Coordination Compounds — NEET & JEE PYQ Vault

How to use this. Attempt the question first with the solution closed, then open it. If you got it right but your reasoning differed from the Baby Steps, that gap is worth noting — it usually means a lucky guess. Log every wrong answer by type: didn't know the concept, misread the question, or ran out of time. Exam labels indicate the pattern the question follows; several appear in more than one year and in both papers.

Werner's Theory & Ionisation

06 Q

[ 01 ]^A NEET / AIPMTWerner

1 mol each of CoCl₃·6NH₃, CoCl₃·5NH₃, CoCl₃·4NH₃ and CoCl₃·3NH₃ are treated with excess AgNO₃. The moles of AgCl precipitated are respectively:

(a)3, 2, 1, 0
(b)0, 1, 2, 3
(c)3, 3, 3, 3
(d)1, 2, 3, 4

Answer — (a) 3, 2, 1, 0

Given

Four cobalt(III) chloride–ammonia complexes, 1 mol each, excess AgNO₃.

Asked

Moles of AgCl precipitated from each.

Concept

Only ionisable Cl⁻ (outside the square bracket, primary valence) reacts with Ag⁺. Cl⁻ inside the coordination sphere is non-ionisable and gives no precipitate. Secondary valence of Co³⁺ is fixed at 6.

Formula

moles AgCl = moles of Cl⁻ outside [ ]

Baby steps

Co³⁺ needs 6 groups inside the bracket. Fill with NH₃ first, then Cl⁻.
6NH₃ → [Co(NH₃)₆]Cl₃ → 3 free Cl⁻ → 3 AgCl
5NH₃ → [Co(NH₃)₅Cl]Cl₂ → 2 free Cl⁻ → 2 AgCl
4NH₃ → [Co(NH₃)₄Cl₂]Cl → 1 free Cl⁻ → 1 AgCl
3NH₃ → [Co(NH₃)₃Cl₃] → 0 free Cl⁻ → 0 AgCl

ShortcutAgCl = (number of NH₃) − 3 for CoCl₃·xNH₃. So x = 6,5,4,3 gives 3,2,1,0 instantly.

[ 02 ]^A NEETWerner

How many ions are produced in solution from the complex CoCl₃·6NH₃?

(a)6
(b)4
(c)3
(d)2

Answer — (b) 4

Given

CoCl₃·6NH₃, i.e. [Co(NH₃)₆]Cl₃.

Asked

Total number of ions on dissociation.

Concept

The coordination sphere behaves as one single ion; counter ions dissociate separately.

Formula

[Co(NH₃)₆]Cl₃ → [Co(NH₃)₆]³⁺ + 3Cl⁻

Baby steps

Write the Werner formula: all 6 NH₃ inside, all 3 Cl⁻ outside.
Complex cation = 1 ion.
Chloride counter ions = 3 ions.
Total = 1 + 3 = 4 ions (1:3 electrolyte).

ShortcutIons = 1 + (charge on complex ion). Charge +3 → 4 ions.

[ 03 ]^A NEETWerner

Which of the following will not give a precipitate with AgNO₃ solution?

(a)[Co(NH₃)₆]Cl₃
(b)[Co(NH₃)₅Cl]Cl₂
(c)[Co(NH₃)₄Cl₂]Cl
(d)[Co(NH₃)₃Cl₃]

Answer — (d) [Co(NH₃)₃Cl₃]

Given

Four cobalt–ammine chlorides.

Asked

The one with no ionisable chloride.

Concept

Precipitation requires free Cl⁻. A neutral complex has zero counter ions.

Formula

No counter ion ⇒ no AgCl

Baby steps

Check what lies outside the square bracket in each option.
Options A, B, C have 3, 2 and 1 Cl⁻ outside respectively.
[Co(NH₃)₃Cl₃] has nothing outside — it is a neutral, non-electrolyte complex.
Hence option D gives no precipitate.

ShortcutNeutral complex (no charge shown, nothing outside brackets) → zero conductivity, zero precipitate.

[ 04 ]^A JEE MainWerner

The secondary valence and the oxidation number of the central metal in PtCl₄·2HCl are respectively:

(a)6 and +4
(b)4 and +2
(c)6 and +2
(d)4 and +4

Answer — (a) 6 and +4

Given

PtCl₄·2HCl; 0 mol AgCl on adding AgNO₃.

Asked

Secondary valence (= coordination number) and oxidation number of Pt.

Concept

Secondary valence = number of groups directly attached = coordination number. Primary valence = oxidation number, satisfied by ionisable anions.

Formula

x + (charge of ligands) = charge on complex ion

Baby steps

No AgCl means all 6 Cl⁻ sit inside the sphere → formula is H₂[PtCl₆].
Number of donor atoms bonded to Pt = 6 → secondary valence 6.
Charge on complex ion = −2. So x + 6(−1) = −2.
x = +4 → Pt(IV).

ShortcutCount total ligands in the bracket = secondary valence. Then balance charges for oxidation number.

[ 05 ]^A NEETWerner

Which statement is not a postulate of Werner's theory?

(a)Primary valences are ionisable and satisfied by negative ions
(b)Secondary valences are non-ionisable and equal the coordination number
(c)Secondary valences have directional character giving definite geometry
(d)Primary valence is fixed for a metal and secondary valence is variable

Answer — (d) Primary valence is fixed for a metal and secondary valence is variable

Given

Statements about Werner's coordination theory.

Asked

The incorrect postulate.

Concept

Werner: primary valence = oxidation state (variable for a metal), secondary valence = coordination number (fixed for a given metal ion).

Formula

—

Baby steps

Primary valence is ionisable, satisfied by anions → A is correct.
Secondary valence is non-ionisable, equals CN → B is correct.
Secondary valences point in fixed directions giving octahedral/tetrahedral shapes → C is correct.
Werner said secondary valence is fixed, not variable. D is the wrong statement.

ShortcutRemember the pair: primary = ionisable & variable; secondary = non-ionisable & fixed + directional.

[ 06 ]^A NEETWerner

FeSO₄ mixed with (NH₄)₂SO₄ in 1:1 molar ratio gives the test of Fe²⁺, but CuSO₄ mixed with aqueous NH₃ in 1:4 molar ratio does not give the test of Cu²⁺. This is because:

(a)Mohr's salt is a double salt; the copper species is a complex
(b)Both are complexes but Fe²⁺ is more reactive
(c)Both are double salts
(d)Cu²⁺ is reduced to Cu⁺

Answer — (a) Mohr's salt is a double salt; the copper species is a complex

Given

Mohr's salt FeSO₄·(NH₄)₂SO₄·6H₂O and [Cu(NH₃)₄]SO₄.

Asked

Reason for the difference in ionic tests.

Concept

Double salts dissociate completely into their simple ions in water; complexes retain the coordination sphere and release almost no free metal ion.

Formula

Double salt → free ions; Complex ion → low dissociation (high K_stability)

Baby steps

Mohr's salt exists only in the solid state; in water it gives Fe²⁺, NH₄⁺ and SO₄²⁻ freely → Fe²⁺ test positive.
CuSO₄ + 4NH₃ → [Cu(NH₃)₄]SO₄, a true complex.
The tetraammine ion does not dissociate appreciably, so free Cu²⁺ is negligible.
Hence no Cu²⁺ test.

ShortcutIf the compound exists only in solid state → double salt. If it survives in solution → complex.

Nomenclature

08 Q

[ 07 ]^B NEETNomenclature

The IUPAC name of K₃[Fe(CN)₆] is:

(a)Potassium hexacyanidoferrate(III)
(b)Potassium hexacyanidoiron(III)
(c)Potassium hexacyanoferrate(II)
(d)Tripotassium hexacyanidoferrate(III)

Answer — (a) Potassium hexacyanidoferrate(III)

Given

K₃[Fe(CN)₆].

Asked

IUPAC name.

Concept

Cation named first. Anionic complex → metal takes the -ate suffix and, for Fe, the Latin stem ferrate. Never state the number of counter ions.

Formula

x + 6(−1) = −3

Baby steps

Cation is K⁺ → write ‘potassium’ first.
Ligand CN⁻ = cyanido; six of them = hexacyanido.
Complex is an anion → ferrate (Latin name + ate).
Oxidation state: x − 6 = −3 ⇒ x = +3 → (III).
Name = potassium hexacyanidoferrate(III).

ShortcutLatin-name metals in anions: Fe→ferrate, Cu→cuprate, Ag→argentate, Pb→plumbate, Sn→stannate, Au→aurate.

[ 08 ]^B NEETNomenclature

The IUPAC name of [Pt(NH₃)₂Cl(NH₂CH₃)]Cl is:

(a)Diamminechlorido(methanamine)platinum(II) chloride
(b)Diamminechloridomethylaminoplatinum(IV) chloride
(c)Bisamminechlorido(methanamine)platinum(II) chloride
(d)Chloridodiammine(methanamine)platinum(II) chloride

Answer — (a) Diamminechlorido(methanamine)platinum(II) chloride

Given

[Pt(NH₃)₂Cl(NH₂CH₃)]Cl.

Asked

IUPAC name.

Concept

Ligands are named alphabetically (ignoring the multiplying prefix), then the metal with its oxidation state in Roman numerals, then the counter anion.

Formula

x + 0 + 0 + (−1) = +1

Baby steps

Alphabetical order of ligand names: ammine, chlorido, methanamine.
Prefix di- for two NH₃ → diammine.
Charge on complex = +1 (one Cl⁻ outside). x − 1 = +1 ⇒ x = +2.
Name = diamminechlorido(methanamine)platinum(II) chloride.

ShortcutAlphabetise by the ligand's own name, not by di/tri. ‘a’ of ammine beats ‘c’ of chlorido.

[ 09 ]^B JEE MainNomenclature

The formula of the compound potassium trioxalatoaluminate(III) is:

(a)K₃[Al(C₂O₄)₃]
(b)K[Al(C₂O₄)₃]
(c)K₃[Al(C₂O₄)₂]
(d)Al[K(C₂O₄)₃]

Answer — (a) K₃[Al(C₂O₄)₃]

Given

Name: potassium trioxalatoaluminate(III).

Asked

Chemical formula.

Concept

Write metal first inside the bracket, then ligands; balance the total charge with counter ions.

Formula

Charge on complex = (oxidation no.) + Σ(ligand charges)

Baby steps

Central metal Al in +3; three oxalate ligands, each C₂O₄²⁻.
Charge on complex = (+3) + 3(−2) = −3.
Three K⁺ needed to neutralise.
Formula = K₃[Al(C₂O₄)₃].

Shortcut‘-ate’ ending always means the complex is the anion, so K goes outside and in front.

[ 10 ]^B NEETNomenclature

The formula of hexaamminecobalt(III) sulphate is:

(a)[Co(NH₃)₆]₂(SO₄)₃
(b)[Co(NH₃)₆]SO₄
(c)[Co(NH₃)₆]₃(SO₄)₂
(d)Co₂(NH₃)₆(SO₄)₃

Answer — (a) [Co(NH₃)₆]₂(SO₄)₃

Given

Complex cation [Co(NH₃)₆]³⁺ with SO₄²⁻ counter ion.

Asked

Formula of the salt.

Concept

NH₃ is neutral, so the charge on the complex ion equals the oxidation state of the metal. Then cross-balance charges.

Formula

Total positive charge = total negative charge

Baby steps

Co(III) with six neutral NH₃ → [Co(NH₃)₆]³⁺.
Counter ion SO₄²⁻.
LCM of 3 and 2 is 6 → two cations and three sulphates.
[Co(NH₃)₆]₂(SO₄)₃.

