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Tamil Nadu HSE Second Year Chemistry study materials
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Tamil Nadu HSE Second Year Chemistry practice questions

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Question 1Metallurgy

In the extraction of aluminium by the Hall–Héroult process, cryolite is added to the electrolyte mainly to

  1. convert alumina into aluminium carbide before electrolysis
  2. lower the melting point of alumina and increase its electrical conductivity Correct answer.
  3. prevent the graphite anode from burning away in oxygen
  4. oxidise the alumina so that aluminium metal is released at the anode

Explanation

Pure alumina melts only near 2323 K, which is impractical, so cryolite (Na3AlF6) is added to dissolve alumina and bring the working temperature down to about 1223 K. The molten cryolite–alumina bath also conducts electricity better than pure alumina would, so the cell operates efficiently.

Question 2Metallurgy

Which of the following ores is a sulphide ore of copper?

  1. Chalcocite Correct answer.
  2. Cuprite
  3. Cuprite after roasting
  4. Malachite

Explanation

Chalcocite is Cu2S, a copper sulphide, and is concentrated by froth flotation before roasting. Cuprite (Cu2O) and malachite (CuCO3·Cu(OH)2) are oxide and carbonate ores respectively.

Question 3Metallurgy

In the froth flotation process, pine oil is used because it

  1. dissolves the gangue particles so they sink
  2. stabilises the froth in which the wetted ore particles rise Correct answer.
  3. reduces the metal ion to the free metal during flotation
  4. acts as a collector that makes sulphide particles water-avid

Explanation

Pine oil is a froth stabiliser: it produces a persistent foam so that the air bubbles carrying the sulphide particles can be skimmed off. Collectors such as xanthates are separate reagents that make the ore particles water-repellent.

Question 4Metallurgy

Which statement correctly describes the role of the slag in a smelting furnace?

  1. It is the impure metal that is later refined by electrolysis.
  2. It is a volatile product that escapes with the flue gases.
  3. It is a lighter molten layer that floats above the molten metal and protects it from oxidation.
  4. It is a fusible waste layer that removes gangue as it forms and can be tapped off separately. Correct answer.

Explanation

A flux is added to combine with the infusible gangue and produce a fusible slag, which is a waste product. Because the slag is less dense than the molten metal, it forms a separate upper layer that is tapped off, carrying away impurities.

Question 5Metallurgy

In the extraction of copper, the 'blister copper' obtained after converting matte is refined by

  1. electrolytic refining using acidified copper sulphate solution Correct answer.
  2. Mond's process with carbon monoxide
  3. liquation on a sloping hearth
  4. zone refining under an inert atmosphere

Explanation

Blister copper is purified by electrolysis in acidified copper sulphate, with impure copper as anode and pure copper as cathode; the anode dissolves and pure copper deposits on the cathode. Mond's process and zone refining are used for nickel and silicon-type purifications, not for copper.

Question 6Metallurgy

During the electrolytic refining of copper, the anode mud that collects at the bottom contains mainly

  1. slaked lime used to neutralise the electrolyte
  2. pure copper that has fallen from the anode
  3. unattacked noble metals such as silver, gold and platinum Correct answer.
  4. copper sulphate crystals that have crystallised out

Explanation

The less reactive impurities present in blister copper, such as silver, gold and platinum, are not oxidised at the anode, so they simply fall as anode mud. This mud is a valuable source of precious metals, while the more reactive impurities dissolve into the electrolyte.

Question 7Metallurgy

In the Mond process for refining nickel, the intermediate compound formed and later decomposed is

  1. Ni(OH)2
  2. NiS
  3. Ni(CO)4 Correct answer.
  4. NiCl2

Explanation

Impure nickel is heated to about 330 K in a stream of carbon monoxide to form volatile nickel tetracarbonyl, Ni(CO)4. This vapour is then decomposed by heating to about 450 K, depositing pure nickel and releasing carbon monoxide for reuse.

Question 8Metallurgy

The chief reason that the reduction of zinc oxide by carbon is carried out above 1100 K is that

  1. the slag becomes insoluble in water below 1100 K
  2. carbon monoxide cannot form below 1100 K
  3. zinc oxide decomposes into zinc and oxygen only above 1100 K
  4. zinc metal produced is a gas above this temperature and distils away, shifting the equilibrium Correct answer.

