Group 13 Elements Explained: Boron to Thallium Properties and Trends

Chapter 13 — The Group 13 Elements: Boron, Aluminium, Gallium, Indium & Thallium

A Complete Competitive Exam Guide | Based on Housecroft & Sharpe, Inorganic Chemistry, 4th Edition

JEE Advanced NEET IIT-JAM GATE CSIR-NET TGT/PGT BITSAT

Group 13 spans from a non-metal (Boron) through a classic amphoteric metal (Aluminium) to the heavy post-transition metals (Gallium, Indium, Thallium). No other group in the p-block shows such a dramatic swing in character — making it one of the most exam-favourite chapters in inorganic chemistry.

1. Introduction — The Big Picture of Group 13

The five elements of Group 13 — Boron (B, Z=5), Aluminium (Al, Z=13), Gallium (Ga, Z=31), Indium (In, Z=49), and Thallium (Tl, Z=81) — share the outer valence configuration ns²np¹. Yet their chemistry is strikingly different, and understanding why is the heart of this chapter.

Boron is a hard, black/brown non-metal with high melting point (2453 K for β-rhombohedral form), extreme hardness, and essentially no simple cation chemistry. It forms an enormous variety of electron-deficient cluster compounds — boranes — that have no parallel elsewhere in the periodic table. Aluminium is the most abundant metal in the Earth's crust and is classically amphoteric. Gallium is famous for its unusually long liquid range (303–2477 K); you can melt it in your hand. Indium emits a high-pitched metallic "cry" when bent — a consequence of crystal twinning. Thallium is highly toxic, and unlike the others it prefers the +1 oxidation state over +3, giving it an alkali-metal-like character in some respects.

The diagonal relationship between Al and Be is important: both form amphoteric oxides, both react with alkali to liberate H₂ (Al) or dissolve (Be), and both have strong tendency to form complex halide anions. This diagonal relationship appears frequently in competitive exams.

The dominant oxidation state for the group is +3. However, the +1 state becomes increasingly stable going down the group, and for Thallium it is the thermodynamically preferred state. This is explained by the thermodynamic 6s inert pair effect, a concept fundamental to understanding the chemistry of heavy p-block metals.

A common misconception is that GaCl₂ represents gallium in the +2 oxidation state. In reality, crystallographic and magnetic data confirm that it is a mixed-valence compound: Ga⁺[GaCl₄]⁻. Similarly, TlI₃ is not Tl³⁺ with I⁻ — it is Tl⁺[I₃]⁻. Be careful with apparent +2 or +3 states for the heavier elements!

2. Occurrence, Extraction and Applications

2.1 Boron

Boron is not found as a free element. Its chief ores are borax (Na₂[B₄O₅(OH)₄]·8H₂O) and kernite (Na₂[B₄O₅(OH)₄]·2H₂O), with massive deposits in the Mojave Desert, California. Boron of commercial purity is made from its oxide:

Na₂[B₄O₅(OH)₄]·8H₂O + H₂SO₄ → 4B(OH)₃ + Na₂SO₄ + 5H₂O (13.1)

2B(OH)₃ → B₂O₃ + 3H₂O (Δ, red heat) (13.2)

Pure boron is obtained by vapour-phase reduction of BBr₃ with H₂, or by pyrolysis of B₂H₆ or BI₃. It has at least four allotropes, the standard state being β-rhombohedral boron.

2.2 Aluminium — The Bayer Process and Hall–Héroult Process

Aluminium is the most abundant metal in the Earth's crust (~8%). Its principal ore is bauxite (hydrated Al₂O₃), which is purified by the Bayer process:

Crude bauxite is dissolved in hot, pressurised aqueous NaOH. Al₂O₃ dissolves (forming [Al(OH)₄]⁻), while Fe₂O₃ is filtered off as "red mud".
The solution is seeded with Al₂O₃·3H₂O crystals (or treated with CO₂) to precipitate crystalline α-Al(OH)₃ (gibbsite).
α-Al(OH)₃ is calcined at ~1300 K to give anhydrous Al₂O₃ (alumina).
Molten alumina is electrolysed in cryolite (Na₃[AlF₆]) at ~1220 K (Hall–Héroult process). Aluminium deposits at the cathode.

In the Hall–Héroult cell, cryolite lowers the melting point of Al₂O₃ from 2345 K to a workable 1220 K. The cryolite is consumed in the process, so synthetic cryolite is manufactured commercially (Eq. 13.46). This energy-intensive step is why aluminium production is linked to hydroelectric power.

2.3 Gallium, Indium and Thallium

Gallium is found in trace amounts in bauxite (associated with Al) and in residues from the zinc-processing industry. The dramatic rise in Ga production since 1980 parallels the growth of the electronics industry — GaAs is the dominant application (LEDs, laser diodes, solar cells, integrated circuits). Gallium nitride (GaN) is the material behind blue and white LEDs and is used in DVD laser diodes.

Indium is recovered as a byproduct of ZnS smelting (sphalerite), since In³⁺ and Zn²⁺ have similar ionic radii and indium substitutes for zinc. The major use of indium worldwide is in indium–tin oxide (ITO) thin films for LCD displays and touchscreens. Thallium is obtained from smelting of Cu, Zn and Pb ores. Its uses are limited due to high toxicity, but include semiconducting materials and Tl-activated sodium iodide in scintillation detectors.

3. Physical Properties and Electronic Structure

3.1 Electronic Configurations

All group 13 elements have the outer electron configuration ns²np¹. However, the relationship to the preceding noble gas is more complex for Ga, In and Tl because d-block elements intervene:

Table 1: Ground State Electronic Configurations of Group 13 Elements
Element	Configuration	After removing 3 valence e⁻
B	[He]2s²2p¹	[He] (noble gas)
Al	[Ne]3s²3p¹	[Ne] (noble gas)
Ga	[Ar]3d¹⁰4s²4p¹	[Ar]3d¹⁰ (not noble gas)
In	[Kr]4d¹⁰5s²5p¹	[Kr]4d¹⁰
Tl	[Xe]4f¹⁴5d¹⁰6s²6p¹	[Xe]4f¹⁴5d¹⁰

For B and Al, IE₄ involves removing an electron from a noble gas configuration — a colossal energy jump (IE₄ = 25,030 kJ mol⁻¹ for B, 11,580 kJ mol⁻¹ for Al). This explains why B and Al have no +4 state. For Ga, In and Tl, the gap between IE₃ and IE₄ is much smaller because the d-electrons screen nuclear charge poorly.

3.2 Ionization Energies and the d-Block Contraction

A close look at IE₁ values in Table 13.1 reveals something strange: IE₁ of Ga (578.8 kJ mol⁻¹) is almost equal to that of Al (577.5 kJ mol⁻¹), and IE₁ of Tl (589.4 kJ mol⁻¹) is actually slightly higher than that of In (558.3 kJ mol⁻¹). This violates the expected decrease down a group.

Why does IE₁ not smoothly decrease down Group 13? The 3d electrons (in Ga) and 4f electrons (in Tl) have low shielding ability (screening power is d < p < s). As nuclear charge increases across the d-block, 4s and 4p electrons are pulled in more effectively than expected. This leads to the d-block contraction at Ga and the f-block (lanthanoid) contraction at Tl. For Tl specifically, relativistic effects additionally stabilize the 6s electrons (relativistic contraction of 6s orbital), contributing to the thermodynamic 6s inert pair effect.