ShortcutCriss-cross the charges: 3+ and 2− → subscripts 2 and 3.

[ 11 ]^B NEETNomenclature

The IUPAC name of the complex Fe₄[Fe(CN)₆]₃ is:

(a)Iron(III) hexacyanidoferrate(II)
(b)Iron(II) hexacyanidoferrate(III)
(c)Tetrairon trihexacyanidoferrate(II)
(d)Iron(III) hexacyanoiron(II)

Answer — (a) Iron(III) hexacyanidoferrate(II)

Given

Fe₄[Fe(CN)₆]₃ (Prussian blue).

Asked

IUPAC name.

Concept

Cation is named first with its oxidation state; the anionic complex takes -ate. Both metals here are iron in different oxidation states.

Formula

4a + 3b = 0 with b = charge of complex ion

Baby steps

Inside the complex: x + 6(−1) = charge. For hexacyanidoferrate(II), x = +2 → charge −4.
Three such anions give −12, so four cations must give +12 → each Fe is +3.
Cation = iron(III); anion = hexacyanidoferrate(II).
Name = iron(III) hexacyanidoferrate(II).

ShortcutWhen the same metal appears twice, the outer one is named as the element, the inner one as -ate.

[ 12 ]^B NEETNomenclature

The IUPAC name of [Ni(CO)₄] is:

(a)Tetracarbonylnickel(0)
(b)Tetracarbonylnickelate(0)
(c)Nickel tetracarbonyl(II)
(d)Tetracarbonylnickel(IV)

Answer — (a) Tetracarbonylnickel(0)

Given

[Ni(CO)₄], a neutral homoleptic carbonyl.

Asked

IUPAC name.

Concept

CO is a neutral ligand named carbonyl. A neutral complex molecule is named like a complex cation (no -ate).

Formula

x + 4(0) = 0 ⇒ x = 0

Baby steps

CO carries no charge, complex carries no charge.
Therefore oxidation state of Ni = 0, written as (0).
Four CO → tetracarbonyl.
Name = tetracarbonylnickel(0).

ShortcutSpecial neutral ligand names to memorise: H₂O = aqua, NH₃ = ammine, CO = carbonyl, NO = nitrosyl.

[ 13 ]^B JEE MainNomenclature

Which of the following is the correct formula for pentaamminenitrito-O-cobalt(III) chloride?

(a)[Co(NH₃)₅(ONO)]Cl₂
(b)[Co(NH₃)₅(NO₂)]Cl₂
(c)[Co(NH₃)₅(ONO)]Cl
(d)[Co(NH₃)₅(NO₃)]Cl₂

Answer — (a) [Co(NH₃)₅(ONO)]Cl₂

Given

Name specifies nitrito-O linkage on Co(III) with five ammine ligands.

Asked

Formula.

Concept

NO₂⁻ is ambidentate. Bonding through O is written –ONO (nitrito-O, red form); bonding through N is written –NO₂ (nitrito-N, yellow form).

Formula

x + 5(0) + (−1) = charge on complex

Baby steps

Co is +3, five neutral NH₃, one NO₂⁻.
Charge on complex = +3 − 1 = +2 → two Cl⁻ outside.
O-bonded form is written ONO.
[Co(NH₃)₅(ONO)]Cl₂.

ShortcutWrite the donor atom first: O-bonded = ONO, N-bonded = NO₂; S-bonded = SCN, N-bonded = NCS.

[ 14 ]^B NEETNomenclature

The IUPAC name of [Co(en)₃]Cl₃ is:

(a)Tris(ethane-1,2-diamine)cobalt(III) chloride
(b)Triethylenediaminecobalt(III) chloride
(c)Tris(ethane-1,2-diamine)cobaltate(III) chloride
(d)Tri(ethane-1,2-diamine)cobalt(III) trichloride

Answer — (a) Tris(ethane-1,2-diamine)cobalt(III) chloride

Given

[Co(en)₃]Cl₃, en = ethane-1,2-diamine.

Asked

IUPAC name.

Concept

When the ligand name already contains a numerical prefix (di-, tri-), use bis, tris, tetrakis and enclose the ligand in parentheses.

Formula

x + 3(0) = +3

Baby steps

en is neutral and didentate; three of them → use tris, not tri.
Ligand name goes in brackets: tris(ethane-1,2-diamine).
Complex is a cation, so metal is named cobalt, oxidation state +3.
Tris(ethane-1,2-diamine)cobalt(III) chloride.

ShortcutLigand name already has di/tri inside it → switch to bis/tris/tetrakis. Applies to en, bipy, PPh₃.

Oxidation State, CN & EAN

07 Q

[ 15 ]^C NEETOx. state

The oxidation number of cobalt in [Co(H₂O)(CN)(en)₂]²⁺ is:

(a)+2
(b)+3
(c)+1
(d)+4

Answer — (b) +3

Given

[Co(H₂O)(CN)(en)₂]²⁺.

Asked

Oxidation number of Co.

Concept

Sum of the metal's oxidation number and all ligand charges equals the overall charge. H₂O and en are neutral, CN⁻ is −1.

Formula

x + Σ(ligand charges) = net charge

Baby steps

H₂O = 0, en = 0 each, CN = −1.
x + 0 + (−1) + 2(0) = +2.
x − 1 = 2 ⇒ x = +3.

ShortcutOnly count charged ligands. Neutral ones (H₂O, NH₃, en, CO, py, PPh₃) can be skipped.

[ 16 ]^C NEETCoordination no.

The coordination number of the central metal ion in [Cr(C₂O₄)₃]³⁻ and [Co(en)₂Cl₂]⁺ are respectively:

(a)3 and 4
(b)6 and 6
(c)6 and 4
(d)3 and 6

Answer — (b) 6 and 6

Given

Two complexes with didentate ligands (oxalate, en).

Asked

Coordination numbers.

Concept

Coordination number counts donor atoms (sigma bonds), not ligand molecules. Pi bonds are not counted.

Formula

CN = Σ(number of ligands × denticity)

Baby steps

Oxalate C₂O₄²⁻ is didentate → 3 × 2 = 6.
en is didentate → 2 × 2 = 4; plus 2 Cl⁻ unidentate = 2.
Total for second complex = 4 + 2 = 6.

ShortcutDidentate ligands to memorise: en, ox, gly, bipy, phen, C₂O₄²⁻. EDTA⁴⁻ is hexadentate.

[ 17 ]^C JEE MainEAN

The effective atomic number (EAN) of the metal in K₄[Fe(CN)₆] is: (Z of Fe = 26)

(a)36
(b)34
(c)38
(d)26

Answer — (a) 36

Given

K₄[Fe(CN)₆], Z_Fe = 26.

Asked

EAN of iron.

Concept

EAN = atomic number − oxidation state + 2 × (coordination number). It counts the total electrons around the metal after ligand donation; 36 (Kr) signals extra stability.

Formula

EAN = Z − (oxidation state) + 2 × CN

Baby steps

Oxidation state of Fe: x + 6(−1) = −4 ⇒ x = +2.
Coordination number = 6.
EAN = 26 − 2 + 2(6) = 24 + 12.
EAN = 36, the krypton configuration.

ShortcutEAN = Z − ox. state + 2CN. If it lands on 36, 54 or 86, the complex is unusually stable.

[ 18 ]^C NEETEAN

For which of the following is the EAN of the central metal not equal to 36?

(a)[Ni(CO)₄] (Z = 28)
(b)[Fe(CN)₆]⁴⁻ (Z = 26)
(c)[Cr(CO)₆] (Z = 24)
(d)[Cu(NH₃)₄]²⁺ (Z = 29)

Answer — (d) [Cu(NH₃)₄]²⁺ (Z = 29)

Given

Four complexes with atomic numbers supplied.

Asked

The complex whose EAN ≠ 36.

Concept

Apply EAN = Z − oxidation state + 2CN to each.

Formula

EAN = Z − ox + 2CN

Baby steps

Ni(CO)₄: 28 − 0 + 8 = 36.
[Fe(CN)₆]⁴⁻: 26 − 2 + 12 = 36.
[Cr(CO)₆]: 24 − 0 + 12 = 36.
[Cu(NH₃)₄]²⁺: 29 − 2 + 8 = 35 ≠ 36.

ShortcutCu(II) ammine is the classic exception in EAN questions — d⁹ can never reach the noble gas count.

[ 19 ]^C NEETOx. state

The oxidation states of the metal in [Fe(CO)₅], [Ni(CN)₄]²⁻ and [Cr(NH₃)₃Cl₃] are respectively:

(a)0, +2, +3
(b)+2, +2, +3
(c)0, 0, +3
(d)0, +2, +6

Answer — (a) 0, +2, +3

Given

Three coordination entities.

Asked

Oxidation state of the central metal in each.

Concept

Neutral ligands contribute zero; anionic ligands contribute their own charge. Balance to the net charge.

Formula

x + Σ(ligand charge) = net charge

Baby steps

[Fe(CO)₅]: CO neutral, complex neutral ⇒ x = 0.
[Ni(CN)₄]²⁻: x − 4 = −2 ⇒ x = +2.
[Cr(NH₃)₃Cl₃]: x + 0 − 3 = 0 ⇒ x = +3.

ShortcutAny homoleptic neutral carbonyl (Ni(CO)₄, Fe(CO)₅, Cr(CO)₆) always has the metal in zero oxidation state.

[ 20 ]^C JEE MainCoordination no.

The number of donor atoms in EDTA⁴⁻ that bind to a metal ion is:

(a)4
(b)5
(c)6
(d)2

Answer — (c) 6

Given

Ethylenediaminetetraacetate ion, EDTA⁴⁻.

Asked

Denticity.

Concept

Denticity is the number of ligating groups a single ligand uses to bind one metal ion.

Formula

—

Baby steps

EDTA⁴⁻ has two nitrogen atoms in the ethylenediamine backbone.
It also has four carboxylate oxygen donors.
Total = 2 N + 4 O = 6 donor atoms → hexadentate.
It therefore forms a very stable chelate, used for water hardness and lead poisoning.

ShortcutEDTA = 2N + 4O = 6. It saturates an octahedral metal single-handedly.

[ 21 ]^C NEETOx. state

In which of the following complexes is the central metal in the +2 oxidation state?

(a)K₃[Fe(CN)₆]
(b)[Cr(NH₃)₆]Cl₃
(c)[Ni(NH₃)₆]Cl₂
(d)K₃[Co(C₂O₄)₃]

Answer — (c) [Ni(NH₃)₆]Cl₂

Given

Four coordination compounds.

Asked

The one with metal in +2.

Concept

Set metal charge x, add ligand charges, equate to the charge on the complex ion deduced from the counter ions.

Formula

x + Σ(ligand charge) = charge on complex ion

Baby steps

K₃[Fe(CN)₆]: x − 6 = −3 ⇒ +3.
[Cr(NH₃)₆]Cl₃: x = +3.
[Ni(NH₃)₆]Cl₂: x = +2.
K₃[Co(ox)₃]: x − 6 = −3 ⇒ +3.

ShortcutWith all-neutral ligands, the oxidation state equals the number of counter anions' total charge.

Isomerism

14 Q

[ 22 ]^D NEETGeometrical

The number of geometrical isomers possible for [Co(NH₃)₃Cl₃] is:

(a)2
(b)3
(c)4
(d)0

Answer — (a) 2

Given

Octahedral complex of the type [Ma₃b₃].

Asked

Number of geometrical isomers.

Concept

[Ma₃b₃] octahedral complexes show facial (fac) and meridional (mer) isomerism. Fac: the three identical ligands occupy one triangular face. Mer: they lie around a meridian.

