Transition Metals and Coordination Chemistry Questions and Answers

Answer: D

The rates vary as H^- \(\gt\) CH₃^- \(\gt\) Ph^- \(\gt\) Cl^-.

In the square-planar complex, L sits trans to the chloride that pyridine replaces (the two PEt₃ groups are trans to each other). The rate of substitution is therefore governed by the trans effect of L: a strong σ-donor weakens the Pt-Cl bond opposite it and stabilises the five-coordinate transition state, so the chloride leaves faster.

Among these ligands, hydride is the strongest σ-donor, then methyl, then phenyl, and chloride is the weakest. This is the classic order found for trans-[PtLCl(PEt₃)₂], where the hydride complex reacts many thousands of times faster than the chloro complex.

Each other option puts chloride, the weakest trans-director here, ahead of one or more of the strong σ-donors.

Answer: B

The green complex is tetrahedral, with 2 unpaired electrons.

Ni(II) is d⁸. In a square planar field the d_x²-y² orbital is pushed far above the rest, all eight electrons pair in the four lower orbitals, and the complex is diamagnetic: that is the red form obtained at low temperature.

With PPh₃ and bromide, both bulky and the bromide weak-field, the tetrahedral isomer is favoured at room temperature. In a tetrahedral field the d⁸ configuration is e⁴t₂⁴, leaving two unpaired electrons in the t₂ set, so the green form is paramagnetic.

A square planar d⁸ complex has no unpaired electrons, which rules out both square planar options, and one unpaired electron would need an odd electron count.

Answer: B

The rate depends on the concentration of [Co(CN)₅Cl]³⁻ only.

Substitution at this inert, low-spin Co(III) centre is dissociative: the slow step is loss of chloride to give the five-coordinate [Co(CN)₅]²⁻, which then picks up hydroxide quickly. So rate = k[Co(CN)₅Cl³⁻], independent of [OH⁻].

The hydroxide dependence familiar from ammine complexes such as [Co(NH₃)₅Cl]²⁺ comes from the conjugate-base (S_N1CB) mechanism, in which OH⁻ removes an N-H proton. Cyanide ligands have no such protons, so that pathway is closed and the options involving [OH⁻] do not apply.

Answer: C

The order is [Cr(CN)₆]³⁻ \(\gt\) [Cr(H₂O)₆]³⁺ \(\gt\) [CrF₆]³⁻.

With the same metal ion, Cr(III), the octahedral splitting follows the spectrochemical series of the ligands: F⁻ \(\lt\) H₂O \(\lt\) CN⁻. Cyanide is a strong-field ligand because it is a good π-acceptor; fluoride is a π-donor and lies at the weak end; water sits between them.

Every other option places fluoride or water above cyanide, or fluoride above water, against the series.

Answer: B

The answer is dichromate ion, +6 and +6, charge transfer.

Acid shifts the chromate-dichromate equilibrium to the right:

\[ \begin{aligned} &2\,\mathrm{CrO_{4}^{2-}} + 2\,\mathrm{H^{+}} \\ &\rightleftharpoons \mathrm{Cr_{2}O_{7}^{2-}} + \mathrm{H_{2}O} \end{aligned} \]

No electrons are transferred, so both chromium atoms stay +6. Cr(VI) is d⁰ and has no d electrons for a d-d transition; the intense colour comes from ligand-to-metal charge transfer, an electron moving from oxygen to chromium.

Chromium(III) chloride would need a reduction that HCl in a few drops does not cause, perchlorate contains no chromium and is colourless, and chromic acid is not the red-orange species formed here.

Answer: B

The increasing order is [Mn(H₂O)₆]²⁺ \(\lt\) [Ti(H₂O)₆]³⁺ \(\lt\) [NiCl₄]²⁻ \(\lt\) [CrO₄]²⁻.

