When the alkali metals burn in excess oxygen at atmospheric pressure, which pairing of metal and main product is correct?

Potassium – KO₂, a superoxide

Lithium – Li₂O₂, a peroxide

Sodium – NaO₂, a superoxide

Potassium – KO₂, a superoxide

Caesium – Cs₂O, a normal oxide

Which statement about compounds of the group 1 and group 2 metals is correct?

BeO dissolves in both acids and alkalis, because it is amphoteric

Sulfate solubility increases down group 2, from BeSO₄ to BaSO₄

Thermal stability of alkali metal hydrides increases from LiH to CsH

Solid BeCl₂ is an ionic lattice made of discrete Be²⁺ and Cl⁻ ions

Towards a common base such as NH₃, the Lewis acid strength of the boron trihalides increases in the order BF₃ < BCl₃ < BBr₃. Which explanation is accepted?

Back-donation from halogen p into boron’s empty 2p is greatest for F and lost in the adduct

Fluorine, the most electronegative halogen, leaves boron the most electron-poor in BF₃

Larger halogens let boron use vacant 3d orbitals to accept the lone pair of the base

Bulky bromine atoms force boron to turn pyramidal before the base even arrives

Back-donation from halogen p into boron’s empty 2p is greatest for F and lost in the adduct

Which statement about the structure of diborane, B₂H₆, is correct?

Two hydrogen atoms bridge the borons through three-centre two-electron bonds

Each boron is sp² hybridised and the whole molecule is planar, like ethene

Two hydrogen atoms bridge the borons through three-centre two-electron bonds

It has a B–B single bond and six terminal B–H bonds, just like ethane

All six B–H bonds are equal in length, since the electrons are fully delocalised

SiCl₄ fumes in moist air and is rapidly hydrolysed, whereas CCl₄ is unaffected by water. What is the main reason?

The larger Si atom can bind an attacking water molecule as a fifth ligand; small C cannot

C–Cl bonds are far more polar than Si–Cl bonds, so water is repelled from the carbon

SiCl₄ is an ionic solid of Si⁴⁺ and Cl⁻ ions, which water hydrates and splits at once

Hydrolysis of CCl₄ would absorb heat, whereas hydrolysis of SiCl₄ releases a lot of it

Which of the following statements about oxides and oxoacids of the p-block elements are correct?

N₂O₅ is the acid anhydride of nitric acid

H₃PO₃ is dibasic because one of its three hydrogens is bonded directly to phosphorus

N₂O₅ is the acid anhydride of nitric acid

NO is an acidic oxide that dissolves in water to give nitrous acid

H₃PO₃ is dibasic because one of its three hydrogens is bonded directly to phosphorus

Al₂O₃ is a purely basic oxide that dissolves in acids but not in alkalis

The permanganate ion, MnO₄⁻, is intensely purple although its manganese is in the +7 oxidation state. What is the origin of its colour?

Ligand-to-metal charge transfer, an electron moving from oxygen to an empty Mn d orbital

d–d transitions of the three 3d electrons left on Mn⁷⁺, split by the tetrahedral field

Unpaired electrons on Mn⁷⁺, which make the ion paramagnetic and hence coloured

Ligand-to-metal charge transfer, an electron moving from oxygen to an empty Mn d orbital

Light emitted by Mn⁷⁺ ions as water slowly reduces them to the Mn²⁺ state

Which of the following statements about the 3d elements are correct?

Zinc is not regarded as a transition element, because neither Zn nor Zn²⁺ has an incomplete d subshell

Manganese shows the highest oxidation state in the 3d series, +7

Aqueous Sc³⁺ is coloured because of d–d transitions

Zinc is not regarded as a transition element, because neither Zn nor Zn²⁺ has an incomplete d subshell

Atomic radii increase steadily from Sc to Zn as d electrons are added

Manganese shows the highest oxidation state in the 3d series, +7

[CoF₆]³⁻ has a magnetic moment of about 4.9 BM, whereas [Co(NH₃)₆]³⁺ is diamagnetic. Both contain octahedral Co³⁺ (d⁶). Which interpretation is correct?

F⁻ is weak-field, so Δo P gives t₂g⁶

The fluoride complex is really tetrahedral, and the small Δt leaves four electrons unpaired

F⁻ is weak-field, so Δo P gives t₂g⁶

Fluoride reduces cobalt to Co²⁺, and the d⁷ Co²⁺ ion carries four unpaired electrons

The ammine complex is square planar dsp², which pairs every electron, unlike the fluoride

[Ni(CN)₄]²⁻ is diamagnetic while [NiCl₄]²⁻ is paramagnetic with two unpaired electrons. Which assignment of geometry and hybridisation fits these facts?

[Ni(CN)₄]²⁻ square planar and dsp²; [NiCl₄]²⁻ tetrahedral and sp³

Both tetrahedral and sp³; CN⁻ pairs the electrons simply through its stronger field

[Ni(CN)₄]²⁻ tetrahedral and sp³; [NiCl₄]²⁻ square planar and dsp²

Both square planar and dsp²; Cl⁻ is too weak a ligand to pair any electrons

[Ni(CN)₄]²⁻ square planar and dsp²; [NiCl₄]²⁻ tetrahedral and sp³

Two complexes have the formulas [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅(SO₄)]Br. Which observation distinguishes them, and what type of isomers are they?

