JEE Mains Chemistry · The p-Block Elements

Boron and Aluminium Compounds

Borax, boric acid, diborane and borazine, and the boron and aluminium halides: electron-deficient compounds that accept electron pairs, with boron limited to four bonds and aluminium able to reach six.

All The p-Block Elements notes JEE Mains Chemistry strategy

Why this matters

Twenty-one PYQs, twenty of them multiple choice, and one from 2026. Nine are about borax and boric acid: the true formula of borax, the borax bead colours and why boric acid is a weak Lewis acid; seven about diborane and borazine, their bonds, shapes and preparation; five about boron and aluminium halides as Lewis acids: back-bonding in BF₃, the covalency of boron and the octahedral aluminium ion in water.

Concept 1 of 3: Borax, the borax bead test and boric acid

Borax is usually written

\mathrm{Na_2B_4O_7\cdot 10H_2O}

, but its true formula is

\mathrm{Na_2[B_4O_5(OH)_4]\cdot 8H_2O}

: two of the four borons are tetrahedral and two are trigonal. On strong heating it melts to a clear glass of sodium metaborate and boric anhydride. The boric anhydride combines with a metal oxide to give a coloured metaborate, which is the borax bead test. Boric acid looks like a triprotic acid but is not a proton donor at all: boron has an empty p orbital, so

\mathrm{B(OH)_3}

takes a hydroxide ion from water, and the water left behind releases the

\mathrm{H^{+}}

Definition

True formula of borax: $\mathrm{Na_2[B_4O_5(OH)_4]\cdot 8H_2O}$ .
Borax in water is alkaline: $\mathrm{Na_2B_4O_7 + 7H_2O \rightarrow 2NaOH + 4H_3BO_3}$ ; a strong base with a weak acid.
Borax bead: $\mathrm{Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 + B_2O_3}$ , then $\mathrm{B_2O_3 + MO \rightarrow M(BO_2)_2}$ .
Copper: in the non-luminous (oxidising) flame, blue-green copper(II) metaborate $\mathrm{Cu(BO_2)_2}$ ; in the luminous (reducing) flame, colourless copper(I) metaborate $\mathrm{CuBO_2}$ or red copper metal.
Cobalt: blue $\mathrm{Co(BO_2)_2}$ .
Boric acid is a weak, monobasic Lewis acid. In the solid, planar $\mathrm{B(OH)_3}$ units are joined into layers by hydrogen bonds, which is why it is a solid while $\mathrm{BF_3}$ is a gas.

Borax bead and boric acid

\mathrm{Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 + B_2O_3} \qquad \mathrm{B(OH)_3 + 2H_2O \rightarrow [B(OH)_4]^{-} + H_3O^{+}}

Worked example

A borax bead is touched to a trace of copper sulphate and heated first in the non-luminous flame and then in the luminous flame. Write the reactions and give the colour and the oxidation state of copper each time.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 13 April 2023 · Q144Moderate

Example 1 · The p-Block Elements · Boron and Aluminium Compounds

If the formula of Borax is

Na_{2}{\text{ }B}_{4}O_{x}(OH)_{y}\cdot zH_{2}O

, then

x + y + z =

___

Cupric metaborate is blue-green, not colourless

In the oxidising flame copper gives blue-green copper(II) metaborate,

\mathrm{Cu(BO_2)_2}

. The colourless one is copper(I) metaborate,

\mathrm{CuBO_2}

, formed in the reducing (luminous) flame.

Boric acid is monobasic, not tribasic

The three OH groups in

\mathrm{B(OH)_3}

do not ionise. Boric acid accepts one hydroxide ion from water, so it releases one

\mathrm{H^{+}}

per molecule and is a weak acid.

Drill 8 more on borax, the borax bead test and boric acid

Concept 2 of 3: Structure and preparation of diborane and borazine

Diborane,

\mathrm{B_2H_6}

, has only twelve valence electrons, too few for eight ordinary bonds. It makes do with four normal B–H bonds at the ends and two hydrogens that each bridge both borons, holding them with one pair of electrons spread over three atoms. These three-centre two-electron bonds are the 'banana bonds'. Borazine,

\mathrm{B_3N_3H_6}

, made by heating diborane with ammonia, has no such bonds: it is a flat ring of alternating B and N with delocalised electrons, called inorganic benzene.

