Coordination Compounds

Werner's two valencies. Primary valency is the oxidation state, satisfied by the ions written outside the square bracket. Secondary valency is the coordination number, satisfied by ligands inside it.

Only species outside the bracket ionise, so only those precipitate with silver nitrate. Chloride bound inside the sphere is invisible to AgNO₃.

Cobalt(III) has a coordination number of 6 — assume this unless a question forces otherwise. It is the single most-used fact in this chapter.

1. 2 mol AgCl means 2 chlorides outside. So the formula is [Co(NH₃)_xCl_y−2]Cl₂.
2. Check the ion count: 1 complex cation + 2 chloride anions = 3 ions. ✓ That confirms the split rather than adding new information.
3. Charge balance: the cation must be 2+ to pair with two Cl⁻. With Co in the +3 state, +3 − (y−2) = +2, so y − 2 = 1 and y = 3.
4. Fill the coordination sphere to 6: x + (y−2) = 6 → x + 1 = 6 → x = 5.

Secondary valency = coordination number = the number of donor atoms inside the square bracket, whatever they happen to be.

The procedure is mechanical: the AgCl count tells you how many chlorides are outside; everything else in the formula must be inside; then count the donor atoms inside.

The trap in IV is forgetting that the chlorides left over after the ionisable ones are removed are still ligands, and still count.

1. Compound I, PdCl₂·4NH₃: 2 AgCl, so both chlorides are outside. Structure is [Pd(NH₃)₄]Cl₂.
2. Count inside for I: 4 NH₃, nothing else. Secondary valency = 4.
3. Compound IV, CoCl₃·4NH₃: only 1 AgCl, so 1 chloride outside and 2 chlorides inside. Structure is [Co(NH₃)₄Cl₂]Cl.
4. Count inside for IV: 4 NH₃ + 2 Cl = 6. So the pair is 4 and 6.

Assertion. The two chlorides can only be adjacent (cis) or opposite (trans). There is no third arrangement — on an octahedron, any two positions are either 90° apart or 180° apart. So the count is 2, not 3, and A is false.

Where the 3 comes from: the cis form is chiral and exists as a d/l pair. Add those to trans and you get 3 stereoisomers. But optical isomers are not geometrical isomers, and A says geometrical.

Reason. Co(III) with a coordination number of 6 is octahedral. True, and independently so.

1. Judge A alone. Place the first Cl anywhere; the second is either at 90° (cis) or 180° (trans). Two options. A claims three — false.
2. Judge R alone. Co(III), coordination number 6 → octahedral. True.
3. A false, R true → the option that says exactly that.
4. The explanation clause never comes into play: once A is false, whether R explains it is not asked.

A ligand is symmetrical when its donor atoms are equivalent — whichever end binds, the result is the same.

Oxalate donates through two identical oxygens. en donates through two identical nitrogens. EDTA's donor set is symmetric about its centre.

Thiocyanate is the odd one: S–C≡N. It can bind through sulphur (thiocyanato, –SCN) or through nitrogen (isothiocyanato, –NCS), and those give genuinely different compounds. Two different donor atoms means unsymmetrical — and it is the same property that makes it ambidentate.

1. Oxalate: two O donors, identical → symmetrical.
2. en: two N donors, identical → symmetrical.
3. EDTA: 2 N + 4 O, arranged symmetrically about the backbone → symmetrical.
4. SCN⁻: donor is S at one end, N at the other → unsymmetrical.

Coordination number counts donor atoms, not ligands. This is the single most common slip in the chapter. Three ligands, all bidentate, gives six donor atoms.

Oxidation number comes from charge balance. One chloride sits outside, so the complex ion carries +1. Inside, en contributes 0 and oxalate contributes −2.

1. Denticities: en is bidentate, oxalate is bidentate.
2. CN = (2 × 2) + (1 × 2) = 6. Six donor atoms, from three ligands.
3. Charge on the complex ion: one Cl⁻ outside means +1 inside.
4. x + 0 + 0 + (−2) = +1 → x = +3. Sum = 6 + 3 = 9.

A chelate is a ligand that grips the metal at two or more points and thereby closes a ring. Two conditions, and both are needed: polydentate, and actually forming the ring.

Charge is irrelevant. Carbonate carries 2− just like oxalate, but at NCERT level it is treated as monodentate — one oxygen bound, no ring.

1. en — two N donors, closes a five-membered ring. Chelate.
2. Oxalate — two O donors, closes a five-membered ring. Chelate.
3. H₂O — one donor atom. Cannot chelate.
4. EDTA⁴⁻ — six donors (2 N, 4 O), the textbook chelator. Chelate. CO₃²⁻ is taken as monodentate. Total = 3.

Check each type against a specific structural feature. Do not judge them as a group.