Table 2: Key Physical Data for Group 13 Elements
Property	B	Al	Ga	In	Tl
Atomic number Z	5	13	31	49	81
Melting point (K)	2453	933	303	430	576.5
IE₁ (kJ mol⁻¹)	800.6	577.5	578.8	558.3	589.4
IE₃ (kJ mol⁻¹)	3660	2745	2963	2704	2878
Cov. radius (pm)	88	130	122	150	155
E°(M³⁺/M) (V)	—	−1.66	−0.55	−0.34	+0.72

3.3 The Thermodynamic 6s Inert Pair Effect

For Tl (and analogously for Pb in Group 14, Bi in Group 15), the most stable compounds are those in which the element exhibits the lower oxidation state (+1 for Tl, +2 for Pb, +3 for Bi). This stability of the lower state arises from a combination of:

Relativistic contraction of 6s orbital: As Z increases to 81, electrons in the 1s orbital move with speeds approaching the speed of light (Z/137 ≈ 0.59 for Tl). Relativistic mass increase causes orbital contraction. This extra s-contraction propagates to the 6s orbital, increasing its binding energy.
Poor shielding by intervening d and f electrons: The 4f and 5d electrons have low shielding power, so the effective nuclear charge experienced by 6s electrons in Tl is larger than expected by simple extrapolation.
Thermochemical consequences: The high IE₂ + IE₃ for Tl means the energy cost of forming Tl³⁺ is not recovered by the lattice energy gained, unless the anion is F⁻ (which gives the largest lattice energy due to small size). Hence TlF₃ is the only stable simple ionic trihalide of Tl.

The worked example on thermochemistry of TlF and TlF₃ (Worked Example 13.1 in Housecroft) is a high-probability numerical for IIT-JAM and GATE. The Born-Haber cycle for TlF(s) + F₂(g) → TlF₃(s) gives ΔH° = −297 kJ mol⁻¹, confirming TlF₃ is stable. The key inputs: IE₂ + IE₃ of Tl, ΔₐH°(F,g), ΔEA(F,g), and lattice energies of TlF and TlF₃.

Fig. 1 — Variation in first, second, and third ionization energies across Group 13. Note the discontinuities at Ga and Tl due to poor d/f-electron shielding and relativistic effects.

4. The Elements — Structures and Reactivity

4.1 Allotropes of Boron

Boron is unique in the p-block for its cluster-based solid-state chemistry. All allotropes are built on one fundamental unit: the icosahedral B₁₂ unit, where each boron is connected to five others within the icosahedron. The bonding inside each B₁₂ cage is delocalized — there are simply not enough valence electrons for classical 2-centre bonds to every neighbour. This electron deficiency is the defining characteristic of boron chemistry.

α-Rhombohedral boron: The simplest allotrope to understand conceptually. The structure consists of B₁₂ icosahedra linked by B–B covalent bonds to form an infinite 3D lattice. One way to picture it: the B₁₂ units are the "spheres" in a cubic close-packed (ccp) arrangement.
β-Rhombohedral boron (standard state): More complex, built from B₈₄ units. Each B₈₄ unit contains a central B₁₂ icosahedron, with a B atom bonded to each of the 12, plus an outer B₆₀ cage. The B₆₀ sub-unit has the same structure as fullerene C₆₀ — a beautiful structural analogy! The β-form melts at 2453 K.

Fig. 2 — Schematic representation of the icosahedral B₁₂ unit. Each boron atom is bonded to 5 others within the icosahedron. The bonding is delocalized due to electron deficiency.

4.2 Reactivity of the Group 13 Elements

Boron is chemically inert at room temperature (only F₂ attacks it). At high temperatures, it reacts with most non-metals, metals, and NH₃. The key contrast is with the heavier elements:

2Al + 3H₂SO₄(dilute, aq) → Al₂(SO₄)₃ + 3H₂ (13.3)

2Al + 2MOH + 6H₂O → 2M[Al(OH)₄] + 3H₂↑ (M = Na, K) (13.4)

2Al + Fe₂O₃ → Al₂O₃ + 2Fe (Thermite Reaction, highly exothermic) (13.5)

The thermite reaction (Eq. 13.5) is a favourite for numerical problems. ΔrH° = −847.6 kJ per mol of Fe₂O₃ (calculated from ΔfH° values: Al₂O₃ = −1675.7, Fe₂O₃ = −824.2 kJ mol⁻¹). The heat liberated is sufficient to melt the iron produced (ΔfusH(Fe) = 13.8 kJ mol⁻¹). Applications: welding railway tracks (Goldschmidt process).

A particularly important reactivity trend: only Ga liberates H₂ from aqueous alkali among the heavier Group 13 metals. In and Tl dissolve in most acids to give In(III) and Tl(I) respectively, but do not react with alkali solutions in the same way as Al. Also, when Tl reacts with halogens, the products depend on the halogen:

2Tl + 2Br₂ → Tl[TlBr₄] (mixed-valence: Tl⁺ and Tl³⁺) (13.6)

3Tl + 2I₂ → Tl₃I₄ (13.7)

5. Simple Hydrides

5.1 Diborane (B₂H₆) — The Paradigm of Electron-Deficient Bonding

With three valence electrons, boron might form BH₃, but this monomer dimerizes spontaneously. The stable species is diborane(6), B₂H₆. Its structure cannot be explained by classical 2-centre, 2-electron (2c-2e) bonds; instead it uses 3-centre, 2-electron (3c-2e) bonds for the bridging B–H–B interactions.

Fig. 3 — Structure of diborane (B₂H₆). Yellow atoms are bridging hydrogens involved in 3c-2e bonds (banana bonds). Terminal B–H bonds are shorter (119 pm) than bridging B–H bonds (133 pm).

The key features of diborane structure worth memorising for exams:

Molecular formula: B₂H₆; exists as a colourless gas (bp 180.5 K).
4 terminal H atoms (2 per B), 2 bridging H atoms forming 3c-2e bonds.
B–H_terminal = 119 pm; B–H_bridge = 133 pm (bridging bonds are longer and weaker).
∠H_b–B–H_b ≈ 97°; the bridging plane is perpendicular to the plane of terminal atoms.
ΔfH° = +36 kJ mol⁻¹ (endothermic formation — kinetically stable, thermodynamically borderline).

