Why is boric acid monobasic and not tribasic?

Boric acid does not release its own hydroxyl protons. The electron-deficient B atom accepts a hydroxide ion from water: B(OH)3 + 2H2O gives [B(OH)4]- + H3O+. Only one H3O+ is produced per H3BO3, so the acid is monobasic and behaves as a Lewis acid.

What is the diagonal relationship between B and Si?

B and Si show diagonal similarity: both form covalent easily hydrolysed halides, both form weakly acidic oxo-acids (H3BO3 and H4SiO4), and both form acidic oxides (B2O3 and SiO2) that react with alkali to give borates and silicates.

Group13 Elements: The Boron Family

Q: What is the inert pair effect and how does it affect Group 13 chemistry?

The inert pair effect is the reluctance of the ns2 electrons of heavy p-block atoms to participate in bonding. In Group 13 this means the +1 state becomes more stable down the group. By thallium, Tl+ is more stable than Tl3+, and Tl2O3 acts as a strong oxidising agent.

Q: What are 3-centre 2-electron bonds in diborane?

Diborane has 12 valence electrons but needs 8 two-electron bonds, so it is electron-deficient. The two bridging B-H-B links are held by a single pair of electrons shared across three atoms. These are called 3-centre 2-electron or banana bonds. Each boron is sp3 hybridised.

Q: Why does borax give a coloured bead with transition metal oxides?

On heating, borax forms transparent 2NaBO2 + B2O3. When a transition metal oxide is added, B2O3 forms a coloured metaborate M(BO2)x. The specific colour is a signature of the metal ion and its oxidation state.

Q: Why is AlCl3 dimeric in the vapour state while BCl3 is monomeric?

In BCl3 the empty 2p orbital of B is stabilised by weak back-donation from Cl and BCl3 remains monomeric. In AlCl3 no such back-bonding is possible, so Al is electron-deficient and two units dimerise into Al2Cl6 with two bridging Cl atoms.

Q: What is borazine and why is it called inorganic benzene?

Borazine B3N3H6 is a planar six-membered ring with alternating B and N atoms each bearing one H. It is isoelectronic and isostructural with benzene, has a delocalised pi system formed by N lone pair donation into empty B p orbitals, and is prepared by heating diborane with ammonia.

Q: Why is Ga liquid at slightly above room temperature?

In solid gallium the atoms form discrete Ga2 dimers linked by weak van der Waals forces rather than a fully-connected metallic lattice. Very little thermal energy is needed to break them up, giving Ga an unusually low melting point of 302.9 K.

ChemistryThe p-Block Elements: Group 13 & 14For JEE aspirants

The Boron family or Group 13 consists of five elements: boron (B), aluminium (Al), gallium (Ga), indium (In) and thallium (Tl). All have the general valence-shell configuration $n s^{2} n p^{1}$ , giving them three valence electrons and a common $+ 3$ oxidation state. Boron is the only non-metal; the rest are metals. Down the group, the $+ 1$ state becomes progressively more stable owing to the inert pair effect, and Ga shows an unusually low melting point (about $302.9$ K) that makes it liquid on a warm palm.

Key Formulas & Facts - Quick Reference

General valence configuration: $n s^{2} n p^{1}$ (n = 2 to 6)
Common oxidation states: $+ 3$ (all), $+ 1$ (Ga, In, Tl - stability increases due to inert pair effect)
Diborane synthesis: $2 B F_{3} + 6 N a H 450 K B_{2} H_{6} + 6 N a F$
Orthoboric acid on heating: $H_{3} B O_{3} 370 K H B O_{2} red heat B_{2} O_{3}$
Lewis acid order of boron trihalides: $B F_{3} < B C l_{3} < B B r_{3} < B I_{3}$
Borax formula: $N a_{2} B_{4} O_{7} \cdot 10 H_{2} O$ or $N a_{2} [B_{4} O_{5} (O H)_{4}] \cdot 8 H_{2} O$
Borax with dil. HCl: $N a_{2} B_{4} O_{7} \cdot 10 H_{2} O + 2 H C l \to 4 H_{3} B O_{3} + 2 N a C l + 5 H_{2} O$
Borazine ("inorganic benzene"): $B_{3} N_{3} H_{6}$

1. Introduction and Electronic Configuration

Group 13 elements sit just after the $s$ -block. Because they carry a $p$ -electron in the outermost shell, they open the $p$ -block. Boron is a hard, high-melting non-metal (metalloid); Al, Ga, In, Tl are soft metals.