Explanation

Zinc boils at about 1180 K, so above roughly 1100 K the zinc formed is volatile and is carried off as vapour, preventing the reverse reaction. The removal of the product drives the reduction of ZnO by carbon or CO forward.

Question 9Metallurgy

In the extraction of iron in a blast furnace, limestone is added mainly to

  1. oxidise the iron to iron(II) oxide for easier reduction
  2. provide carbon for reducing iron(III) oxide
  3. lower the melting point of the iron produced
  4. remove the acidic silica gangue by forming calcium silicate slag Correct answer.

Explanation

Limestone decomposes to calcium oxide, which acts as a basic flux and combines with acidic silica gangue to form calcium silicate slag. The slag floats on the molten iron and is drawn off separately, removing the impurity.

Question 10Metallurgy

Which of the following concentration methods is most suitable for a low-grade ore in which the ore particles are magnetic?

  1. Liquation on a sloping hearth
  2. Leaching with a suitable solvent
  3. Magnetic separation using a magnetic roller Correct answer.
  4. Froth flotation

Explanation

Magnetic separation exploits a difference in magnetic behaviour: on a belt passing over a magnetic roller, magnetic ore particles cling to the roller and fall nearer, while non-magnetic gangue is thrown farther. It is therefore the natural choice when the valuable mineral is magnetic, as in magnetic iron ore.

Question 11Metallurgy

In the extraction of aluminium, the carbon anodes are gradually consumed and must be replaced. The consumption is chiefly caused by

  1. aluminium metal dissolving into the graphite lattice
  2. the anodes melting because the bath is above 2000 K
  3. cryolite reacting with graphite to give sodium carbide
  4. the anodes reacting with the oxygen liberated at them to form carbon dioxide Correct answer.

Explanation

At the anode, oxide ions are discharged and the liberated oxygen attacks the hot carbon electrode, forming CO and CO2 and slowly burning it away. This is why the anode is consumed and periodically replaced in the Hall–Héroult cell.

Question 12Metallurgy

In the cyanide process for extracting silver, the metal is finally obtained from the soluble complex by

  1. electrolysing the solution with a silver anode
  2. passing hydrogen sulphide to precipitate silver sulphide
  3. adding dilute sulphuric acid to precipitate silver sulphate
  4. displacing it with zinc, which is more electropositive Correct answer.

Explanation

The silver dissolves as the complex ion [Ag(CN)2]−, and zinc, being more electropositive, displaces silver from this solution. The displaced silver is then collected and purified further.

Question 13Metallurgy

In the Ellingham diagram, the curve for the formation of carbon monoxide slopes downward as temperature rises. This implies that

  1. carbon monoxide is more stable at low temperature than at high temperature
  2. carbon becomes a stronger reducing agent at high temperature for oxides whose curves lie above it Correct answer.
  3. carbon can never reduce any metal oxide at high temperature
  4. the entropy change for CO formation is negative

Explanation

A downward-sloping curve means the free energy of formation of CO becomes more negative as temperature rises, so carbon is increasingly able to reduce oxides whose curves lie above the CO line. This is why coke reduction works well at the high temperatures of the blast furnace.

Question 14Metallurgy

A metal oxide can be reduced by carbon only if the free energy of formation of carbon monoxide at that temperature is more negative than that of the metal oxide. Using this principle, which situation best explains why alumina is not reduced by coke in a blast furnace?

  1. Aluminium has a higher melting point than iron, so it cannot be displaced by carbon.
  2. Alumina is a covalent compound and therefore cannot react with carbon at all.
  3. Coke reacts with alumina to form aluminium carbide, which then decomposes to alumina.
  4. The free energy of formation of alumina is more negative than that of carbon monoxide over the practical furnace range. Correct answer.

Explanation

The Ellingham curves show that the Al2O3 line lies well below the CO line at normal furnace temperatures, so carbon cannot reduce alumina there. Aluminium is therefore obtained by electrolytic reduction of alumina dissolved in cryolite instead.

Question 15p-Block Elements

Which of the following is the correct general electronic configuration of the valence shell of group 17 elements (halogens)?

  1. ns²np⁵ Correct answer.
  2. ns²np⁶
  3. ns²np⁴
  4. ns²np³

Explanation

Halogens belong to group 17, which lies one group before the noble gases. Each halogen atom has seven electrons in its valence shell, arranged as ns²np⁵, and needs one more electron to attain the stable octet.