B₂H₆ + 6H₂O → 2B(OH)₃ + 6H₂ (rapid hydrolysis) (13.11)

B₂H₆ + 3O₂ → B₂O₃ + 3H₂O ΔrH° = −2138 kJ mol⁻¹ (13.12)

5.2 Preparation of B₂H₆

3Na[BH₄] + 4Et₂O·BF₃ → 2B₂H₆ + 3Na[BF₄] + 4Et₂O (laboratory method) (13.8)

2BF₃ + 6NaH → B₂H₆ + 6NaF (450 K, industrial method) (13.10)

5.3 Lewis Acid Behaviour of BH₃ and BX₃

The monomer BH₃ (formed transiently from B₂H₆) is an excellent Lewis acid. The important series for exam problems: Lewis acid strength of boron trihalides towards typical Lewis bases is BF₃ < BCl₃ < BBr₃. This order is counter-intuitive based on electronegativity alone (F is most electronegative → most electron withdrawing → strongest Lewis acid expected for BF₃). The real explanation involves π-bonding:

Why BF₃ is a weaker Lewis acid than BBr₃: In BX₃ (trigonal planar, sp² B), there is partial π-bonding from filled X lone pair p-orbitals into the empty B 2p orbital. This back-donation is greatest for F (2p–2p overlap, smaller radius, better energy match) and least for I. When a Lewis base donates to BX₃ and forces the geometry to change from trigonal planar to tetrahedral, this π-bonding is lost (the 2p orbital on B is now used in the fourth bond). The reorganization energy (cost of breaking π-character) is greatest for BF₃. The net stabilization in adduct formation is least favourable for BF₃ despite its higher Lewis acid character in the absence of back-donation. The adduct stability order: L·BF₃ < L·BCl₃ < L·BBr₃.

Supporting evidence: B–F bond length increases from 130 pm in BF₃ to 145 pm in [BF₄]⁻ — consistent with loss of B–F π-character.

Exception: For very weak Lewis bases like CO, negligible geometric change occurs. In this case, the polarity of BX₃ dominates, and OC·BF₃ > OC·BCl₃ in stability. Be alert to such subtleties in CSIR-NET and IIT-JAM.

5.4 The [MH₄]⁻ Ions — Tetrahydridoborates and Aluminates

The [BH₄]⁻ and [AlH₄]⁻ ions are the most important hydridic reducing agents in chemistry:

Na[BH₄]: White, non-volatile, ionic solid (NaCl-type structure). Stable in dry air and kinetically stable in water. Selective reducing agent — reduces ketones/aldehydes but not esters. Prepared by treating Na with B and H₂ under pressure.
Li[AlH₄]: Powerful reducing agent, prepared by reaction of LiH + AlCl₃ in Et₂O, or directly from Li + Al + H₂ at 250 bar. Violently decomposed by water: Li[AlH₄] + 4H₂O → LiOH + Al(OH)₃ + 4H₂.

For [BH₄]⁻ and [AlH₄]⁻ comparison questions: NaBH₄ is ionic (NaCl structure), kinetically stable in water (selective reducing agent). LiAlH₄ is also ionic but much more reactive with water — reduced rapidly. The order of reducing power: LiAlH₄ > NaBH₄. This appears frequently in NEET and JEE.

6. Halides and Complex Halides

6.1 Boron Trihalides (BX₃)

The boron trihalides BF₃, BCl₃, BBr₃ and BI₃ are all monomeric, trigonal planar molecules (point group D₃h). Physical state at room temperature: BF₃ is a gas (bp 172 K), BCl₃ and BBr₃ are liquids, BI₃ is a white solid (mp 316 K). They are all strong Lewis acids and are hydrolysed by water:

BX₃ + 3H₂O → B(OH)₃ + 3HX (X = Cl, Br, I) (13.36)

B₂O₃ + 3CaF₂ + 3H₂SO₄(conc) → 2BF₃ + 3CaSO₄ + 3H₂O (synthesis of BF₃) (13.28)

BF₃ is commercially available as its diethyl ether adduct, Et₂O·BF₃ (compound 13.11) — a liquid at 298 K, making it a convenient Lewis acid catalyst for Friedel–Crafts reactions, polymerisation, and esterification.

6.2 Tetrafluoroborate [BF₄]⁻ — An "Innocent" Anion

4BF₃ + 6H₂O → 3[H₃O]⁺ + 3[BF₄]⁻ + B(OH)₃ (13.29)

B(OH)₃ + 4HF → [H₃O]⁺ + [BF₄]⁻ + 2H₂O (13.30)

The [BF₄]⁻ ion is tetrahedral (Td symmetry) and coordinates very weakly to metal centres. It is widely used as a "non-coordinating" counter-ion in organometallic and coordination chemistry to precipitate cationic complexes without interfering with their coordination sphere.

IR activity of [BF₄]⁻: Td symmetry. It has only 2 IR-active vibrational modes (both T₂), corresponding to the stretching (ν₃, T₂) and bending (ν₄, T₂) modes. The symmetric stretching (A₁) and the two E modes are IR-inactive. This is a common question in GATE and CSIR-NET spectroscopy sections.

6.3 Al(III) Halides — The Friedel–Crafts Catalysts

AlCl₃, AlBr₃ and AlI₃ are obtained by direct combination of elements. Unlike BX₃, they are dimeric in the vapour phase (Al₂X₆) and often in solution. In the solid state, AlCl₃ has a layer structure with octahedral Al. In the vapour, Al₂Cl₆ contains tetrahedral Al with X→Al coordinate bonds from Cl lone pairs.

Fig. 4 — Structure of Al₂Cl₆ dimer. Orange (bridging) Cl atoms donate lone pairs to Al via coordinate (dative) bonds. Terminal Al–Cl bonds (206 pm) are shorter than bridging bonds (221 pm).

AlCl₃ is the premier catalyst in Friedel–Crafts alkylation and acylation. The reaction involves formation of a carbocation (or acylium ion) and the [AlCl₄]⁻ anion:

RCl + AlCl₃ → R⁺ + [AlCl₄]⁻ (alkylation catalyst step) (13.47)

2[AlCl₄]⁻ ⇌ [Al₂Cl₇]⁻ + Cl⁻ (in molten salt systems) (13.48)

6.4 Thallium Halides — Stability Inversion

Tl(III) halides are unstable compared to Tl(I) halides. TlCl₃ and TlBr₃ spontaneously decompose:

TlBr₃ → TlBr + Br₂ (spontaneous decomposition of Tl³⁺ halides) (13.49)

The compound "TlI₃" is actually thallium(I) triiodide — it is isomorphous with alkali metal triiodides (M⁺[I₃]⁻). This is a critical distinction for exam questions on thallium chemistry.

7. Oxides, Hydroxides and Oxoacids

Acid-Base trend in Group 13 oxides: B₂O₃ is exclusively acidic → Al₂O₃ and Ga₂O₃ are amphoteric → In₂O₃ and Tl₂O₃ are basic. This group trend (increasing basic character going down) is a fundamental pattern across the p-block and is tested repeatedly.

7.1 Boron Oxides and Boric Acid

B₂O₃ is obtained by dehydrating boric acid at red heat (see Eq. 13.2). It is a vitreous solid with a 3D covalent structure of planar BO₃ units (B–O = 138 pm) linked by shared oxygen atoms. Under high pressure (>50 kbar) and at 803 K, a denser polymorph forms with tetrahedral BO₄ units. B₂O₃ is an important component of borosilicate glass (Pyrex), reducing thermal expansion and lowering the working temperature from 1983 K (for pure silica) to ~1093 K.