Element	Symbol	Z	Electronic Configuration	Character
Boron	B	5	$[H e] 2 s^{2} 2 p^{1}$	Non-metal (metalloid)
Aluminium	Al	13	$[N e] 3 s^{2} 3 p^{1}$	Metal
Gallium	Ga	31	$[A r] 3 d^{10} 4 s^{2} 4 p^{1}$	Metal
Indium	In	49	$[K r] 4 d^{10} 5 s^{2} 5 p^{1}$	Metal
Thallium	Tl	81	$[X e] 4 f^{14} 5 d^{10} 6 s^{2} 6 p^{1}$	Metal

Occurrence

Boron: not found free; principal ores are borax $N a_{2} B_{4} O_{7} \cdot 10 H_{2} O$ , kernite $N a_{2} B_{4} O_{7} \cdot 4 H_{2} O$ and colemanite $C a_{2} B_{6} O_{11} \cdot 5 H_{2} O$ .
Aluminium: most abundant metal in earth's crust ( $\sim 8.3%$ ); chief ore is bauxite $A l_{2} O_{3} \cdot 2 H_{2} O$ .
Ga, In, Tl: found only in traces, mostly as by-products of Zn and Al extraction.

2. Trends in Physical Properties

Figure 1: Trends across Group 13. Atomic radius (left) rises from B to Tl but dips at Ga because the

3 d^{10}

electrons screen poorly. First ionization enthalpy (right) zig-zags:

G a > A l

(

d

-block contraction) and

T l > I n

(poor

4 f^{14}

shielding, lanthanoid contraction).

Atomic and ionic radii

Both radii generally increase from B to Tl as new shells are added. However, Ga is smaller than Al. Between Al and Ga the ten $3 d$ electrons are filled; their poor screening lets the effective nuclear charge on Ga's outer electrons become larger than expected.

Ionization enthalpy

The first ionization enthalpy generally decreases from B to Al but then shows irregularities: $B > A l < G a > I n < T l$ . The higher IE of Ga (over Al) is the $d$ -block contraction; the higher IE of Tl (over In) is caused by poor $4 f^{14}$ shielding (lanthanoid contraction).

Density

Density increases down the group. Aluminium ( $2.70$ g cm $^{- 3}$ ) is unusually light for a metal - one reason it dominates aerospace materials.

Melting and boiling points

Melting points do not follow a simple down-the-group trend. B melts very high ( $2349$ K) due to its covalent icosahedral $B_{12}$ units, and Ga melts anomalously low ( $302.9$ K) because its solid contains discrete $G a_{2}$ dimers held by weak van der Waals forces.

Figure 2: Melting points do not follow a smooth trend. B is highest because of its covalent icosahedral

B_{12}

network; Ga is anomalously low (

302.9

K) because solid gallium packs as weakly bound

G a_{2}

dimers.

Oxidation states and the inert pair effect

All members show the $+ 3$ state. Ga, In and Tl additionally show $+ 1$ . The $+ 1$ state becomes more stable on descending the group, and by Tl it is the more stable state. The inert pair effect explains this: the two $n s^{2}$ electrons of the heavier atoms are held tightly by the nucleus (through poor $d$ and $f$ shielding) and become chemically inert, leaving only the $n p^{1}$ electron for bonding.

Figure 3: The inert pair effect. In Al both

3 s^{2}

and

3 p^{1}

electrons bond, so

A l^{3 +}

dominates. Down the group the

n s^{2}

pair is held so tightly (poor

d

and

f

shielding) that in Tl only the

6 p^{1}

electron is used, making

T l^{+}

the stable ion.

Stability of $+ 1$ vs $+ 3$ : in Tl,

T l^{+}

is more stable than

T l^{3 +}

;

T l^{3 +}

compounds like

T l_{2} O_{3}

are strong oxidising agents that readily revert to

T l^{+}

Solved Example 1

Explain why the

+ 1

oxidation state of Tl is more stable than

+ 3

, while for Al the opposite is true.