Definition

Lab preparation: $\mathrm{2NaBH_4 + I_2 \rightarrow B_2H_6 + 2NaI + H_2}$ .
Other routes: $\mathrm{4BF_3 + 3LiAlH_4 \rightarrow 2B_2H_6 + 3LiF + 3AlF_3}$ ; industrially $\mathrm{2BF_3 + 6NaH \xrightarrow{450\,K} B_2H_6 + 6NaF}$ .
Lithium aluminium hydride: $\mathrm{4LiH + AlCl_3 \rightarrow LiAlH_4 + 3LiCl}$ (aluminium chloride is the dimer $\mathrm{Al_2Cl_6}$ ).
Diborane is a Lewis acid: with trimethylamine it gives $\mathrm{Me_3N \rightarrow BH_3}$ , where boron is tetrahedral.
Borazine: $\mathrm{3B_2H_6 + 6NH_3 \rightarrow 3[BH_2(NH_3)_2]^{+}[BH_4]^{-} \xrightarrow{\Delta} 2B_3N_3H_6 + 12H_2}$ .

Feature	Diborane, B₂H₆	Borazine, B₃N₃H₆
Shape	Non-planar: the two $\mathrm{BH_2}$ ends lie in one plane, the two bridging H above and below it	Planar six-membered ring of alternating B and N
Bonds	Four terminal 2-centre-2-electron B–H bonds and two bridging 3-centre-2-electron B–H–B bonds	Only ordinary 2-centre-2-electron bonds, with π electrons delocalised round the ring Banana bonds belong to diborane, never to borazine.
Hybridisation of boron	About $sp^3$	$sp^2$
Bond angles and lengths	Terminal H–B–H 122°, bridge H–B–H 97°; terminal B–H 119 pm, bridging B–H 134 pm	All six B–N bonds equal in length
With water	$\mathrm{B_2H_6 + 6H_2O \rightarrow 2B(OH)_3 + 6H_2}$	$\mathrm{B_3N_3H_6 + 9H_2O \rightarrow 3B(OH)_3 + 3NH_3 + 3H_2}$
Acid-base nature	Lewis acid; split by bases such as $\mathrm{NMe_3}$	Polar B–N bonds make it more reactive than benzene

The terminal H–B–H angle is wider than the bridge angle, so the terminal bonds have more s character and less p character.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 13 April 2023 · Q35Moderate

Example 2 · The p-Block Elements · Boron and Aluminium Compounds

The incorrect statement from the following for borazine is:

Diborane has two 3-centre bonds, not four

Of the eight B–H links in

\mathrm{B_2H_6}

, four are ordinary terminal bonds. The four bridging links make just two 3-centre-2-electron bonds, one for each bridging hydrogen.

Diborane is not planar and its boron is not sp²

The two bridging hydrogens sit above and below the plane of the four terminal hydrogens, and each boron is roughly

sp^3

. The flat,

sp^2

molecule is borazine.

BH₃ is a Lewis acid, not a Lewis base

Boron in

\mathrm{BH_3}

has only six electrons and an empty p orbital, so it accepts an electron pair, for example from

\mathrm{NMe_3}

. It has no lone pair to donate.

Drill 6 more on structure and preparation of diborane and borazine

Concept 3 of 3: Boron and aluminium halides as Lewis acids: back-bonding and maximum covalency

A boron trihalide has only six electrons round boron, so it accepts a lone pair: it is a Lewis acid. A filled p orbital on each halogen can also push electron density back into boron's empty 2p orbital. This back-bonding works best when the orbitals are the same size, 2p with 2p, so it is strongest in

\mathrm{BF_3}

and weakens as the halogen grows. Boron has no d orbitals, so it can form at most four bonds. Aluminium can use its 3d orbitals and reach six, as in

\mathrm{[AlF_6]^{3-}}

and

\mathrm{[Al(H_2O)_6]^{3+}}

Definition

Back-bonding $p\pi$ – $p\pi$ is strongest in $\mathrm{BF_3}$ : $\mathrm{BF_3 > BCl_3 > BBr_3 > BI_3}$ .
Lewis acid strength runs the other way, $\mathrm{BF_3 < BCl_3 < BBr_3 < BI_3}$ , because back-bonding fills boron's empty orbital.
Maximum covalency of boron is 4: its valence shell has only 2s and 2p orbitals. So $\mathrm{BF_6^{3-}}$ and $\mathrm{[B(H_2O)_6]^{3+}}$ do not exist.
Group 13 trihalides are covalent and hydrolyse in water. $\mathrm{BCl_3}$ gives boric acid and $\mathrm{[B(OH)_4]^{-}}$ ; $\mathrm{AlCl_3}$ in acidified water gives the octahedral ion $\mathrm{[Al(H_2O)_6]^{3+}}$ , in which aluminium is $sp^3d^2$ .