Boric acid, B(OH)₃: A white, layered solid. The layers are held together by hydrogen bonds (O–H···O, O···O = 270 pm), giving boric acid its slippery, soapy feel (used as a lubricant). In aqueous solution, boric acid behaves as a Lewis acid (not Brønsted — it doesn't donate a proton directly):

B(OH)₃(aq) + 2H₂O(l) ⇌ [B(OH)₄]⁻(aq) + [H₃O]⁺(aq) pKa = 9.1 (13.52)

Boric acid is a Lewis acid — it accepts OH⁻ from water to complete its octet (forming the tetrahedral [B(OH)₄]⁻), thereby generating H⁺. It does NOT donate a proton directly. This mechanistic distinction is frequently tested in NEET and JEE. The acidity increases in the presence of 1,2-diols (due to chelate formation), which is why boric acid is used in buffer solutions with polyhydroxy compounds.

7.2 Borate Anions — The Boron–Oxygen World

Boron–oxygen chemistry is rich in structural diversity. Key structural principle: in planar BO₃ units, B–O ≈ 136 pm (shorter, π-bonding present); in tetrahedral BO₄ units, B–O ≈ 148 pm (longer, π-bonding lost on going to sp³). Important borate minerals and anions:

Table 3: Important Borate Anions and Minerals
Species	Structure	Occurrence
[BO₃]³⁻	Trigonal planar	Orthoborate
[B₂O₅]⁴⁻ (pyroborate)	Two BO₃ sharing one vertex	B–O–B angle varies 131–153°
[B₄O₅(OH)₄]²⁻	2 trigonal + 2 tetrahedral B	Borax, Kernite
[B₃O₃(OH)₃]	6-membered ring (metaboric acid)	B₃O₃(OH)₃

Fig. 5 — Comparison of trigonal planar BO₃ (present in e.g. boric acid, borax) and tetrahedral BO₄ (formed when BO₃ accepts OH⁻ or bridges to another B). The elongation of B–O reflects loss of π-character.

7.3 Aluminium Oxide — Polymorphs and Amphoterism

Al₂O₃ occurs in two main forms:

α-Al₂O₃ (corundum): hcp array of O²⁻ ions with Al³⁺ in two-thirds of octahedral sites. Extremely hard (corundum is harder than all minerals except diamond), density 4.0 g cm⁻³, resistant to acids. Gemstones (ruby = Cr³⁺-doped corundum; sapphire = Fe²⁺/Ti⁴⁺-doped). Made by dehydrating Al(OH)₃ at ~1300 K.
γ-Al₂O₃ (activated alumina): Defect spinel structure, density 3.5 g cm⁻³, large surface area. Used as a catalyst and stationary phase in chromatography (acidic, neutral, and basic forms commercially available). Made by dehydrating γ-AlO(OH) at <720 K.

Both Al(OH)₃ and Al₂O₃ are amphoteric:

Al(OH)₃ + [OH]⁻ → [Al(OH)₄]⁻ (dissolves in base) (13.57)

Al(OH)₃ + 3[H₃O]⁺ → [Al(OH₂)₆]³⁺ (dissolves in acid) (13.58)

γ-Al₂O₃ + 3H₂O + 2[OH]⁻ → 2[Al(OH)₄]⁻ (13.55)

8. Compounds Containing Nitrogen — The B≡N Analogy

The BN unit is isoelectronic with C₂ (C≡C). One B and one N together have the same total valence electrons as two carbons. This leads to profound structural analogies: hexagonal BN (like graphite), cubic BN (like diamond), and borazine (B₃N₃H₆, like benzene). These "inorganic analogues of hydrocarbons" are high-frequency exam topics.

8.1 Boron Nitride (BN)

BN exists in three main polymorphs:

Table 4: Polymorphs of Boron Nitride (BN)
Form	Structure	Carbon analogue	Hardness	Conductivity
α-BN (hexagonal)	Layered hexagonal rings; B over N in adjacent layers (eclipsed)	Graphite (but graphite has staggered layers)	Soft (lubricant)	Insulator (white)
β-BN (cubic, borazon)	Zinc blende structure; B–N = 157 pm	Diamond	~Diamond hardness	Insulator
γ-BN (wurtzite)	Wurtzite structure; formed at ~12 kbar	Lonsdaleite	Very hard	Insulator

Fig. 6 — Layer structure of α-BN (hexagonal boron nitride). B atoms in one layer sit directly above N atoms in the adjacent layer (eclipsed arrangement), unlike graphite where alternate layers are staggered. Interlayer distance = 330 pm (van der Waals). B–N within layers = 144.6 pm.

α-BN vs. graphite comparison (favourite exam question): Both are white/dark layered lubricants. But α-BN is WHITE and an INSULATOR (B–N polarity opens a large band gap), whereas graphite is BLACK and a CONDUCTOR (delocalised π-electrons give metallic character). The eclipsed vs staggered layer arrangement is the key structural distinction.

8.2 Borazine — "Inorganic Benzene"

Borazine, (HBNH)₃, is a colourless liquid (mp 215 K, bp 328 K) with an aromatic odour. Its physical properties closely resemble those of benzene. The B₃N₃ ring is planar with equal B–N bonds (144 pm, same as in α-BN), suggesting some π-electron delocalization from N lone pairs into empty B 2p orbitals.

Synthesis of borazine:

NH₄Cl + Na[BH₄] → H₃N·BH₃ → (HBNH)₃ (−NaCl, −H₂) (13.61)

BCl₃ + 3NH₄Cl → (ClBNH)₃ + 9HCl (420 K, chlorobenzene) (13.62)

Despite isoelectronic structure with benzene, borazine is NOT aromatic in the same sense. The polarity of the B–N bond (χ_B = 2.0, χ_N = 3.0) makes the ring electronically non-uniform: B atoms are electron-poor (susceptible to nucleophilic attack), N atoms are electron-rich (susceptible to electrophilic attack). Therefore, borazine undergoes addition reactions (not substitution), unlike benzene:

(HBNH)₃ + 3HCl → (ClHBNH₂)₃ (addition, not substitution!) (13.63)

(HBNH)₃ + 3H₂O → {H(HO)BNH₂}₃ (addition of water) (13.64)

9. Alums, Aqua Ions and Coordination Chemistry

9.1 Alums

Alums have the general formula M⁺M³⁺(SO₄)₂·12H₂O. The monovalent cation M⁺ is typically K⁺, Rb⁺, Cs⁺, [NH₄]⁺, but Na⁺, Li⁺ and even Tl⁺ alums exist. The trivalent cation M³⁺ can be Al, Ga, In (not Tl — too large), Ti, V, Cr, Mn, Fe or Co (requires similar ionic radius to Al³⁺). Alums form beautiful octahedral crystals; the colour arises from the M³⁺ ion (e.g., KFe(SO₄)₂·12H₂O is purple due to [Fe(OH₂)₆]³⁺).

9.2 The M³⁺ Aqua Ions and Hydrolysis

In acidic solution, Group 13 metal ions form octahedral hexaaqua complexes [M(OH₂)₆]³⁺. These are Brønsted acids:

[M(OH₂)₆]³⁺ ⇌ [M(OH₂)₅(OH)]²⁺ + H⁺ (acidity increases down the group)

As pH increases, a cascade of multinuclear hydrolysis species forms before Al(OH)₃ precipitates. At low pH: [Al₂(OH)₂]⁴⁺; at higher pH: [Al₇(OH)₁₆]⁵⁺; then Al(OH)₃; in alkali: [Al(OH)₄]⁻ (tetrahedral), [Al(OH)₆]³⁻ (octahedral).

