Group 13 (Boron family) — anomaly of B, diagonal B-Si; BX₃ Lewis acidity; diborane B₂H₆ (3c-2e bond), borazine

1. Why is Boron anomalous? (WHAT & WHY)

WHY each property follows:

Property	Reason	Consequence
Non-metallic, hard	Tiny size, high IE → won't lose e⁻ as ions	B is a metalloid; rest are metals
Only covalent compounds	Sum of first 3 IEs is huge	No simple $B^{3+}$ ion exists
Max covalency 4	No d-orbitals → octet can't expand	$BF_4^-$ exists, $BF_6^{3-}$ does not

2. Diagonal relationship: B ↔ Si (WHY similar)

Shared properties (evidence):

Both form giant covalent / acidic oxides ( $B_2O_3$ , $SiO_2$ ) that are glassy/weakly acidic.
Both form volatile, hydrolysable halides ( $BCl_3 + 3H_2O \to H_3BO_3 + 3HCl$ ; $SiCl_4 + 2H_2O \to SiO_2 + 4HCl$ ).
Both form electron-precise/covalent hydrides that are flammable (boranes, silanes).
Both form borates / silicates with chains and rings.

3. Lewis acidity of BX₃ (DERIVE the trend)

Boron in $BX_3$ has only 6 electrons around it (an empty $2p_z$ orbital) → it accepts a lone pair → Lewis acid: $BF_3 + :NH_3 \longrightarrow F_3B\!\leftarrow\!NH_3$

Derivation of the trend (HOW): The empty $2p_z$ on B can be filled by back-bonding ( $p\pi$ – $p\pi$ ) from a halogen lone pair. Stronger back-bonding → B is less electron-deficient → weaker Lewis acid.

F is in period 2, same size as B → $2p$ – $2p$ overlap is excellent → strong back-donation → B least acidic.
Going F→Cl→Br→I, halogen p-orbitals get larger and more diffuse → overlap with B's small $2p$ gets worse → back-bonding weakens → B stays electron-hungry → acidity rises.

4. Diborane $B_2H_6$ — the 3c–2e bond (DERIVE the structure)

Structure (HOW the electrons distribute):

4 terminal B–H bonds → normal 2c–2e → uses $4 \times 2 = 8$ electrons.
2 bridging B–H–B bonds → 3c–2e → uses $2 \times 2 = 4$ electrons.
Total $= 8 + 4 = 12$ electrons ✔ (matches!)

Figure — Group 13 (Boron family) — anomaly of B, diagonal B-Si; BX₃ Lewis acidity; diborane B₂H₆ (3c-2e bond), borazine

Preparation & key reactions: $3LiAlH_4 + 4BF_3 \to 2B_2H_6 + 3LiF + 3AlF_3$ $B_2H_6 + 6H_2O \to 2H_3BO_3 + 6H_2$ $B_2H_6 + 2NMe_3 \to 2Me_3N\!\cdot\!BH_3 \quad (\text{symmetric cleavage})$ $B_2H_6 + 3O_2 \to B_2O_3 + 3H_2O \quad (\Delta H = -2165\,kJ/mol,\ \text{very exothermic})$

5. Borazine $B_3N_3H_6$ — "inorganic benzene"

Flashcards

Why does no free

B^{3+}

ion exist?

Sum of its first three ionisation enthalpies is too large to be compensated by lattice/hydration energy → B forms only covalent bonds.

Maximum covalency of boron and why?

4; because it has no d-orbitals and cannot expand its octet (e.g.

BF_4^-

exists,

BF_6^{3-}

does not).

State the diagonal relationship of boron.

B resembles Si (same group as Al but diagonally placed) due to similar electronegativity and charge/size ratio.

Give one piece of evidence for B–Si diagonal relationship.

Both have acidic covalent oxides (

B_2O_3

SiO_2

) and form volatile, easily hydrolysed halides.

Order of Lewis acidity of boron halides?

BF_3 < BCl_3 < BBr_3 < BI_3

Why is

BF_3

the weakest Lewis acid?

Strong

2p

–

2p

back-bonding from F fills B's empty

2p

orbital, reducing electron deficiency.

How many valence electrons in

B_2H_6

12 (

2\times3 + 6\times1

Why is

B_2H_6

electron-deficient?

8 normal 2c–2e B–H bonds would need 16 electrons but only 12 are available.

