electrons to pair. They stay spread out following Hund's rule, which leaves the inner 3d orbitals full and forces the complex to use outer 4d orbitals — giving sp³d², high spin.

With d⁴ unpaired, that is four unpaired electrons.

4  Formula or rule
weak field → high spin → sp³d² (outer orbital)
strong field → low spin → d²sp³ (inner orbital)
5  Baby steps
  1. Oxidation state: x − 6 = −3, so Mn³⁺.
  2. Electronic configuration of Mn³⁺ = 3d⁴.
  3. Cl⁻ is weak field, so no pairing occurs: the four electrons occupy four separate d orbitals → 4 unpaired.
  4. With the 3d orbitals occupied, the complex must use 4d for bonding → sp³d².
  5. sp³d², paramagnetic with four unpaired electrons.
6  Diagram
Inner orbital or outer orbital?STRONG field ligandCN⁻, CO, NO₂⁻, enelectrons PAIR upd²sp³ (inner)low spinWEAK field ligandCl⁻, F⁻, Br⁻, H₂Oelectrons stay UNPAIREDsp³d² (outer)high spin[MnCl₆]³⁻: Mn³⁺ is d⁴, and Cl⁻ is a WEAK field ligandso nothing pairs → sp³d², FOUR unpaired electronsd²sp³ and low spin always travel together
7  Shortcuts
The hybridisation and the spin state always travel together, so getting one right gives you the other:

sp³d² ⇔ outer orbital ⇔ high spin ⇔ maximum unpaired
d²sp³ ⇔ inner orbital ⇔ low spin ⇔ minimum unpaired

The chosen option pairs sp³d² with only two unpaired electrons, which is internally contradictory — sp³d² means nothing paired up.

Spectrochemical series (weak → strong): I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO.
Q 90Left blankSpin-only magnetic moment

The correct decreasing order of spin only magnetic moment values (BM) of Cu⁺, Cu²⁺, Cr²⁺ and Cr³⁺ ions is

Correct answer (left blank)Cr²⁺ > Cr³⁺ > Cu²⁺ > Cu⁺
1  Given
  • Cu = [Ar]3d¹⁰4s¹; Cr = [Ar]3d⁵4s¹.
  • Cu⁺ = 3d¹⁰; Cu²⁺ = 3d⁹.
  • Cr²⁺ = 3d⁴; Cr³⁺ = 3d³.
2  Asked

The decreasing order of spin-only magnetic moments.

3  Concept

μ = √(n(n+2)) BM, where n is the number of unpaired electrons. Since μ rises steadily with n, ranking by magnetic moment is exactly the same as ranking by n — no square roots need ever be evaluated.

4  Formula or rule
μ = √(n(n+2)) BM  ·  n = 4, 3, 2, 1, 0 → 4.90, 3.87, 2.83, 1.73, 0
5  Baby steps
  1. Cr²⁺ = d⁴ → 4 unpaired → 4.90 BM.
  2. Cr³⁺ = d³ → 3 unpaired → 3.87 BM.
  3. Cu²⁺ = d⁹ → 1 unpaired → 1.73 BM.
  4. Cu⁺ = d¹⁰ → 0 unpaired → 0 BM (diamagnetic).
  5. Order: Cr²⁺ > Cr³⁺ > Cu²⁺ > Cu⁺.
6  Diagram
Unpaired electrons set the magnetic momentCr²⁺d⁴4 unpaired4.90 BMCr³⁺d³3 unpaired3.87 BMCu²⁺d⁹1 unpaired1.73 BMCu⁺d¹⁰0 unpaired0 BMμ = √(n(n+2)) BM — ranking by μ is ranking by nCr²⁺ > Cr³⁺ > Cu²⁺ > Cu⁺
7  Shortcuts
Count unpaired electrons and stop there. The order of μ is the order of n, so the square-root formula is only needed if an actual value is asked for.

Two configurations worth knowing on sight: d¹⁰ is always diamagnetic (μ = 0) and d⁵ gives the maximum for a first-row ion (5 unpaired, 5.92 BM).

Removing electrons: for transition metals the 4s electrons leave first, then the 3d. So Cu⁺ is d¹⁰ (not d⁹4s¹) — that order is what makes Cu⁺ diamagnetic.

Weak areas — Coordination Compounds

Six of the eleven were left blank and five were attempted and missed — but the errors here are unusually concentrated. Coordination Compounds supplied ten of the eleven questions, and almost all of them come back to one idea: what is inside the square bracket and what is outside it.

QWhat it testedSettled by
46, 73formula from the namecharge on the complex ion, then cross-multiply
47primary vs secondary valencycount inside and outside the bracket
52minimum conductancenothing outside the bracket → a non-electrolyte
69Werner's theorysecondary valencies are directional and non-ionisable

Q46 and Q73 are the same question, asked twice on one paper — tris(ethane-1,2-diamine)cobalt(III) sulphate — and both were left blank. That is 8 marks lost to a single formula, and the second appearance would have been free had the first been worked out.

Two internal contradictions worth spotting. In Q81 the chosen option pairs sp³d² with only two unpaired electrons, but sp³d² means nothing paired up — the option contradicts itself. In Q80 the reason genuinely is the mechanism behind the assertion, so the “not the explanation” option cannot be right.

The one table to build: hybridisation, spin state and orbital type always travel together. d²sp³ = inner = low spin = minimum unpaired; sp³d² = outer = high spin = maximum unpaired. Get one and you have all four.