9.3 Coordination Complexes of M³⁺ Ions

The Group 13 M³⁺ ions (especially Al³⁺, Ga³⁺, In³⁺) form a wide range of coordination complexes. Octahedral is the most common geometry:

[M(acac)₃] (M = Al, Ga, In): Octahedral, tris-acetylacetonate; used in NMR shift reagents
[M(ox)₃]³⁻ (M = Al, Ga, In): Octahedral, tris-oxalate; chiral
[Al(quinolinate)₃]: Used in gravimetric analysis of Al³⁺; green fluorescence makes it useful in OLEDs

The Ga³⁺ and In³⁺ ions form very stable complexes with macrocyclic ligands containing pendant carboxylate groups (log K ≈ 20). These are used in radiopharmaceuticals: ⁶⁷Ga (γ-emitter, t½ = 3.2 days), ⁶⁸Ga (β⁺, for PET scanning, t½ = 68 min), and ¹¹¹In (γ-emitter, t½ = 2.8 days) — tumour-seeking agents for cancer imaging and therapy.

9.4 Redox Chemistry in Aqueous Solution

The standard reduction potentials (E°(M³⁺/M)) follow the order Al³⁺ ≪ Ga³⁺ < In³⁺ < Tl³⁺:

Al³⁺(aq): E° = −1.66 V (most difficult to reduce; Al is an excellent reducing agent)
Ga³⁺(aq): E° = −0.55 V
In³⁺(aq): E° = −0.34 V; E°(In³⁺/In⁺) = −0.44 V; E°(In⁺/In) = −0.14 V
Tl³⁺(aq): E° = +0.72 V (powerful oxidising agent)

The dramatic increase in E°(M³⁺/M) from Al to Tl (−1.66 to +0.72 V) is NOT simply due to atomic size. A major contributor is the large increase in the sum of the first three ionization energies on going from Al to Ga (due to the d-block contraction and poor 3d shielding). The Gibbs energy cycle shows: the larger IE₁+IE₂+IE₃ for Ga relative to Al means Ga³⁺ is more easily reduced back.

10. Metal Borides

Solid-state metal borides are characteristically extremely hard, involatile, high-melting and chemically inert. They are industrially critical in rocket cones, turbine blades, and refractory applications. Their structural diversity is enormous and reflects the unusual bonding capabilities of boron clusters.

Table 5: Classification of Metal Borides by Boron Arrangement
Boron Organisation	Examples	Key Properties
Isolated B atoms	Ni₃B, Pd₅B₂	Metal-like
B–B pairs	Cr₅B₃	—
Chains of B	CrB, FeB, HfB	Hard
Sheets (2D)	MgB₂, TiB₂, CrB₂	MgB₂: superconductor Tc = 39 K
B₆ octahedra (linked)	CaB₆, LaB₆, CeB₆	LaB₆/CeB₆: electron emitters
B₁₂ icosahedra	ZrB₁₂, UB₁₂	—

MgB₂ (discovered in 2001) is a superconductor with Tc = 39 K — the highest Tc for any binary intermetallic compound. The boron atoms form graphite-like sheets (2D hexagonal layers), and superconductivity arises primarily from phonons coupling to electrons in the boron σ-bands. This is a landmark discovery featured in CSIR-NET and GATE questions.

TiO₂ + B₂O₃ → TiB₂ (Na, high temperature) (13.79)

11. Borane and Carbaborane Clusters — Wade's Rules

This section is perhaps the most conceptually demanding in the chapter, but once understood, it is one of the most reliably answered topics in CSIR-NET, IIT-JAM and GATE. The key insight: boron clusters are electron-deficient — they have too few valence electrons for all-localized 2c-2e bonds, so MO theory (and Wade's empirical rules) must be used.

11.1 Classification of Borane Clusters

Table 6: Classification of Borane Clusters (Wade's Framework)
Type	General Formula	Cage Description	Electron pairs	Example
closo	[BₙHₙ]²⁻	Closed deltahedral cage, n vertices	(n+1)	[B₆H₆]²⁻ (octahedron)
nido	BₙHₙ₊₄	Open cage, (n−1) vertices of parent closo n	(n+1)	B₅H₉ (square pyramid)
arachno	BₙHₙ₊₆	Open cage, (n−2) vertices of parent closo n	(n+1)	B₄H₁₀
hypho	BₙHₙ₊₈	Open cage, (n−3) vertices of parent closo n	(n+1)	Rare

11.2 Applying Wade's Rules — Step-by-Step

Procedure for Wade's Rules electron count:

Assume every B carries a terminal H. Each {BH} unit contributes 2 electrons to cluster bonding (3 valence e⁻ − 1 for terminal B–H bond = 2).
Each additional H (beyond the terminal set) contributes 1 electron.
Each {CH} unit contributes 3 electrons (in carbaboranes).
Add electrons from any overall charge (−1 charge adds 1 e⁻; −2 charge adds 2 e⁻).
Total electrons ÷ 2 = number of electron pairs = (n+1), from which n = number of vertices in parent deltahedron.
Count how many vertices are occupied. Vacant vertices tell you the cluster type: 0 vacant = closo; 1 vacant = nido; 2 vacant = arachno; 3 vacant = hypho.

Worked Example 1: [B₆H₆]²⁻

6 {BH} units → 6 × 2 = 12 electrons. Charge 2⁻ → +2 electrons. Total = 14 electrons = 7 pairs = (n+1) → n = 6. Six-vertex deltahedron = octahedron. All 6 vertices occupied → closo structure. ✓

Worked Example 2: B₅H₉

5 {BH} units → 5 × 2 = 10 electrons. Additional H atoms = 9 − 5 = 4 → +4 electrons. Total = 14 electrons = 7 pairs = (n+1) → n = 6. Parent = octahedron (6 vertices). Only 5 vertices occupied → 1 vacant → nido. Structure = square-based pyramid ✓

Worked Example 3: B₄H₁₀

4 {BH} units → 8 electrons. Additional H = 10 − 4 = 6 → +6 electrons. Total = 14 = 7 pairs → n = 6. Only 4 vertices occupied → 2 vacant → arachno. ✓

Fig. 7 — The closo → nido → arachno relationship in borane clusters. All three have 7 electron pairs for cluster bonding (n+1 = 7), but progressively fewer vertices are occupied. Note the increasing openness from left to right.

11.3 Carbaboranes — When CH Replaces BH

When a CH unit replaces a BH unit in a borane cage, each such replacement adds one extra electron (C has 4 valence electrons vs B with 3). Carbaboranes are rationalized by the same Wade's rules, with {CH} contributing 3 electrons per vertex instead of 2.

Example: C₂B₄H₆ (formed from B₄H₁₀ + ethyne):

4 {BH} units → 4 × 2 = 8 e⁻; 2 {CH} units → 2 × 3 = 6 e⁻; no extra H, no charge.
Total = 14 = 7 pairs → n = 6. Six-vertex deltahedron = octahedron. All 6 vertices occupied → closo.
Two isomers (1,2 and 1,6 positions for the two C atoms).