Formula

[Ma₃b₃] → 2 geometrical isomers (fac, mer)

Baby steps

Identify the type: three NH₃ and three Cl⁻ → Ma₃b₃.
Place the three Cl⁻ on adjacent corners of one face → fac isomer.
Place them around the meridian (all in one plane through the metal) → mer isomer.
Neither is optically active (both have a plane of symmetry). Answer = 2.

ShortcutMemorise the octahedral counts: Ma₄b₂ → 2 (cis/trans); Ma₃b₃ → 2 (fac/mer); Ma₂b₂c₂ → 6 stereoisomers.

[ 23 ]^D NEETGeometrical

The number of geometrical isomers of [Cr(C₂O₄)₃]³⁻ is:

(a)0
(b)2
(c)3
(d)4

Answer — (a) 0

Given

Tris-chelate octahedral complex with three identical didentate oxalate ligands.

Asked

Number of geometrical isomers.

Concept

[M(AA)₃] with three identical symmetrical didentate ligands has only one geometrical arrangement, but it is chiral — it shows optical isomerism (d and l), not geometrical.

Formula

[M(AA)₃] → 0 geometrical, 2 optical

Baby steps

All three ligands are identical and symmetrical, so no cis/trans distinction is possible.
Geometrical isomers = 0.
However the propeller-shaped ion has no plane of symmetry.
So it exists as a pair of non-superimposable mirror images (2 optical isomers).

ShortcutTris-chelates like [Co(en)₃]³⁺ and [Cr(ox)₃]³⁻: always 0 geometrical, always 2 optical.

[ 24 ]^D NEET / JEE MainOptical

The total number of stereoisomers of [CoCl₂(en)₂]⁺ is:

(a)2
(b)3
(c)4
(d)6

Answer — (b) 3

Given

Octahedral [MX₂(AA)₂] type complex.

Asked

Total stereoisomers (geometrical + optical).

Concept

[MX₂(AA)₂] gives cis and trans geometrical isomers. The trans form has a plane of symmetry (achiral); the cis form is chiral and splits into d and l.

Formula

Total = 1 (trans) + 2 (cis d, l) = 3

Baby steps

Draw trans: the two Cl⁻ are 180° apart → has a plane of symmetry → optically inactive → 1 isomer.
Draw cis: the two Cl⁻ are adjacent (90°) → no plane of symmetry → chiral.
Cis therefore exists as d and l → 2 isomers.
Total = 1 + 2 = 3 stereoisomers.

ShortcutFor [MX₂(AA)₂]: only the cis form is optically active. Same result for [PtCl₂(en)₂]²⁺.

[ 25 ]^D NEETGeometrical

How many geometrical isomers are possible for the square planar complex [Pt(NH₃)(Br)(Cl)(py)]?

(a)2
(b)3
(c)4
(d)6

Answer — (b) 3

Given

Square planar MABCD type complex.

Asked

Number of geometrical isomers.

Concept

In a square planar MABCD complex, fixing one ligand and cycling which ligand sits trans to it gives three distinct arrangements. Square planar complexes have a plane of symmetry (the molecular plane), so none are optically active.

Formula

Square planar MABCD → 3 geometrical isomers, 0 optical

Baby steps

Fix NH₃ at one corner.
Put Br trans to it → isomer 1.
Put Cl trans to it → isomer 2.
Put py trans to it → isomer 3. Total = 3, none optically active.

ShortcutNumber of square planar MABCD isomers = 3 (one for each ligand placed trans to a chosen reference).

[ 26 ]^D NEETGeometrical

Which of the following complexes does not show geometrical isomerism?

(a)[Pt(NH₃)₂Cl₂] (square planar)
(b)[Co(NH₃)₄Cl₂]⁺
(c)[Ni(NH₃)₂Cl₂] (tetrahedral)
(d)[CoCl₂(en)₂]⁺

Answer — (c) [Ni(NH₃)₂Cl₂] (tetrahedral)

Given

Four heteroleptic complexes of different geometry.

Asked

The one with no geometrical isomerism.

Concept

In a tetrahedral complex all four positions are mutually adjacent — every ligand is at 109.5° to every other. There is no cis/trans distinction.

Formula

Tetrahedral ⇒ no geometrical isomerism

Baby steps

Square planar MA₂B₂ → cis and trans exist.
Octahedral MA₄B₂ → cis and trans exist.
Octahedral [MX₂(AA)₂] → cis and trans exist.
Tetrahedral MA₂B₂ → all positions equivalent → no geometrical isomerism.

ShortcutTetrahedral = never geometrical isomerism. It can show optical isomerism only with four different ligands (rare).

[ 27 ]^D NEETLinkage

Which of the following ligands can give rise to linkage isomerism?

(a)NH₃
(b)en
(c)NO₂⁻
(d)Cl⁻

Answer — (c) NO₂⁻

Given

Four ligands.

Asked

The one causing linkage isomerism.

Concept

Linkage isomerism needs an ambidentate ligand — one having two different donor atoms, only one of which coordinates at a time.

Formula

Ambidentate ligands: NO₂⁻, SCN⁻, CN⁻

Baby steps

NH₃ has only one donor (N) → unidentate, no choice.
en has two N donors but both bind simultaneously → chelating, not ambidentate.
Cl⁻ has one donor atom type.
NO₂⁻ can bind through N (–NO₂, nitrito-N, yellow) or O (–ONO, nitrito-O, red) → linkage isomerism.

ShortcutAmbidentate short list: NO₂⁻ (N or O), SCN⁻ (S or N), CN⁻ (C or N). Two donors but only one binds.

[ 28 ]^D NEETIonisation

[Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br exhibit which type of isomerism?

(a)Linkage
(b)Coordination
(c)Ionisation
(d)Solvate

Answer — (c) Ionisation

Given

Two salts with the same molecular formula but interchanged ligand and counter ion.

Asked

Type of isomerism.

Concept

Ionisation isomerism arises when the counter ion is itself a potential ligand and swaps places with a ligand inside the sphere. The two isomers give different ions in solution.

Formula

Test: BaCl₂ → BaSO₄; AgNO₃ → AgBr

Baby steps

First salt releases free SO₄²⁻ → gives white BaSO₄ with Ba²⁺, no AgBr.
Second salt releases free Br⁻ → gives pale yellow AgBr with Ag⁺, no BaSO₄.
Different ions in solution from the same formula → ionisation isomerism.

ShortcutLigand and counter ion swap → ionisation. If the swapping species is water, call it solvate (hydrate) isomerism.

[ 29 ]^D NEETCoordination

[Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆] are examples of:

(a)Ionisation isomers
(b)Coordination isomers
(c)Linkage isomers
(d)Geometrical isomers

Answer — (b) Coordination isomers

Given

Two salts in which both cation and anion are complex ions.

Asked

Type of isomerism.

Concept

Coordination isomerism arises from the interchange of ligands between the cationic and anionic complexes of two different metal ions.

Formula

—

Baby steps

In the first, NH₃ is on Co³⁺ and CN⁻ is on Cr³⁺.
In the second, the ligand sets have swapped metals.
Both the cation and anion are complex → requirement for coordination isomerism.
Answer = coordination isomerism.

ShortcutBoth ions inside brackets → coordination isomerism. One ion inside, one bare → ionisation isomerism.

[ 30 ]^D NEETSolvate

[Cr(H₂O)₆]Cl₃ (violet) and [Cr(H₂O)₅Cl]Cl₂·H₂O (grey-green) are:

(a)Optical isomers
(b)Linkage isomers
(c)Hydrate (solvate) isomers
(d)Coordination isomers

Answer — (c) Hydrate (solvate) isomers

Given

Two chromium(III) chloride hydrates of the same composition, different colours.

Asked

Type of isomerism.

Concept

Solvate isomerism (hydrate isomerism when the solvent is water) differs in whether a solvent molecule is coordinated to the metal or merely held in the crystal lattice.

Formula

Moles AgCl on AgNO₃ = number of Cl⁻ outside

Baby steps

Violet form has all six H₂O coordinated and 3 free Cl⁻ → 3 mol AgCl.
Grey-green form has one Cl⁻ inside and one lattice H₂O → only 2 mol AgCl.
Same formula, different distribution of water → hydrate/solvate isomerism.

ShortcutThe number of AgCl moles distinguishes hydrate isomers experimentally — a favourite follow-up question.

[ 31 ]^D NEETOptical

Which of the following is optically active?

(a)trans-[CrCl₂(ox)₂]³⁻
(b)cis-[CrCl₂(ox)₂]³⁻
(c)trans-[Co(NH₃)₄Cl₂]⁺
(d)[Ni(CO)₄]

Answer — (b) cis-[CrCl₂(ox)₂]³⁻

Given

Various octahedral and tetrahedral complexes.

Asked

The chiral (optically active) species.

Concept

A complex is optically active if it has no plane and no centre of symmetry, so that it is non-superimposable on its mirror image.

Formula

Chiral ⇒ no plane of symmetry

Baby steps

Trans forms place identical ligands opposite each other → a plane of symmetry always exists → achiral.
In cis-[CrCl₂(ox)₂]³⁻ the two chelate rings twist like a propeller → no symmetry plane.
[Ni(CO)₄] is tetrahedral with four identical ligands → achiral.
Answer = cis-[CrCl₂(ox)₂]³⁻.

ShortcutIn exams, trans is almost never optically active. Look for cis + chelate ring.

[ 32 ]^D JEE MainOptical

The number of optical isomers of [Co(en)₃]³⁺ is:

(a)0
(b)2
(c)3
(d)4

Answer — (b) 2

Given

Tris(ethane-1,2-diamine)cobalt(III) ion.

Asked

Number of optical isomers.

Concept

Three didentate chelate rings around an octahedral centre create a three-bladed propeller with no symmetry plane → a pair of enantiomers (d and l, or Δ and Λ).

Formula

[M(AA)₃] → 2 enantiomers

Baby steps

Draw the octahedron with three en rings spanning cis positions.
Check for a plane of symmetry — there is none.
The mirror image cannot be superimposed by rotation.
Hence 2 optical isomers (dextro and laevo).

ShortcutAny [M(AA)₃] tris-chelate: 2 optical isomers, no geometrical isomers, no meso form.

[ 33 ]^D NEETIsomerism

The type of isomerism shown by [Co(NH₃)₅(NO₂)]Cl₂ is:

(a)Linkage and ionisation
(b)Only optical
(c)Only geometrical
(d)Coordination and solvate

Answer — (a) Linkage and ionisation

Given

[Co(NH₃)₅(NO₂)]Cl₂.

Asked

Types of isomerism possible.

Concept

NO₂⁻ is ambidentate → linkage isomerism. Cl⁻ outside can exchange with NO₂⁻ inside → ionisation isomerism.

Formula

—

Baby steps

Ambidentate NO₂⁻ can bind as –NO₂ (yellow) or –ONO (red) → linkage.
Swap: [Co(NH₃)₅Cl](NO₂)Cl → ionisation.
MA₅B type has only one spatial arrangement → no geometrical isomerism.
Five identical NH₃ give a symmetry plane → no optical activity.

ShortcutMA₅B octahedral complexes never show geometrical or optical isomerism — only structural types.

[ 34 ]^D JEE MainGeometrical

The total number of stereoisomers possible for an octahedral complex of the type [Ma₂b₂c₂] is:

(a)3
(b)5
(c)6
(d)8

Answer — (c) 6

Given

Octahedral complex with three pairs of different unidentate ligands.

Asked

Total stereoisomers.

Concept

[Ma₂b₂c₂] gives 5 geometrical arrangements; the all-cis one lacks a symmetry plane and splits into a d,l pair.