Question 16p-Block Elements

The anomalous behaviour of nitrogen compared with the other group 15 elements is mainly due to its

  1. presence of vacant d-orbitals in the valence shell
  2. low electronegativity and low ionisation enthalpy
  3. small size and absence of d-orbitals Correct answer.
  4. large atomic radius and high metallic character

Explanation

Nitrogen is the first member of group 15 and has a very small atomic radius with no d-orbitals available for bonding. It therefore cannot expand its covalency beyond four, forms strong multiple bonds and exists as a diatomic gas, unlike the heavier members.

Question 17p-Block Elements

Which noble gas is used to provide an inert atmosphere during the welding of metals such as aluminium and stainless steel?

  1. Radon
  2. Neon
  3. Helium
  4. Argon Correct answer.

Explanation

Argon is comparatively cheap, readily available from liquid air and chemically inert, so it is used as a shielding gas in arc welding. It prevents oxidation of the hot metal by atmospheric oxygen and nitrogen.

Question 18p-Block Elements

Which of the following oxides of group 16 elements is amphoteric in nature?

  1. SO₂
  2. TeO₂ Correct answer.
  3. SeO₂
  4. SO₃

Explanation

In group 16 the acidic character of the dioxides decreases down the group as metallic character increases. Tellurium dioxide reacts both with acids and with bases, so it is amphoteric, whereas the oxides of sulphur and selenium are acidic.

Question 19p-Block Elements

The bond angle in the hydrides of group 16 elements changes in the order H₂O > H₂S > H₂Se > H₂Te. The best explanation for this trend is that

  1. the hybridisation of the central atom changes from sp³ to sp² down the group
  2. the electronegativity of the central atom decreases down the group, so bond pairs move away from it Correct answer.
  3. the number of lone pairs on the central atom increases down the group
  4. the bond length decreases down the group, forcing the angle to close

Explanation

As the central atom becomes larger and less electronegative down the group, the bonding electron pairs lie farther from it and closer to hydrogen. Greater repulsion between bond pairs relative to lone pairs is lost, so the angle falls from about 104.5° in water towards 90° in H₂Te.

Question 20p-Block Elements

In the industrial manufacture of sulphuric acid by the contact process, the catalyst used in the conversion of SO₂ to SO₃ is

  1. MnO₂
  2. Fe₂O₃
  3. V₂O₅ Correct answer.
  4. Pt asbestos only

Explanation

The contact process oxidises sulphur dioxide to sulphur trioxide over vanadium(V) oxide, which is cheaper and less easily poisoned than platinum. The reaction is carried out at about 700 K and 2 bar to give a good equilibrium yield.

Question 21p-Block Elements

Bleaching powder is manufactured by the action of chlorine on dry slaked lime. The formula usually assigned to the active constituent of bleaching powder is

  1. Ca(OCl)Cl Correct answer.
  2. Ca(OCl)₂·2Ca(OH)₂
  3. CaCl₂
  4. Ca(ClO₃)₂

Explanation

Chlorine reacts with dry slaked lime to give a mixture in which calcium hypochlorite and calcium chloride share a formula unit written as Ca(OCl)Cl. Its bleaching action comes from the hypochlorite ion releasing nascent oxygen.

Question 22p-Block Elements

Which interhalogen compound has a bent (angular) shape?

  1. BrF₅
  2. ICl
  3. ClF
  4. ClF₃ Correct answer.

Explanation

ClF₃ is of the type AX₃E₂, so the three bonding pairs and two lone pairs give a T-shaped molecule, which is angular rather than linear. The other compounds listed are linear diatomic species or, in the case of BrF₅, square pyramidal.

Question 23p-Block Elements

Which property of helium makes it particularly suitable for use in gas thermometers and in filling party balloons that rise in air?

  1. Its high solubility in water
  2. Its high density compared with air
  3. Its very low density and very low boiling point Correct answer.
  4. Its ability to form stable compounds with fluorine

Explanation

Helium has the lowest boiling point of any substance and a density far below that of air. These properties give it a wide liquid range for low-temperature thermometry and enough buoyancy for balloons.

Question 24p-Block Elements

Among the hydrides of group 15 elements, the one with the lowest boiling point is

  1. SbH₃
  2. AsH₃
  3. PH₃ Correct answer.
  4. NH₃

Explanation

Boiling point normally rises with molar mass down group 15, but ammonia is exceptional because of strong intermolecular hydrogen bonding. Phosphine, with only weak van der Waals forces and no hydrogen bonding, has the lowest boiling point of the group.

