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Coordination Compounds · Sections 5.1–5.3

Werner · Terms
& IUPAC Names

Thirty NEET-pattern questions on the first three sections of the chapter, each worked through Given, Asked, Concept, Formula, Baby Steps and a Shortcut. Attempt with solutions closed, then open to check the reasoning — not just the answer.

30Questions
105.1
105.2
105.3
30Shortcuts

Suggested timing. These three sections are fact-and-rule based, so speed matters more than depth. Target 35–40 seconds per question, giving about 20 minutes for the full set. Anything slower than a minute on a nomenclature question means the alphabetical-order rule or the Latin-name list still needs revision.

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5.1 Werner01–10
5.2 Terms11–20
5.3 IUPAC21–30
5.1

Werner's Theory of Coordination Compounds

10 Q
[ 01 ]W NEET styleAgCl count

1 mol of PtCl4·5NH3 on treatment with excess AgNO3 gives 3 mol of AgCl. The correct formulation of the compound is:

  • (a)[Pt(NH3)5Cl]Cl3
  • (b)[Pt(NH3)5Cl3]Cl
  • (c)[Pt(NH3)4Cl2]Cl2
  • (d)[Pt(NH3)5]Cl4

Answer — (a) [Pt(NH3)5Cl]Cl3

Given
PtCl4·5NH3, 1 mol; AgCl obtained = 3 mol. Pt(IV) has secondary valence 6.
Asked
Werner formulation showing which chlorides lie inside the coordination sphere.
Concept
Only ionisable chloride outside the square bracket precipitates with Ag+. The coordination number of Pt(IV) is fixed at 6, so the bracket must contain exactly six donor groups.
Formula
moles AgCl = number of Cl− outside [ ]
Baby steps
  1. 3 mol AgCl ⇒ 3 Cl− are outside the bracket.
  2. Total Cl in the compound = 4, so 4 − 3 = 1 Cl− sits inside.
  3. Inside the bracket: 5 NH3 + 1 Cl− = 6 groups, matching the secondary valence of 6.
  4. Formula = [Pt(NH3)5Cl]Cl3.
ShortcutFill the bracket to the coordination number using NH3 first, then top up with Cl−. Whatever is left over goes outside.
[ 02 ]W NEET styleConductivity

Which of the following cobalt(III) complexes behaves as a 1:1 electrolyte in aqueous solution?

  • (a)CoCl3·6NH3
  • (b)CoCl3·5NH3
  • (c)CoCl3·4NH3
  • (d)CoCl3·3NH3

Answer — (c) CoCl3·4NH3

Given
Series of cobalt(III) chloride–ammonia complexes; secondary valence of Co(III) = 6.
Asked
The complex giving a 1:1 electrolyte.
Concept
The electrolyte ratio equals (1 complex ion) : (number of counter ions). Conductivity rises with the total number of ions released.
Formula
1:n electrolyte ⇒ complex ion + n counter ions
Baby steps
  1. 6NH3 → [Co(NH3)6]Cl3 → 1:3 electrolyte, 4 ions.
  2. 5NH3 → [Co(NH3)5Cl]Cl2 → 1:2 electrolyte, 3 ions.
  3. 4NH3 → [Co(NH3)4Cl2]Cl → 1:1 electrolyte, 2 ions.
  4. 3NH3 → [Co(NH3)3Cl3] → non-electrolyte, 0 ions.
ShortcutElectrolyte ratio = charge on the complex ion. Charge +1 → 1:1, charge +3 → 1:3.
[ 03 ]W NEET styleValences

The primary and secondary valences of cobalt in [Co(NH3)5Cl]Cl2 are respectively:

  • (a)2 and 6
  • (b)3 and 6
  • (c)3 and 5
  • (d)6 and 3

Answer — (b) 3 and 6

Given
[Co(NH3)5Cl]Cl2.
Asked
Primary valence and secondary valence of Co.
Concept
Primary valence = oxidation number, ionisable, satisfied by negative ions. Secondary valence = coordination number, non-ionisable, satisfied by neutral molecules or negative ions.
Formula
Primary = oxidation state; Secondary = total donor groups inside [ ]
Baby steps
  1. Charge on the complex ion = +2 (two Cl− outside).
  2. Oxidation state: x + 5(0) + (−1) = +2 ⇒ x = +3 → primary valence 3.
  3. Groups inside the bracket = 5 NH3 + 1 Cl− = 6.
  4. Secondary valence 6.
ShortcutCareful: primary valence counts all three chlorides (one inside, two outside), because oxidation state does not care about position.
[ 04 ]W NEET styleFormulation