Only [Co(NH₃)₅Br]SO₄ gives a white precipitate with BaCl₂; they are ionisation isomers

Only [Co(NH₃)₅(SO₄)]Br gives a white precipitate with BaCl₂; they are linkage isomers

Both give a pale-yellow precipitate with AgNO₃; they are coordination isomers

Only [Co(NH₃)₅Br]SO₄ gives a white precipitate with BaCl₂; they are ionisation isomers

Only [Co(NH₃)₅Br]SO₄ gives a pale-yellow precipitate with AgNO₃; they are hydrate isomers

Which of the following statements about isomerism in coordination complexes are correct?

cis-[PtCl₂(NH₃)₂] is the anticancer drug cisplatin

[Co(NH₃)₃Cl₃] exists as fac and mer isomers

A tetrahedral complex [MA₂B₂] exists as cis and trans isomers

cis-[PtCl₂(NH₃)₂] is the anticancer drug cisplatin

[Co(NH₃)₃Cl₃] exists as fac and mer isomers

trans-[Co(en)₂Cl₂]⁺ is chiral and can be resolved into enantiomers

s, p and d Block Elements: Oxides, Halides and Hydrides, the 3d Series and Coordination Complexes

This chapter answers Section XL0.3 of XL0 Chemistry, the part of the Life Sciences (XL) paper that every XL candidate answers: “Oxides, halides and hydrides of alkali, alkaline earth metals, B, Al, Si, N, P, and S. General characteristics of 3d elements. Coordination complexes: valence bond and crystal field theory, colour, geometry, magnetic properties and isomerism.” It first walks through the oxides, halides and hydrides of the named s- and p-block elements, explaining each trend rather than listing it, then sets out what makes the 3d metals distinctive, and finally treats coordination complexes by valence-bond and crystal-field theory, with their colour, geometry, magnetism and isomers. The numericals are gas volumes from hydrides, spin-only magnetic moments, crystal field stabilisation energies, Δo from an absorption wavelength and counts of isomers.

1. Alkali and alkaline earth metals: oxides, hydrides and halides

Oxides. Burnt in excess oxygen, lithium gives the normal oxide Li₂O, sodium mainly the peroxide Na₂O₂ (O₂²⁻), and potassium, rubidium and caesium the superoxides MO₂ (O₂⁻, paramagnetic). A large cation stabilises a large anion in the lattice, so the bigger the metal ion, the bigger the oxygen anion it can hold. All are strongly basic: Na₂O + H₂O → 2NaOH, peroxides give H₂O₂, and superoxides give H₂O₂ and O₂; KO₂ regenerates breathing air because 4KO₂ + 2CO₂ → 2K₂CO₃ + 3O₂. Group 2 metals burn to MO (barium also gives BaO₂). Basic character rises down group 2: BeO is amphoteric, MgO weakly basic, and CaO, SrO and BaO strongly basic. Down group 2 the hydroxides become more soluble while the sulfates become less soluble, so BaSO₄ is insoluble enough to be swallowed as an X-ray contrast agent.

Hydrides and halides. The alkali metals and Ca, Sr and Ba form ionic (saline) hydrides containing H⁻, which react with water to release hydrogen: NaH + H₂O → NaOH + H₂ and CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂. Thermal stability falls from LiH to CsH as the lattice weakens. BeH₂ and MgH₂ are covalent, polymeric solids; BeH₂ is held together by Be–H–Be bridges. The halides are mostly high-melting ionic solids, but the small, polarising Li⁺ and Be²⁺ give covalent character (Fajans): LiCl dissolves in organic solvents, and BeCl₂ is a chain polymer of Cl-bridged, tetrahedral Be in the solid, a Cl-bridged dimer in the vapour and a linear monomer at high temperature. Its aqueous solution is acidic because the small Be²⁺ polarises its hydration water. These similarities give the diagonal relationships Li–Mg and Be–Al: Be and Al both have amphoteric oxides and covalent, bridged chlorides that act as Lewis acids.

⚠️ Opposite solubility trends in group 2

Hydroxide solubility rises from Be(OH)₂ to Ba(OH)₂, but sulfate solubility falls from BeSO₄ to BaSO₄. A small anion such as OH⁻ is held most strongly in the lattice by small cations, while for the large sulfate ion the lattice energy changes little down the group and the falling hydration energy of the cation decides the trend.

2. Boron, aluminium and silicon

Boron and aluminium. B₂O₃ is acidic and dissolves in water to give boric acid, B(OH)₃, a weak monobasic Lewis acid: it does not release its own protons but accepts OH⁻, B(OH)₃ + H₂O ⇌ [B(OH)₄]⁻ + H⁺. Al₂O₃ and Al(OH)₃ are amphoteric, dissolving in acid as Al³⁺ and in NaOH as [Al(OH)₄]⁻. The trigonal planar BX₃ halides are electron-deficient Lewis acids, and towards a common base their strength runs BF₃ < BCl₃ < BBr₃, against the electronegativity order. The reason is back-bonding: a filled 2p orbital on F donates into the empty 2p orbital of B, an overlap that is best for the small F atom and is lost when the boron turns pyramidal in an adduct. BCl₃ and AlCl₃ hydrolyse readily, and AlCl₃, a Lewis acid catalyst, exists as the Cl-bridged dimer Al₂Cl₆ in the vapour at moderate temperatures.