Formula

5 geometrical + 1 extra enantiomer = 6 stereoisomers

Baby steps

Arrangements: all three pairs trans (1); one pair cis and two trans, choosing which pair is cis (3); all three pairs cis (1). Total geometrical = 5.
The all-cis isomer is chiral → contributes 2 forms instead of 1.
Total = 4 + 2 = 6 stereoisomers.

Shortcut[Ma₂b₂c₂] = 6 stereoisomers (5 geometrical, one of which is a d/l pair). Worth memorising outright.

[ 35 ]^D NEETGeometrical

cis-platin, an anticancer drug, is:

(a)trans-[Pt(NH₃)₂Cl₂]
(b)cis-[Pt(NH₃)₂Cl₂]
(c)cis-[Pt(NH₃)₄Cl₂]
(d)[Pt(NH₃)₂Cl₄]

Answer — (b) cis-[Pt(NH₃)₂Cl₂]

Given

Anticancer platinum complex.

Asked

Its structure.

Concept

Only the cis isomer of diamminedichloridoplatinum(II) is biologically active; it binds to two adjacent guanine bases of DNA. The trans isomer is inactive because its Cl⁻ ligands are too far apart.

Formula

—

Baby steps

Pt(II) with CN 4 → square planar (dsp²).
Type MA₂B₂ → cis and trans isomers exist.
The cis form has both Cl⁻ adjacent, at the right distance to chelate DNA.
Answer = cis-[Pt(NH₃)₂Cl₂].

ShortcutGeometry decides biology: cis works, trans does not. A one-line application question that repeats often.

Valence Bond Theory

14 Q

[ 36 ]^E NEETVBT

The hybridisation and magnetic nature of [Fe(CN)₆]³⁻ are: (Z of Fe = 26)

(a)sp³d², paramagnetic with 5 unpaired e⁻
(b)d²sp³, paramagnetic with 1 unpaired e⁻
(c)d²sp³, diamagnetic
(d)dsp², paramagnetic

Answer — (b) d²sp³, paramagnetic with 1 unpaired e⁻

Given

[Fe(CN)₆]³⁻, Z_Fe = 26.

Asked

Hybridisation and magnetic behaviour.

Concept

CN⁻ is a strong field ligand: it forces pairing of the 3d electrons, freeing two inner 3d orbitals for d²sp³ hybridisation (inner orbital / low spin complex).

Formula

μ = √n(n+2) BM

Baby steps

Oxidation state: x − 6 = −3 ⇒ Fe³⁺.
Fe³⁺ = [Ar]3d⁵, i.e. five unpaired electrons in the free ion.
CN⁻ is strong field → pair up to t_2g⁵ → only 1 unpaired electron.
Two vacant 3d orbitals + 4s + 4p → d²sp³, octahedral, paramagnetic, μ = √3 = 1.73 BM.

ShortcutInner d used → d²sp³ (low spin). Outer d used → sp³d² (high spin). CN⁻, CO, NO₂⁻ → always inner.

[ 37 ]^E NEETVBT

[FeF₆]³⁻ is strongly paramagnetic whereas [Fe(CN)₆]³⁻ is weakly paramagnetic because:

(a)F⁻ is a weak field ligand and does not pair the 3d electrons, while CN⁻ does
(b)Fe has different oxidation states in the two
(c)F⁻ causes greater splitting than CN⁻
(d)[FeF₆]³⁻ is tetrahedral

Answer — (a) F⁻ is a weak field ligand and does not pair the 3d electrons, while CN⁻ does

Given

Two Fe(III) octahedral complexes with different ligands.

Asked

Reason for the difference in paramagnetism.

Concept

Both have Fe³⁺ (d⁵). The ligand field strength decides whether the electrons pair. Weak field → high spin (5 unpaired); strong field → low spin (1 unpaired).

Formula

μ = √n(n+2)

Baby steps

[FeF₆]³⁻: F⁻ weak field → no pairing → n = 5 → μ = √35 = 5.92 BM, sp³d², outer orbital.
[Fe(CN)₆]³⁻: CN⁻ strong field → pairing → n = 1 → μ = 1.73 BM, d²sp³, inner orbital.
Hence the fluoride complex is strongly, the cyanide complex weakly, paramagnetic.

ShortcutSame metal, same oxidation state, different μ ⇒ the question is testing ligand field strength, nothing else.

[ 38 ]^E NEETVBT

[NiCl₄]²⁻ is paramagnetic while [Ni(CN)₄]²⁻ is diamagnetic, though both have coordination number 4. The hybridisations are respectively:

(a)sp³ and dsp²
(b)dsp² and sp³
(c)sp³ and sp³
(d)d²sp³ and dsp²

Answer — (a) sp³ and dsp²

Given

Two Ni(II) complexes with CN = 4. Ni²⁺ = 3d⁸.

Asked

Hybridisation of each.

Concept

A weak field ligand leaves the two unpaired 3d electrons alone → only 4s and 4p are available → sp³, tetrahedral, paramagnetic. A strong field ligand pairs them, vacating one 3d orbital → dsp², square planar, diamagnetic.

Formula

μ = √n(n+2)

Baby steps

Ni²⁺ is 3d⁸ → two unpaired electrons in the free ion.
Cl⁻ weak field → no pairing → hybridise 4s + 4p → sp³, tetrahedral, μ = 2.83 BM.
CN⁻ strong field → the two electrons pair into one 3d orbital, emptying another.
One 3d + 4s + two 4p → dsp², square planar, n = 0, diamagnetic.

ShortcutCN = 4 with a strong ligand → square planar dsp², diamagnetic. CN = 4 with a weak ligand → tetrahedral sp³, paramagnetic.

[ 39 ]^E NEETMagnetic moment

The spin-only magnetic moment of [Cr(H₂O)₆]³⁺ is: (Z of Cr = 24)

(a)1.73 BM
(b)2.83 BM
(c)3.87 BM
(d)5.92 BM

Answer — (c) 3.87 BM

Given

[Cr(H₂O)₆]³⁺, Z_Cr = 24.

Asked

Spin-only magnetic moment.

Concept

Only unpaired electrons contribute. For d¹–d³, weak and strong field give the same count because the t_2g orbitals fill singly under Hund's rule either way.

Formula

μ = √n(n+2) BM

Baby steps

Cr³⁺ = [Ar]3d³.
Three electrons occupy t_2g singly → n = 3 (regardless of ligand strength).
μ = √3(3+2) = √15.
μ = 3.87 BM.

ShortcutMemorise the ladder: n = 1→1.73, 2→2.83, 3→3.87, 4→4.90, 5→5.92 BM.

[ 40 ]^E NEETMagnetic moment

Amongst the following, the ion with the highest magnetic moment value is:

(a)[Cr(H₂O)₆]³⁺
(b)[Fe(H₂O)₆]²⁺
(c)[Zn(H₂O)₆]²⁺
(d)[Ni(H₂O)₆]²⁺

Answer — (b) [Fe(H₂O)₆]²⁺

Given

Four hexaaqua complexes. H₂O is a weak field ligand throughout.

Asked

The highest magnetic moment.

Concept

With the same weak ligand, the complex with the most unpaired electrons wins.

Formula

μ = √n(n+2)

Baby steps

Cr³⁺ = d³ → n = 3 → 3.87 BM.
Fe²⁺ = d⁶, high spin (t_2g⁴e_g²) → n = 4 → 4.90 BM.
Zn²⁺ = d¹⁰ → n = 0 → diamagnetic.
Ni²⁺ = d⁸ → n = 2 → 2.83 BM. Highest is [Fe(H₂O)₆]²⁺.

ShortcutWith H₂O (weak field), just count unpaired electrons the free-ion way — no pairing happens.

[ 41 ]^E NEETVBT

The spin-only magnetic moment of [MnBr₄]²⁻ is 5.9 BM. The geometry of the complex ion is:

(a)Square planar
(b)Tetrahedral
(c)Octahedral
(d)Trigonal bipyramidal

Answer — (b) Tetrahedral

Given

[MnBr₄]²⁻, μ = 5.9 BM, CN = 4.

Asked

Geometry.

Concept

With CN = 4 the two options are tetrahedral (sp³) and square planar (dsp²). The measured moment identifies the number of unpaired electrons and hence which one it is.

Formula

μ = √n(n+2) ⇒ 5.9 = √35 ⇒ n = 5

Baby steps

Mn²⁺ = 3d⁵.
μ = 5.9 BM ⇒ n(n+2) = 35 ⇒ n = 5 unpaired electrons.
All five d electrons remain unpaired, so no 3d orbital is free for dsp².
Hybridisation must be sp³ → tetrahedral.

Shortcutμ = 5.9 BM → n = 5 → no d orbital vacated → cannot be square planar. Geometry follows from μ.

[ 42 ]^E NEETVBT

The number of unpaired electrons in the square planar [Pt(CN)₄]²⁻ ion is:

(a)0
(b)1
(c)2
(d)4

Answer — (a) 0

Given

Square planar [Pt(CN)₄]²⁻.

Asked

Number of unpaired electrons.

Concept

Square planar geometry requires dsp² hybridisation, which needs one empty (n−1)d orbital. That orbital can only be emptied by pairing all the d electrons.

Formula

dsp² ⇒ all d electrons paired

Baby steps

Pt²⁺ has a d⁸ configuration (5d⁸).
Square planar → dsp² hybridisation is compulsory.
To free one 5d orbital, the two unpaired electrons must pair.
Unpaired electrons = 0; the ion is diamagnetic.

ShortcutEvery square planar d⁸ complex is diamagnetic — [Ni(CN)₄]²⁻, [Pt(CN)₄]²⁻, [PtCl₄]²⁻.

[ 43 ]^E NEETVBT

[NiCl₄]²⁻ is paramagnetic while [Ni(CO)₄] is diamagnetic, though both are tetrahedral. The reason is:

(a)Ni is in +2 state in the first and 0 in the second; CO pairs the electrons, Cl⁻ does not
(b)[Ni(CO)₄] is square planar
(c)CO is a weak field ligand
(d)Cl⁻ is a strong field ligand

Answer — (a) Ni is in +2 state in the first and 0 in the second; CO pairs the electrons, Cl⁻ does not

Given

Two tetrahedral nickel complexes with different magnetic behaviour.

Asked

Explanation.

Concept

Oxidation state and ligand strength both change. In Ni(CO)₄, nickel is in the zero oxidation state (3d⁸4s²) and the strong field CO forces the 4s electrons into 3d, giving 3d¹⁰.

Formula

μ = √n(n+2)

Baby steps

[NiCl₄]²⁻: Ni²⁺ = 3d⁸; Cl⁻ weak → 2 unpaired → sp³, paramagnetic (μ = 2.83 BM).
[Ni(CO)₄]: Ni(0) = 3d⁸4s².
CO is very strong field → 4s electrons shift to 3d → 3d¹⁰4s⁰, all paired.
Empty 4s and three 4p → sp³, tetrahedral, diamagnetic.

ShortcutNi(CO)₄ is the standard ‘same geometry, different magnetism’ pair. Key is Ni(0) → 3d¹⁰.

[ 44 ]^E NEETVBT

[Co(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺ is an outer orbital complex because:

(a)Co³⁺ is d⁶ and can vacate two 3d orbitals on pairing; Ni²⁺ is d⁸ and cannot
(b)NH₃ is stronger towards Ni than Co
(c)Ni²⁺ is smaller than Co³⁺
(d)Co³⁺ uses 4d orbitals

Answer — (a) Co³⁺ is d⁶ and can vacate two 3d orbitals on pairing; Ni²⁺ is d⁸ and cannot

Given

Two hexaammine complexes with the same ligand.