1 mol of CrCl3·6H2O gives 2 mol of AgCl with excess AgNO3. The correct formula of the compound is:

  • (a)[Cr(H2O)6]Cl3
  • (b)[Cr(H2O)5Cl]Cl2·H2O
  • (c)[Cr(H2O)4Cl2]Cl·2H2O
  • (d)[Cr(H2O)3Cl3]·3H2O

Answer — (b) [Cr(H2O)5Cl]Cl2·H2O

Given
CrCl3·6H2O; 2 mol AgCl per mol; Cr(III) coordination number 6.
Asked
Correct hydrate formulation.
Concept
This is hydrate (solvate) isomerism. Water can either be coordinated to the metal or sit free in the crystal lattice, and the number of ionisable chlorides changes accordingly.
Formula
moles AgCl = number of free Cl−
Baby steps
  1. 2 mol AgCl ⇒ 2 Cl− outside, so 1 Cl− is inside the bracket.
  2. Coordination number must total 6: 1 Cl− + 5 H2O = 6.
  3. The sixth water molecule is left in the lattice, written after a dot.
  4. Formula = [Cr(H2O)5Cl]Cl2·H2O (grey-green form).
ShortcutViolet form = [Cr(H2O)6]Cl3 → 3 AgCl. Grey-green = 2 AgCl. Dark green = [Cr(H2O)4Cl2]Cl·2H2O → 1 AgCl.
[ 05 ]W NEET styleIons

The total number of ions produced in solution by PtCl4·2KCl, which gives no precipitate with AgNO3, is:

  • (a)2
  • (b)3
  • (c)4
  • (d)6

Answer — (b) 3

Given
PtCl4·2KCl; 0 mol AgCl with AgNO3.
Asked
Number of ions on dissociation.
Concept
Zero precipitate means every chloride is locked inside the coordination sphere, leaving only the potassium ions free.
Formula
Total ions = 1 complex ion + counter ions
Baby steps
  1. No AgCl ⇒ all 6 Cl− are inside the bracket.
  2. Formula = K2[PtCl6].
  3. It dissociates as K2[PtCl6] → 2K+ + [PtCl6]2−.
  4. Total ions = 2 + 1 = 3 (a 2:1 electrolyte).
ShortcutThe whole coordination sphere always counts as one ion, no matter how many ligands it holds.
[ 06 ]W NEET styleDouble salt

Which of the following is a double salt and not a coordination compound?

  • (a)K4[Fe(CN)6]
  • (b)KAl(SO4)2·12H2O
  • (c)[Cu(NH3)4]SO4
  • (d)[Ni(CO)4]

Answer — (b) KAl(SO4)2·12H2O

Given
Four compounds formed by combining simpler compounds.
Asked
The double salt.
Concept
Both double salts and complexes form in stoichiometric ratios, but a double salt dissociates completely into its simple ions in water, whereas a complex retains its coordination sphere.
Formula
Double salt → all simple ions; Complex → complex ion survives
Baby steps
  1. Potash alum, KAl(SO4)2·12H2O, gives K+, Al3+ and SO42− freely in water.
  2. It therefore answers the tests for both K+ and Al3+ → double salt.
  3. K4[Fe(CN)6] gives [Fe(CN)6]4−, not free Fe2+ → complex.
  4. The copper ammine and nickel carbonyl are also true complexes.
ShortcutDouble salts to memorise: carnallite KCl·MgCl2·6H2O, Mohr's salt FeSO4·(NH4)2SO4·6H2O, potash alum. They exist only in the solid state.
[ 07 ]W NEET styleIsomers

The green and violet forms of CoCl3·4NH3 both give 1 mol of AgCl and behave as 1:1 electrolytes. Werner explained this by proposing that they are:

  • (a)Different oxidation states of cobalt
  • (b)Isomers with different spatial arrangements of the same groups
  • (c)Double salts of different composition
  • (d)Compounds with different coordination numbers