Species	Covalency of the central atom	Shape	Why
$\mathrm{BF_3}$	3	Trigonal planar	Electron deficient; back-bonding from F partly fills boron's empty p orbital
$\mathrm{[BF_4]^{-}}$	4 (oxidation state still +3)	Tetrahedral	Fluoride donates a pair into boron's empty orbital
$\mathrm{BF_6^{3-}}$	Would need 6	Does not exist	Boron has no d orbitals, so four bonds is its limit The reason NCERT gives is the missing d orbitals.
$\mathrm{[AlF_6]^{3-}}$	6	Octahedral	Aluminium uses its 3d orbitals
$\mathrm{[Al(H_2O)_6]^{3+}}$	6	Octahedral, $sp^3d^2$	Formed when aluminium chloride dissolves in acidified water
$\mathrm{Al_2Cl_6}$	4	Two tetrahedra sharing an edge of two bridging Cl	Each aluminium completes its octet through a chlorine lone pair

Covalency counts the bonds round the atom; it is not the oxidation state. Boron is +3 in both BF₃ and [BF₄]⁻.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 20 · Q34Moderate

Example 3 · The p-Block Elements · Boron and Aluminium Compounds

In which one of the following molecules strongest back donation of an electron pair from halide to boron is expected?

Strongest back-bonding means weakest Lewis acid

Back-bonding is strongest in

\mathrm{BF_3}

because boron's 2p and fluorine's 2p orbitals are the same size. That filling of boron's empty orbital makes

\mathrm{BF_3}

the WEAKEST Lewis acid of the boron trihalides, not the strongest.

Covalency 4 does not mean oxidation state +4

\mathrm{[BF_4]^{-}}

boron forms four bonds, but its oxidation state is

x + 4(-1) = -1

, so

x = +3

Drill 4 more on boron and aluminium halides as lewis acids: back-bonding and maximum covalency

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Borax, the borax bead test and boric acid
Borax bead and boric acid
$\mathrm{Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 + B_2O_3} \qquad \mathrm{B(OH)_3 + 2H_2O \rightarrow [B(OH)_4]^{-} + H_3O^{+}}$

Reference tables (2)

Structure and preparation of diborane and borazine6 rows

Feature	Diborane, B₂H₆	Borazine, B₃N₃H₆
Shape	Non-planar: the two $\mathrm{BH_2}$ ends lie in one plane, the two bridging H above and below it	Planar six-membered ring of alternating B and N
Bonds	Four terminal 2-centre-2-electron B–H bonds and two bridging 3-centre-2-electron B–H–B bonds	Only ordinary 2-centre-2-electron bonds, with π electrons delocalised round the ring Banana bonds belong to diborane, never to borazine.
Hybridisation of boron	About $sp^3$	$sp^2$
Bond angles and lengths	Terminal H–B–H 122°, bridge H–B–H 97°; terminal B–H 119 pm, bridging B–H 134 pm	All six B–N bonds equal in length
With water	$\mathrm{B_2H_6 + 6H_2O \rightarrow 2B(OH)_3 + 6H_2}$	$\mathrm{B_3N_3H_6 + 9H_2O \rightarrow 3B(OH)_3 + 3NH_3 + 3H_2}$
Acid-base nature	Lewis acid; split by bases such as $\mathrm{NMe_3}$	Polar B–N bonds make it more reactive than benzene

The terminal H–B–H angle is wider than the bridge angle, so the terminal bonds have more s character and less p character.

Boron and aluminium halides as Lewis acids: back-bonding and maximum covalency6 rows

Species	Covalency of the central atom	Shape	Why
$\mathrm{BF_3}$	3	Trigonal planar	Electron deficient; back-bonding from F partly fills boron's empty p orbital
$\mathrm{[BF_4]^{-}}$	4 (oxidation state still +3)	Tetrahedral	Fluoride donates a pair into boron's empty orbital
$\mathrm{BF_6^{3-}}$	Would need 6	Does not exist	Boron has no d orbitals, so four bonds is its limit The reason NCERT gives is the missing d orbitals.
$\mathrm{[AlF_6]^{3-}}$	6	Octahedral	Aluminium uses its 3d orbitals
$\mathrm{[Al(H_2O)_6]^{3+}}$	6	Octahedral, $sp^3d^2$	Formed when aluminium chloride dissolves in acidified water
$\mathrm{Al_2Cl_6}$	4	Two tetrahedra sharing an edge of two bridging Cl	Each aluminium completes its octet through a chlorine lone pair

Covalency counts the bonds round the atom; it is not the oxidation state. Boron is +3 in both BF₃ and [BF₄]⁻.

Watch out for (7)

Cupric metaborate is blue-green, not colourless
→ Borax, the borax bead test and boric acid
Boric acid is monobasic, not tribasic
→ Borax, the borax bead test and boric acid
Diborane has two 3-centre bonds, not four
→ Structure and preparation of diborane and borazine
Diborane is not planar and its boron is not sp²
→ Structure and preparation of diborane and borazine
BH₃ is a Lewis acid, not a Lewis base
→ Structure and preparation of diborane and borazine
Strongest back-bonding means weakest Lewis acid
→ Boron and aluminium halides as Lewis acids: back-bonding and maximum covalency
Covalency 4 does not mean oxidation state +4
→ Boron and aluminium halides as Lewis acids: back-bonding and maximum covalency

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