Wade's Rules shortcut for exams: Count total cluster-bonding electron pairs (SEPs). For closo = n+1; nido = n+2 (where n = actual number of vertices occupied); arachno = n+3. Alternatively, total electrons / 2 = n+1 (closo). Memorize this: The number of SEPs is always ONE MORE than the number of vertices in the parent deltahedron.

12. Comprehensive Exam-Focused Summary

12.1 Most Important Reactions for JEE / NEET

1. Thermite: 2Al + Fe₂O₃ → Al₂O₃ + 2Fe (ΔrH° = −847.6 kJ mol⁻¹)

2. Al in NaOH: 2Al + 2NaOH + 6H₂O → 2Na[Al(OH)₄] + 3H₂↑

3. Boric acid as Lewis acid: B(OH)₃ + 2H₂O ⇌ [B(OH)₄]⁻ + H₃O⁺ (pKa = 9.1)

4. Hydrolysis of B₂H₆: B₂H₆ + 6H₂O → 2B(OH)₃ + 6H₂

5. Al amphoterism: Al(OH)₃ + OH⁻ → [Al(OH)₄]⁻ AND Al(OH)₃ + 3H⁺ → Al³⁺ + 3H₂O

6. Tl(III) instability: TlBr₃ → TlBr + Br₂ (Tl³⁺ → Tl⁺)

7. Cryolite synthesis: Al(OH)₃ + 6HF + 3NaOH → Na₃[AlF₆] + 6H₂O

8. LiAlH₄ decomposition: LiAlH₄ + 4H₂O → LiOH + Al(OH)₃ + 4H₂

12.2 Key Trends to Memorise

Table 7: Critical Trends in Group 13 for Competitive Exams
Property	Trend down the group	Key Exception
Metallic character	Non-metal (B) → Metal (Al, Ga, In, Tl)	Al is amphoteric
Stability of +3 state	Decreases; +1 becomes more stable	Tl prefers +1
Oxide character	Acidic → Amphoteric → Basic	—
Lewis acidity of MX₃	BX₃ > AlX₃ (B compounds) — but within BX₃: BF₃ < BCl₃ < BBr₃	Counter-intuitive π-bond explanation
IE₁	Non-monotonic: B>Al≈Ga>In<Tl	Ga ≈ Al (d-contraction)
Halides	BX₃ monomeric (trigonal); AlX₃ dimeric (via Cl bridging) in vapour	BX₃ has π-bonding
Aqueous basicity	Increases	TlOH is as strong a base as KOH

12.3 Mnemonics and Memory Tricks

Group 13 Elements: B Al Ga In Tl → "BAGInTl" → "Bag it'll" (carry the group in a bag). Or: Brilliant Aluminium Glows In Tlight (slight misspelling for memory).

closo-nido-arachno-hypho: CNAH → "Canada" (minus the ad). Each step = one fewer vertex, two more H, same number of SEPs.

Lewis acidity of BX₃: BF₃ < BCl₃ < BBr₃ → "The lighter the halogen, the more π-bonding back-donation, the weaker the Lewis acid." Counterintuitive: Fluorine's 2p orbitals match boron 2p best → strongest π-donation → most reorganization cost → weakest Lewis acid for adduct formation.

12.4 NMR Spectroscopy of Group 13 — Exam Calculations

All Group 13 elements have NMR-active nuclei. Most important for exams:

¹¹B NMR (I = 3/2, 80.4% abundance): Routinely used to characterize boron compounds. Trigonal planar B appears at higher δ (less shielded, ~10–70 ppm) than tetrahedral B (~0–(−20) ppm). In B₂H₆: one ¹¹B environment, which couples to 2 terminal H (giving a triplet) and 2 bridging H (giving another triplet) → spectrum is a triplet of triplets.
²⁷Al NMR (I = 5/2, 100%): [Al(OH₂)₆]³⁺ appears near δ 0 ppm (reference). Tetrahedral Al appears at higher field. Used to probe Al speciation in solution.
²⁰⁵Tl NMR (I = 1/2, 70.5%): Used as a substitute for Na⁺ and K⁺ in biological systems (since Tl⁺ has similar radius to K⁺), allowing study of K⁺ channels and Na⁺-dependent enzymes.

¹H NMR of B₂H₆: Two types of H — terminal (4H) and bridging (2H). Terminal H couples to ¹¹B (I = 3/2, four spin states ±3/2, ±1/2) → 1:1:1:1 quartet. Bridging H couples to TWO ¹¹B nuclei → combined spin states give 1:2:3:4:3:2:1 septet. Ratio of integration: terminal:bridge = 4:2 = 2:1. This is Worked Example 13.2 in Housecroft — frequently appears in CSIR-NET and IIT-JAM.

13. Spinel Structures — A Bonus Framework Topic

Spinels (general formula AB₂X₄, e.g. MgAl₂O₄, FeCr₂O₄, Fe₃O₄) are a large and important family of mixed metal oxides. They are built on a cubic close-packed array of O²⁻ ions with:

Normal spinel: A²⁺ in 1/8 of the tetrahedral holes; B³⁺ in 1/2 of the octahedral holes. (δ = 0)
Inverse spinel: Half the B³⁺ in tetrahedral holes, other half of B³⁺ and all A²⁺ in octahedral holes. (δ = 0.5)

Examples: MgAl₂O₄ (normal, δ = 0); Fe₃O₄ (inverse, δ = 0.5 — equal Fe²⁺ and Fe³⁺ in octahedral sites, Fe³⁺ in tetrahedral sites; this is why magnetite is a ferrimagnet). The factor δ ranges 0–0.5 depending on the crystal field stabilization energy preference of the metal ions for each type of site.

In CSIR-NET and GATE, spinel questions often ask: "Classify FeCr₂O₄ as normal or inverse." — Normal (Cr³⁺ strongly prefers octahedral by CFSE), so Fe²⁺ in tetrahedral and Cr³⁺ in octahedral → normal spinel. Magnetite (Fe₃O₄) is inverse because Fe²⁺ has no CFSE preference but Fe³⁺ in tetrahedral gives overall stability when Fe²⁺ joins Fe³⁺ in octahedral.