Answer — (b) Isomers with different spatial arrangements of the same groups

Given
Two compounds with identical empirical formula CoCl3·4NH3 but different colours and identical conductivity.
Asked
Werner's explanation for the difference.
Concept
Identical formula and identical ionisation behaviour but different physical properties points to isomerism — the same groups arranged differently in space around the metal.
Formula
Both = [Co(NH3)4Cl2]Cl
Baby steps
  1. Both give 1 mol AgCl ⇒ both are [Co(NH3)4Cl2]Cl.
  2. Same formula, same conductivity, so composition cannot be the difference.
  3. The two Cl− ligands inside can be adjacent (cis, violet) or opposite (trans, green).
  4. Hence they are geometrical isomers.
ShortcutThis pair is historically important: it let Werner argue for the octahedral shape, since a planar hexagon would have given three isomers, not two.
[ 08 ]W NEET stylePostulates

According to Werner's theory, the secondary valences are satisfied by:

  • (a)Only negative ions
  • (b)Only neutral molecules
  • (c)Neutral molecules or negative ions
  • (d)Only positive ions

Answer — (c) Neutral molecules or negative ions

Given
Werner's postulates.
Asked
What satisfies the secondary valence.
Concept
Secondary valence is the coordination number. Any species with a lone pair can act as the ligand — neutral (NH3, H2O, CO) or anionic (Cl−, CN−). Primary valence, in contrast, is satisfied only by negative ions.
Formula
—
Baby steps
  1. In [Co(NH3)6]3+ all six secondary valences are satisfied by neutral NH3.
  2. In [Fe(CN)6]4− all six are satisfied by anionic CN−.
  3. In [Co(NH3)4Cl2]+ both types appear together.
  4. So the answer is neutral molecules or negative ions.
ShortcutPrimary → negative ions only. Secondary → neutral or negative. That single contrast answers most postulate questions.
[ 09 ]W NEET styleSecondary valence

1 mol of PdCl2·4NH3 gives 2 mol of AgCl with excess AgNO3. The secondary valence of palladium is:

  • (a)2
  • (b)4
  • (c)6
  • (d)8

Answer — (b) 4

Given
PdCl2·4NH3; 2 mol AgCl per mol.
Asked
Secondary valence (coordination number) of Pd.
Concept
Count the groups that remain bonded to the metal after removing everything that precipitated as AgCl.
Formula
Secondary valence = number of donor groups inside [ ]
Baby steps
  1. 2 mol AgCl ⇒ both chlorides are outside the bracket.
  2. Nothing but the four NH3 molecules remains inside.
  3. Formula = [Pd(NH3)4]Cl2.
  4. Secondary valence = 4.
ShortcutPd(II) and Pt(II) are coordination number 4 (square planar); Pt(IV) and Co(III) are coordination number 6.
[ 10 ]W NEET styleConductivity

The correct order of increasing molar conductivity of the following complexes in aqueous solution is:

  • (a)[Co(NH3)3Cl3] < [Co(NH3)4Cl2]Cl < [Co(NH3)5Cl]Cl2 < [Co(NH3)6]Cl3
  • (b)[Co(NH3)6]Cl3 < [Co(NH3)5Cl]Cl2 < [Co(NH3)4Cl2]Cl < [Co(NH3)3Cl3]
  • (c)All four have the same molar conductivity
  • (d)[Co(NH3)4Cl2]Cl < [Co(NH3)3Cl3] < [Co(NH3)6]Cl3 < [Co(NH3)5Cl]Cl2

Answer — (a) [Co(NH3)3Cl3] < [Co(NH3)4Cl2]Cl < [Co(NH3)5Cl]Cl2 < [Co(NH3)6]Cl3

Given
Four cobalt(III) ammine chlorides already written in Werner form.
Asked
Increasing order of molar conductivity.
Concept
Molar conductivity depends on the total number of ions released per formula unit. More ions and higher charges give higher conductivity.
Formula
Λm ∝ number of ions produced
Baby steps
  1. [Co(NH3)3Cl3] → 0 ions, non-electrolyte, lowest.
  2. [Co(NH3)4Cl2]Cl → 2 ions (1:1).
  3. [Co(NH3)5Cl]Cl2 → 3 ions (1:2).
  4. [Co(NH3)6]Cl3 → 4 ions (1:3), highest. Order = option A.
ShortcutConductivity order always follows the AgCl order. More free Cl− → more ions → higher Λm.
5.2

Definitions of Important Terms

10 Q
[ 11 ]D NEET styleDenticity

Which of the following is a hexadentate ligand?