Describe bonding in

B_2H_6

4 terminal 2c–2e B–H bonds (8 e⁻) + 2 bridging 3c–2e B–H–B bonds (4 e⁻); no direct B–B bond.

What is a 3c–2e bond?

One electron pair shared among three atoms (e.g. B–H–B bridge).

Bridging vs terminal B–H bond length in diborane?

Bridging ≈ 133 pm (longer/weaker), terminal ≈ 119 pm.

Product of diborane hydrolysis?

H_3BO_3

(boric acid) +

H_2

What is borazine and its formula?

"Inorganic benzene",

B_3N_3H_6

; planar ring of alternating B and N.

Why is borazine more reactive than benzene?

Its

\pi

ring is polar (

B^{\delta+}

N^{\delta-}

), so it readily adds reagents like HCl.

Recall Feynman: explain to a 12-year-old

Boron is a tiny atom with only three "hands" (electrons) but four empty pockets to hold hands. So it's always trying to grab extra hands — that makes it a "Lewis acid" (a hand-grabber). When two boron-hydrogen molecules meet, they don't have enough hands, so two hydrogens stand between two borons and let three atoms share a single handshake — a special "3-friends-one-handshake" bond. With fluorine, the F atom shares its spare hands so nicely that boron stops grabbing — that's why $BF_3$ is the least grabby. And because boron acts so much like silicon (its diagonal neighbour), they make similar glassy oxides and smelly halides.

Connections

p-Block Overview
Lewis Acids and Bases
Back-bonding (pπ–pπ)
Ionisation Enthalpy Trends
Boron Trihalides
Boric Acid H3BO3
Benzene Aromaticity (compare with borazine)
Group 14 — Silicon (diagonal partner)

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Group 13 ka pehla member Boron apne hi group ka rebel hai. Reason simple hai: woh bahut chhota hai aur uske paas sirf 3 valence electrons hain, lekin 4 empty pockets (orbitals) hain. Is "electron-deficiency" ki wajah se Boron hamesha electron pair maangta rehta hai — isliye woh ek Lewis acid hai. Aur kyunki uske paas d-orbital nahi hai, woh octet expand nahi kar sakta, max covalency sirf 4 hai (isliye $BF_4^-$ banta hai par $BF_6^{3-}$ nahi). $B^{3+}$ ion bhi nahi banta — kyunki itne chhote atom se 3 electron nikaalne ki energy bahut zyada hai.

Diagonal relationship ka funda: period mein right jaao to electronegativity badhti hai, group mein neeche jaao to ghatti hai — ye dono effects diagonal pe cancel ho jaate hain, isliye Boron, Silicon jaisa behave karta hai (dono ke acidic oxide, dono ke hydrolysable halides).

BX₃ ki Lewis acidity ka order ulta lagta hai: $BF_3 < BCl_3 < BBr_3 < BI_3$ . Yaad rakho — ye electronegativity se nahi, back-bonding se decide hota hai. Fluorine apna lone pair Boron ke empty $2p$ orbital mein bahut achhe se daal deta hai (size match perfect), to Boron satisfied ho jaata hai, kam acidic. Iodine bada aur diffuse hai, back-bonding weak, Boron bhookha rehta hai — sabse strong acid.

Diborane $B_2H_6$ ka asli maza — sirf 12 electrons hain, normal bonds ke liye 16 chahiye, to 4 short. Solution: do hydrogen bridge banate hain jahan 2 electron teen atoms ko jodte hain — ye hai 3c–2e (banana) bond. 4 terminal normal bonds + 2 bridge bonds. Koi seedha B–B bond nahi hota! Aur borazine ( $B_3N_3H_6$ ) ko "inorganic benzene" bolte hain kyunki B aur N milke average 4 electron dete hain, bilkul carbon jaisa, par ring polar hone ki wajah se benzene se zyada reactive hai.

Group 13 (Boron family) — anomaly of B, diagonal B-Si; BX₃ Lewis acidity; diborane B₂H₆ (3c-2e bond), borazine

1. Why is Boron anomalous? (WHAT & WHY)

2. Diagonal relationship: B ↔ Si (WHY similar)

3. Lewis acidity of BX₃ (DERIVE the trend)

4. Diborane $B_2H_6$ — the 3c–2e bond (DERIVE the structure)

5. Borazine $B_3N_3H_6$ — "inorganic benzene"

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — p-Block

Connections