Diborane, B₂H₆, has only 12 valence electrons, too few for eight ordinary bonds. Each boron holds two terminal H by normal two-centre two-electron bonds, and the two borons are joined by two B–H–B bridges, each a three-centre two-electron bond; the four terminal H lie in one plane and the two bridging H above and below it, and the bridge B–H bonds are longer than the terminal ones. Diborane burns in air and hydrolyses: B₂H₆ + 6H₂O → 2B(OH)₃ + 6H₂. The complex hydrides NaBH₄ and LiAlH₄ deliver H⁻ and reduce aldehydes and ketones to alcohols; LiAlH₄ is stronger and reacts violently with water, whereas NaBH₄ can be used in water or alcohol.

Silicon. CO₂ is a gas of small linear molecules, but SiO₂ is a hard, high-melting network solid in which every Si is bonded tetrahedrally to four O and every O bridges two Si. Silicon is too large for good pπ–pπ overlap with oxygen, so it makes four Si–O single bonds instead of two double bonds. SiO₂ is acidic: fused with NaOH or Na₂CO₃ it gives silicates, and HF attacks it (SiO₂ + 4HF → SiF₄ + 2H₂O), which is why HF etches glass. Silicates are built from SiO₄ tetrahedra sharing corners into chains, sheets (micas, clays) and frameworks (quartz, feldspars, zeolites), and diatoms build their cell walls from hydrated silica. SiCl₄ fumes in moist air and hydrolyses completely to hydrated silica and HCl, whereas CCl₄ is unaffected by water: the larger Si atom can take a water molecule as a fifth ligand in a five-coordinate intermediate, but the small, crowded, four-coordinate carbon cannot. Silane, SiH₄, ignites spontaneously in air, unlike methane, because Si–H and Si–Si bonds are weaker than C–H and C–C, which also limits silicon’s catenation.

3. Nitrogen, phosphorus and sulfur

Key oxides, hydrides and halides of N, P and S
Element	Oxides	Hydride	Halides and hydrolysis
N	N₂O (+1) and NO (+2) neutral; NO is an odd-electron radical. N₂O₃ (+3) gives HNO₂; NO₂ (+4) is brown, paramagnetic and dimerises to N₂O₄; N₂O₅ (+5) is the anhydride of HNO₃	NH₃: pyramidal, 107°, hydrogen-bonded (bp −33 °C), a good base and ligand	NCl₃ + 3H₂O → NH₃ + 3HOCl; NF₃ resists hydrolysis; no NCl₅, since N cannot exceed four bonds
P	P₄O₆ and P₄O₁₀: P₄ cages with an O inserted in each of the six P–P edges; P₄O₁₀ also has a terminal P=O on each P. P₄O₆ gives H₃PO₃, P₄O₁₀ gives H₃PO₄ and is a powerful drying agent	PH₃: about 93.5°, no significant hydrogen bonding, a much weaker base than NH₃, toxic	PCl₃ + 3H₂O → H₃PO₃ + 3HCl; PCl₅ is trigonal bipyramidal as a gas, [PCl₄]⁺[PCl₆]⁻ as a solid, and water converts it via POCl₃ to H₃PO₄
S	SO₂: bent, acidic (gives H₂SO₃), mainly a reducing agent, used to preserve foods; SO₃: trigonal planar, the anhydride of H₂SO₄	H₂S: bent, about 92°, weak diprotic acid, bp about −60 °C, highly toxic	SF₆: octahedral and kinetically inert, its S shielded by six F; SF₄: see-saw, hydrolyses readily (SF₄ + 2H₂O → SO₂ + 4HF)

Three patterns run through the table. The acidity of oxides rises with the oxidation state of the element (NO neutral, N₂O₅ strongly acidic; SO₂ gives a weak acid, SO₃ a strong one). Basicity falls from NH₃ to PH₃ because phosphorus’s lone pair sits in a larger, more diffuse orbital. And the basicity of a phosphorus oxoacid equals the number of P–OH groups, since H bonded directly to P is not ionisable: H₃PO₄ is tribasic, H₃PO₃ (one P–H) dibasic and H₃PO₂ (two P–H) monobasic, and the P–H bonds make the last two reducing agents.