  • (a)Ethane-1,2-diamine
  • (b)Oxalate
  • (c)EDTA4−
  • (d)Glycinate

Answer — (c) EDTA4−

Given
Four common ligands.
Asked
The hexadentate one.
Concept
Denticity is the number of donor atoms a single ligand uses to bind one metal ion simultaneously.
Formula
Denticity = number of ligating donor atoms
Baby steps
  1. en (H2NCH2CH2NH2) has two N donors → didentate.
  2. Oxalate C2O42− has two O donors → didentate.
  3. Glycinate has one N and one O → didentate.
  4. EDTA4− binds through 2 N + 4 O = 6 donor atoms → hexadentate.
ShortcutEDTA alone can fill an entire octahedral coordination sphere — that is why its complexes top every stability list.
[ 12 ]D NEET styleAmbidentate

Which pair of ligands is ambidentate?

  • (a)NH3 and H2O
  • (b)NO2− and SCN−
  • (c)en and ox
  • (d)Cl− and Br−

Answer — (b) NO2− and SCN−

Given
Four pairs of ligands.
Asked
The ambidentate pair.
Concept
An ambidentate ligand has two different donor atoms but coordinates through only one at a time. This is different from a didentate ligand, which uses both donors simultaneously.
Formula
Ambidentate → two possible donors, one used
Baby steps
  1. NH3 and H2O each have a single donor atom → unidentate.
  2. en and ox have two donors that bind together → didentate chelating.
  3. NO2− binds through N (–NO2) or O (–ONO).
  4. SCN− binds through S (thiocyanato-S) or N (thiocyanato-N). Both are ambidentate.
ShortcutAmbidentate → causes linkage isomerism. Didentate → causes chelation. Do not confuse the two.
[ 13 ]D NEET styleHomo/heteroleptic

Which of the following is a homoleptic complex?

  • (a)[Co(NH3)4Cl2]+
  • (b)[Co(NH3)6]3+
  • (c)[Pt(NH3)2Cl2]
  • (d)[Cr(NH3)3(H2O)3]3+

Answer — (b) [Co(NH3)6]3+

Given
Four coordination entities.
Asked
The homoleptic one.
Concept
Homoleptic: the metal is bound to only one kind of donor group. Heteroleptic: more than one kind of donor group is present.
Formula
One ligand type ⇒ homoleptic
Baby steps
  1. [Co(NH3)4Cl2]+ has NH3 and Cl− → heteroleptic.
  2. [Co(NH3)6]3+ has only NH3 → homoleptic.
  3. [Pt(NH3)2Cl2] has two ligand types → heteroleptic.
  4. [Cr(NH3)3(H2O)3]3+ has two ligand types → heteroleptic.
ShortcutOnly heteroleptic complexes can show geometrical isomerism — a homoleptic complex has nothing to arrange differently.
[ 14 ]D NEET styleCoordination no.

The coordination number of the central metal ion is determined by:

  • (a)The total number of sigma bonds formed by the ligands with the metal
  • (b)The total number of sigma and pi bonds together
  • (c)The number of ligand molecules only
  • (d)The oxidation state of the metal

Answer — (a) The total number of sigma bonds formed by the ligands with the metal

Given
Definition of coordination number.
Asked
What is counted.
Concept
Coordination number counts ligand donor atoms bonded by sigma bonds only. Pi bonds, if formed between ligand and metal, are not counted.
Formula
CN = number of sigma-bonded donor atoms
Baby steps
  1. Each donor atom donates a lone pair into a metal orbital → one sigma bond.
  2. A didentate ligand such as en therefore contributes 2, not 1.
  3. In metal carbonyls, the M→CO pi back bond is extra and is not counted.
  4. So Ni(CO)4 has CN = 4, not 8.
ShortcutCount donor atoms, never ligand molecules. [Co(en)3]3+ has 3 ligands but CN = 6.
[ 15 ]D NEET styleChelate

Which of the following complexes contains a chelate ring?