14. Real-World Applications — Exam Context

Table 8: Applications of Group 13 Elements and Compounds
Compound/Element	Application	Why it works
Al metal	Packaging, aerospace, overhead wires	Low density (2.7 g cm⁻³), high strength after alloying, corrosion resistance via oxide layer
GaAs, GaN	LEDs, laser diodes, solar cells, 5G	Direct band gap semiconductors; tunable emission wavelength
ITO (In₂O₃/SnO₂)	LCD touchscreens, solar cells	Transparent and electrically conducting
Al₂O₃ (α, corundum)	Abrasive, gemstones, laser crystals (ruby)	Extreme hardness (Mohs 9); Cr³⁺ doping → red emission
Al(OH)₃	Water purification, mordant for dyes	Flocculation of colloidal particles; adsorption of dye molecules
Na[BH₄]	Selective reducing agent in organic synthesis	Reduces ketones/aldehydes, not esters or carboxylic acids
BN (hexagonal)	High-temperature lubricant, mould release	Low shear strength between layers (like graphite), thermally stable
BN (cubic)	Cutting tool abrasive (second to diamond)	Diamond-like hardness, better thermal stability in air than diamond
B₁₀H₁₄ (decaborane)	Boron neutron capture therapy (BNCT)	¹⁰B captures slow neutrons (high cross-section); delivers localised radiation to tumour
AlCl₃	Friedel–Crafts catalyst	Powerful Lewis acid; generates carbocations/acylium ions
MgB₂	Superconducting wires (Tc = 39 K)	Cheaper than cuprate superconductors; usable with liquid H₂ cooling
Borax	Washing powder, glazes, buffer solutions	Hydrolyses to H₂O₂ (bleach); flux in brazing; buffer with boric acid

Boron in Biology: Boron is an essential plant micronutrient (since 1923). Deficiency causes die-back of terminal buds, hollow hearts in vegetables, and failure of grain to set. Boron's role is in cross-linking rhamnogalacturonan II (RG-II) in plant cell walls via borate-diol diester bridges. This cross-linking is necessary for normal plant growth. Borate fertilisers (borax) are applied to crops, but excess boron is toxic.

15. Final Exam Strategy and High-Value Topics

Topics most likely to appear in JEE Advanced / IIT-JAM / CSIR-NET:

Thermodynamic 6s inert pair effect — Born-Haber cycle numericals for TlF/TlF₃
Lewis acidity order of BX₃ (with mechanistic explanation)
Wade's Rules — electron counting and structure prediction for boranes/carbaboranes
Structure and bonding in B₂H₆ (3c-2e bonds, NMR prediction)
α-BN vs. graphite and cubic BN vs. diamond comparisons
Borazine vs. benzene: similarities AND differences
Spinel vs. inverse spinel classification
Boric acid as Lewis acid (not Brønsted); mechanism
Amphoteric behaviour of Al₂O₃ and Al(OH)₃ with equations
Extraction of Al: Bayer process + Hall–Héroult electrolysis

For NEET specifically: Focus on physical properties (melting points, reactivity), industrial processes (Bayer, thermite), amphoteric nature of Al, applications of boron compounds (borax, boric acid, borosilicate glass), and the basic reactions of Al with acid and base. Avoid getting lost in MO theory and Wade's rules — those are more relevant for GATE/JAM/CSIR.

For GATE / CSIR-NET / IIT-JAM specifically: Wade's rules (including carbaboranes), spinel structures, relativistic effects and inert pair effect numericals, NMR prediction (especially ¹¹B and ¹H of B₂H₆), and structure/bonding in BN allotropes are the highest-value topics. Master Worked Examples 13.1–13.9 from the textbook.

16. Relativistic Effects in Detail — Theory for Advanced Exams

Relativistic effects become significant for elements with atomic number Z > 70. For Thallium (Z = 81), the innermost 1s electrons move at about 58% of the speed of light (v/c = Z/137 ≈ 0.59). According to special relativity, the mass of an electron increases as: m = m₀/(1−v²/c²)^½. This relativistic mass increase causes the Bohr orbit radius r = n²a₀/Z to contract (since r is inversely proportional to m). The result is that 1s orbitals, and by extension all s-orbitals, undergo relativistic contraction — up to ~20% for Z = 80.

The consequences cascade through the electronic structure:

s-orbital contraction: 6s orbitals in Tl (and Au, Hg, Pb, Bi) are stabilized — their energy drops and they become harder to remove. This raises IE(6s) far above what simple extrapolation from In would predict.
p-orbital contraction: Less affected (low electron density near nucleus), but still noticeable for 6p orbitals.
d and f orbital expansion: As s and p orbitals contract and shield the nucleus more effectively, d and f electrons "see" less nuclear charge and undergo relativistic expansion. This is why gold has anomalous optical properties (golden colour rather than silver).
Inert pair effect: Relativistically stabilized 6s² pair resists ionization → lower oxidation state (+1 for Tl, +2 for Pb, +3 for Bi) is preferred over +3, +4, +5 respectively.

Fig. 8 — Schematic showing how relativistic contraction stabilises (lowers the energy of) the 6s orbital in heavy elements like Tl. The large gap between 6s² and 6p¹ means the 6s pair is "inert" — high energy cost to ionize. This is the origin of the 6s inert pair effect.

17. ¹¹B NMR Spectroscopy — A Complete Analytical Tool

Boron-11 (¹¹B) NMR spectroscopy is one of the most powerful diagnostic tools in boron chemistry. Because ¹¹B has spin I = 3/2, it couples to protons in a distinctive way that reveals coordination environment and connectivity:

Table 9: ¹¹B NMR Chemical Shift Ranges for Common Boron Environments
Boron Environment	Coordination	δ(¹¹B) / ppm	Key Example
Tricoordinate (sp²)	3	+10 to +70	BF₃: δ ≈ +10; BCl₃: δ ≈ +47; BR₃
Tetracoordinate (sp³)	4	−5 to −25	[BF₄]⁻: δ ≈ −1; [BH₄]⁻: δ ≈ −41
Terminal BH in boranes	3 (terminal)	−5 to +15	B₂H₆ terminal BH₂
Cluster B (borane)	4–6 (in cage)	−5 to −30	[B₆H₆]²⁻: δ ≈ −16
BN in borazine	3	+28 to +32	(HBNH)₃

Coupling patterns in ¹H NMR coupled to ¹¹B (I = 3/2, 4 spin states):

A proton coupled to one ¹¹B (spin 3/2) gives a 1:1:1:1 quartet (four equally intense lines).
A proton coupled to two ¹¹B nuclei gives a 1:2:3:4:3:2:1 septet (seven lines).
In proton-decoupled (¹¹B{¹H}) NMR, all B–H couplings are removed, giving singlets — useful for counting distinct B environments.

18. Practical Chemistry — Handling Air-Sensitive Group 13 Compounds

Most hydrides and organometallic compounds of the group 13 elements are extraordinarily reactive with air and moisture. Diborane (B₂H₆) burns spontaneously in air; Ga₂H₆ decomposes above 243 K; Li[AlH₄] reacts violently with water. These compounds must be handled using high-vacuum techniques with all-glass apparatus, or in a Schlenk line (vacuum/inert gas manifold) or an inert atmosphere glove box (argon or N₂).

The Schlenk technique involves a double-manifold glass line connected to a vacuum pump on one side and an inert gas cylinder on the other. Reactions are conducted in round-bottomed flasks connected via thick-walled tubing. Solutions are transferred by a cannula (flexible tube with needle) under positive inert gas pressure, or by syringe. This technique is standard in modern inorganic synthesis and is the subject of whole textbooks (e.g., Shriver & Drezdon, "The Manipulation of Air-Sensitive Compounds").