Linkage needs an ambidentate ligand. NO₂⁻ binds through N (nitro) or O (nitrito). ✓

Ionisation needs an exchangeable pair — something inside that could swap with the counter ion. Cl⁻ outside can trade with NO₂⁻ inside, giving [Co(NH₃)₄(NO₂)Cl]NO₂. ✓

Geometrical needs an MA₄B₂ octahedron. Cis and trans both exist. ✓

Optical needs no plane of symmetry anywhere. With four identical ammines, even the cis form has a mirror plane, so it is achiral. ✗

1. Look for an ambidentate ligand: NO₂⁻ is there. Linkage ✓
2. Look outside the bracket: Cl⁻ can exchange with an inner NO₂⁻. Ionisation ✓
3. Classify the octahedron: MA₄B₂, so cis/trans exist. Geometrical ✓
4. Test for chirality: with four identical NH₃, both cis and trans have a mirror plane. Optical ✗. That leaves exactly one option.

Nuclearity is just the number of metal atoms. Read the formula and count. Fe appears once, so mononuclear.

The five carbonyls are a distraction — they set the coordination number (5, trigonal bipyramidal), not the nuclearity.

For contrast, the other carbonyls NCERT names: Co₂(CO)₈ is dinuclear, Fe₂(CO)₉ is dinuclear, and Fe₃(CO)₁₂ or Co₄(CO)₁₂ are poly-nuclear.

1. Count the metal atoms in Fe(CO)₅: one.
2. One metal atom → mononuclear.
3. Do not let the subscript 5 pull you toward another answer — it counts ligands.
4. Shape, for completeness: trigonal bipyramidal, dsp³, diamagnetic.

Statement I. Ti in [TiF₆]³⁻ is +3, so d¹. That single electron goes into t_2g, leaving two d orbitals free for d²sp³ hybridisation — inner orbital. One unpaired electron, so paramagnetic. Both claims correct.

Note that F⁻ being a weak-field ligand does not matter here. Weak versus strong only decides how electrons distribute when there is a choice, and with d¹ there is none.

Statement II. Co(III) is d⁶ and NH₃ is strong-field, so all six electrons pair in t_2g. That gives d²sp³ — inner orbital, not outer — and zero unpaired electrons, so diamagnetic, not paramagnetic. Wrong on both counts.

1. Statement I: Ti(III) → d¹ → one unpaired → paramagnetic; two empty d orbitals available → inner orbital. Correct.
2. Statement II: Co(III) → d⁶; NH₃ strong field → t_2g⁶e_g⁰.
3. That configuration is inner orbital (d²sp³) and has n = 0, so diamagnetic. The statement says outer and paramagnetic — both wrong.
4. I correct, II incorrect.

On an octahedron there are only two possible angles between two ligands: 90° (adjacent) or 180° (opposite). No third value exists, which is exactly why only cis and trans exist.

90° = adjacent = cis. 180° = opposite = trans. The question is a vocabulary check dressed as geometry.

1. Recognise the geometry: six ligands around Co(III) → octahedral.
2. List the only two possible ligand–ligand angles: 90° and 180°.
3. 90° means the two chlorides are next to each other.
4. Adjacent identical ligands → cis.

Read an IUPAC name as a set of instructions, left to right:

tetrammine = 4 NH₃, a ligand, so inside the bracket. diaqua = 2 H₂O, also a ligand, also inside. copper(II) = Cu²⁺, the central metal. hydroxide is named after the metal as a separate word — that makes it the counter ion, outside the bracket.

The rule that decides this question: anything named as a separate word after the metal is outside the coordination sphere. Ligands are prefixes joined to the metal name.

Charge check: Cu is +2, both ligands are neutral, so the complex ion is 2+ and needs two OH⁻ outside.

1. Collect the ligands: 4 NH₃ and 2 H₂O, both inside the bracket → [Cu(NH₃)₄(H₂O)₂].
2. “Hydroxide” is a separate word after the metal, so it is the counter ion, outside.
3. Balance the charge: Cu²⁺ with neutral ligands gives a 2+ ion, needing 2 OH⁻.
4. [Cu(NH₃)₄(H₂O)₂](OH)₂. Note the distractors write water as ·2H₂O, which is water of crystallisation, a different thing entirely from an aqua ligand.

In a square plane every position is equivalent to every other by rotation, so isomerism only appears when there is a genuine choice of relative placement.

With one odd ligand (Ma₃b or Mab₃), wherever you put it the other three fill the rest — rotate the molecule and every arrangement looks the same. One structure only.

With two of each, the pair can sit adjacent (cis) or opposite (trans), and no rotation converts one into the other. Two distinct compounds.

1. Ma₄: all four identical, nothing to arrange. One structure.
2. Ma₃b: the single b has no distinguishable position. One structure.
3. Mab₃: same argument with the roles swapped. One structure.
4. Ma₂b₂: the two b's are either adjacent (cis) or opposite (trans) → two isomers. Cisplatin and transplatin are exactly this case.