  • (a)[Co(NH3)6]3+
  • (b)[Ni(CO)4]
  • (c)[Cr(C2O4)3]3−
  • (d)[CoCl4]2−

Answer — (c) [Cr(C2O4)3]3−

Given
Four coordination entities.
Asked
The one containing a chelate ring.
Concept
A chelate forms when a di- or polydentate ligand uses two or more donor atoms to grip the same metal ion, closing a ring.
Formula
Chelate ⇒ ring formed by one ligand + metal
Baby steps
  1. NH3, CO and Cl− are all unidentate → no ring possible.
  2. Oxalate C2O42− is didentate, binding through two oxygen atoms.
  3. Each oxalate plus the Cr centre closes a five-membered ring.
  4. So [Cr(C2O4)3]3− is the chelate complex.
ShortcutChelate complexes are always more stable than comparable unidentate ones — that is the chelate effect, driven mainly by entropy.
[ 16 ]D NEET styleCoordination sphere

In the complex K4[Fe(CN)6], the coordination sphere and the counter ion are respectively:

  • (a)K+ and [Fe(CN)6]4−
  • (b)[Fe(CN)6]4− and K+
  • (c)Fe2+ and CN−
  • (d)CN− and Fe2+

Answer — (b) [Fe(CN)6]4− and K+

Given
K4[Fe(CN)6].
Asked
Identify the coordination sphere and the counter ion.
Concept
The coordination sphere is the central atom together with its attached ligands, enclosed in square brackets. Ionisable groups written outside the bracket are counter ions.
Formula
[ metal + ligands ] = coordination sphere; outside = counter ion
Baby steps
  1. Inside the bracket: Fe2+ with six CN− ligands.
  2. Charge: x + 6(−1) = −4 ⇒ x = +2, so the sphere is [Fe(CN)6]4−.
  3. Outside the bracket: four K+ ions.
  4. Coordination sphere = [Fe(CN)6]4−; counter ion = K+.
ShortcutThe coordination sphere does not dissociate in water — which is why this compound gives no Fe2+ test.
[ 17 ]D NEET styleLewis acid

In a coordination entity, the central metal atom or ion acts as a:

  • (a)Lewis base
  • (b)Lewis acid
  • (c)Bronsted acid
  • (d)Reducing agent

Answer — (b) Lewis acid

Given
Metal–ligand interaction in a coordination entity.
Asked
Role of the central metal.
Concept
A ligand donates a lone pair (electron pair donor = Lewis base); the metal accepts it into a vacant orbital (electron pair acceptor = Lewis acid).
Formula
M (vacant orbital) ← :L (lone pair)
Baby steps
  1. Ligands such as NH3 and CN− carry lone pairs.
  2. The metal ion has vacant d, s and p orbitals available for hybridisation.
  3. The lone pair is donated into these vacant orbitals, forming a coordinate bond.
  4. Metal = electron pair acceptor = Lewis acid; ligand = Lewis base.
ShortcutEvery ligand is a Lewis base, every central metal ion is a Lewis acid. No exceptions in this chapter.
[ 18 ]D NEET stylePolyhedron

The coordination polyhedra of [Co(NH3)6]3+, [Ni(CO)4] and [PtCl4]2− are respectively:

  • (a)Octahedral, tetrahedral, square planar
  • (b)Tetrahedral, octahedral, square planar
  • (c)Octahedral, square planar, tetrahedral
  • (d)Square planar, tetrahedral, octahedral

Answer — (a) Octahedral, tetrahedral, square planar

Given
Three coordination entities with coordination numbers 6, 4 and 4.
Asked
Their coordination polyhedra.
Concept
The coordination polyhedron is the spatial arrangement of the donor atoms around the metal. CN 6 gives octahedral; CN 4 gives either tetrahedral or square planar depending on the metal and ligand.
Formula
CN 6 → octahedral; CN 4 → tetrahedral or square planar
Baby steps
  1. [Co(NH3)6]3+: CN 6 → octahedral.
  2. [Ni(CO)4]: CN 4, Ni(0) with d10 → sp3 → tetrahedral.
  3. [PtCl4]2−: CN 4, Pt(II) d8 → dsp2 → square planar.
  4. Order = octahedral, tetrahedral, square planar.
ShortcutFor CN 4: d8 metals (Ni2+, Pd2+, Pt2+) with strong ligands go square planar; everything else goes tetrahedral.
[ 19 ]D NEET styleCoordination no.