Solution:

Down Group 13, the two $n s^{2}$ electrons of the heavier atoms become progressively harder to unpair and use in bonding because they are held tightly by the nucleus (poorly shielded $d$ and $f$ electrons let the effective nuclear charge on $6 s$ be very large). This is the inert pair effect. In Al, no filled $d$ or $f$ shells are involved, so both $3 s$ and $3 p$ electrons participate in bonding and $A l^{3 +}$ dominates. In Tl, the $6 s^{2}$ pair stays inert and only the $6 p^{1}$ electron is lost, giving $T l^{+}$ as the stable ion.

3. Anomalous Behaviour of Boron and Diagonal Relationship with Silicon

Boron differs sharply from the rest of Group 13 due to its very small size, high ionization enthalpy, high electronegativity ( $2.0$ ) and absence of $d$ -orbitals in its valence shell.

Points of anomaly

B is a hard, high-melting non-metal; all others are soft, low-melting metals.
B forms only covalent compounds; the others form ionic compounds in the $+ 3$ state (with covalent exceptions for small ions).
Boron trihalides ( $B F_{3}$ , $B C l_{3}$ ) are strong Lewis acids and stay monomeric; $A l C l_{3}$ dimerises to $A l_{2} C l_{6}$ .
Maximum covalency of B is $4$ (no $d$ -orbitals); Al and beyond can reach $6$ (e.g. $[A l F_{6}]^{3 -}$ ).
Boron hydrides (boranes) are electron-deficient with $3 c$ - $2 e$ bonds; aluminium hydride is a polymeric solid $(A l H_{3})_{n}$ .

Figure 4: In the vapour,

A l C l_{3}

dimerises to

A l_{2} C l_{6}

. Two chlorine atoms bridge the aluminium centres, each donating a lone pair (dashed dative bond) so both Al atoms reach an octet and become

s p^{3}

tetrahedral.

B C l_{3}

cannot do this because boron is too small.

Diagonal relationship: B and Si

Because of similar charge-to-size ratio, boron resembles silicon (Group 14):

Both are non-metals of similar electronegativity and form giant covalent networks or clusters.
Both form volatile, covalent, easily hydrolysed halides: $B C l_{3} + 3 H_{2} O \to H_{3} B O_{3} + 3 H C l$ ; $S i C l_{4} + 4 H_{2} O \to H_{4} S i O_{4} + 4 H C l$ .
Both form acidic oxides: $B_{2} O_{3}$ and $S i O_{2}$ react with alkali to give borates and silicates.
Both form weak oxo-acids: $H_{3} B O_{3}$ (boric acid) and $H_{4} S i O_{4}$ (silicic acid).
Hydrides of both catch fire in air (boranes and silanes are pyrophoric).

Figure 5: Boron sits diagonally above-left of silicon. Their similar charge-to-size ratio makes B mimic Si: both form covalent, easily hydrolysed halides, weakly acidic oxo-acids (

H_{3} B O_{3}

H_{4} S i O_{4}

) and acidic oxides (

B_{2} O_{3}

S i O_{2}

4. Chemical Trends of Group 13

Reactivity with air and water

Amorphous B is fairly reactive; crystalline B is inert. Al forms a protective oxide film. Ga, In and Tl behave as typical metals.

$4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$ (thin protective coat)

$2 A l + 6 H_{2} O \to 2 A l (O H)_{3} + 3 H_{2}$ (only when oxide film is removed)

Reactivity with acids and alkalis

Boron does not react with dilute acids but reacts with hot concentrated $H N O_{3}$ :

$B + 3 H N O_{3} \to H_{3} B O_{3} + 3 N O_{2}$

Boron also reacts with alkalis above $773$ K:

$2 B + 6 N a O H Δ 2 N a_{3} B O_{3} + 3 H_{2}$

Aluminium is amphoteric - reacts with both acids and alkalis (details in the Aluminium concept).

Nature of oxides and hydroxides

Acidic character decreases and basic character increases down the group:

B_{2} O_{3}

(acidic)

>

A l_{2} O_{3}

G a_{2} O_{3}

(amphoteric)

>

I n_{2} O_{3}

T l_{2} O_{3}

(basic)

Figure 6: Acidic character of the

M_{2} O_{3}

oxides falls and basic character rises down Group 13:

B_{2} O_{3}

is acidic,

A l_{2} O_{3}

and

G a_{2} O_{3}

are amphoteric, while

I n_{2} O_{3}

and

T l_{2} O_{3}

are basic.