Band intensity follows the selection rules:

• [Mn(H₂O)₆]²⁺, high-spin d⁵: every d-d transition is both spin-forbidden and Laporte-forbidden, so it is almost colourless.
• [Ti(H₂O)₆]³⁺, d¹ octahedral: spin-allowed but Laporte-forbidden, weak.
• [NiCl₄]²⁻, tetrahedral: no centre of symmetry, so d-p mixing relaxes the Laporte rule and the d-d bands are much stronger.
• [CrO₄]²⁻, d⁰: its colour is an allowed ligand-to-metal charge transfer, the most intense of all.

Any order putting the manganese ion above the titanium ion, or a d-d band above the charge-transfer band, breaks these rules.

Answer: B

The order is \(k_{22} \gt k_{11} \gt k_{33}\).

In outer-sphere self-exchange the main barrier is the reorganisation needed to make the two partners' bond lengths equal before the electron jumps.

• [Fe(bpy)₃]^2+/3+: both are low spin (t_2g⁶ and t_2g⁵); the electron moves between nonbonding t_2g orbitals, the Fe-N distances barely change, and the large aromatic ligands help electronic coupling. Self-exchange is very fast, of the order of 10⁸ M^-1 s^-1.
• [Fe(H₂O)₆]^2+/3+: both are high spin and the electron comes from the antibonding e_g set, so the Fe-O bonds change noticeably; the rate is only a few M^-1 s^-1.
• [Co(NH₃)₆]^2+/3+: high-spin Co(II), t_2g⁵e_g², must become low-spin Co(III), t_2g⁶, a change of spin state with a large change in Co-N distance. Self-exchange is extremely slow, about 10^-6 M^-1 s^-1 or less.

Any order that does not put the bipyridyl couple first and the cobalt ammine couple last is therefore wrong.

Answer: D

The species is Cu²⁺.

Copper(II) and hexacyanoferrate(II) give the classic chocolate-brown precipitate of copper(II) hexacyanoferrate(II), Cu₂[Fe(CN)₆], which does not dissolve in dilute acids (it dissolves in ammonia). This is the standard confirmatory test for Cu²⁺, and it is the reaction the question describes.

Fe³⁺ gives the deep blue precipitate Prussian blue, not a brown one, and Th⁴⁺ gives a white precipitate. The uranyl ion, UO₂²⁺, also gives a brown hexacyanoferrate(II); the key treats the brown precipitate insoluble in dilute acid as the copper test, which is the textbook reading of this description.

Answer: B

The match is P-III, Q-IV, R-II, S-I.

Octahedral Ni(II), d⁸, has a ³A_2g ground state and three spin-allowed bands, in order of increasing energy: ³T_2g(F), ³T_1g(F) and ³T_1g(P). Their energies follow the wavelengths in reverse.

• P, at 1000 nm, is the lowest energy band: ³T_2g(F), whose energy equals 10Dq (III).
• R, at 630 nm: ³T_1g(F) (II).
• S, at 375 nm, is the highest energy band: ³T_1g(P) (I).
• Q, at 770 nm with ε only 8, is far weaker than the rest: the spin-forbidden transition to ¹E_g(D) (IV), which borrows intensity from the nearby spin-allowed band.

The other options either assign the weak band to a spin-allowed transition or reverse the energy order of the three triplet bands.

Answer: D

Formation of [M(en)₃]²⁺ is least favoured for Fe.

For high-spin divalent ions of the first transition series with the same ligands, complex stability follows the Irving-Williams series:

\[ \begin{aligned} &\mathrm{Mn^{2+} \lt Fe^{2+} \lt Co^{2+}} \\ &\mathrm{\lt Ni^{2+} \lt Cu^{2+} \gt Zn^{2+}} \end{aligned} \]

The order follows the decreasing ionic radius across the series and the growing ligand-field stabilisation up to Ni²⁺, with Cu²⁺ gaining extra Jahn-Teller stabilisation. Of the four metals offered, Fe²⁺ stands lowest, so replacing its six waters by three ethylenediamines is least favourable.

Co²⁺ and Ni²⁺ form [M(en)₃]²⁺ with much larger overall stability constants, and Cu²⁺, though its third en binds only weakly because of Jahn-Teller distortion, still has the highest overall affinity for en of the four.