Question 25p-Block Elements

Which of the following statements about the oxoacids of sulphur is correct?

  1. Oleum is a mixture of H₂SO₄ and H₂S₂O₇
  2. Sulphurous acid contains sulphur in the +6 oxidation state
  3. Sulphuric acid contains two S–O–H bonds and is therefore monobasic
  4. The S–O–H bond is present in all oxoacids of sulphur and gives them their acidic character Correct answer.

Explanation

Every oxoacid of sulphur has at least one S–O–H linkage, and the hydrogen of that group can be released as a proton. The number of such groups fixes the basicity of the acid.

Question 26p-Block Elements

The inert pair effect is best illustrated by which observation for group 13 elements?

  1. Thallium forms Tl⁺ compounds more readily than Tl³⁺ compounds Correct answer.
  2. Aluminium forms Al³⁺ compounds readily
  3. Boron forms only covalent compounds
  4. Gallium has a higher melting point than indium

Explanation

The reluctance of the ns² pair to take part in bonding increases down group 13, so the +1 state becomes more stable than +3 for the heaviest members. Thallium(I) compounds are therefore more stable than thallium(III) compounds.

Question 27p-Block Elements

Carbon shows a maximum covalency of four, whereas silicon shows a covalency of six. The reason is that

  1. silicon is more electronegative than carbon
  2. carbon has a larger atomic radius than silicon
  3. silicon has vacant 3d orbitals available for bonding while carbon has none Correct answer.
  4. carbon has vacant p-orbitals that cannot be used

Explanation

Carbon's valence shell is n = 2 and contains only 2s and 2p orbitals, so it can accommodate at most eight electrons. Silicon has empty 3d orbitals that can accept electron pairs, allowing it to expand its octet to form species such as SiF₆²⁻.

Question 28p-Block Elements

In the reaction of chlorine with cold dilute sodium hydroxide, the oxidation states of chlorine in the products are

  1. 0 in Cl₂ and −1 in NaCl only
  2. −1 in NaCl and +1 in NaOCl Correct answer.
  3. +1 in NaCl and −1 in NaOCl
  4. −1 in NaCl and +5 in NaClO₃

Explanation

Cold dilute alkali gives sodium chloride and sodium hypochlorite through disproportionation of chlorine. Chlorine is reduced to −1 in the chloride and oxidised to +1 in the hypochlorite.

Question 29p-Block Elements

Which of the following correctly relates the acid strength of the oxoacids of chlorine?

  1. HClO > HClO₂ > HClO₃ > HClO₄
  2. HClO₄ > HClO₃ > HClO₂ > HClO Correct answer.
  3. HClO₂ > HClO₄ > HClO > HClO₃
  4. HClO₃ > HClO₄ > HClO > HClO₂

Explanation

Acid strength increases as the number of terminal oxygen atoms attached to chlorine increases, because these oxygens withdraw electron density and stabilise the conjugate base. Perchloric acid, with three terminal oxygens, is the strongest of the four.

Question 30p-Block Elements

A mixture of two gases is passed through water, and one component dissolves to give an acidic solution while the other does not. Which pair of noble gases best fits this behaviour when the soluble gas forms no oxoacid?

  1. Helium and neon
  2. Xenon and radon
  3. Argon and krypton
  4. This behaviour is not shown by any noble gas; noble gases are chemically inert under ordinary conditions Correct answer.

Explanation

Under ordinary conditions the noble gases are monatomic and chemically inert, so none of them dissolves in water to give an acidic solution. Their very low solubility in water does not produce acidic species.

Question 31p-Block Elements

When excess chlorine is passed over hot concentrated sodium hydroxide solution, the products are sodium chloride, sodium chlorate and water. The number of moles of chlorine that disproportionate per mole of chlorate formed is

  1. 2
  2. 3 Correct answer.
  3. 6
  4. 1

Explanation

The balanced equation is 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O. One chlorine atom is oxidised from 0 to +5 while five are reduced from 0 to −1, so three moles of Cl₂ are consumed per mole of chlorate.

Question 32p-Block Elements

Which of the following statements about the relative strengths of hydrogen halides as reducing agents is correct?