Halides and Lewis acidity of $B X_{3}$

All Group 13 trihalides are Lewis acids because the central atom has an empty $p$ -orbital. Naively, the more electronegative F should make $B F_{3}$ the strongest acid, but the observed order is:

Lewis acid strength:

B F_{3} < B C l_{3} < B B r_{3} < B I_{3}

This is explained by $p π$ - $p π$ back-bonding: in $B F_{3}$ the small F atom donates lone-pair electron density into the empty $p$ -orbital of B, partially filling it and reducing the electron deficiency. Back-bonding is strongest for F (best size match with B $2 p$ ) and weakest for I (large size mismatch), so $B I_{3}$ is the strongest Lewis acid.

Figure 7: A filled halogen

p

-orbital donates lone-pair density into the empty

2 p

-orbital of boron (

p π

p π

back-bonding), partly quenching its electron deficiency. This overlap is best for small F, so

B F_{3}

is the weakest Lewis acid and the order is

B F_{3} < B C l_{3} < B B r_{3} < B I_{3}

JEE Advanced

Bond order of B-F in $B F_{3}$ : counting only the $σ$ bond, the B-F bond order is $1$ . But because back-donation delocalises electron density from the three F lone pairs into B's empty $p$ -orbital, each B-F bond gains an additional $1/3$ of $π$ -character. The effective bond order therefore becomes:

$B-F bond order = 1 + \frac{1}{3} = 1.33$

This is consistent with the observed B-F bond length in $B F_{3}$ ( $130$ pm) being shorter than a normal single B-F bond in the $B F_{4}^{-}$ anion ( $143$ pm, where no back-bonding is possible because B is already saturated).

5. Important Compounds of Boron

5.1 Diborane, $B_{2} H_{6}$

Diborane is the simplest borane and the parent of a rich family. It is a colourless, pyrophoric gas with an unpleasant odour.

Preparation:

Laboratory: $2 N a B H_{4} + I_{2} \to B_{2} H_{6} + 2 N a I + H_{2}$

Industrial: $2 B F_{3} + 6 N a H 450 K B_{2} H_{6} + 6 N a F$

From boron trifluoride and $L i A l H_{4}$ : $4 B F_{3} + 3 L i A l H_{4} \to 2 B_{2} H_{6} + 3 L i F + 3 A l F_{3}$

Structure: Diborane has $12$ valence electrons but $8$ bonds appear necessary to hold it together, so it is electron-deficient. Its two B-H-B bridges are held by three-centre two-electron (3c-2e) "banana bonds": one pair of electrons is shared across two B atoms and one H atom. Each B is $s p^{3}$ hybridised.

Figure 8: Diborane

B_{2} H_{6}

has four normal terminal

B

H

bonds and two three-centre two-electron

B

H

B

bridges (amber clouds). These "banana" bonds share one electron pair over three atoms, holding the electron-deficient molecule together. Each boron is

s p^{3}

hybridised.

Reactions:

Combustion: $B_{2} H_{6} + 3 O_{2} \to B_{2} O_{3} + 3 H_{2} O$ ( $Δ H = - 1976$ kJ mol $^{- 1}$ , highly exothermic; once tested as a rocket fuel).
Hydrolysis: $B_{2} H_{6} + 6 H_{2} O \to 2 H_{3} B O_{3} + 6 H_{2}$
With Lewis bases (symmetric cleavage): $B_{2} H_{6} + 2 N M e_{3} \to 2 M e_{3} N \cdot B H_{3}$
With $N H_{3}$ (unsymmetric cleavage): $B_{2} H_{6} + 2 N H_{3} \to [B H_{2} (N H_{3})_{2}]^{+} [B H_{4}]^{-}$ , which on heating gives borazine.

JEE Advanced

Borazine, $B_{3} N_{3} H_{6}$ ("inorganic benzene"): when the diborane-ammonia ammoniate is heated to $\sim 450$ K, it loses hydrogen to give a six-membered $B_{3} N_{3}$ ring isoelectronic and isostructural with benzene.