4. General characteristics of the 3d elements

Configuration. Sc to Zn are [Ar] 3dⁿ 4s² (Cr 3d⁵4s¹, Cu 3d¹⁰4s¹). A transition element has an incomplete d subshell in the atom or a common ion, so Sc (3d¹) qualifies but Zn, 3d¹⁰ in the atom and in Zn²⁺, does not.
Metallic properties and size. Hard, dense metals with high melting points, because d electrons join the metallic bonding. Radii shrink at first and then stay nearly constant across the series, since added 3d electrons shield the outer 4s electrons poorly; this similarity in size lets them form alloys and interstitial compounds (C, N or H in the lattice, as in steel).
Variable oxidation states, usually differing by one, because 4s and 3d are close in energy. The maximum rises to +7 at Mn (MnO₄⁻) and then falls; Sc shows only +3 and Zn only +2. Mn²⁺ and Fe³⁺ (d⁵) are especially stable, and high states are stabilised by O and F (CrO₄²⁻, Cr₂O₇²⁻, MnO₄⁻, all strong oxidants).
Colour, magnetism, catalysis and complexes. Partly filled d subshells give coloured ions (d–d transitions) and paramagnetism (unpaired electrons); d⁰ Sc³⁺ and Ti⁴⁺ and d¹⁰ Zn²⁺ are colourless and diamagnetic. Variable oxidation states and active surfaces make them catalysts: Fe in ammonia synthesis, V₂O₅ in the oxidation of SO₂ to SO₃, Ni in the hydrogenation of oils. Small, highly charged ions with vacant d orbitals form many complexes, and Fe, Cu, Zn, Mn and Co are essential to life.

⚠️ Not every colour is d–d

MnO₄⁻ (Mn⁷⁺, d⁰) and Cr₂O₇²⁻ (Cr⁶⁺, d⁰) have no d electrons, yet they are intensely coloured. Their colour comes from ligand-to-metal charge transfer: an electron jumps from an oxygen orbital to an empty metal d orbital. Such bands are allowed and therefore far more intense than the weak, Laporte-forbidden d–d bands of, say, [Ti(H₂O)₆]³⁺.

5. Coordination complexes: valence-bond and crystal-field theory, geometry, colour and magnetism

A complex has a central metal ion bound to ligands, Lewis bases donating electron pairs; the number of donor atoms is the coordination number. A ligand with several donor atoms (ethylenediamine, en, bidentate; EDTA⁴⁻, hexadentate) forms rings and is a chelate. Valence-bond theory gives the metal hybrid orbitals to receive the ligand pairs: sp³ tetrahedral ([NiCl₄]²⁻, two unpaired electrons), dsp² square planar ([Ni(CN)₄]²⁻, diamagnetic, its electrons paired to free one 3d orbital), d²sp³ inner-orbital octahedral ([Co(NH₃)₆]³⁺, diamagnetic) and sp³d² outer-orbital octahedral ([CoF₆]³⁻, four unpaired electrons). VBT rationalises geometry and magnetism but cannot explain colour, cannot say in advance which ligands force pairing, and gives no energies.

Crystal-field theory treats ligands as point charges that split the five d orbitals. In an octahedral field the e_g pair (d_x²−y², d_z²), pointing at the ligands, rises by 0.6Δo and the t₂g trio (d_xy, d_xz, d_yz), pointing between them, falls by 0.4Δo. In a tetrahedral field the order is inverted (e below t₂) and Δt ≈ (4/9)Δo, so tetrahedral complexes are almost always high-spin. A square planar field raises d_x²−y² far above the rest, so d⁸ ions with strong-field ligands (Ni²⁺ with CN⁻, and Pd²⁺ and Pt²⁺ generally) are square planar and diamagnetic. Δo grows along the spectrochemical series I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO, with the metal’s charge (M³⁺ > M²⁺) and down a group (4d and 5d ≫ 3d).

For d⁴ to d⁷ octahedral ions there is a choice. If Δo is smaller than the pairing energy P, electrons occupy e_g before pairing (high spin); if Δo > P they pair in t₂g (low spin). The crystal field stabilisation energy is CFSE = (−0.4 n(t₂g) + 0.6 n(e_g))Δo, plus P for each pair forced beyond those in the free ion; stems state whether to include P. Spin-only magnetic moment: μ = √(n(n + 2)) BM for n unpaired electrons, giving 1.73, 2.83, 3.87, 4.90 and 5.92 BM for n = 1 to 5. Worked example: Fe³⁺ is d⁵. With weak-field F⁻, [FeF₆]³⁻ is t₂g³e_g², five unpaired, spin-only μ = 5.92 BM, CFSE = 3(−0.4) + 2(0.6) = 0. With strong-field CN⁻, [Fe(CN)₆]³⁻ is t₂g⁵, one unpaired, spin-only μ = 1.73 BM, CFSE = −2.0Δo (plus 2P). A measured moment thus reveals the spin state. Octahedral d⁹ Cu²⁺ is also Jahn–Teller distorted, its two axial bonds lengthened.

Colour. A d–d transition absorbs light of energy Δo, and the complex shows the complementary colour. Δo per mole = N_A hc/λ: a band at 500 nm (20 000 cm⁻¹) means Δo = 6.022 × 10²³ × 6.626 × 10⁻³⁴ × 3.00 × 10⁸/(500 × 10⁻⁹) = 239 kJ mol⁻¹. The d¹ ion [Ti(H₂O)₆]³⁺ absorbs near 500 nm, in the green, and looks purple. Replacing water by the stronger-field NH₃ raises Δo and shifts absorption to shorter wavelength, which is why [Cu(NH₃)₄]²⁺ is a deeper blue-violet than hydrated Cu²⁺.