  1. HF > HCl > HBr > HI
  2. HBr > HI > HCl > HF
  3. HI > HBr > HCl > HF Correct answer.
  4. HCl > HBr > HI > HF

Explanation

The reducing power of hydrogen halides increases down the group as the H–X bond becomes weaker and the halide ion more easily oxidised. Thus HI is the strongest reducing agent and HF the weakest, since the H–F bond is very strong.

Question 33Transition and Inner Transition Elements

In the 3d transition series, the atomic radius decreases from Sc to Cr and then remains almost constant up to Cu. Which statement best explains this trend?

  1. Nuclear charge increases while the added 3d electrons shield each other poorly, so the net attraction on the outer 4s electrons increases until pairing begins. Correct answer.
  2. Successive elements have identical effective nuclear charge, so the radius cannot change at all from Sc to Cu.
  3. The 4s orbital contracts steadily because it has no shielding from the 3d subshell throughout the series.
  4. The decrease stops because the 3d subshell becomes completely filled at chromium.

Explanation

Across the series the nuclear charge rises by one unit per element, but the extra electron enters the inner 3d subshell, where shielding of the 4s electrons is imperfect. The resulting increase in effective nuclear charge contracts the atom until electron pairing in 3d orbitals produces repulsive interactions that offset further contraction. Hence the radius falls up to Cr and then levels off.

Question 34Transition and Inner Transition Elements

Which ion is expected to be colourless in aqueous solution because it has no partially filled d subshell?

  1. Mn2+
  2. Sc3+ Correct answer.
  3. Ni2+
  4. Cu2+

Explanation

Scandium(III) has the electronic configuration [Ar]3d0, so there is no partially filled d subshell available for d-d electronic transitions. Colour in aqueous transition metal ions arises mainly from d-d absorption, so Sc3+ is colourless. The other ions listed all retain unpaired or partially filled 3d electrons.

Question 35Transition and Inner Transition Elements

The spin-only magnetic moment of a high-spin Mn2+ ion (3d5) is closest to which value?

  1. 5.92 BM Correct answer.
  2. 3.87 BM
  3. 2.83 BM
  4. 1.73 BM

Explanation

High-spin Mn2+ has five unpaired 3d electrons, so n = 5. Using the spin-only formula μ = √(n(n+2)) BM gives √(5 × 7) = √35 ≈ 5.92 BM. The value follows directly from the number of unpaired electrons.

Question 36Transition and Inner Transition Elements

Which statement correctly describes the oxidation states shown by manganese in its compounds?

  1. Manganese shows a continuous range from +2 to +7, with +2 being the most stable in aqueous solution.
  2. Manganese never attains the +7 state because that would require removal of all 3d and 4s electrons.
  3. Manganese can show +2, +3, +4, +6 and +7, and MnO4- is a strong oxidising agent in acidic medium. Correct answer.
  4. Manganese shows only the +2 state in all its compounds.

Explanation

Manganese has the configuration [Ar]3d5 4s2 and can lose the 4s electrons and any number of 3d electrons, giving states from +2 up to +7. The +2 state is the most stable in aqueous solution, while permanganate (MnO4-, Mn in +7) is a powerful oxidising agent in acid. The +6 state occurs in manganate, MnO4^2-.

Question 37Transition and Inner Transition Elements

Which of the following is the correct electronic configuration of the Cr3+ ion?

  1. [Ar]3d3 Correct answer.
  2. [Ar]3d4 4s1
  3. [Ar]3d1 4s2
  4. [Ar]3d5

Explanation

Chromium has the configuration [Ar]3d5 4s1, and the ion Cr3+ is formed by removing three electrons. The single 4s electron and two 3d electrons are lost, leaving [Ar]3d3. This half-filled preference does not survive in the trivalent ion.

Question 38Transition and Inner Transition Elements

Why is Zn2+ not regarded as a typical transition metal ion even though zinc is placed in the 3d series?

  1. Its 3d subshell is completely empty, so it cannot show any d-d transitions.
  2. It cannot form any coordinate complexes with ligands such as NH3.
  3. It has a single unpaired electron in the 4s orbital, which makes it paramagnetic.
  4. Its 3d subshell is completely filled ([Ar]3d10), so it has no partially filled d orbitals and its compounds are generally colourless and diamagnetic. Correct answer.

Explanation

A typical transition metal ion has an incomplete d subshell in at least one of its common oxidation states. Zn2+ is [Ar]3d10, with a completely filled d subshell, so d-d transitions are impossible and its compounds are colourless and diamagnetic. It also lacks the variable oxidation states characteristic of transition metals.