$3 B_{2} H_{6} + 6 N H_{3} Δ 2 B_{3} N_{3} H_{6} + 12 H_{2}$

Figure 9 (JEE Advanced): Borazine is a planar six-membered ring of alternating B and N atoms, each carrying one H. It is isoelectronic and isostructural with benzene, hence "inorganic benzene"; the N lone pairs delocalise into empty B

p

-orbitals to give a

π

system.

5.2 Borax, $N a_{2} B_{4} O_{7} \cdot 10 H_{2} O$

Borax is the most important compound of boron. Its correct structural formula is $N a_{2} [B_{4} O_{5} (O H)_{4}] \cdot 8 H_{2} O$ ; it contains two 3-coordinated (trigonal) and two 4-coordinated (tetrahedral) boron atoms linked through oxygen bridges.

Figure 10: The tetraborate anion

[B_{4} O_{5} (OH)_{4}]^{2 -}

in borax has two tetrahedral (dark) and two trigonal (orange) boron atoms joined by five bridging oxygens, with the central oxygen linking the two tetrahedral borons. Each boron carries one OH group, and each tetrahedral boron holds one negative charge. Two

N a^{+}

and eight waters complete

N a_{2} [B_{4} O_{5} (OH)_{4}] \cdot 8 H_{2} O

Preparation (from boric acid):

$4 H_{3} B O_{3} + N a_{2} C O_{3} \to N a_{2} B_{4} O_{7} + 6 H_{2} O + C O_{2}$

JEE Advanced

Preparation from colemanite ore, $C a_{2} B_{6} O_{11} \cdot 5 H_{2} O$ : the powdered ore is boiled with sodium carbonate solution.

$C a_{2} B_{6} O_{11} + 2 N a_{2} C O_{3} \to N a_{2} B_{4} O_{7} + 2 C a C O_{3} ↓ + 2 N a B O_{2}$

The insoluble $C a C O_{3}$ is filtered off. The filtrate is boiled with excess $C O_{2}$ to convert unwanted sodium metaborate to borax and precipitate sodium bicarbonate:

$4 N a B O_{2} + C O_{2} \to N a_{2} B_{4} O_{7} + N a_{2} C O_{3}$

Reactions:

With water (hydrolysis): $N a_{2} B_{4} O_{7} + 7 H_{2} O \to 2 N a O H + 4 H_{3} B O_{3}$ (alkaline solution).
Action of heat: $N a_{2} B_{4} O_{7} \cdot 10 H_{2} O Δ N a_{2} B_{4} O_{7} Δ 2 N a B O_{2} + B_{2} O_{3}$ (transparent glassy bead).
With dilute HCl: $N a_{2} B_{4} O_{7} \cdot 10 H_{2} O + 2 H C l \to 4 H_{3} B O_{3} + 2 N a C l + 5 H_{2} O$

Borax bead test: the molten

2 N a B O_{2} + B_{2} O_{3}

bead dissolves coloured metal oxides to give characteristic coloured metaborates - blue for Co, green for Cr, brown for Ni, deep blue for Cu - used in qualitative analysis to identify transition metal ions.

Example (borax bead with copper sulphate):

$C u S O_{4} Δ C u O + S O_{3}$ ; then $C u O + B_{2} O_{3} \to C u (B O_{2})_{2}$ (blue bead)

5.3 Orthoboric acid, $H_{3} B O_{3}$

Boric acid is a white, waxy, soapy solid with a layered structure held by hydrogen bonds. It is a weak monobasic Lewis acid but does not donate a proton directly; instead it accepts $O H^{-}$ from water:

$B (O H)_{3} + 2 H_{2} O ⇌ [B (O H)_{4}]^{-} + H_{3} O^{+}$

Figure 11: Orthoboric acid

H_{3} B O_{3}

is a layered solid. Each planar

B (OH)_{3}

molecule is linked to three neighbours by six

O - H \dots O

hydrogen bonds (dashed), because every oxygen donates one and accepts one. These bonds build flat sheets that stack through weak forces, which is why the solid feels soft and soapy.