6. Isomerism in coordination complexes

Kinds of isomerism, with examples
Type	What differs	Example and test
Ionisation	Which anion is coordinated and which is free	[Co(NH₃)₅Br]SO₄ gives BaSO₄ with Ba²⁺; [Co(NH₃)₅(SO₄)]Br gives AgBr with Ag⁺
Hydrate	Water inside versus outside the coordination sphere	[Cr(H₂O)₆]Cl₃ (violet), [Cr(H₂O)₅Cl]Cl₂·H₂O, [Cr(H₂O)₄Cl₂]Cl·2H₂O: AgNO₃ precipitates 3, 2 and 1 Cl⁻ per formula
Linkage	Donor atom of an ambidentate ligand	M–NO₂ (nitro, N-bound) and M–ONO (nitrito, O-bound); SCN⁻ by S or N
Coordination	Ligands swapped between cationic and anionic complexes	[Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆]
Geometrical	Relative positions of ligands	Square planar MA₂B₂: cis and trans; octahedral MA₄B₂: cis and trans; MA₃B₃: fac and mer. Tetrahedral MA₂B₂ has none
Optical	Non-superimposable mirror images (enantiomers)	[Co(en)₃]³⁺ as Δ and Λ; cis-[Co(en)₂Cl₂]⁺ is chiral, trans is not

Counting stereoisomers means listing geometrical isomers, then checking each for a mirror plane. [Co(en)₂Cl₂]⁺ has a trans form, which has mirror planes, and a cis form, which has none and so exists as two enantiomers: three stereoisomers in all. [MA₂B₂C₂] has five geometrical isomers, one of which (all three pairs cis) is chiral, giving six stereoisomers. Geometry matters biologically: cisplatin, cis-[PtCl₂(NH₃)₂], is an anticancer drug that cross-links neighbouring guanines in DNA, while its trans isomer is clinically inactive.

Key takeaways

Li gives Li₂O, Na gives Na₂O₂ and K, Rb, Cs give superoxides MO₂; group 2 oxides grow more basic downward (BeO amphoteric). Saline hydrides contain H⁻ and release H₂ with water; Be and Li compounds are the most covalent, and BeCl₂ is a bridged polymer.
Boron compounds are electron-deficient: B(OH)₃ is a monobasic Lewis acid, BF₃ < BCl₃ < BBr₃ in Lewis acidity owing to back-bonding, and diborane has two 3c–2e B–H–B bridges. Al₂O₃ is amphoteric and AlCl₃ dimerises.
SiO₂ is a network of corner-sharing SiO₄ tetrahedra, acidic and attacked by HF; SiCl₄ hydrolyses but CCl₄ does not; SiH₄ ignites in air.
Oxide acidity rises with oxidation state (NO and N₂O neutral, N₂O₅ and SO₃ strongly acidic); NH₃ is more basic than PH₃ and has a wider angle; the basicity of a P oxoacid equals its number of P–OH groups.
3d metals show variable oxidation states (up to +7 at Mn), colour, paramagnetism, catalysis and complex formation; d⁰ and d¹⁰ ions are colourless unless charge transfer colours them.
Octahedral splitting puts t₂g at −0.4Δo and e_g at +0.6Δo; Δo versus P decides high or low spin; μ = √(n(n + 2)) BM; Δo = N_A hc/λ. Isomers are ionisation, hydrate, linkage, coordination, geometrical and optical; [Co(en)₂Cl₂]⁺ has three stereoisomers.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

