RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Chemistry · 11 questions · 26 marks
Why is one cobalt complex orange and another violet when both contain the same metal and the same number of ligands? Coordination chemistry answers that with Werner's theory of primary and secondary valencies and with crystal field theory, which explains colour, magnetism and stability from the splitting of d orbitals. You will also learn to name complexes by IUPAC rules and to spot the several kinds of isomerism they show.
The IUPAC name of K₃[Fe(C₂O₄)₃] is:
Answer
Potassium trioxalatoferrate(III). The complex ion carries a 3− charge and three oxalate ligands each carry 2−, so iron is +3 (x − 6 = −3). Because the complex ion is an anion, the metal takes the Latin stem with the -ate ending, giving ferrate(III), and the cation is named first.
Which complex is expected to be diamagnetic?
Answer
[Co(NH₃)₆]³⁺. Co³⁺ is d⁶, and NH₃ is a strong-field ligand, so Δ₀ exceeds the pairing energy and all six electrons pair up in the t₂g set, giving t₂g⁶ e_g⁰ with no unpaired electrons. Fluoride is a weak-field ligand so [CoF₆]³⁻ is high spin with four unpaired electrons, and both [Fe(H₂O)₆]²⁺ and tetrahedral [NiCl₄]²⁻ are paramagnetic.
One mole of a cobalt(III) complex with the empirical formula CoCl₃·5NH₃ gives two moles of AgCl on treatment with excess silver nitrate. The correct formula is:
Answer
[Co(NH₃)₅Cl]Cl₂. Only chloride outside the coordination sphere is free to be precipitated as AgCl, so two moles of precipitate means two ionisable chlorides. The third chloride must be coordinated to cobalt, giving a coordination number of 6 with five ammonia molecules and one chloride inside the sphere.
Which of the following pairs are linkage isomers?
Answer
[Co(NH₃)₅(NO₂)]Cl₂ and [Co(NH₃)₅(ONO)]Cl₂. Linkage isomerism arises when an ambidentate ligand binds through a different donor atom — here nitrite bonds through nitrogen in one isomer and through oxygen in the other. The second pair is ionisation isomerism, the third hydrate isomerism, and the fourth geometrical isomerism.
Assertion (A): [Ni(CN)₄]²⁻ is square planar and diamagnetic while [NiCl₄]²⁻ is tetrahedral and paramagnetic. Reason (R): Cyanide is a strong-field ligand that forces pairing of the 3d electrons of Ni²⁺, freeing a d orbital for dsp² hybridisation, whereas the weak-field chloride leaves the electrons unpaired so only sp³ hybridisation is possible.
Answer
Both A and R are true and R is the correct explanation of A. Ni²⁺ is 3d⁸ with two unpaired electrons. With CN⁻ the large splitting pairs them into one orbital, emptying a 3d orbital that combines with 4s and two 4p orbitals to give dsp² hybridisation and a square planar, diamagnetic ion. With Cl⁻ the splitting is too small to pay the pairing energy, so the 3d electrons stay as they are and the metal uses 4s and 4p only, giving sp³ hybridisation, a tetrahedral shape and two unpaired electrons.
Write the IUPAC names of (i) [Pt(NH₃)₂Cl₂] and (ii) [Cr(H₂O)₄Cl₂]Cl, and state the oxidation number and coordination number of the metal in each.
Answer
(i) Diamminedichloridoplatinum(II). Platinum: oxidation number +2 (the two chlorides carry the whole charge of the neutral complex), coordination number 4. (ii) Tetraaquadichloridochromium(III) chloride. Chromium: oxidation number +3 (x − 2 = +1 for the complex cation), coordination number 6. In both names the ligands are listed alphabetically ignoring the multiplying prefixes: ammine before chlorido, and aqua before chlorido.
Using crystal field theory, explain the difference in colour and magnetic behaviour between [Fe(H₂O)₆]³⁺ and [Fe(CN)₆]³⁻.
Answer
Iron is +3 in both, so the metal ion is d⁵ in each case; the difference comes entirely from the ligand field strength. Water is a weak-field ligand, so Δ₀ is smaller than the pairing energy P. The electrons occupy all five orbitals singly, giving t₂g³ e_g², five unpaired electrons and μ = √35 = 5.92 BM — a strongly paramagnetic, pale yellow-brown ion. Cyanide is a strong-field ligand near the top of the spectrochemical series, so Δ₀ is greater than P. Four electrons pair up to give t₂g⁵ e_g⁰, only one unpaired electron and μ = √3 = 1.73 BM — weakly paramagnetic. The colours differ because the energy of the d–d transition equals Δ₀; the larger gap in the cyanido complex absorbs shorter-wavelength light, so the transmitted colour differs from that of the aqua complex.
Draw and name the geometrical isomers possible for [Co(en)₂Cl₂]⁺, and state which of them is optically active and why.
Answer
The ion is octahedral with two bidentate ethane-1,2-diamine ligands and two chlorides. Two geometrical isomers exist: the cis isomer, in which the two chlorides occupy adjacent positions at 90°, and the trans isomer, in which they lie opposite each other at 180°. Only the cis isomer is optically active. It has no plane of symmetry and no centre of symmetry, so it is chiral and exists as a pair of non-superimposable mirror images (d and l forms) that rotate plane-polarised light in opposite directions. The trans isomer possesses a plane of symmetry passing through the two chlorides, so it is superimposable on its mirror image and is optically inactive.