19. Isoelectronic Relationships — A Unifying Principle

The concept of isoelectronic species ties together many apparently unrelated structures across the chapter and is a powerful tool in exams:

Table 10: Key Isoelectronic Pairs and Series in Group 13 Chemistry
Boron species	Isoelectronic carbon species	Structural relationship
[BN]ₙ unit (BN)	[CC]ₙ (carbon)	Hexagonal BN ↔ graphite; cubic BN ↔ diamond
Borazine (B₃N₃H₆)	Benzene (C₆H₆)	Isoelectronic planar 6-membered ring; but BN ≠ CC in reactivity
[BN₂]³⁻	CO₂ (or [OCO])	Both linear, D∞h symmetry
[BN₃]⁶⁻	[CO₃]²⁻	Both trigonal planar; B⁻ isoelectronic with C
BPO₄ lattice	SiO₂ lattice	Alternating B/P replace Si; same total valence electrons
[B(OMe)₄]⁻	Si(OMe)₄	Both tetrahedral; B⁻ isoelectronic with Si
{CH} in carbaborane	{BH}⁻ (isoelectronic)	CH contributes 3 e⁻, BH contributes 2 e⁻ to cluster bonding

Isoelectronic reasoning: B⁻ and C are isoelectronic (both have 6 electrons total, same as carbon's neutral atom). P⁺ and Si are also isoelectronic. Therefore BP, like SiC and C₂, forms a zinc-blende structure. In BPO₄, replacing two Si atoms by one B and one P preserves the total electron count → same crystal structure as SiO₂ (cristobalite). These isoelectronic deductions are standard in IIT-JAM and CSIR-NET.

20. Worked Numerical Problems — Exam Practice

Problem 1 (JEE/IIT-JAM level): Born-Haber cycle for TlF₃

Given: Lattice energies: TlF(s) = −845 kJ mol⁻¹; TlF₃(s) = −5493 kJ mol⁻¹. IE₂(Tl) = +1971, IE₃(Tl) = +2878 kJ mol⁻¹. ΔₐH°(F,g) = +79 kJ mol⁻¹. ΔEA(F,g) = −328 kJ mol⁻¹.

Calculate: ΔH° for the reaction TlF(s) + F₂(g) → TlF₃(s)

TlF(s) + F₂(g) → TlF₃(s) Using Hess's Law thermochemical cycle: ΔlatticeH°(TlF) + ΔH° = IE₂ + IE₃ + 2ΔₐH°(F,g) + 2ΔEA(F,g) + Δlattice H°(TlF₃) ΔH° = IE₂ + IE₃ + 2ΔₐH°(F) + 2ΔEA(F) + Δlattice H°(TlF₃) − Δlattice H°(TlF) = 1971 + 2878 + (2×79) − (2×328) − 5493 + 845 = 1971 + 2878 + 158 − 656 − 5493 + 845 = −297 kJ mol⁻¹ (negative: TlF₃ is thermodynamically stable)

Problem 2 (CSIR-NET level): Wade's Rules for an Unknown Borane

Question: An anionic borane has the formula [B₅H₈]⁻. Apply Wade's rules to predict its structure.

[B₅H₈]⁻: — 5 {BH} units → 5×2 = 10 electrons — additional H = 8 − 5 = 3 H → 3 electrons — charge 1⁻ → 1 electron Total = 10 + 3 + 1 = 14 electrons = 7 pairs n+1 = 7 → n = 6 → parent deltahedron = OCTAHEDRON (6-vertex) Occupied vertices: 5 (one vacant) → NIDO structure Structure: Square-based pyramid; bridging H atoms on the 3 B–B edges of the open square face.

Problem 3 (GATE level): Lewis Acid Strength Interpretation

Question: The enthalpy of formation of adducts L·BX₃ from the reaction: pyridine(soln) + BX₃(g) → py·BX₃(soln) is −75 kJ mol⁻¹ (BF₃), −82 kJ mol⁻¹ (BCl₃), −88 kJ mol⁻¹ (BBr₃). Explain why BF₃ gives the least stable adduct despite F being the most electronegative halogen.

Model Answer: In BX₃ (trigonal planar, sp² B), partial π-bonding exists from filled X lone pair p-orbitals into the empty B 2p orbital. This back-donation is: BF₃ > BCl₃ > BBr₃ (because 2p–2p overlap with F is best). On adduct formation, the geometry changes from trigonal planar to tetrahedral at B, destroying the π-bond. The reorganization energy (endothermic) is greatest for BF₃ (loses most π-bonding) and least for BBr₃. The net exothermic stabilization from L→B σ-bond is offset most in BF₃, giving the least negative ΔH° for adduct formation. Hence, Lewis acid strength order for adduct stability: BF₃ < BCl₃ < BBr₃.

Problem 4 (JEE Advanced level): Identifying GaCl₂ Oxidation State

Question: A sample of "GaCl₂" is reported to be diamagnetic. Explain what this means about the true nature of GaCl₂.

Answer: If gallium were genuinely in the +2 state, it would have the configuration [Ar]3d¹⁰4s¹ — a single unpaired electron — making it paramagnetic. Since "GaCl₂" is diamagnetic, there are no unpaired electrons. This is consistent with the formulation Ga⁺[GaCl₄]⁻, a mixed-valence compound containing Ga⁺ ([Ar]3d¹⁰, no unpaired e⁻) and Ga³⁺ as [GaCl₄]⁻ (empty d subshell, no unpaired e⁻). Both Ga⁺ and Ga³⁺ are diamagnetic → the bulk compound is diamagnetic. This experiment elegantly confirms the non-existence of Ga(II) in this compound.

21. Quick-Revision Flashcard Summary

One-line summaries for last-minute revision:

B is non-metal; Al is amphoteric metal; Ga, In, Tl are metals (Tl behaves like alkali metals in +1 state).
All Group 13 elements have ns²np¹ outer configuration. +3 is characteristic; +1 becomes stable for Tl.
IE₁ trend non-monotonic: Ga ≈ Al (d-contraction); Tl > In (relativistic + f-contraction).
Boric acid is a Lewis acid (accepts OH⁻ from water), NOT a Brønsted acid. pKa = 9.1.
BX₃ Lewis acid order: BF₃ < BCl₃ < BBr₃ (π-bonding loss on adduct formation).
B₂H₆: 4 terminal H + 2 bridging H; 3c-2e bonds; colourless gas, bp 180.5 K.
α-BN: eclipsed layers (unlike graphite's staggered); insulator (white); lubricant.
β-BN (borazon): zinc blende structure; diamond-hard; insulator; abrasive.
Borazine: isoelectronic with benzene but undergoes addition (not substitution) due to B-N polarity.
Wade's rules: SEPs = (n+1) for n-vertex closo, (n+2) for nido, (n+3) for arachno.
MgB₂: superconductor Tc = 39 K (highest for binary compound, discovered 2001).
TlI₃ = Tl⁺[I₃]⁻ (NOT Tl³⁺ with I⁻); GaCl₂ = Ga⁺[GaCl₄]⁻ (NOT Ga²⁺).
Normal spinel: A²⁺ tetrahedral, B³⁺ octahedral (δ=0); Inverse: B³⁺ tetrahedral + (A²⁺ + B³⁺) octahedral (δ=0.5).
Alum: M⁺M³⁺(SO₄)₂·12H₂O; M³⁺ is octahedral [M(H₂O)₆]³⁺ in crystal.
ITO: indium-tin oxide; transparent conductor → LCDs, touchscreens.

Group 13 & 14 Elements Explained: Boron to Thallium Properties and Trends