Question 39Transition and Inner Transition Elements

Which property of transition metals is directly responsible for their ability to act as heterogeneous catalysts?

  1. The availability of variable oxidation states and partially filled d orbitals that allow adsorption and electron transfer at the surface Correct answer.
  2. Their complete d10 configuration in the elemental state
  3. Their inability to form alloys with other metals
  4. Their low melting points compared with alkali metals

Explanation

Transition metal surfaces can adsorb reactant molecules and use their variable oxidation states and partially filled d orbitals to weaken bonds and transfer electrons. This provides a low-energy pathway for reactions, which is why metals such as Fe, Ni, Pt and V2O5 are used as catalysts. The unfilled d orbitals are central to this behaviour.

Question 40Transition and Inner Transition Elements

In the reaction of acidified KMnO4 with oxalic acid at room temperature, what is the role of the small amount of Mn2+ formed initially?

  1. It acts as an autocatalyst, accelerating the reaction after an initial slow stage. Correct answer.
  2. It acts as an inhibitor and stops the reaction completely.
  3. It is oxidised to MnO2, which is the actual oxidising agent.
  4. It has no effect on the rate and is simply a spectator ion.

Explanation

Permanganate oxidation of oxalic acid in acid is slow at first, but the Mn2+ produced catalyses further reaction, so the rate increases as the reaction proceeds. This is a classic example of autocatalysis. Adding a little Mn2+ at the start removes the induction period.

Question 41Transition and Inner Transition Elements

Which set correctly matches the lanthanoid property with its usual explanation?

  1. Lanthanoid contraction - removal of 4f electrons as the atomic number increases
  2. Lanthanoid contraction - poor shielding of the 4f electrons, causing a gradual decrease in size across the series Correct answer.
  3. Lanthanoid contraction - increasing shielding of 4f electrons, causing a gradual increase in size
  4. Lanthanoid contraction - expansion of the 5d subshell as the 4f subshell fills

Explanation

As the 4f subshell fills across the lanthanoids, the 4f electrons shield the outer electrons poorly. The effective nuclear charge therefore rises steadily and the atomic and ionic radii decrease gradually, a trend called the lanthanoid contraction. It explains the close similarity in size of the heavier 4d and 5d elements.

Question 42Transition and Inner Transition Elements

The most common oxidation state of the lanthanoids is +3. Which statement explains why Ce4+ and Eu2+ are also relatively accessible?

  1. Both ions have empty 4f subshells.
  2. Both ions have completely filled 4f subshells.
  3. Both ions contain unpaired electrons only in the 5d subshell.
  4. Ce4+ has the stable [Xe] configuration and Eu2+ has the stable half-filled 4f7 configuration. Correct answer.

Explanation

Cerium loses its single 4f electron and a 6s electron to give Ce4+, which has the noble gas configuration [Xe]. Europium in the +2 state is [Xe]4f7, a half-filled subshell that confers extra stability. These favourable configurations make the +4 and +2 states of these elements unusually accessible.

Question 43Transition and Inner Transition Elements

Which of the following statements about the actinoids is correct?

  1. Actinoids show a wider range of oxidation states than lanthanoids because the 5f, 6d and 7s orbitals are close in energy. Correct answer.
  2. Actinoid elements have completely filled 5f subshells in all their compounds.
  3. Actinoids show only the +3 oxidation state, exactly like the lanthanoids.
  4. Actinoids are lighter than lanthanoids and occur before them in the periodic table.

Explanation

In actinoids the 5f, 6d and 7s orbitals have comparable energies, so electrons from all three can participate in bonding. This allows oxidation states ranging from +3 to +7, as seen in uranium compounds such as UO2^2+. Lanthanoids are more restricted, mainly to +3.

Question 44Transition and Inner Transition Elements

Which comparison of the 3d, 4d and 5d transition series is correct?

  1. The 4d and 5d metals generally show fewer oxidation states and are less dense than the 3d metals.
  2. The 4d and 5d metals show a greater tendency to form complexes with higher coordination numbers and are more dense than the 3d metals. Correct answer.
  3. The 4d and 5d metals have smaller atomic radii than the corresponding 3d metals in every group.
  4. The 3d metals form the strongest metal-metal bonds and have the highest enthalpies of atomisation of all three series.