Asked

Reason for inner vs outer orbital character.

Concept

Inner orbital (d²sp³) needs two empty (n−1)d orbitals. Pairing can free them only if the d-electron count allows it.

Formula

d²sp³ (inner) vs sp³d² (outer)

Baby steps

Co³⁺ = 3d⁶. Pairing gives t_2g⁶, leaving two empty 3d orbitals → d²sp³, diamagnetic, inner.
Ni²⁺ = 3d⁸. Even with maximum pairing, at most one 3d orbital can be freed.
Two inner d orbitals are unavailable, so it must use outer 4d → sp³d², outer orbital, paramagnetic with 2 unpaired electrons.

Shortcutd⁸ octahedral is always outer orbital (sp³d²), no matter how strong the ligand.

[ 45 ]^E JEE MainMagnetic moment

A metal ion in an octahedral complex shows a spin-only magnetic moment of 4.90 BM. The number of unpaired electrons is:

(a)2
(b)3
(c)4
(d)5

Answer — (c) 4

Given

μ = 4.90 BM.

Asked

Number of unpaired electrons.

Concept

Invert the spin-only formula.

Formula

μ = √n(n+2) ⇒ n² + 2n − μ² = 0

Baby steps

Square both sides: n(n+2) = (4.90)² ≈ 24.
n² + 2n − 24 = 0.
(n + 6)(n − 4) = 0.
n = 4 (reject the negative root).

ShortcutInstead of solving the quadratic, match against the memorised ladder: 4.90 BM is always n = 4.

[ 46 ]^E NEETVBT

Which of the following is an inner orbital (low spin) complex?

(a)[CoF₆]³⁻
(b)[MnCl₆]³⁻
(c)[Mn(CN)₆]³⁻
(d)[FeF₆]³⁻

Answer — (c) [Mn(CN)₆]³⁻

Given

Four octahedral complexes of first-series metals.

Asked

The inner orbital complex.

Concept

Only strong field ligands (CN⁻, CO, NO₂⁻, en, NH₃ for +3 ions) cause pairing and inner (n−1)d participation.

Formula

Strong field ⇒ d²sp³

Baby steps

F⁻ and Cl⁻ sit at the weak end of the spectrochemical series → outer orbital sp³d².
CN⁻ sits near the strong end.
In [Mn(CN)₆]³⁻, Mn³⁺ = d⁴ pairs to t_2g⁴ → 2 unpaired, two 3d orbitals free.
Hence d²sp³, inner orbital.

ShortcutHalide ligand in the formula → assume outer orbital / high spin. CN⁻ or CO → inner orbital / low spin.

[ 47 ]^E JEE MainVBT

The hybridisation of the central metal in [Fe(H₂O)₆]²⁺ and the number of unpaired electrons are:

(a)d²sp³, 0
(b)sp³d², 4
(c)d²sp³, 4
(d)sp³d², 2

Answer — (b) sp³d², 4

Given

[Fe(H₂O)₆]²⁺.

Asked

Hybridisation and unpaired electron count.

Concept

H₂O is a weak field ligand, so the d⁶ ion stays high spin and must use outer 4d orbitals.

Formula

μ = √n(n+2)

Baby steps

Fe²⁺ = 3d⁶.
H₂O weak field → no extra pairing → t_2g⁴e_g² → 4 unpaired electrons.
No vacant 3d orbitals remain, so 4d must be used.
Hybridisation = sp³d², octahedral, μ = 4.90 BM.

ShortcutWeak field octahedral → sp³d² and free-ion unpaired count. Strong field → d²sp³ and maximum pairing.

[ 48 ]^E NEETVBT

Which one of the following complexes is diamagnetic?

(a)[CoF₆]³⁻
(b)[Co(NH₃)₆]³⁺
(c)[Fe(H₂O)₆]²⁺
(d)[NiCl₄]²⁻

Answer — (b) [Co(NH₃)₆]³⁺

Given

Four complexes with known ligand strengths.

Asked

The diamagnetic one.

Concept

Diamagnetic means zero unpaired electrons. For d⁶ with a strong field ligand, all six electrons pair into t_2g.

Formula

n = 0 ⇒ μ = 0

Baby steps

[CoF₆]³⁻: F⁻ weak → t_2g⁴e_g² → 4 unpaired, paramagnetic.
[Co(NH₃)₆]³⁺: NH₃ strong enough for Co(III) → t_2g⁶e_g⁰ → 0 unpaired, diamagnetic.
[Fe(H₂O)₆]²⁺: 4 unpaired.
[NiCl₄]²⁻: 2 unpaired.

ShortcutLow spin d⁶ is the classic diamagnetic case: [Co(NH₃)₆]³⁺, [Fe(CN)₆]⁴⁻, [Co(C₂O₄)₃]³⁻.

[ 49 ]^E NEETVBT

[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]²⁻ is diamagnetic. This is because:

(a)Cr³⁺ is d³ with three unpaired electrons that cannot pair; Ni²⁺ in a strong square planar field pairs completely
(b)NH₃ is a weaker ligand than CN⁻ for every metal
(c)Cr³⁺ is larger than Ni²⁺
(d)[Ni(CN)₄]²⁻ is tetrahedral

Answer — (a) Cr³⁺ is d³ with three unpaired electrons that cannot pair; Ni²⁺ in a strong square planar field pairs completely

Given

Two complexes with different magnetic behaviour.

Asked

Correct explanation.

Concept

A d³ ion has three electrons in three separate t_2g orbitals; no amount of ligand field strength can pair them. A d⁸ ion in a square planar strong field pairs into four orbitals leaving none unpaired.

Formula

μ = √n(n+2)

Baby steps

Cr³⁺ = 3d³ → t_2g³, three unpaired regardless of ligand → μ = 3.87 BM, paramagnetic, d²sp³.
Ni²⁺ = 3d⁸; CN⁻ strong → electrons pair, one 3d empties.
dsp² square planar with all electrons paired → diamagnetic.

Shortcutd¹, d², d³ and d⁸(tetrahedral) can never be made diamagnetic by any ligand.

Crystal Field Theory

14 Q

[ 50 ]^F NEETCFT

The relationship between the crystal field splitting in tetrahedral and octahedral fields for the same metal, same ligands and same metal–ligand distance is:

(a)Δ_t = (4/9)Δ_o
(b)Δ_t = (9/4)Δ_o
(c)Δ_t = (2/5)Δ_o
(d)Δ_t = Δ_o

Answer — (a) Δ_t = (4/9)Δ_o

Given

Same metal, same ligands, same M–L distance in two geometries.

Asked

Relation between Δ_t and Δ_o.

Concept

A tetrahedral field has only 4 ligands (not 6) and none of them points directly at any d orbital, so the splitting is much smaller and the pattern is inverted (e below t₂).

Formula

Δ_t = (4/9)Δ_o ≈ 0.44Δ_o

Baby steps

Factor 2/3 from having 4 ligands instead of 6.
Factor 2/3 again because no ligand points along a d orbital lobe.
Net = (2/3)(2/3) = 4/9.
Because Δ_t is small, it is almost always less than the pairing energy P.

ShortcutΔ_t < P always ⇒ tetrahedral complexes are essentially always high spin. Never look for low spin tetrahedral in an exam.

[ 51 ]^F NEETSpectrochemical

The correct order of ligands in the spectrochemical series (increasing field strength) is:

(a)I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O < NH₃ < en < CN⁻ < CO
(b)CO < CN⁻ < en < NH₃ < H₂O < F⁻ < I⁻
(c)F⁻ < I⁻ < H₂O < CN⁻ < NH₃ < CO
(d)H₂O < I⁻ < NH₃ < CO < CN⁻

Answer — (a) I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O < NH₃ < en < CN⁻ < CO

Given

Common ligands.

Asked

Correct increasing order of crystal field strength.

Concept

The spectrochemical series is an experimentally determined order based on absorption spectra, not on ligand charge. Anionic ligands sit at the weak end — a fact CFT itself cannot explain.

Formula

I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < EDTA⁴⁻ < NH₃ < en < CN⁻ < CO

Baby steps

Halides are weakest, decreasing in strength down the group (I⁻ weakest).
Neutral O-donors (H₂O) come next, then N-donors (NH₃, en).
C-donors (CN⁻, CO) are strongest because of pi back bonding.
Option A follows this order correctly.

ShortcutDonor atom trend: I < Br < Cl < F < O < N < C. Remember the mnemonic order and the two ends: I⁻ weakest, CO strongest.

[ 52 ]^F NEETCFSE

The crystal field stabilisation energy (CFSE) for a d⁶ octahedral complex with a strong field ligand is:

(a)−0.4Δ_o
(b)−1.6Δ_o
(c)−2.4Δ_o
(d)−1.2Δ_o

Answer — (c) −2.4Δ_o

Given

Octahedral d⁶, strong field (low spin).

Asked

CFSE (ignoring the pairing energy term).

Concept

Each t_2g electron is stabilised by 0.4Δ_o; each e_g electron is destabilised by 0.6Δ_o.

Formula

CFSE = [−0.4(n_t2g) + 0.6(n_eg)]Δ_o

Baby steps

Strong field d⁶ → t_2g⁶e_g⁰.
n_t2g = 6, n_eg = 0.
CFSE = −0.4(6) + 0.6(0) = −2.4.
CFSE = −2.4Δ_o — the maximum possible, which is why low spin d⁶ complexes are exceptionally stable.

ShortcutUse −0.4 per t_2g and +0.6 per e_g. Low spin d⁶ gives the maximum CFSE of −2.4Δ_o.

[ 53 ]^F JEE MainCFSE

The CFSE of the high spin octahedral complex [Fe(H₂O)₆]²⁺ is:

(a)−0.4Δ_o
(b)−1.2Δ_o
(c)−2.4Δ_o
(d)0

Answer — (a) −0.4Δ_o

Given

Fe²⁺ = d⁶, H₂O weak field → high spin.

Asked

CFSE.

Concept

In a weak field the electrons follow Hund's rule first, filling all five orbitals singly before pairing.

Formula

CFSE = [−0.4 n_t2g + 0.6 n_eg]Δ_o

Baby steps

High spin d⁶ → t_2g⁴e_g².
CFSE = −0.4(4) + 0.6(2).
= −1.6 + 1.2 = −0.4.
CFSE = −0.4Δ_o.

ShortcutFor high spin d⁵ the CFSE is exactly zero (−0.4×3 + 0.6×2 = 0). Useful checkpoint.

[ 54 ]^F NEETCFT

The electronic configuration of the d orbitals in [CoF₆]³⁻ is:

(a)t_2g⁶e_g⁰
(b)t_2g⁴e_g²
(c)t_2g³e_g³
(d)t_2g⁵e_g¹

Answer — (b) t_2g⁴e_g²

Given

[CoF₆]³⁻; Co³⁺ = d⁶; F⁻ weak field.

Asked

t_2g/e_g configuration.

Concept

When Δ_o < P, electrons prefer to occupy the higher e_g level singly rather than pay the pairing energy.

Formula

Δ_o < P ⇒ high spin

Baby steps

Co³⁺ has 6 d electrons.
F⁻ is a weak field ligand → Δ_o < P → high spin.
Fill all five orbitals singly (5 electrons), then pair the sixth in t_2g.
Result: t_2g⁴e_g², 4 unpaired, μ = 4.90 BM, paramagnetic.

ShortcutHigh spin: fill singly across all 5 orbitals first, then pair. Low spin: fill t_2g completely first.