The coordination number of cobalt in [Co(en)2(C2O4)]+ is:

  • (a)3
  • (b)4
  • (c)6
  • (d)8

Answer — (c) 6

Given
[Co(en)2(C2O4)]+; en and oxalate are both didentate.
Asked
Coordination number of Co.
Concept
Multiply each ligand by its denticity and add. Never count ligand molecules directly.
Formula
CN = Σ(number of ligands × denticity)
Baby steps
  1. Two en ligands, each didentate → 2 × 2 = 4 donor atoms.
  2. One oxalate, didentate → 1 × 2 = 2 donor atoms.
  3. Total donor atoms = 4 + 2 = 6.
  4. So Co has CN 6 and an octahedral polyhedron, even though only three ligand molecules are present.
ShortcutThree chelating ligands almost always mean CN = 6. Count donors, not molecules.
[ 20 ]D NEET styleOxidation no.

The oxidation number of the central atom is defined as the charge it would carry if:

  • (a)All the ligands are removed along with the electron pairs shared with the metal
  • (b)All the ligands are removed leaving their electron pairs on the metal
  • (c)Only the anionic ligands are removed
  • (d)The counter ions are removed

Answer — (a) All the ligands are removed along with the electron pairs shared with the metal

Given
Definition of oxidation number in a coordination entity.
Asked
The correct definition.
Concept
By convention the coordinate bond electrons are assigned back to the ligand, since the ligand donated them. Whatever charge is then left on the metal is its oxidation number.
Formula
Oxidation number written as a Roman numeral after the metal name
Baby steps
  1. Strip away each ligand together with the lone pair it donated.
  2. The metal is left with only its own electrons.
  3. The residual charge is the oxidation number.
  4. Example: in [Cu(CN)4]3−, removing four CN− leaves Cu+, written as Cu(I).
ShortcutOxidation number can be zero or even negative (Ni(CO)4 is Ni(0)). Coordination number never can.
5.3

IUPAC Nomenclature

10 Q
[ 21 ]N NEET styleNaming

The IUPAC name of [Cr(NH3)3(H2O)3]Cl3 is:

  • (a)Triaquatriamminechromium(III) chloride
  • (b)Triamminetriaquachromium(III) chloride
  • (c)Triamminetriaquachromate(III) chloride
  • (d)Trichloridotriamminechromium(III) hydrate

Answer — (b) Triamminetriaquachromium(III) chloride

Given
[Cr(NH3)3(H2O)3]Cl3.
Asked
IUPAC name.
Concept
Ligands are named in alphabetical order of their ligand names, ignoring the multiplying prefixes. The complex here is a cation, so the metal keeps its ordinary English name.
Formula
x + 3(0) + 3(0) = +3
Baby steps
  1. Ligand names: ammine and aqua → compare next letters, m before q, so ammine comes first.
  2. Three of each → triammine, triaqua.
  3. All ligands are neutral and three Cl− lie outside, so the complex is +3 ⇒ Cr(III).
  4. Name = triamminetriaquachromium(III) chloride.
ShortcutAlphabetise the ligand name, not the prefix. Here you must go to the second letter: ammine before aqua.
[ 22 ]N NEET styleFormula

The formula of potassium tetrachloridopalladate(II) is:

  • (a)K[PdCl4]
  • (b)K2[PdCl4]
  • (c)[Pd(KCl)4]
  • (d)K4[PdCl2]

Answer — (b) K2[PdCl4]

Given
Name: potassium tetrachloridopalladate(II).
Asked
Formula.
Concept
The -ate suffix marks an anionic complex, so the potassium is a counter cation written outside and in front. Balance the charges.
Formula
Charge on complex = oxidation number + Σ(ligand charges)
Baby steps
  1. Pd is in +2; four Cl− ligands contribute −4.
  2. Charge on complex = +2 − 4 = −2 → [PdCl4]2−.
  3. Two K+ are needed to neutralise it.
  4. Formula = K2[PdCl4].
ShortcutWhen writing a formula: metal first inside the bracket, then ligands alphabetically. When writing a name, the order reverses.
[ 23 ]N NEET styleNaming

The IUPAC name of [Ag(NH3)2][Ag(CN)2] is:

  • (a)Diamminesilver(I) dicyanidoargentate(I)
  • (b)Diamminesilver(II) dicyanidoargentate(0)
  • (c)Dicyanidoargentate(I) diamminesilver(I)
  • (d)Diamminesilver(I) dicyanidosilver(I)

Answer — (a) Diamminesilver(I) dicyanidoargentate(I)