When the alkali metals burn in excess oxygen at atmospheric pressure, which pairing of metal and main product is correct?
1. Lithium – Li₂O₂, a peroxide
2. Sodium – NaO₂, a superoxide
3. Potassium – KO₂, a superoxide
4. Caesium – Cs₂O, a normal oxide
Show answer
Answer: C — Potassium – KO₂, a superoxide
Lithium gives the normal oxide Li₂O, sodium mainly the peroxide Na₂O₂, and potassium, rubidium and caesium the superoxides MO₂. Larger cations are better at stabilising large anions of low charge density in a lattice, so down the group the product changes from O²⁻ to O₂²⁻ and then to O₂⁻, whose single negative charge is spread over two atoms. Pairing sodium with a superoxide shifts the series up by one metal, the commonest slip.
4.21 g of calcium hydride (molar mass 42.1 g mol⁻¹) reacts completely with water: CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂. Taking the molar volume of a gas at STP (273 K, 1 atm) as 22.4 L mol⁻¹, the volume of hydrogen released, in litres at STP to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 4.48
Moles of CaH₂ = 4.21/42.1 = 0.100 mol. Each mole gives 2 mol H₂, because the two H⁻ ions each take a proton from water, so n(H₂) = 0.200 mol and V = 0.200 × 22.4 = 4.48 L. Forgetting the factor of 2 gives 2.24 L.
Which statement about compounds of the group 1 and group 2 metals is correct?
1. BeO dissolves in both acids and alkalis, because it is amphoteric
2. Sulfate solubility increases down group 2, from BeSO₄ to BaSO₄
3. Thermal stability of alkali metal hydrides increases from LiH to CsH
4. Solid BeCl₂ is an ionic lattice made of discrete Be²⁺ and Cl⁻ ions
Show answer
Answer: A — BeO dissolves in both acids and alkalis, because it is amphoteric
BeO is amphoteric: it gives Be²⁺ salts with acids and beryllate with alkalis, part of beryllium’s diagonal resemblance to aluminium. Sulfate solubility falls down the group (BaSO₄ is insoluble), which is the tempting reversal of the hydroxide trend. Hydride stability falls from LiH to CsH as the lattice weakens, and the small, polarising Be²⁺ makes solid BeCl₂ a covalent chain polymer with chloride bridges, not an ionic lattice.
Towards a common base such as NH₃, the Lewis acid strength of the boron trihalides increases in the order BF₃ < BCl₃ < BBr₃. Which explanation is accepted?
1. Fluorine, the most electronegative halogen, leaves boron the most electron-poor in BF₃
2. Larger halogens let boron use vacant 3d orbitals to accept the lone pair of the base
3. Bulky bromine atoms force boron to turn pyramidal before the base even arrives
4. Back-donation from halogen p into boron’s empty 2p is greatest for F and lost in the adduct
Show answer
Answer: D — Back-donation from halogen p into boron’s empty 2p is greatest for F and lost in the adduct
In planar BX₃ a filled p orbital of X donates into the empty 2p orbital of B. The overlap is best for the small F atom, so BF₃ gains the most π stabilisation, and it loses all of it when boron becomes tetrahedral in an adduct; BF₃ therefore pays the highest price to accept a base. The electronegativity argument predicts the opposite order and is the tempting trap. Boron has no accessible d orbitals, and the halogens do not make boron pyramidal before adduct formation.
Which statement about the structure of diborane, B₂H₆, is correct?
1. Each boron is sp² hybridised and the whole molecule is planar, like ethene
2. Two hydrogen atoms bridge the borons through three-centre two-electron bonds
3. It has a B–B single bond and six terminal B–H bonds, just like ethane
4. All six B–H bonds are equal in length, since the electrons are fully delocalised
Show answer
Answer: B — Two hydrogen atoms bridge the borons through three-centre two-electron bonds
Diborane has only 12 valence electrons. Eight go into four terminal two-centre B–H bonds, and the remaining four form two B–H–B bridges, each a three-centre two-electron bond. The ethane-like structure would need 14 electrons, which diborane does not have. The bridging H atoms lie above and below the plane of the four terminal ones, so the molecule is not planar, and the bridge bonds are longer than the terminal ones.
SiCl₄ fumes in moist air and is rapidly hydrolysed, whereas CCl₄ is unaffected by water. What is the main reason?
1. The larger Si atom can bind an attacking water molecule as a fifth ligand; small C cannot
2. C–Cl bonds are far more polar than Si–Cl bonds, so water is repelled from the carbon
3. SiCl₄ is an ionic solid of Si⁴⁺ and Cl⁻ ions, which water hydrates and splits at once
4. Hydrolysis of CCl₄ would absorb heat, whereas hydrolysis of SiCl₄ releases a lot of it
Show answer
Answer: A — The larger Si atom can bind an attacking water molecule as a fifth ligand; small C cannot
Hydrolysis begins with water attacking the central atom. Silicon is large enough to accept it as a fifth ligand in a five-coordinate intermediate, after which Cl⁻ leaves; carbon is small and its four Cl atoms shield it, so no such intermediate forms. The resistance of CCl₄ is kinetic, since its hydrolysis to CO₂ and HCl is thermodynamically favourable. Si–Cl is in fact more polar than C–Cl (Si is less electronegative than C), and SiCl₄ is a volatile molecular liquid, not an ionic solid.