[ 55 ]^F NEETCFT

For a d⁴ octahedral ion, if Δ_o > P then the configuration is:

(a)t_2g³e_g¹
(b)t_2g⁴e_g⁰
(c)t_2g²e_g²
(d)t_2g¹e_g³

Answer — (b) t_2g⁴e_g⁰

Given

d⁴ octahedral ion, Δ_o > P.

Asked

Electronic configuration.

Concept

The fourth electron chooses between paying the pairing energy P (going into t_2g) or paying Δ_o (going into e_g). It takes the cheaper route.

Formula

Δ_o > P ⇒ low spin; Δ_o < P ⇒ high spin

Baby steps

First three electrons fill t_2g singly.
The fourth faces the choice. Here Δ_o > P, so promotion to e_g is more expensive than pairing.
It pairs in t_2g → t_2g⁴e_g⁰, 2 unpaired, low spin.

ShortcutCompare Δ_o with P like comparing two prices — the electron always picks the cheaper option.

[ 56 ]^F NEETCFT

The hexaaqua manganese(II) ion contains five unpaired electrons while the hexacyano ion contains only one. The reason is:

(a)CN⁻ is a strong field ligand giving Δ_o > P, so pairing occurs
(b)Mn has different oxidation states in the two ions
(c)H₂O causes greater splitting than CN⁻
(d)The cyano complex is tetrahedral

Answer — (a) CN⁻ is a strong field ligand giving Δ_o > P, so pairing occurs

Given

[Mn(H₂O)₆]²⁺ and [Mn(CN)₆]⁴⁻; both Mn(II) = d⁵.

Asked

CFT explanation.

Concept

Same d⁵ ion, two different ligand fields. Weak field keeps all five unpaired; strong field pairs them down to one.

Formula

Δ_o vs P decides spin state

Baby steps

Mn²⁺ = 3d⁵ in both.
H₂O weak → Δ_o < P → t_2g³e_g² → 5 unpaired, μ = 5.92 BM.
CN⁻ strong → Δ_o > P → t_2g⁵e_g⁰ → 1 unpaired, μ = 1.73 BM.

ShortcutHigh spin d⁵ is the maximum-paramagnetism case in the whole 3d series (μ = 5.92 BM).

[ 57 ]^F JEE MainCFT

In an octahedral crystal field, the e_g orbitals are raised and the t_2g orbitals lowered relative to the barycentre by:

(a)0.6Δ_o and 0.4Δ_o respectively
(b)0.4Δ_o and 0.6Δ_o respectively
(c)0.5Δ_o each
(d)Δ_o and Δ_o/2

Answer — (a) 0.6Δ_o and 0.4Δ_o respectively

Given

Octahedral splitting diagram.

Asked

Energy shifts of the two sets.

Concept

The barycentre (centre of gravity of energy) must be conserved: total energy rise of the two e_g orbitals equals the total energy fall of the three t_2g orbitals.

Formula

2 × (0.6Δ_o) = 3 × (0.4Δ_o)

Baby steps

Let e_g rise by x and t_2g fall by y, with x + y = Δ_o.
Barycentre rule: 2x = 3y.
Solving: x = 0.6Δ_o, y = 0.4Δ_o.
So e_g is raised by 3/5 Δ_o and t_2g lowered by 2/5 Δ_o.

ShortcutTwo orbitals up by 3/5, three orbitals down by 2/5. The 2×3 = 3×2 balance is the memory hook.

[ 58 ]^F NEETCFT

Which of the following d orbitals experience the greatest repulsion from ligands in an octahedral field?

(a)d_xy and d_yz
(b)d_x²−y² and d_z²
(c)d_xz and d_yz
(d)d_xy, d_yz and d_xz

Answer — (b) d_x²−y² and d_z²

Given

Octahedral arrangement of six ligands along the ±x, ±y, ±z axes.

Asked

Which orbitals are destabilised most.

Concept

Repulsion is greatest when the orbital lobes point directly at the incoming ligands.

Formula

Axial orbitals → e_g (higher energy)

Baby steps

Ligands approach along the three Cartesian axes.
d_x²−y² lies along x and y; d_z² lies along z → head-on repulsion.
d_xy, d_yz, d_xz point between the axes → less repulsion.
So the e_g set (d_x²−y², d_z²) is raised.

ShortcutOrbital name with only axis letters (x²−y², z²) → points along axes → e_g. Two-letter names (xy, yz, xz) → t_2g. In tetrahedral, it flips.

[ 59 ]^F NEETCFT

The magnitude of Δ_o is expected to be largest for:

(a)[Co(H₂O)₆]³⁺
(b)[Co(NH₃)₆]³⁺
(c)[Co(CN)₆]³⁻
(d)[CoF₆]³⁻

Answer — (c) [Co(CN)₆]³⁻

Given

Four Co(III) octahedral complexes.

Asked

Largest crystal field splitting.

Concept

With the metal and its oxidation state held constant, Δ_o is decided purely by the position of the ligand in the spectrochemical series.

Formula

Δ_o ∝ ligand field strength

Baby steps

Rank the four ligands: F⁻ < H₂O < NH₃ < CN⁻.
CN⁻ is the strongest of the four.
Therefore [Co(CN)₆]³⁻ has the largest Δ_o.
It also absorbs the shortest wavelength (310 nm, ultraviolet) and appears pale yellow.

ShortcutLarger Δ_o → higher energy absorbed → shorter wavelength absorbed. Δ_o and λ_absorbed are inversely related.

[ 60 ]^F JEE MainCFT

Δ_o for a given ligand increases in the order:

(a)3d < 4d < 5d metal ions
(b)5d < 4d < 3d metal ions
(c)4d < 3d < 5d metal ions
(d)It is independent of the metal

Answer — (a) 3d < 4d < 5d metal ions

Given

Same ligand, metals from different transition series.

Asked

Trend of Δ_o down a group.

Concept

Larger, more diffuse 4d and 5d orbitals overlap the ligand orbitals more effectively, producing much stronger splitting. Higher charge on the metal also raises Δ_o.

Formula

Δ_o: 3d < 4d < 5d; also M²⁺ < M³⁺

Baby steps

Going down a group, d orbitals become larger and less shielded.
Metal–ligand overlap improves, so repulsion and splitting increase.
Δ_o rises by roughly 30–50% from 3d to 4d and again from 4d to 5d.
Hence 3d < 4d < 5d.

ShortcutConsequence worth remembering: 4d and 5d complexes are nearly always low spin, whatever the ligand.

[ 61 ]^F NEETCFT

Low spin configurations are rarely observed in tetrahedral complexes because:

(a)Δ_t is small and always less than the pairing energy
(b)Tetrahedral complexes have no d orbitals available
(c)Δ_t is greater than Δ_o
(d)Tetrahedral complexes are always diamagnetic

Answer — (a) Δ_t is small and always less than the pairing energy

Given

Tetrahedral crystal field.

Asked

Reason for the absence of low spin tetrahedral complexes.

Concept

Since Δ_t = (4/9)Δ_o, the splitting energy is too small to overcome the pairing energy, so electrons prefer to spread out singly.

Formula

Δ_t < P

Baby steps

Only four ligands, none pointing directly at a d orbital.
Δ_t comes out at roughly 4/9 of Δ_o.
Pairing energy P almost always exceeds this small splitting.
So electrons occupy the higher t₂ set singly → high spin.

ShortcutAlso note: the ‘g’ subscript is dropped for tetrahedral (e, t₂) because there is no centre of symmetry.

[ 62 ]^F NEETCFT

Which of the following statements about crystal field theory is incorrect?

(a)The metal–ligand bond is treated as purely ionic
(b)Ligands are treated as point charges or point dipoles
(c)It correctly predicts that anionic ligands cause the greatest splitting
(d)It explains the colour and magnetic properties of complexes

Answer — (c) It correctly predicts that anionic ligands cause the greatest splitting

Given

Statements about CFT.

Asked

The incorrect one.

Concept

CFT assumes purely electrostatic interaction, so it predicts that anionic ligands should split most strongly. Experimentally, anionic ligands (I⁻, Br⁻, Cl⁻) sit at the weak end — a known failure of the theory.

Formula

—

Baby steps

CFT is an electrostatic model → A is a correct description.
Ligands as point charges/dipoles → B is correct.
The prediction about anionic ligands fails against experiment → C is the incorrect statement.
CFT does succeed on colour and magnetism → D is correct.

ShortcutCFT's two admitted weaknesses: it ignores covalency, and it wrongly ranks anionic ligands as strong field. LFT/MOT fix these.

[ 63 ]^F JEE MainCFSE

The CFSE of an octahedral complex with configuration t_2g³e_g⁰ is:

(a)−0.4Δ_o
(b)−0.8Δ_o
(c)−1.2Δ_o
(d)−2.0Δ_o

Answer — (c) −1.2Δ_o

Given

d³ octahedral ion, e.g. Cr³⁺.

Asked

CFSE.

Concept

Each electron in the stabilised t_2g set contributes −0.4Δ_o.

Formula

CFSE = [−0.4 n_t2g + 0.6 n_eg]Δ_o

Baby steps

n_t2g = 3, n_eg = 0.
CFSE = −0.4 × 3.
= −1.2Δ_o.
This large stabilisation is why Cr(III) complexes are kinetically inert and very common.

Shortcutd³ and low spin d⁶ have the two largest CFSE values — both give unusually stable, inert complexes.

Colour in Complexes

07 Q

[ 64 ]^G NEETColour

[Ti(H₂O)₆]³⁺ is violet in colour because:

(a)It absorbs in the blue-green region and transmits the complementary violet
(b)It emits violet light
(c)The Ti–O bond is violet
(d)It has a charge transfer transition

Answer — (a) It absorbs in the blue-green region and transmits the complementary violet

Given

[Ti(H₂O)₆]³⁺, Ti³⁺ = 3d¹. Absorption at about 498 nm.

Asked

Reason for the violet colour.

Concept

The colour arises from a d–d transition: the single electron is promoted from t_2g to e_g by absorbing visible light. The observed colour is the complementary colour of the light absorbed.

Formula

Δ_o = hν = hc/λ

Baby steps

Ti³⁺ is 3d¹; ground state is t_2g¹e_g⁰.
Absorbed light promotes the electron: t_2g¹e_g⁰ → t_2g⁰e_g¹.
The energy needed corresponds to blue-green light near 498 nm.
Blue-green is removed, so the transmitted light appears violet.

ShortcutComplementary pairs: absorbs blue-green → looks violet; absorbs yellow → looks violet-blue; absorbs green → looks red.

[ 65 ]^G NEETColour

Which of the following complexes is colourless?

(a)[Ti(H₂O)₆]³⁺
(b)[Cu(H₂O)₄]²⁺
(c)[Zn(NH₃)₄]²⁺
(d)[Co(NH₃)₆]³⁺

Answer — (c) [Zn(NH₃)₄]²⁺

Given

Four complexes of first-series metals.

Asked

The colourless one.

Concept

Colour from d–d transitions requires a partially filled d subshell. A d⁰ or d¹⁰ ion has no possible d–d transition and is colourless.

Formula

Colour ⇒ d¹ to d⁹

Baby steps

Ti³⁺ = d¹ → coloured (violet).
Cu²⁺ = d⁹ → coloured (blue).
Zn²⁺ = d¹⁰ → no vacant d orbital to be promoted into → colourless.
Co³⁺ = d⁶ → coloured (yellow-orange).

ShortcutColourless ions to recognise instantly: Sc³⁺, Ti⁴⁺ (d⁰) and Zn²⁺, Cu⁺, Ag⁺, Cd²⁺ (d¹⁰).