Given
[Ag(NH3)2][Ag(CN)2] — both cation and anion are complex.
Asked
IUPAC name.
Concept
When both ions are complex, name the cation first, then the anion. The anionic complex takes the -ate suffix and, for silver, the Latin stem argentate.
Formula
Cation: x + 2(0) = +1; Anion: x + 2(−1) = −1
Baby steps
  1. Cation [Ag(NH3)2]+: NH3 neutral ⇒ Ag is +1 → diamminesilver(I).
  2. Anion [Ag(CN)2]−: x − 2 = −1 ⇒ Ag is +1.
  3. Anionic silver uses the Latin name → dicyanidoargentate(I).
  4. Name = diamminesilver(I) dicyanidoargentate(I).
ShortcutLatin stems appear only in anions: Fe→ferrate, Cu→cuprate, Ag→argentate, Au→aurate, Pb→plumbate, Sn→stannate.
[ 24 ]N NEET styleFormula

The formula of dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate is:

  • (a)[PtCl2(en)2](NO3)2
  • (b)[PtCl2(en)2]NO3
  • (c)[Pt(en)2](NO3)4
  • (d)K2[PtCl2(en)2]

Answer — (a) [PtCl2(en)2](NO3)2

Given
Name specifies Pt(IV), two Cl−, two en, nitrate counter ion.
Asked
Formula.
Concept
‘Bis’ signals two en ligands (used because the ligand name already contains a numerical prefix). Balance the charge to find the number of nitrates.
Formula
Charge = +4 + 2(−1) + 2(0)
Baby steps
  1. Pt is +4; two Cl− give −2; two neutral en give 0.
  2. Charge on complex = +4 − 2 = +2.
  3. Two NO3− are needed outside.
  4. Formula = [PtCl2(en)2](NO3)2.
Shortcut‘bis’ = 2, ‘tris’ = 3, ‘tetrakis’ = 4 — used only when the ligand name itself contains di/tri (en, bipy, PPh3).
[ 25 ]N NEET styleNaming

The IUPAC name of [Co(NH3)4Cl(NO2)]Cl is:

  • (a)Tetraamminechloridonitrito-N-cobalt(III) chloride
  • (b)Tetraamminechloridonitrito-N-cobalt(II) chloride
  • (c)Chloridotetraamminenitrito-N-cobaltate(III) chloride
  • (d)Tetraamminenitritochloridocobalt(III) chloride

Answer — (a) Tetraamminechloridonitrito-N-cobalt(III) chloride

Given
[Co(NH3)4Cl(NO2)]Cl.
Asked
IUPAC name.
Concept
Alphabetical ligand order, ambidentate NO2− written with its donor atom specified, then the metal with its oxidation state, then the counter anion.
Formula
x + 4(0) + (−1) + (−1) = +1
Baby steps
  1. Alphabetical: ammine, chlorido, nitrito.
  2. NO2− written as NO2 means it is N-bonded → nitrito-N.
  3. Charge on complex = +1 (one Cl− outside). x − 1 − 1 = +1 ⇒ x = +3.
  4. Name = tetraamminechloridonitrito-N-cobalt(III) chloride.
ShortcutWith ambidentate ligands the donor atom must be stated: nitrito-N or nitrito-O; thiocyanato-S or thiocyanato-N.
[ 26 ]N NEET styleFormula

The formula of iron(III) hexacyanidoferrate(II) is:

  • (a)Fe3[Fe(CN)6]4
  • (b)Fe4[Fe(CN)6]3
  • (c)Fe[Fe(CN)6]
  • (d)Fe2[Fe(CN)6]3

Answer — (b) Fe4[Fe(CN)6]3

Given
Cation Fe3+, anion [Fe(CN)6]4−.
Asked
Formula.
Concept
The metal named as the element is the cation; the metal named with -ate is inside the anionic complex. Balance total positive and negative charge.
Formula
Charge of anion: x + 6(−1) = charge, with x = +2
Baby steps
  1. Hexacyanidoferrate(II): +2 − 6 = −4 → [Fe(CN)6]4−.
  2. Cation is iron(III) → Fe3+.
  3. Criss-cross: four Fe3+ (+12) balance three [Fe(CN)6]4− (−12).
  4. Formula = Fe4[Fe(CN)6]3 (Prussian blue).
ShortcutTwo different Roman numerals for the same metal is the clue that one iron is inside the bracket and one is outside.
[ 27 ]N NEET styleNaming