Preparation:

From borax: $N a_{2} B_{4} O_{7} + 2 H C l + 5 H_{2} O \to 2 N a C l + 4 H_{3} B O_{3}$

From diborane: $B_{2} H_{6} + 6 H_{2} O \to 2 H_{3} B O_{3} + 6 H_{2}$

Action of heat:

$H_{3} B O_{3} 370 K H B O_{2} (metaboric) Δ H_{2} B_{4} O_{7} (tetraboric) red heat B_{2} O_{3}$

Uses: mild antiseptic and eyewash (dilute solution), food preservative, borosilicate glass manufacture, glazes and enamels, buffer in nuclear reactors.

5.4 Boron trifluoride, $B F_{3}$

$B F_{3}$ is a colourless, pungent gas with a trigonal planar structure ( $s p^{2}$ boron, bond angle $120°$ ). It is a strong Lewis acid (though the weakest in the $B X_{3}$ series) and forms coordination compounds with Lewis bases.

Figure 12: (Left)

B F_{3}

is trigonal planar with

s p^{2}

boron and

12 0^{\circ}

F-B-F angles. (Right) Accepting a lone pair from

N H_{3}

(or

F^{-}

), boron rehybridises to

s p^{3}

and the adduct

H_{3} N \to B F_{3}

is tetrahedral; the dashed orange line is the dative (coordinate) bond.

Preparation:

$B_{2} O_{3} + 3 C a F_{2} + 3 H_{2} S O_{4} Δ 2 B F_{3} + 3 C a S O_{4} + 3 H_{2} O$

Reactions with Lewis bases:

$B F_{3} + : N H_{3} \to F_{3} B \leftarrow N H_{3}$

$B F_{3} + F^{-} \to [B F_{4}]^{-}$ (tetrafluoroborate anion)

Uses: catalyst for Friedel-Crafts alkylation and acylation, cationic polymerisation (of isobutylene), esterification, and as an ionisation source in mass spectrometry.

Solved Example 2

Although fluorine is more electronegative than chlorine,

B C l_{3}

is a stronger Lewis acid than

B F_{3}

. Explain.

Solution:

Electronegativity alone would predict $B F_{3}$ to be the stronger acid (F pulls electron density more strongly, leaving B more electron-deficient). But there is a second effect: $p π$ - $p π$ back-donation from filled halogen $p$ -orbitals into the empty $p$ -orbital on B. Because F ( $2 p$ ) is exactly the right size to overlap with B ( $2 p$ ), this back-donation is very effective in $B F_{3}$ and partially fills B's empty orbital, reducing its acceptor ability. In $B C l_{3}$ , the Cl $3 p$ orbital is too large to overlap efficiently with B $2 p$ , so back-donation is weak and B remains strongly electron-deficient. Hence $B C l_{3} > B F_{3}$ in Lewis acidity.

Solved Example 3

Write balanced equations for (i) the borax bead test with

C o O

and (ii) the reaction of borax with dilute HCl. State one commercial use of the second product.

Solution:

(i) $N a_{2} B_{4} O_{7} Δ 2 N a B O_{2} + B_{2} O_{3}$ ; then $C o O + B_{2} O_{3} \to C o (B O_{2})_{2}$ (blue bead).

(ii) $N a_{2} B_{4} O_{7} \cdot 10 H_{2} O + 2 H C l \to 4 H_{3} B O_{3} + 2 N a C l + 5 H_{2} O$ . Orthoboric acid $H_{3} B O_{3}$ is used as a mild antiseptic (eyewash), a food preservative and in the manufacture of borosilicate (Pyrex) glass.

JEE Advanced

Complex hydrides - reducing power of $L i A l H_{4}$ vs $N a B H_{4}$ : both are complex hydrides where a metal cation is paired with a tetrahedral $[M H_{4}]^{-}$ anion. $L i A l H_{4}$ is a much stronger reducing agent than $N a B H_{4}$ because:

Al is less electronegative than B, so the Al-H bond is more polar (H bears more hydridic $H^{-}$ character).
The Al-H bond is weaker (larger orbitals, poorer overlap) and thus more easily broken to deliver hydride.

Consequence: $L i A l H_{4}$ reduces esters, carboxylic acids, amides and even nitro groups. $N a B H_{4}$ is mild - it reduces only aldehydes and ketones, leaving esters, acids and nitro groups untouched. This selectivity is exploited routinely in organic synthesis.