Every one of these is Co(III), octahedral, d⁶. Metal, oxidation state and geometry are all identical, so the ligand alone decides Δ₀. That is the question telling you which variable to look at.

The spectrochemical series ranks ligands by field strength. The order to hold in memory, weak to strong:

I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO

CN⁻ sits at the strong end, second only to CO.

1. Confirm what is constant: Co(III), octahedral, d⁶ in all four options.
2. Place each ligand on the series: oxalate < H₂O < NH₃ < CN⁻.
3. The strongest field gives the largest splitting.
4. [Co(CN)₆]³⁻ has the highest Δ₀.

Do not match by elimination. Work out shape and moment for each complex independently, write them down, and only then look at List-II.

A [Ni(CO)₄] — CO is neutral, so Ni is 0, 3d¹⁰4s². CO is a strong field ligand: the 4s electrons shift into 3d, giving 3d¹⁰ — completely filled, so sp³ tetrahedral, 0 BM.

B [Ni(CN)₄]²⁻ — Ni(II), d⁸. CN⁻ is strong, so the two unpaired d electrons pair up, freeing one 3d orbital: dsp² square planar, 0 BM.

C [NiCl₄]²⁻ — also Ni(II), d⁸, but Cl⁻ is weak, so no pairing. Two unpaired electrons: sp³ tetrahedral, μ = √8 = 2.83 → 2.8 BM.

D [MnBr₄]²⁻ — Mn(II), d⁵, Br⁻ weak. Five unpaired: tetrahedral, μ = √35 = 5.92 → 5.9 BM.

1. Get the oxidation state and d-count for each: Ni(0) d¹⁰, Ni(II) d⁸, Ni(II) d⁸, Mn(II) d⁵.
2. Classify each ligand as strong or weak: CO strong, CN⁻ strong, Cl⁻ weak, Br⁻ weak.
3. Count unpaired electrons: 0, 0, 2, 5. Convert with √(n(n+2)): 0, 0, 2.83, 5.92 BM.
4. Now match. The two 0 BM entries are separated by shape — A is tetrahedral (III), B is square planar (II). So A-III, B-II, C-I, D-IV.

Diamagnetic means zero unpaired electrons. Three things decide it, in this order: the oxidation state (giving the d-count), the ligand field strength, and then the filling.

A Mn(III), d⁴, CN⁻ strong → t_2g⁴: one orbital holds a lone electron, and a second holds another. 2 unpaired, paramagnetic.

B Co(III), d⁶, NH₃ strong → t_2g⁶e_g⁰. 0 unpaired, diamagnetic. ✓

C Fe(II), d⁶, CN⁻ strong → same filling. 0 unpaired, diamagnetic. ✓

D Co(III), d⁶, but H₂O and F⁻ are both weak → high spin t_2g⁴e_g². 4 unpaired, paramagnetic.

1. Fix the oxidation state and d-count for each: Mn(III) d⁴, Co(III) d⁶, Fe(II) d⁶, Co(III) d⁶.
2. Classify each ligand set: CN⁻ strong, NH₃ strong, CN⁻ strong, H₂O/F⁻ weak.
3. Only d⁶ with a strong field reaches zero unpaired — that rules A out (wrong d-count) and D out (wrong field).
4. B and C both qualify.

This is straight NCERT recall of the applications section, and the four compounds pair with four different metals:

• EDTA — lead poisoning. Given as Ca-EDTA; the calcium is displaced by the more strongly bound Pb²⁺, which is then excreted.
• D-penicillamine — copper poisoning, and Wilson's disease.
• Desferrioxime B — iron overload.
• Cis-platin — not a chelating agent at all; it is an anti-tumour drug that binds DNA.

The principle behind all three chelators is the same: a ligand that binds the toxic metal far more strongly than the body's own ligands do, forming a stable water-soluble complex that the kidneys can clear.

1. Recognise cis-platin as the odd one out: an anti-cancer drug, not a chelator. Eliminate.
2. Desferrioxime B is the iron chelator. Eliminate.
3. D-penicillamine is the copper chelator. Eliminate.
4. EDTA is left, and is indeed the lead antidote.

The charge on a coordination entity is simply the metal's oxidation state plus the charges of its ligands. Nothing constrains that sum to a sign.

One example of each is enough to kill the three "always" options:

• Cationic — [Co(NH₃)₆]³⁺: +3 with six neutral ligands.
• Anionic — [Fe(CN)₆]⁴⁻: +2 − 6 = −4.
• Neutral — [Ni(CO)₄]: metal at 0 with neutral ligands. Also [Pt(NH₃)₂Cl₂], +2 − 2 = 0.

1. Test “always neutral” against [Co(NH₃)₆]³⁺. Fails.
2. Test “always anionic” against the same example. Fails.
3. Test “always cationic” against [Fe(CN)₆]⁴⁻. Fails.
4. All three sign restrictions are refuted, so the inclusive option is correct.