The IUPAC name of K2[Zn(OH)4] is:

  • (a)Potassium tetrahydroxidozincate(II)
  • (b)Potassium tetrahydroxidozinc(II)
  • (c)Dipotassium tetrahydroxidozincate(II)
  • (d)Potassium tetrahydroxozincium(II)

Answer — (a) Potassium tetrahydroxidozincate(II)

Given
K2[Zn(OH)4].
Asked
IUPAC name.
Concept
Anionic complex → metal takes -ate. OH− is named hydroxido. The number of counter ions is never stated in the name of an ionic compound.
Formula
x + 4(−1) = −2
Baby steps
  1. Cation named first: potassium.
  2. Four OH− → tetrahydroxido.
  3. Complex is an anion → zincate; oxidation state x − 4 = −2 ⇒ x = +2.
  4. Name = potassium tetrahydroxidozincate(II).
ShortcutNever write ‘dipotassium’ or ‘trichloride’ in a coordination compound name — counter ion counts are always omitted.
[ 28 ]N NEET styleRules

While writing the formula of a coordination entity, the correct sequence is:

  • (a)Ligands first in alphabetical order, then the central atom
  • (b)Central atom first, then ligands in alphabetical order, all enclosed in square brackets
  • (c)Central atom first, then anionic ligands followed by neutral ligands
  • (d)Ligands in order of increasing charge, then the central atom

Answer — (b) Central atom first, then ligands in alphabetical order, all enclosed in square brackets

Given
IUPAC rules for writing formulas.
Asked
The correct sequence.
Concept
Formula writing and name writing follow opposite orders. In the formula the metal leads; in the name the ligands lead.
Formula
[ M(ligands alphabetical) ]charge
Baby steps
  1. The central atom is listed first inside the bracket.
  2. Ligands follow in alphabetical order, regardless of their charge.
  3. Polyatomic ligands and abbreviations are enclosed in parentheses.
  4. The whole entity is enclosed in square brackets, with the charge as a right superscript.
ShortcutFormula: metal → ligands. Name: ligands → metal. This reversal is tested directly almost every year.
[ 29 ]N NEET styleNaming

The IUPAC name of [Pt(NH3)2Cl(NO2)] is:

  • (a)Diamminechloridonitrito-N-platinum(II)
  • (b)Diamminechloridonitrito-N-platinate(II)
  • (c)Diamminechloridonitrito-O-platinum(IV)
  • (d)Chloridodiamminenitrito-N-platinum(II)

Answer — (a) Diamminechloridonitrito-N-platinum(II)

Given
[Pt(NH3)2Cl(NO2)], a neutral complex.
Asked
IUPAC name.
Concept
A neutral complex molecule is named like a complex cation — the metal keeps its ordinary name with no -ate suffix and no counter ion is mentioned.
Formula
x + 2(0) + (−1) + (−1) = 0
Baby steps
  1. Alphabetical ligand order: ammine, chlorido, nitrito.
  2. NO2 written with N first ⇒ N-bonded → nitrito-N.
  3. Complex is neutral: x − 2 = 0 ⇒ x = +2.
  4. Name = diamminechloridonitrito-N-platinum(II).
ShortcutNo square-bracket charge and nothing written outside ⇒ neutral complex ⇒ no -ate, no counter ion in the name.
[ 30 ]N NEET styleRules

According to the 2004 IUPAC recommendations, the chloride ligand in a complex is named as:

  • (a)Chloro
  • (b)Chlorido
  • (c)Chlorate
  • (d)Chlorine

Answer — (b) Chlorido

Given
Modern IUPAC convention for anionic ligands.
Asked
Correct name for Cl− as a ligand.
Concept
Anionic ligand names now end in –ido rather than the older –o. So chloro becomes chlorido, cyano becomes cyanido, hydroxo becomes hydroxido.
Formula
Anionic ligand → ends in –ido
Baby steps
  1. Older texts used chloro, bromo, cyano, hydroxo.
  2. The 2004 draft changed these to chlorido, bromido, cyanido, hydroxido.
  3. NCERT follows the newer form throughout Unit 5.
  4. Answer = chlorido.
ShortcutUpdated names: Cl− chlorido, Br− bromido, CN− cyanido, OH− hydroxido, H− hydrido. Neutral names are unchanged: aqua, ammine, carbonyl, nitrosyl.