Common Mistakes to Avoid

Watch out

Do not write $B F_{3}$ as the strongest Lewis acid in the $B X_{3}$ series; the electronegativity argument is overridden by $p π$ - $p π$ back-bonding, giving $B I_{3} > B B r_{3} > B C l_{3} > B F_{3}$ .
Diborane bridges are not normal covalent bonds. They are 3-centre 2-electron bonds. Drawing a straight $B - H - B$ line with a lone pair is wrong; use the curved "banana bond" convention.
Boric acid is a Lewis acid, not a Bronsted acid. It does not donate its own $H^{+}$ ; it accepts $O H^{-}$ from water and releases $H^{+}$ from that water.
Borax is not simply $N a_{2} B_{4} O_{7} \cdot 10 H_{2} O$ - the structural formula is $N a_{2} [B_{4} O_{5} (O H)_{4}] \cdot 8 H_{2} O$ , with $2$ trigonal and $2$ tetrahedral B atoms in the anion.
Do not confuse the $+ 1$ oxidation state of Tl (Group 13 inert pair) with that of Pb (Group 14). Both are inert pair effects but they leave different electrons behind.
Ga is smaller than Al because of $d$ -block contraction, not because it lies higher in the group.
Never write B in diborane as $s p^{2}$ - it is $s p^{3}$ . The four hybrid orbitals form two terminal $σ$ bonds and contribute to the two 3c-2e bridges.

Frequently Asked Questions

Q1. Why is boron a non-metal while the rest of Group 13 are metals?

Boron has a very small atomic size and a very high ionization enthalpy, which prevents it from losing electrons to form $B^{3 +}$ . It also has a high electronegativity ( $2.0$ ), close to that of silicon. As a result, it can only form covalent compounds and behaves as a non-metal. The heavier members have larger atoms and lower ionization enthalpies, so they lose electrons more easily and show metallic character.

Q2. What is the inert pair effect and how does it affect Group 13 chemistry?

The inert pair effect is the reluctance of the $n s^{2}$ electrons of heavy $p$ -block atoms to participate in bonding, because they are pulled tightly by the nucleus through poor $d$ and $f$ shielding. In Group 13 this means the $+ 1$ state (loss of just the $n p^{1}$ electron) becomes more stable down the group. By thallium, $T l^{+}$ is more stable than $T l^{3 +}$ , and $T l_{2} O_{3}$ acts as a strong oxidising agent.

Q3. What are 3-centre 2-electron bonds in diborane?

Diborane has $12$ valence electrons but appears to need $8$ two-electron bonds, so it is electron-deficient. The two bridging $B - H - B$ links are held by a single pair of electrons shared across three atoms (two B and one H). These are called 3-centre 2-electron (3c-2e) or "banana" bonds and are the hallmark of electron-deficient boranes. Each boron is $s p^{3}$ hybridised.

Q4. Why is boric acid $H_{3} B O_{3}$ a monobasic and not a tribasic acid?

Boric acid does not release its own three hydroxyl protons. Instead, the electron-deficient B atom accepts a hydroxide ion from a water molecule: $B (O H)_{3} + 2 H_{2} O ⇌ [B (O H)_{4}]^{-} + H_{3} O^{+}$ . Only one $H_{3} O^{+}$ is generated per $H_{3} B O_{3}$ , so the acid is effectively monobasic and behaves as a Lewis acid.

Q5. Why does borax give a coloured bead with transition metal oxides?

On heating, borax forms transparent glassy $2 N a B O_{2} + B_{2} O_{3}$ . When a transition metal oxide is added, $B_{2} O_{3}$ combines with it to form a coloured metaborate $M (B O_{2})_{x}$ . The specific colour (blue for Co, green for Cr, brown for Ni, blue for Cu) is a signature of the metal ion and its oxidation state, so the borax bead test is used in qualitative inorganic analysis.

Q6. Why is $A l C l_{3}$ dimeric in the vapour state while $B C l_{3}$ is monomeric?

In $B C l_{3}$ , the empty $2 p$ orbital of B is stabilised by $p π$ - $p π$ back-donation from Cl (even a weak overlap helps), so B keeps a trigonal planar geometry and stays monomeric. In $A l C l_{3}$ , the vacant orbitals on Al are $3 p$ , which cannot back-bond effectively with Cl $3 p$ ; Al is still electron-deficient, so two $A l C l_{3}$ units dimerise to $A l_{2} C l_{6}$ , in which two Cl atoms bridge and donate lone pairs, completing octets on both Al atoms.