(i) State the main postulates of Werner's coordination theory. (ii) A cobalt(III) chloride–ammonia compound has the composition CoCl₃·4NH₃; one mole of it gives one mole of AgCl with excess AgNO₃, and the freshly prepared solution has an electrical conductivity corresponding to two ions. Deduce its structural formula, write its IUPAC name and predict the number of geometrical isomers. (iii) State two limitations of valence bond theory.
Answer
(i) Werner proposed that a metal exhibits two kinds of valency. The primary valency is ionisable, is satisfied by negative ions, and equals the oxidation state of the metal. The secondary valency is non-ionisable, is satisfied by neutral molecules or negative ions called ligands, and equals the coordination number. The secondary valencies are directed in space towards fixed positions, which fixes the geometry of the complex; for a coordination number of six the positions are octahedral, and for four they are tetrahedral or square planar. Species inside the coordination sphere are written in square brackets and do not ionise in solution. (ii) Only one chloride is precipitated, so only one is outside the coordination sphere; a conductivity of two ions confirms a 1:1 electrolyte. The other two chlorides must be bonded to cobalt along with the four ammonia molecules, giving a coordination number of 6. Structural formula: [Co(NH₃)₄Cl₂]Cl. IUPAC name: tetraamminedichloridocobalt(III) chloride. Being an octahedral MA₄B₂ type, it has two geometrical isomers: cis (the two chlorides adjacent, violet) and trans (the two chlorides opposite, green). Neither is optically active, since both possess a plane of symmetry. (iii) Valence bond theory cannot explain the colour of complexes, since it says nothing about the energy gap between d orbitals. It also gives no quantitative account of thermodynamic or kinetic stability, does not predict whether a given complex will be inner- or outer-orbital in advance, and offers no explanation for the order of the spectrochemical series.
(i) Explain crystal field splitting in an octahedral complex, defining Δ₀ and the crystal field stabilisation energy. (ii) On this basis, work out the electron configuration, number of unpaired electrons and spin state of [Mn(H₂O)₆]²⁺ and [Mn(CN)₆]⁴⁻. (iii) Explain why tetrahedral complexes are almost never low spin.
Answer
(i) In a free gaseous ion the five d orbitals are degenerate. When six ligands approach along the x, y and z axes, their lone pairs repel electrons in the d orbitals that point directly at them — d_x²−y² and d_z², the e_g set — raising their energy, while d_xy, d_yz and d_zx (the t₂g set) point between the ligands and are lowered. The energy gap between the two sets is the crystal field splitting energy Δ₀. Relative to the barycentre, each t₂g orbital is lowered by 0.4Δ₀ and each e_g orbital raised by 0.6Δ₀. The crystal field stabilisation energy is the net lowering, CFSE = (−0.4n(t₂g) + 0.6n(e_g))Δ₀. Whether electrons pair depends on the comparison of Δ₀ with the pairing energy P: if Δ₀ > P the complex is low spin, if Δ₀ < P it is high spin. (ii) Mn²⁺ is [Ar]3d⁵ in both. With water, a weak-field ligand, Δ₀ < P, so the configuration is t₂g³ e_g², five unpaired electrons, high spin, μ = √35 = 5.92 BM. With cyanide, a strong-field ligand, Δ₀ > P, so electrons pair in the lower set: t₂g⁵ e_g⁰, one unpaired electron, low spin, μ = √3 = 1.73 BM. The change from five unpaired electrons to one, with no change in oxidation state, is direct evidence for crystal field splitting. (iii) In a tetrahedral field only four ligands approach, and they approach between the axes rather than along them, so the repulsion is weaker and the orbital sets are less well matched to the ligand directions. The result is Δ_t ≈ (4/9)Δ₀ for the same metal and ligands. Such a small splitting is almost always less than the pairing energy, so electrons prefer to occupy separate orbitals with parallel spins, and tetrahedral complexes are therefore high spin in practice.
A student prepares two cobalt(III) complexes. Complex P is [Co(NH₃)₆]Cl₃, an orange-yellow solid. Complex Q is K₃[CoF₆], a blue solid. Both are octahedral and cobalt is +3 in each. (a) Write the IUPAC name of each complex. (b) State the d-electron configuration of Co³⁺ and, using the spectrochemical series, predict the t₂g/e_g arrangement in P and in Q. (c) Calculate the spin-only magnetic moment of each complex. (d) Explain why the two complexes differ in colour even though both contain cobalt(III).
Answer
(a) P is hexaamminecobalt(III) chloride. Q is potassium hexafluoridocobaltate(III). (b) Cobalt(III) is [Ar]3d⁶. Ammonia lies high in the spectrochemical series, so in P the splitting Δ₀ exceeds the pairing energy and the arrangement is t₂g⁶ e_g⁰ (low spin, inner orbital, d²sp³). Fluoride is a weak-field ligand, so in Q the splitting is smaller than the pairing energy and the arrangement is t₂g⁴ e_g² (high spin, outer orbital, sp³d²). (c) In P there are no unpaired electrons, so n = 0 and μ = 0 BM; the complex is diamagnetic. In Q, t₂g⁴ e_g² leaves four unpaired electrons, so μ = √(4 × 6) = √24 = 4.90 BM and the complex is strongly paramagnetic. (d) Colour comes from the absorption of visible light in a d–d transition whose energy equals Δ₀, and the eye sees the complementary colour of what is absorbed. Because NH₃ produces a much larger Δ₀ than F⁻, complex P absorbs higher-energy, shorter-wavelength light than complex Q. The two therefore transmit different colours — orange-yellow for P and blue for Q — despite having the same metal in the same oxidation state.
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