[ 66 ]^G NEETColour

The correct order of the wavelengths of absorption in the visible region for [Ni(NO₂)₆]⁴⁻, [Ni(NH₃)₆]²⁺ and [Ni(H₂O)₆]²⁺ is:

(a)[Ni(H₂O)₆]²⁺ > [Ni(NH₃)₆]²⁺ > [Ni(NO₂)₆]⁴⁻
(b)[Ni(NO₂)₆]⁴⁻ > [Ni(NH₃)₆]²⁺ > [Ni(H₂O)₆]²⁺
(c)[Ni(NH₃)₆]²⁺ > [Ni(H₂O)₆]²⁺ > [Ni(NO₂)₆]⁴⁻
(d)All absorb at the same wavelength

Answer — (a) [Ni(H₂O)₆]²⁺ > [Ni(NH₃)₆]²⁺ > [Ni(NO₂)₆]⁴⁻

Given

Three Ni(II) octahedral complexes with different ligands.

Asked

Decreasing order of absorbed wavelength.

Concept

Stronger ligand → larger Δ_o → more energy needed → shorter wavelength absorbed. Wavelength order is therefore the reverse of the field-strength order.

Formula

Δ_o = hc/λ ⇒ λ ∝ 1/Δ_o

Baby steps

Field strength: H₂O < NH₃ < NO₂⁻.
So Δ_o: H₂O complex < NH₃ complex < NO₂⁻ complex.
Wavelength is inversely proportional to Δ_o.
λ: H₂O > NH₃ > NO₂⁻.

ShortcutWrite the spectrochemical order, then simply reverse it for wavelength. Weakest ligand absorbs the longest λ.

[ 67 ]^G NEETColour

On progressively adding ethane-1,2-diamine to an aqueous solution of [Ni(H₂O)₆]²⁺, the colour changes from green to:

(a)Pale blue → blue/purple → violet
(b)Yellow → orange → red
(c)Colourless throughout
(d)Red → pink → brown

Answer — (a) Pale blue → blue/purple → violet

Given

[Ni(H₂O)₆]²⁺ with en added in 1:1, 2:1 and 3:1 molar ratios.

Asked

Sequence of colours.

Concept

Each water replaced by the stronger-field en raises Δ_o, shifting absorption to shorter wavelength and moving the observed colour across the spectrum.

Formula

Δ_o increases as en replaces H₂O

Baby steps

[Ni(H₂O)₆]²⁺ is green.
+1 en → [Ni(H₂O)₄(en)]²⁺, pale blue.
+2 en → [Ni(H₂O)₂(en)₂]²⁺, blue/purple.
+3 en → [Ni(en)₃]²⁺, violet.

ShortcutOne en replaces two water molecules each time, because en is didentate. Watch the stoichiometry, not just the colour.

[ 68 ]^G NEETColour

Anhydrous CuSO₄ is white but CuSO₄·5H₂O is blue because:

(a)In the absence of ligands there is no crystal field splitting, so no d–d transition occurs
(b)Water is blue in colour
(c)Cu²⁺ becomes d¹⁰ on hydration
(d)The sulphate ion absorbs visible light

Answer — (a) In the absence of ligands there is no crystal field splitting, so no d–d transition occurs

Given

Anhydrous and hydrated copper(II) sulphate.

Asked

Reason for the colour difference.

Concept

d–d transitions require the d orbitals to be split, and splitting requires a ligand field. Without coordinated ligands the five d orbitals stay degenerate and no visible light is absorbed.

Formula

No ligand ⇒ Δ = 0 ⇒ no d–d transition

Baby steps

Cu²⁺ is d⁹ in both compounds, so the electron count is not the issue.
In anhydrous CuSO₄ there is no aqua ligand field → degenerate d orbitals → white.
In the pentahydrate, [Cu(H₂O)₄]²⁺ forms → d orbitals split.
It absorbs red light near 600 nm → appears blue.

ShortcutSame idea: heating [Ti(H₂O)₆]Cl₃ to drive off water turns it colourless.

[ 69 ]^G JEE MainColour

[Fe(CN)₆]⁴⁻ and [Fe(H₂O)₆]²⁺ have different colours in dilute solution because:

(a)The two ligands produce different Δ_o, so light of different energy is absorbed
(b)Iron is in different oxidation states
(c)One is octahedral and the other tetrahedral
(d)Only one of them undergoes a d–d transition

Answer — (a) The two ligands produce different Δ_o, so light of different energy is absorbed

Given

Two Fe(II) hexacoordinate complexes.

Asked

Reason for the colour difference.

Concept

Colour depends on the energy gap Δ_o, which depends on the ligand. The metal, its oxidation state and the geometry are identical here, so only the ligand can be responsible.

Formula

λ_absorbed = hc/Δ_o

Baby steps

Both contain Fe²⁺ (d⁶) in an octahedral field.
CN⁻ is strong field → large Δ_o → absorbs short wavelength.
H₂O is weak field → small Δ_o → absorbs long wavelength.
Different absorbed wavelengths → different observed colours.

ShortcutIf a question gives the same metal and oxidation state but different colours, the answer is always ‘different Δ_o due to different ligands’.

[ 70 ]^G NEETColour

A complex absorbs light of wavelength 500 nm (blue-green). The colour observed will be:

(a)Blue-green
(b)Red
(c)Yellow
(d)Green

Answer — (b) Red

Given

Absorption maximum at 500 nm.

Asked

Observed colour.

Concept

The observed colour is the complementary colour of the absorbed light, generated by the wavelengths that are transmitted.

Formula

Observed = complementary of absorbed

Baby steps

500 nm falls in the blue-green region.
Blue-green light is removed from the white light passing through.
The complementary colour of blue-green is red.
This matches [Co(NH₃)₅(H₂O)]³⁺, which absorbs at 500 nm and looks red.

ShortcutNCERT Table 5.3 pairs to memorise: 535 nm yellow→violet; 500 blue-green→red; 475 blue→yellow-orange; 600 red→blue.

Metal Carbonyls

05 Q

[ 71 ]^H NEETCarbonyls

The metal–carbon bond in metal carbonyls possesses:

(a)Only sigma character
(b)Only pi character
(c)Both sigma and pi character (synergic bonding)
(d)Purely ionic character

Answer — (c) Both sigma and pi character (synergic bonding)

Given

M–CO bonding in homoleptic carbonyls.

Asked

Nature of the M–C bond.

Concept

Synergic bonding: CO donates its carbon lone pair into an empty metal orbital (sigma, ligand→metal), and the metal donates electron density from a filled d orbital into the vacant π* orbital of CO (pi back bonding, metal→ligand). Each strengthens the other.

Formula

M ← CO (σ) and M → CO (π*)

Baby steps

Carbon's lone pair enters a vacant metal orbital → sigma bond.
Filled metal d electrons flow into the antibonding π* of CO → pi bond.
The two donations reinforce each other → synergic effect.
Result: the M–C bond strengthens and the C–O bond weakens.

ShortcutSynergic bonding is a two-way street. Note the outcome: M–C bond order rises, C–O bond order falls.

[ 72 ]^H JEE MainCarbonyls

As a result of pi back bonding in metal carbonyls, the C–O bond:

(a)Becomes stronger and shorter
(b)Becomes weaker and longer
(c)Remains unchanged
(d)Becomes fully ionic

Answer — (b) Becomes weaker and longer

Given

Synergic bonding in M–CO.

Asked

Effect on the C–O bond.

Concept

Back-donated electron density enters the antibonding π* orbital of CO. Filling an antibonding orbital reduces the bond order.

Formula

Electron density in π* ⇒ bond order decreases

Baby steps

Free CO has a bond order of 3 (triple bond).
Metal d electrons populate the CO π* orbital.
Antibonding population lowers the C–O bond order below 3.
The bond becomes weaker and longer; the IR stretching frequency drops from ~2143 cm⁻¹.

ShortcutMore back bonding → lower C–O stretching frequency. Anionic carbonyls like [Fe(CO)₄]²⁻ back bond most.

[ 73 ]^H NEETCarbonyls

The geometries of Ni(CO)₄, Fe(CO)₅ and Cr(CO)₆ are respectively:

(a)Tetrahedral, trigonal bipyramidal, octahedral
(b)Square planar, tetrahedral, octahedral
(c)Tetrahedral, square pyramidal, octahedral
(d)Octahedral, trigonal bipyramidal, tetrahedral

Answer — (a) Tetrahedral, trigonal bipyramidal, octahedral

Given

Three homoleptic metal carbonyls.

Asked

Their geometries.

Concept

Geometry follows the coordination number: 4 → tetrahedral (sp³), 5 → trigonal bipyramidal (sp³d), 6 → octahedral (d²sp³).

Formula

CN 4 / 5 / 6 → Td / TBP / Oh

Baby steps

Ni(CO)₄: CN 4, sp³ → tetrahedral.
Fe(CO)₅: CN 5, sp³d → trigonal bipyramidal.
Cr(CO)₆: CN 6, d²sp³ → octahedral.
All three are diamagnetic and obey the 18-electron (EAN = 36) rule.

ShortcutAll simple neutral carbonyls are diamagnetic — CO always pairs everything up.

[ 74 ]^H JEE MainCarbonyls

Which of the following carbonyls contains a metal–metal bond with no bridging CO groups?

(a)[Co₂(CO)₈]
(b)[Mn₂(CO)₁₀]
(c)[Ni(CO)₄]
(d)[Fe(CO)₅]

Answer — (b) [Mn₂(CO)₁₀]

Given

Four metal carbonyls.

Asked

The one with an M–M bond and no bridging CO.

Concept

Polynuclear carbonyls may hold together through a direct metal–metal bond, bridging CO ligands, or both.

Formula

—

Baby steps

Ni(CO)₄ and Fe(CO)₅ are mononuclear → no M–M bond at all.
[Co₂(CO)₈] has a Co–Co bond plus two bridging CO groups.
[Mn₂(CO)₁₀] is two square pyramidal Mn(CO)₅ units joined only by a direct Mn–Mn bond.
Answer = [Mn₂(CO)₁₀].

ShortcutMn₂(CO)₁₀ = M–M only. Co₂(CO)₈ = M–M + 2 bridging CO. This exact contrast is the question.

[ 75 ]^H NEETCarbonyls

In the Mond process, impure nickel is purified by:

(a)Forming Ni(CO)₄ which is then decomposed at higher temperature
(b)Electrolysis of nickel sulphate
(c)Reduction with carbon
(d)Distillation of nickel

Answer — (a) Forming Ni(CO)₄ which is then decomposed at higher temperature

Given

Purification of nickel.

Asked

Basis of the Mond process.

Concept

A volatile coordination compound is formed selectively from the impure metal and then thermally decomposed to give the pure metal — a classic industrial use of complex formation.

Formula

Ni + 4CO ⟶^{330 K} Ni(CO)₄ ⟶^{450–470 K} Ni + 4CO

Baby steps

Impure nickel is warmed with CO at about 330 K.
Only nickel forms the volatile Ni(CO)₄; impurities are left behind.
The vapour is passed to a hotter zone (450–470 K).
Ni(CO)₄ decomposes, depositing pure nickel and releasing CO for reuse.

ShortcutMond's process = Ni. Van Arkel = Ti and Zr (iodide). Do not mix the two up.

Stability & Chelate Effect

05 Q

[ 76 ]^I NEETStability

Amongst the following, the most stable complex is:

(a)[Fe(H₂O)₆]³⁺
(b)[Fe(NH₃)₆]³⁺
(c)[Fe(C₂O₄)₃]³⁻
(d)[FeCl₆]³⁻

Answer — (c) [Fe(C₂O₄)₃]³⁻

Given

Four Fe(III) octahedral complexes.