Q7. What is borazine and why is it called "inorganic benzene"?

Borazine $B_{3} N_{3} H_{6}$ is a planar six-membered ring with alternating B and N atoms, each bearing one H. It is isoelectronic (same total electron count) and isostructural (same shape) with benzene $C_{6} H_{6}$ : same $D_{3 h}$ symmetry, similar bond lengths and a delocalised $π$ system (N lone pairs donate into empty B $p$ -orbitals). This close mimicry earned it the popular name "inorganic benzene". It is prepared by heating diborane with ammonia: $3 B_{2} H_{6} + 6 N H_{3} \to 2 B_{3} N_{3} H_{6} + 12 H_{2}$ .

Q8. Why is Ga liquid at slightly above room temperature?

In the solid state, gallium exists as discrete $G a_{2}$ dimers rather than a fully-connected metallic lattice. The forces between these dimers are weak van der Waals type, so very little thermal energy is needed to break them up. This gives Ga an unusually low melting point of about $302.9$ K ( $29.8°$ C), while it still boils at $2673$ K - the widest liquid range of any element.

Q9. State the diagonal relationship between B and Si with two examples.

Because B and Si have similar charge-to-size ratio and comparable electronegativity, they show diagonal similarity: (i) both form covalent, easily hydrolysed halides - $B C l_{3} + 3 H_{2} O \to H_{3} B O_{3} + 3 H C l$ and $S i C l_{4} + 4 H_{2} O \to H_{4} S i O_{4} + 4 H C l$ ; (ii) both form weakly acidic oxo-acids ( $H_{3} B O_{3}$ and $H_{4} S i O_{4}$ ) and acidic oxides ( $B_{2} O_{3}$ , $S i O_{2}$ ) that react with alkali to give borates and silicates respectively.

Previous year questions on Group13 Elements: The Boron Family

7 questions from past papers, each with a step-by-step solution.

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1. Introduction and Electronic Configuration

Occurrence

2. Trends in Physical Properties

Atomic and ionic radii

Ionization enthalpy

Density

Melting and boiling points

Oxidation states and the inert pair effect

3. Anomalous Behaviour of Boron and Diagonal Relationship with Silicon

Points of anomaly

Diagonal relationship: B and Si

4. Chemical Trends of Group 13

Reactivity with air and water

Reactivity with acids and alkalis

Nature of oxides and hydroxides

Halides and Lewis acidity of BX3​

5. Important Compounds of Boron

5.1 Diborane, B2​H6​

5.2 Borax, Na2​B4​O7​⋅10H2​O

5.3 Orthoboric acid, H3​BO3​

5.4 Boron trifluoride, BF3​

Common Mistakes to Avoid

Frequently Asked Questions

Q1. Why is boron a non-metal while the rest of Group 13 are metals?

Q2. What is the inert pair effect and how does it affect Group 13 chemistry?

Q3. What are 3-centre 2-electron bonds in diborane?

Q4. Why is boric acid H3​BO3​ a monobasic and not a tribasic acid?

Q5. Why does borax give a coloured bead with transition metal oxides?

Q6. Why is AlCl3​ dimeric in the vapour state while BCl3​ is monomeric?

Q7. What is borazine and why is it called "inorganic benzene"?

Q8. Why is Ga liquid at slightly above room temperature?

Q9. State the diagonal relationship between B and Si with two examples.

Previous year questions on Group13 Elements: The Boron Family

Ready to master The p-Block Elements: Group 13 & 14?

Halides and Lewis acidity of $B X_{3}$

5.1 Diborane, $B_{2} H_{6}$

5.2 Borax, $N a_{2} B_{4} O_{7} \cdot 10 H_{2} O$

5.3 Orthoboric acid, $H_{3} B O_{3}$

5.4 Boron trifluoride, $B F_{3}$

Q4. Why is boric acid $H_{3} B O_{3}$ a monobasic and not a tribasic acid?

Q6. Why is $A l C l_{3}$ dimeric in the vapour state while $B C l_{3}$ is monomeric?