Your Answer : (Not Answered)

Correct Answer : B

The two lanthanoids are cerium and europium, recovered as Ce⁴⁺ and Eu²⁺.

Almost all lanthanoids are stable only as Ln³⁺, with very similar radii, which is why they are so hard to separate. Cerium can be oxidised to Ce⁴⁺ (an empty 4f⁰ shell), whose chemistry differs sharply from Ln³⁺: it hydrolyses and precipitates or is extracted selectively. Europium can be reduced to Eu²⁺ (a half-filled 4f⁷ shell), which behaves like an alkaline-earth ion and can be removed, for example as insoluble EuSO₄.

Pr⁴⁺, Tb⁴⁺, Nd⁴⁺ and Dy⁴⁺ exist only in a few solid fluorides and oxides and are not usable in solution, and 'Ln' is the general symbol for any lanthanoid, so options A, C and D do not describe a practical separation.

Your Answer : (Not Answered)

Correct Answer : A

The γ-ray source for ⁵⁷Fe Mössbauer spectroscopy is ⁵⁷Co.

⁵⁷Co (half-life about 272 days), usually diffused into a rhodium matrix, decays by electron capture to an excited state of ⁵⁷Fe, which reaches the 14.4 keV level and emits the 14.4 keV γ-ray that the ⁵⁷Fe nuclei in the sample absorb resonantly. The source must contain the parent of the same nucleus being studied.

^119mSn is the source for ¹¹⁹Sn Mössbauer spectroscopy, not for iron. ^121mSn is the parent used for ¹²¹Sb work, and ¹²⁹Tc is not a Mössbauer source.

Your Answer : (Not Answered)

Correct Answer : B

Cu-Zn superoxide dismutase produces H₂O₂ and O₂.

The enzyme catalyses the disproportionation of the superoxide radical anion, a toxic by-product of respiration:

\[ \mathrm{2O_{2}^{-} + 2H^{+} \rightarrow H_{2}O_{2} + O_{2}} \]

The copper cycles between Cu(II) and Cu(I), while zinc plays a structural role. One superoxide is oxidised to O₂ and the other reduced to H₂O₂, which catalase later breaks down. Water and the hydroxyl radical are not the products of this step.

Your Answer : (Not Answered)

Correct Answer : B

The incorrect statement is it contains a nucleophilic carbene centre.

A Fischer carbene, such as (CO)₅Cr=C(OMe)Me, has a metal in a low oxidation state with π-acceptor ligands, a formal M=C bond, and a heteroatom (O or N) on the carbene carbon. The carbene carbon is electron-poor and electrophilic: nucleophiles such as amines attack it, and the heteroatom stabilises it by donating its lone pair.

A nucleophilic carbene carbon is the mark of Schrock (alkylidene) carbenes, with high-oxidation-state metals. The other three statements are standard features of Fischer carbenes.

Your Answer : (Not Answered)

Correct Answer : D

The extra moment comes from spin-orbit coupling, an orbital contribution added to the spin-only value.

In K₃[Fe(CN)₆] iron is Fe(III), d⁵, and cyanide is a strong-field ligand, so the complex is low spin, t_2g⁵, with one unpaired electron. The spin-only moment for one electron is

\[ \mu_{s} = \sqrt{n(n+2)} = \sqrt{3} \approx 1.73\ \mu_{B} \]

The observed 2.3 µB is larger because the ground term is an orbitally degenerate ²T_2g, so the orbital angular momentum is not quenched and spin-orbit coupling adds to the moment.

It is not the spin-only value of one electron (that is 1.73), and spin-only values come only in fixed steps (1.73, 2.83 and so on), so 'between one and two' is not a spin-only value at all. Ferromagnetic coupling needs interacting paramagnetic centres, which a magnetically dilute salt like this does not have.

Your Answer : (Not Answered)

Correct Answer : A

The reason is that the energy required for LMCT is higher for ReO₄^- than for MnO₄^-.

Both ions are d⁰ (Mn(VII) and Re(VII)), so they have no d-d transitions at all. The intense purple of permanganate comes from ligand-to-metal charge transfer: an electron moves from oxygen lone pairs to the empty metal d orbitals. Mn(VII) is strongly oxidising and its 3d orbitals lie low, so this absorption falls in the visible region. Re(VII) is much harder to reduce and its 5d orbitals lie higher, so the charge transfer moves into the ultraviolet and perrhenate is colourless.

The colour is not MLCT, since the metal has no d electrons to give, and both d-d options fail because a d⁰ ion has no d-d transitions.

Your Answer : (Not Answered)

Correct Answer : B

The correct order is H₂O \(\gt\) NH₃ \(\gt\) en \(\gt\) NCS⁻ \(\gt\) CN⁻.