Asked

The most stable one.

Concept

Chelate effect: complexes with didentate or polydentate ligands forming rings are far more stable than comparable complexes with unidentate ligands, mainly because chelation increases entropy.

Formula

Stability: unidentate < didentate < polydentate

Baby steps

H₂O, NH₃ and Cl⁻ are all unidentate.
Oxalate is didentate and forms five-membered chelate rings.
Three oxalate ligands release six water molecules, giving a large positive ΔS.
Hence [Fe(C₂O₄)₃]³⁻ is the most stable.

ShortcutSpot the chelating ligand (en, ox, EDTA, gly) in the options — that option is almost always the stable one.

[ 77 ]^I NEETStability

The chelate effect refers to the fact that:

(a)Chelate complexes are less stable than those with unidentate ligands
(b)Complexes with polydentate ligands are more stable than similar complexes with unidentate ligands
(c)Chelates are always coloured
(d)Chelation always increases the coordination number

Answer — (b) Complexes with polydentate ligands are more stable than similar complexes with unidentate ligands

Given

Definition question.

Asked

Meaning of the chelate effect.

Concept

A polydentate ligand binds through several donor atoms at once, forming rings. Even if a ring-opening step occurs, the ligand stays attached at the other end, so dissociation is far less likely.

Formula

ΔG = ΔH − TΔS; chelation makes ΔS strongly positive

Baby steps

[Ni(NH₃)₆]²⁺ + 3en → [Ni(en)₃]²⁺ + 6NH₃.
Four particles become seven → entropy increases substantially.
A positive ΔS makes ΔG more negative → higher stability constant.
Hence chelate complexes are more stable.

ShortcutChelate effect is mainly an entropy effect. Five- and six-membered rings are the most stable ring sizes.

[ 78 ]^I NEETStability

When excess aqueous KCN is added to copper sulphate solution and H₂S is then passed, no CuS is precipitated because:

(a)[Cu(CN)₄]³⁻ is very stable and gives too few Cu⁺ ions to exceed the K_sp of CuS
(b)CuS is soluble in water
(c)H₂S does not react with Cu²⁺
(d)KCN oxidises Cu²⁺ to Cu³⁺

Answer — (a) [Cu(CN)₄]³⁻ is very stable and gives too few Cu⁺ ions to exceed the K_sp of CuS

Given

CuSO₄ + excess KCN, then H₂S gas.

Asked

Reason no CuS precipitate forms.

Concept

A very high stability constant means the complex barely dissociates, so the free metal ion concentration stays below the level needed to reach the solubility product of the sulphide.

Formula

[Cu⁺][S²⁻] < K_sp(CuS) ⇒ no precipitate

Baby steps

Excess CN⁻ reduces Cu²⁺ to Cu⁺ and forms [Cu(CN)₄]³⁻.
This complex has a very large stability constant.
Free Cu⁺ in solution is therefore extremely low.
The ionic product never exceeds K_sp of CuS → no precipitate.

ShortcutSame logic explains why K₄[Fe(CN)₆] gives no Fe²⁺ test: strong complex → negligible free ion.

[ 79 ]^I JEE MainStability

The stability of the following complexes increases in the order:

(a)[Ni(NH₃)₆]²⁺ < [Ni(en)₃]²⁺ < [Ni(EDTA)]²⁻
(b)[Ni(EDTA)]²⁻ < [Ni(en)₃]²⁺ < [Ni(NH₃)₆]²⁺
(c)[Ni(en)₃]²⁺ < [Ni(NH₃)₆]²⁺ < [Ni(EDTA)]²⁻
(d)All are equally stable

Answer — (a) [Ni(NH₃)₆]²⁺ < [Ni(en)₃]²⁺ < [Ni(EDTA)]²⁻

Given

Three Ni(II) complexes with unidentate, didentate and hexadentate ligands.

Asked

Increasing order of stability.

Concept

Stability rises with denticity because a single polydentate ligand forms more chelate rings and gives a larger entropy gain.

Formula

Denticity: NH₃ = 1, en = 2, EDTA = 6

Baby steps

NH₃ is unidentate → no chelate rings → least stable.
en is didentate → three five-membered rings → more stable.
EDTA is hexadentate → five rings from one ligand → most stable.
Order: NH₃ < en < EDTA complex.

ShortcutHigher denticity → higher stability constant. EDTA complexes top almost every stability list.

[ 80 ]^I NEETStability

Aqueous copper sulphate gives a bright green solution with aqueous KCl but a green precipitate with aqueous KF. This is because:

(a)[CuCl₄]²⁻ is a soluble complex, while F⁻ forms insoluble CuF₂
(b)Cl⁻ is a stronger field ligand than F⁻
(c)Copper is reduced by KCl
(d)Both form the same complex

Answer — (a) [CuCl₄]²⁻ is a soluble complex, while F⁻ forms insoluble CuF₂

Given

CuSO₄(aq) treated separately with KF and KCl.

Asked

Explanation of the two observations.

Concept

Whether a complex forms depends on the ligand. Cl⁻, being larger and more polarisable, displaces water to form a soluble tetrachloridocuprate(II); the small, hard F⁻ instead gives a sparingly soluble simple salt.

Formula

[Cu(H₂O)₄]²⁺ + 4Cl⁻ → [CuCl₄]²⁻ + 4H₂O

Baby steps

In water, copper(II) exists as blue [Cu(H₂O)₄]²⁺.
With KCl, water is replaced by chloride giving bright green [CuCl₄]²⁻ in solution.
With KF, no stable fluoro complex forms; instead green CuF₂ precipitates.
Hence one gives a solution and the other a precipitate.

ShortcutPrecipitate = simple salt formed. Coloured solution = complex ion formed. That distinction answers the question.

Importance & Applications

06 Q

[ 81 ]^J NEETApplications

The metal ions present in chlorophyll, haemoglobin and vitamin B₁₂ are respectively:

(a)Mg, Fe, Co
(b)Fe, Mg, Co
(c)Co, Fe, Mg
(d)Mg, Co, Fe

Answer — (a) Mg, Fe, Co

Given

Three biologically important coordination compounds.

Asked

Their central metal ions.

Concept

Biological coordination compounds are built around specific metals, each tuned to its function.

Formula

—

Baby steps

Chlorophyll, the green photosynthetic pigment → magnesium.
Haemoglobin, the red oxygen carrier of blood → iron.
Vitamin B₁₂ (cyanocobalamine), the anti-pernicious anaemia factor → cobalt.
Order = Mg, Fe, Co.

ShortcutAdd two more to the list: carbonic anhydrase and carboxypeptidase A both contain zinc.

[ 82 ]^J NEETApplications

EDTA is used in the treatment of:

(a)Iron deficiency
(b)Lead poisoning
(c)Cancer
(d)Diabetes

Answer — (b) Lead poisoning

Given

Chelate therapy.

Asked

Medical use of EDTA.

Concept

Chelate therapy removes toxic metal ions from the body by binding them into stable, water-soluble complexes that can be excreted.

Formula

Pb²⁺ + EDTA⁴⁻ → [Pb(EDTA)]²⁻

Baby steps

Lead ions in the body bind to enzymes and cause toxicity.
EDTA⁴⁻ is hexadentate and forms an extremely stable complex with Pb²⁺.
The complex is water soluble and is excreted in urine.
Hence EDTA treats lead poisoning.

ShortcutOther chelating drugs: D-penicillamine and desferrioxime B remove excess copper and iron. cis-platin treats tumours.

[ 83 ]^J NEETApplications

Wilkinson's catalyst, used for the hydrogenation of alkenes, is:

(a)[(Ph₃P)₃RhCl]
(b)[Ni(CO)₄]
(c)K[PtCl₃(C₂H₄)]
(d)[Co(NH₃)₆]Cl₃

Answer — (a) [(Ph₃P)₃RhCl]

Given

An industrial homogeneous catalyst.

Asked

Its formula.

Concept

Coordination compounds act as homogeneous catalysts by reversibly binding substrates at vacant coordination sites.

Formula

—

Baby steps

Wilkinson's catalyst is a rhodium(I) complex.
Its ligands are three triphenylphosphine groups and one chloride.
Formula = [(Ph₃P)₃RhCl].
It catalyses the hydrogenation of alkenes at ordinary temperature and pressure.

ShortcutNamed catalysts worth knowing: Wilkinson (Rh, hydrogenation), Ziegler-Natta (Ti/Al, polymerisation), Zeise's salt (Pt–alkene).

[ 84 ]^J NEETApplications

In black and white photography, undecomposed AgBr is removed by washing with hypo, forming:

(a)[Ag(NH₃)₂]⁺
(b)[Ag(S₂O₃)₂]³⁻
(c)[Ag(CN)₂]⁻
(d)Ag₂S

Answer — (b) [Ag(S₂O₃)₂]³⁻

Given

Fixing step in photography using sodium thiosulphate (hypo).

Asked

The complex ion formed.

Concept

Thiosulphate acts as a ligand and dissolves the insoluble silver halide by forming a soluble complex — the same principle as complexation in metallurgy.

Formula

AgBr + 2S₂O₃²⁻ → [Ag(S₂O₃)₂]³⁻ + Br⁻

Baby steps

Hypo is Na₂S₂O₃, supplying S₂O₃²⁻ ligands.
Each Ag⁺ binds two thiosulphate ligands.
The resulting [Ag(S₂O₃)₂]³⁻ is water soluble and washes away.
The image is thereby fixed.

ShortcutSilver and gold extraction uses the same trick with cyanide: [Ag(CN)₂]⁻ and [Au(CN)₂]⁻, then displacement by zinc.

[ 85 ]^J NEETApplications

The hardness of water is estimated by titration with:

(a)Na₂EDTA
(b)AgNO₃
(c)KMnO₄
(d)NaOH

Answer — (a) Na₂EDTA

Given

Analytical determination of water hardness.

Asked

The titrant used.

Concept

Ca²⁺ and Mg²⁺ both form stable 1:1 chelates with EDTA. Their stability constants differ enough to allow selective estimation.

Formula

Ca²⁺ + EDTA⁴⁻ → [Ca(EDTA)]²⁻

Baby steps

Hardness is caused by dissolved Ca²⁺ and Mg²⁺.
EDTA⁴⁻ binds each of them in a 1:1 ratio.
The endpoint is detected with an indicator such as Eriochrome Black T.
Titrant = Na₂EDTA.

ShortcutEDTA appears three times in this chapter: water hardness, lead poisoning, and as the standard hexadentate ligand.

[ 86 ]^J NEETApplications

Gold is extracted from its ore by forming which coordination entity, before being displaced by zinc?

(a)[Au(CN)₂]⁻
(b)[AuCl₄]⁻
(c)[Au(NH₃)₂]⁺
(d)[Au(S₂O₃)₂]³⁻

Answer — (a) [Au(CN)₂]⁻

Given

Cyanide process (Mac Arthur–Forrest process) for gold.

Asked

The complex formed.

Concept

Metallurgical leaching uses complex formation to bring an unreactive metal into solution, followed by displacement with a more reactive metal.

Formula

4Au + 8CN⁻ + 2H₂O + O₂ → 4[Au(CN)₂]⁻ + 4OH⁻

Baby steps

Gold is treated with dilute cyanide in the presence of air and water.
It dissolves as the linear dicyanidoaurate(I) ion, [Au(CN)₂]⁻.
Zinc, being more electropositive, displaces gold from this solution.
Metallic gold is recovered.

ShortcutElectroplating uses the same complexes ([Ag(CN)₂]⁻, [Au(CN)₂]⁻) because slow ion release gives a smooth, even deposit.