Explanation

The larger size of 4d and 5d metals allows higher coordination numbers and more extensive complex formation, and their greater mass and closer packing make them denser. The 3d series shows the highest enthalpies of atomisation in several groups because its d orbitals are compact and overlap strongly. The 4d and 5d metals also show more accessible high oxidation states.

Question 45Transition and Inner Transition Elements

Aqueous CuSO4 is blue, but adding excess ammonia gives a deep blue solution. Which change accounts for the deeper colour?

  1. Ammonia oxidises Cu2+ to Cu3+, which is intensely blue.
  2. The sulphate ion absorbs strongly in the blue region after ammonia is added.
  3. Cu2+ is reduced to Cu+ by ammonia, and the d10 ion gives the deep blue colour.
  4. The [Cu(H2O)6]2+ ion is replaced by [Cu(NH3)4(H2O)2]2+, and the stronger NH3 ligands increase the crystal field splitting, shifting absorption to longer wavelength. Correct answer.

Explanation

Ammonia is a stronger field ligand than water, so it increases the splitting of the d orbitals in the copper(II) complex. The d-d absorption shifts to a longer wavelength, and the complementary transmitted colour becomes a deeper blue. The species responsible is the ammine complex [Cu(NH3)4(H2O)2]2+.

Question 46Transition and Inner Transition Elements

Which statement about the magnetic behaviour of transition metal complexes is correct?

  1. A complex with unpaired d electrons is paramagnetic and is attracted into a magnetic field. Correct answer.
  2. A complex with only paired d electrons is paramagnetic and is strongly attracted into a magnetic field.
  3. The magnetic moment depends only on the charge of the central metal ion and not on the number of unpaired electrons.
  4. All octahedral complexes of 3d metals are diamagnetic regardless of the ligands present.

Explanation

Unpaired electrons give a net magnetic moment, so the complex is paramagnetic and is drawn into a magnetic field. The magnitude of the moment depends on the number of unpaired electrons, which in turn depends on the metal, its oxidation state and the strength of the ligand field. Complexes with only paired electrons are diamagnetic.

Question 47Coordination Chemistry

According to Werner's theory, the secondary valency of the central metal ion in [Co(NH3)6]Cl3 is satisfied by which species?

  1. Three ammonia molecules
  2. Three chloride ions
  3. The cobalt(III) ion itself
  4. Six ammonia molecules Correct answer.

Explanation

Werner distinguished primary (ionisable) valency from secondary (non-ionisable) valency. In [Co(NH3)6]Cl3 the three chloride ions satisfy the primary valency, while the six ammonia ligands satisfy the secondary valency of six. Secondary valency is fixed in number and directional, which is why it gives rise to stereochemistry.

Question 48Coordination Chemistry

What is the oxidation state of iron in the complex K4[Fe(CN)6]?

  1. +3
  2. +4
  3. 0
  4. +2 Correct answer.

Explanation

Potassium is +1 and each cyanide ligand carries a charge of -1. Let x be the oxidation state of iron: 4(+1) + x + 6(-1) = 0, giving x = +2. The complex ion [Fe(CN)6]4- therefore contains iron(II).

Question 49Coordination Chemistry

Which of the following is the correct IUPAC name of the complex [Co(NH3)4Cl2]Cl?

  1. Tetraamminedichloridocobaltate(III) chloride
  2. Dichloridotetraamminecobalt(II) chloride
  3. Tetraamminecobalt(III) dichloride
  4. Tetraamminedichloridocobalt(III) chloride Correct answer.

Explanation

Ligands are named alphabetically, so ammine precedes chlorido, and the number of each is shown by the prefixes tetra and di. The cobalt oxidation state is +3, and since the complex cation is positive the metal keeps its ordinary name cobalt. The counter chloride ion is named last as chloride.

Question 50Coordination Chemistry

In the complex [Ni(CN)4]2-, the hybridisation of nickel and the geometry of the complex ion are:

  1. dsp2 and square planar Correct answer.
  2. sp3d2 and octahedral
  3. d2sp3 and octahedral
  4. sp3 and tetrahedral

Explanation

Cyanide is a strong-field ligand, so it forces pairing of the unpaired electrons of Ni(II) and uses one inner 3d orbital. The resulting hybridisation is dsp2, which gives a square planar arrangement of four ligands. Magnetic measurement confirms the complex is diamagnetic, consistent with this description.