The nephelauxetic parameter β is the ratio of the Racah parameter B in the complex to that in the free ion. Ligands that form more covalent bonds expand the d electron cloud more, lower B, and so give a smaller β. The nephelauxetic series runs F⁻ \(\gt\) H₂O \(\gt\) NH₃ \(\gt\) en \(\gt\) NCS⁻ \(\gt\) Cl⁻ \(\gt\) CN⁻ \(\gt\) Br⁻ \(\gt\) I⁻ in decreasing β.

Option A is the same series reversed, which would be the order of increasing nephelauxetic effect, not of β. Options C and D misplace en and CN⁻.

Your Answer : (Not Answered)

Correct Answer : A

Siderite, malachite and calamine are all carbonate ores.

Siderite is iron(II) carbonate, FeCO₃; malachite is basic copper carbonate, CuCO₃·Cu(OH)₂; and calamine is zinc carbonate, ZnCO₃. All three are converted to oxides by calcination before reduction.

Sulphide ores include galena (PbS), zinc blende (ZnS) and copper pyrites (CuFeS₂), and oxide ores include haematite and bauxite. Nitrate ores, such as Chile saltpetre, are not ores of these metals.

Your Answer : (Not Answered)

Correct Answer : C

The formula given for bleaching powder is \(Ca(OCl)_{2}\).

Bleaching powder is made by passing chlorine over dry slaked lime, and is often written \(CaOCl_{2}\) (calcium oxychloride). Its active bleaching and disinfecting component is calcium hypochlorite, \(Ca(OCl)_{2}\), which releases chlorine, and this is the formula the key accepts.

\(MgCO_{3}\) is magnesium carbonate, \(BaCO_{3}\) barium carbonate and \(CaSO_{4}\) calcium sulphate (the basis of gypsum and plaster of Paris); none is a bleaching agent.

Your Answer : (Not Answered)

Correct Answer : B

The hydride ion is the Bronsted-Lowry base of H₂, that is, its conjugate base.

Removing a proton from H₂ leaves H^-, so H₂ and H^- are a conjugate acid-base pair. H^- is an extremely strong base: it takes a proton from water, H^- + H₂O → H₂ + OH^-.

The other statements are false. H^- has two electrons, the same as helium, so it is isoelectronic with He. It is a proton acceptor and an electron-pair donor, so it is neither a Bronsted acid (it has no proton to give to OH^-) nor a Lewis acid.

Your Answer : (Not Answered)

Correct Answer : D

The metals used are Zn and Mg.

In the laboratory, dihydrogen is usually prepared by the action of dilute acid on granulated zinc, \(\mathrm{Zn + 2HCl \rightarrow ZnCl_2 + H_2}\); magnesium reacts the same way. Both react at a steady, controllable rate.

Sodium, calcium and barium react with water and acids so violently that the reaction is hazardous and hard to control. Iron reacts only slowly with dilute acid, and pairing it with sodium in option 3 brings back the hazard.

Your Answer : (Not Answered)

Correct Answer : A

The numbers of shared corners are 0, 1, 2 and 3.

• Orthosilicates (SiO₄⁴⁻, e.g. olivine) contain isolated tetrahedra: no shared corners.
• Pyrosilicates (Si₂O₇⁶⁻) join two tetrahedra through one bridging oxygen: one shared corner per tetrahedron.
• Cyclic silicates (e.g. Si₆O₁₈¹²⁻ in beryl) link tetrahedra in rings, each sharing two corners.
• Sheet silicates (Si₂O₅²⁻, e.g. mica, talc) share three corners per tetrahedron in a two-dimensional net.

The other options scramble this sequence; note that three-dimensional silicates such as quartz share all four corners, one more than the sheets.

Your Answer : (Not Answered)

Correct Answer : A

The configuration is \(d_{xy}^{2} = d_{x^{2}-y^{2}}^{2} \lt d_{z^{2}}^{2} \lt d_{xz}^{0} = d_{yz}^{0}\).

Ferrocene contains Fe(II), d⁶. With the z-axis along the Cp-Fe-Cp line, the d orbitals sort by how they overlap the rings' π orbitals:

• \(d_{xz}\) and \(d_{yz}\) (e_1g) overlap strongly with the filled e₁ ring orbitals. Their antibonding combinations are pushed highest and stay empty.
• \(d_{z^{2}}\) (a_1g) points at the ring centres, where the π system has little density, so it is nearly non-bonding.
• \(d_{xy}\) and \(d_{x^{2}-y^{2}}\) (e_2g) have only a weak δ overlap with the rings and are slightly stabilised by back-donation into empty e₂ ring orbitals, so they lie lowest.