Which of the following statements about oxides and oxoacids of the p-block elements are correct?
1. N₂O₅ is the acid anhydride of nitric acid
2. NO is an acidic oxide that dissolves in water to give nitrous acid
3. H₃PO₃ is dibasic because one of its three hydrogens is bonded directly to phosphorus
4. Al₂O₃ is a purely basic oxide that dissolves in acids but not in alkalis
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Answer: A — N₂O₅ is the acid anhydride of nitric acid; C — H₃PO₃ is dibasic because one of its three hydrogens is bonded directly to phosphorus
N₂O₅ + H₂O → 2HNO₃, so it is the anhydride of nitric acid. H₃PO₃ has two P–OH groups and one P–H bond; only the O–H hydrogens ionise, so it is dibasic. NO, like N₂O, is a neutral oxide; nitrous acid comes from N₂O₃. Al₂O₃ is amphoteric and dissolves in NaOH as [Al(OH)₄]⁻.
The permanganate ion, MnO₄⁻, is intensely purple although its manganese is in the +7 oxidation state. What is the origin of its colour?
1. d–d transitions of the three 3d electrons left on Mn⁷⁺, split by the tetrahedral field
2. Unpaired electrons on Mn⁷⁺, which make the ion paramagnetic and hence coloured
3. Ligand-to-metal charge transfer, an electron moving from oxygen to an empty Mn d orbital
4. Light emitted by Mn⁷⁺ ions as water slowly reduces them to the Mn²⁺ state
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Answer: C — Ligand-to-metal charge transfer, an electron moving from oxygen to an empty Mn d orbital
Manganese ([Ar] 3d⁵ 4s²) has no d electrons left in the +7 state, so d–d transitions are impossible and the ion is diamagnetic. Its colour comes from ligand-to-metal charge transfer: visible light promotes an electron from a filled oxygen orbital into an empty Mn d orbital. Such transitions are allowed, which is why permanganate is far more intensely coloured than a typical d–d complex. Colour does not arise from paramagnetism, and the solution does not glow.
Which of the following statements about the 3d elements are correct?
1. Aqueous Sc³⁺ is coloured because of d–d transitions
2. Zinc is not regarded as a transition element, because neither Zn nor Zn²⁺ has an incomplete d subshell
3. Atomic radii increase steadily from Sc to Zn as d electrons are added
4. Manganese shows the highest oxidation state in the 3d series, +7
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Answer: B — Zinc is not regarded as a transition element, because neither Zn nor Zn²⁺ has an incomplete d subshell; D — Manganese shows the highest oxidation state in the 3d series, +7
Zn is 3d¹⁰ 4s² and Zn²⁺ is 3d¹⁰, so neither has an incomplete d subshell. Manganese reaches +7 in MnO₄⁻, using all seven 3d and 4s electrons; later elements reach lower maxima. Sc³⁺ is d⁰, so it has no d–d transitions and is colourless. Atomic radii decrease at first and then stay nearly constant across the series, because the added 3d electrons shield poorly; they do not rise steadily.
Which of the following aqueous ions is colourless?
1. Zn²⁺
2. Cu²⁺
3. Ni²⁺
4. Cr³⁺
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Answer: A — Zn²⁺
Zn²⁺ is 3d¹⁰: with a full d subshell there is no vacancy to promote an electron into, so no d–d absorption occurs in the visible and the ion is colourless. Cu²⁺ (d⁹, blue), Ni²⁺ (d⁸, green) and Cr³⁺ (d³, violet to green) all have partly filled d subshells and absorb visible light through d–d transitions.
[Fe(H₂O)₆]²⁺ is a high-spin octahedral complex. Using μ = √(n(n + 2)) BM, where n is the number of unpaired electrons, its spin-only magnetic moment, in BM to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 4.90
Fe²⁺ is [Ar] 3d⁶. High spin means t₂g⁴ e_g², with four unpaired electrons. μ = √(4 × 6) = √24 = 4.90 BM. Treating it as low spin (t₂g⁶) would give zero, and taking five unpaired electrons, as for Fe³⁺, would give 5.92 BM.
[Co(NH₃)₆]³⁺ is a low-spin octahedral complex. Using CFSE = (−0.4 n(t₂g) + 0.6 n(e_g))Δo and ignoring pairing energy, its crystal field stabilisation energy is −x Δo. The value of x, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 2.4
Co³⁺ is [Ar] 3d⁶. Low spin puts all six electrons in t₂g (t₂g⁶ e_g⁰), so CFSE = 6 × (−0.4)Δo = −2.4Δo and x = 2.4. The high-spin arrangement t₂g⁴ e_g² would give 4(−0.4) + 2(0.6) = −0.4Δo, the usual wrong answer when the spin state is ignored. NH₃ gives Co³⁺ a Δo larger than the pairing energy, which is why the complex is diamagnetic.
An octahedral d¹ complex shows a single d–d absorption band with its maximum at 490 nm. Taking this band to correspond to Δo, with h = 6.626 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹ and N_A = 6.022 × 10²³ mol⁻¹, Δo in kJ mol⁻¹, to the nearest whole number, is ____.
Numerical answer — type the value.
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Answer: 244
Energy per photon = hc/λ = (6.626 × 10⁻³⁴ × 3.00 × 10⁸)/(490 × 10⁻⁹) = 4.057 × 10⁻¹⁹ J. Per mole: 4.057 × 10⁻¹⁹ × 6.022 × 10²³ = 2.443 × 10⁵ J mol⁻¹ = 244 kJ mol⁻¹. For a d¹ ion the single band is the t₂g → e_g jump, so its energy is Δo directly. Forgetting N_A, or leaving the answer in J mol⁻¹, are the usual slips.
[CoF₆]³⁻ has a magnetic moment of about 4.9 BM, whereas [Co(NH₃)₆]³⁺ is diamagnetic. Both contain octahedral Co³⁺ (d⁶). Which interpretation is correct?
1. The fluoride complex is really tetrahedral, and the small Δt leaves four electrons unpaired
2. F⁻ is weak-field, so Δo < P gives t₂g⁴e_g²; NH₃ is stronger, so Δo > P gives t₂g⁶
3. Fluoride reduces cobalt to Co²⁺, and the d⁷ Co²⁺ ion carries four unpaired electrons
4. The ammine complex is square planar dsp², which pairs every electron, unlike the fluoride
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Answer: B — F⁻ is weak-field, so Δo < P gives t₂g⁴e_g²; NH₃ is stronger, so Δo > P gives t₂g⁶
4.9 BM corresponds to four unpaired electrons (√24). With weak-field F⁻ the splitting is smaller than the pairing energy, so d⁶ spreads as t₂g⁴e_g² (high spin, the VBT outer-orbital sp³d² case). NH₃ lies higher in the spectrochemical series, Δo exceeds P, and all six electrons pair in t₂g (low spin, inner-orbital d²sp³), giving a diamagnetic ion. Six ligands mean octahedral, not tetrahedral or square planar, and F⁻ does not reduce Co³⁺; d⁷ Co²⁺ would in any case have at most three unpaired electrons.
[Ni(CN)₄]²⁻ is diamagnetic while [NiCl₄]²⁻ is paramagnetic with two unpaired electrons. Which assignment of geometry and hybridisation fits these facts?
1. Both tetrahedral and sp³; CN⁻ pairs the electrons simply through its stronger field
2. [Ni(CN)₄]²⁻ tetrahedral and sp³; [NiCl₄]²⁻ square planar and dsp²
3. Both square planar and dsp²; Cl⁻ is too weak a ligand to pair any electrons
4. [Ni(CN)₄]²⁻ square planar and dsp²; [NiCl₄]²⁻ tetrahedral and sp³
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Answer: D — [Ni(CN)₄]²⁻ square planar and dsp²; [NiCl₄]²⁻ tetrahedral and sp³
Ni²⁺ is d⁸. Strong-field CN⁻ pairs the eight electrons into four 3d orbitals, freeing one 3d orbital for dsp² hybridisation: a square planar, diamagnetic ion. Weak-field Cl⁻ leaves two unpaired electrons and the 4s and 4p orbitals give sp³: tetrahedral. A tetrahedral d⁸ ion (e⁴t₂⁴) always has two unpaired electrons, whatever the ligand, so the cyanide complex cannot be tetrahedral; a square planar d⁸ ion is diamagnetic, so the chloride cannot be square planar.
Counting each enantiomer separately, the total number of stereoisomers (geometrical and optical) of the octahedral ion [Co(en)₂Cl₂]⁺, where en is the bidentate ligand ethylenediamine, is ____.
Numerical answer — type the value.
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Answer: 3
There are two geometrical isomers. In trans, the two Cl are opposite and both en rings lie in one plane, which is a mirror plane, so trans is achiral: one isomer. In cis, the Cl are adjacent and the two en rings form a propeller with no mirror plane, so cis exists as a pair of enantiomers: two isomers. Total = 1 + 2 = 3. Answering 2 forgets the optical isomerism of the cis form; answering 4 wrongly makes the trans form chiral.
Two complexes have the formulas [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅(SO₄)]Br. Which observation distinguishes them, and what type of isomers are they?
1. Only [Co(NH₃)₅(SO₄)]Br gives a white precipitate with BaCl₂; they are linkage isomers
2. Both give a pale-yellow precipitate with AgNO₃; they are coordination isomers
3. Only [Co(NH₃)₅Br]SO₄ gives a white precipitate with BaCl₂; they are ionisation isomers
4. Only [Co(NH₃)₅Br]SO₄ gives a pale-yellow precipitate with AgNO₃; they are hydrate isomers
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Answer: C — Only [Co(NH₃)₅Br]SO₄ gives a white precipitate with BaCl₂; they are ionisation isomers
Only ions outside the coordination sphere are free in solution. In [Co(NH₃)₅Br]SO₄ the sulfate is free and gives white BaSO₄ with BaCl₂, while its bromide is bound to cobalt. In [Co(NH₃)₅(SO₄)]Br the bromide is free and gives pale-yellow AgBr with AgNO₃. Isomers that differ in which anion is inside the coordination sphere are ionisation isomers. The other choices attach each test to the wrong compound or name the wrong type of isomerism.
Which of the following statements about isomerism in coordination complexes are correct?
1. A tetrahedral complex [MA₂B₂] exists as cis and trans isomers
2. cis-[PtCl₂(NH₃)₂] is the anticancer drug cisplatin
3. [Co(NH₃)₃Cl₃] exists as fac and mer isomers
4. trans-[Co(en)₂Cl₂]⁺ is chiral and can be resolved into enantiomers
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Answer: B — cis-[PtCl₂(NH₃)₂] is the anticancer drug cisplatin; C — [Co(NH₃)₃Cl₃] exists as fac and mer isomers
Cisplatin is the cis isomer of square planar [PtCl₂(NH₃)₂]; its trans isomer is clinically inactive. An octahedral MA₃B₃ complex has a fac isomer (the three Cl on one face) and a mer isomer (the three Cl on a meridian). In a tetrahedron every position is adjacent to every other, so [MA₂B₂] has no cis–trans isomers. trans-[Co(en)₂Cl₂]⁺ contains a mirror plane and is achiral; it is the cis form that exists as enantiomers.