Six electrons fill e_2g⁴ a_1g², which is the first option. Some treatments place a_1g slightly below e_2g, but the occupancy is the same and the paper's order follows the common textbook diagram. Every other option puts electrons in \(d_{xz}\) and \(d_{yz}\), the strongly antibonding pair, and the third and fourth also leave \(d_{z^{2}}\) empty.

Your Answer : (Not Answered)

Correct Answer : A

The diamagnetic ion is Zn²⁺.

An ion is diamagnetic when all its electrons are paired. Zn²⁺ is 3d¹⁰: a full d subshell with no unpaired electrons.

The others all have unpaired d electrons: Ni²⁺ (3d⁸) has 2, Co²⁺ (3d⁷) has 3, Cu²⁺ (3d⁹) has 1, and Mn²⁺ (3d⁵) has 5, so all four are paramagnetic.

Your Answer : (Not Answered)

Correct Answer : A

The name is Pentaamminenitritocobalt (III) chloride.

The ligand written ONO is nitrite bound through oxygen, named nitrito (nitrito-O). With five ammonia ligands (pentaammine), the complex ion carries a 2+ charge to balance two chloride ions. Ammonia is neutral and nitrite is -1, so cobalt is x - 1 = +2, x = +3. Ligands are named in alphabetical order, ammine before nitrito.

Cobalt (II) gets the oxidation state wrong. 'Nitro' is the name for NO₂ bound through nitrogen, the linkage isomer. Putting nitrito first breaks the alphabetical order, and 'mononitrito' is not used for a single ligand.

Your Answer : (Not Answered)

Correct Answer : E

The ambidentate ligand is nitrite.

An ambidentate ligand has two different donor atoms but uses only one at a time. Nitrite can bind through nitrogen (nitrito-N, –NO₂) or through oxygen (nitrito-O, –ONO), which gives rise to linkage isomerism.

Oxalate and ethylenediamine are didentate (they bind through two atoms at once), and ammonia is monodentate with only one donor atom. Carbon monoxide binds through carbon in metal carbonyls.

Thiocyanate (SCN⁻, bonding through S or N) and cyanide (CN⁻) are other common ambidentate ligands.

Your Answer : (Not Answered)

Correct Answer : B

The common oxidation states of chromium are +3,+6.

Chromium (3d⁵4s¹) is most often found as Cr(III), as in chromium(III) salts and \(Cr_{2}O_{3}\), the most stable state, and as Cr(VI), in chromates and dichromates such as \(K_{2}Cr_{2}O_{7}\).

+2 exists but is strongly reducing; +4 and +5 are rare; +7 is above chromium's maximum, which equals its six valence electrons.

Your Answer : (Not Answered)

Correct Answer : B

The correct statement is: CO and CN^- are good π-accepting ligands.

Both ligands have empty π* orbitals of suitable energy, which accept electron density from filled metal d orbitals (back-bonding). That is why CO and CN^- are strong-field ligands that stabilise metals in low oxidation states.

The other statements are false. BF₃ is a weaker Lewis acid than BI₃, because strong π back-donation from the small fluorine atoms into boron's empty p orbital reduces boron's electron deficiency. cis-Diamminedichloroplatinum(II) (cisplatin) is square planar with both Cl on the same side, so its bond dipoles do not cancel; the trans isomer has zero dipole moment. Boron in BCl₃ is sp² hybridized and trigonal planar.

Your Answer : (Not Answered)

Correct Answer : B

The name is sodium hexachlorophosphate(V).

In Na[PCl₆] the complex ion [PCl₆]^- is an anion. IUPAC names an anionic complex by adding the ending -ate to the central atom's name, giving phosphate, and gives the oxidation state in Roman numerals: six chlorides at -1 and an overall charge of -1 put phosphorus at +5. The cation is named first, so the full name is sodium hexachlorophosphate(V).

'Phosphine' is the name of PH₃ and its derivatives, not of an anionic centre, and the ending '-ite' belongs to oxoanion names (phosphite), not to the nomenclature of complexes.

Your Answer : (Not Answered)

Correct Answer : D

The chelate is that of trifluoroacetylacetone.

Gas chromatography needs a compound that vaporises without decomposing. Trifluoroacetylacetone (Htfa) forms a neutral tris chelate, Al(tfa)₃, in which the fluorinated β-diketonate ligands wrap the metal and leave a non-polar, volatile, thermally stable molecule. Fluorinated β-diketonates are the standard reagents for GC of metal ions such as Al, Be and Cr.

EDTA gives charged, involatile complexes. Ethylene glycol is not a useful chelating agent for Al³⁺ in this sense, and dinonyl phthalate is a common GC stationary phase (a liquid coating), not a chelating ligand.