Reactivity towards Halogens
Back-bonding, Aluminum Dimers, and the $TlI_3$ Anomaly.
Group 13 elements react vigorously with halogens ($X_2 = F_2, Cl_2, Br_2, I_2$) to form trihalides ($MX_3$). Because Group 13 elements have only 3 valence electrons, these trihalides are electron-deficient (hypovalent). How different elements solve this electron deficiency creates some of the most critical concepts for competitive exams.
1. Formation of Trihalides and the $TlI_3$ Trap
All Group 13 elements react directly with halogens at elevated temperatures to form trihalides.
(Where M = B, Al, Ga, In, Tl ; X = F, Cl, Br, I)
The Thallium Exception (JEE/NEET Trap):
Because of the Inert Pair Effect, Thallium prefers the $+1$ oxidation state over the $+3$ state. Furthermore, $Tl^{3+}$ is a strong oxidizing agent, and $I^-$ is a strong reducing agent. Therefore, true Thallium(III) iodide ($Tl^{3+}(I^-)_3$) cannot exist.
2. Boron Trihalides ($BX_3$) & $p\pi-p\pi$ Back-Bonding
Boron trihalides are covalent, planar molecules ($sp^2$ hybridized). The central Boron atom only has 6 electrons in its valence shell, making it an electron-deficient Lewis Acid (it wants to accept a lone pair to complete its octet).
The Lewis Acidity Trend:
Based on electronegativity, Fluorine should pull electrons away from Boron the most, making $BF_3$ the strongest Lewis acid. However, the actual experimental order is completely reversed!
In $BF_3$, Fluorine has a lone pair in its $2p$ orbital, and Boron has an empty $2p$ orbital. Because they are the same size, Fluorine donates its lone pair back to Boron's empty orbital, forming a partial double bond ($p\pi-p\pi$ back-bonding).
This internal electron donation partially satisfies Boron's electron deficiency, making it a weak Lewis acid. As halogen size increases ($Cl=3p, Br=4p, I=5p$), the orbital mismatch with Boron's tiny $2p$ orbital makes back-bonding highly ineffective. Therefore, $BI_3$ gets no internal help and remains a very strong Lewis acid.
3. Aluminum's Solution: The $Al_2Cl_6$ Dimer
Aluminum chloride ($AlCl_3$) is also electron-deficient (6 electrons). However, Aluminum is too large to form effective $p\pi-p\pi$ back-bonds with Chlorine. Instead, it solves its octet problem by dimerizing.
Structure of $Al_2Cl_6$:
Two $AlCl_3$ molecules join together. A Chlorine atom from one molecule donates its lone pair into the empty $3p$ orbital of the Aluminum atom in the other molecule, forming coordinate covalent bonds (dative bonds).
- Hybridization Change: In monomeric $AlCl_3$ (high temp gas), Al is $sp^2$ (planar). In the dimer $Al_2Cl_6$, each Al is surrounded by 4 chlorines, changing to $sp^3$ (tetrahedral).
- Bond Angles & Lengths: The bridging $Al-Cl$ coordinate bonds are longer and weaker than the terminal $Al-Cl$ covalent bonds.
Note: Boron trihalides cannot dimerize because the Boron atom is too small to accommodate four large halogen atoms around it (steric hindrance).
4. Hydrolysis of Trihalides
Because they are Lewis acids with empty p or d orbitals, Group 13 trihalides are readily hydrolyzed by water (acting as a Lewis base).
Hydrolyze completely to yield Orthoboric acid and $HCl$.
$BCl_3 + 3H_2O \rightarrow H_3BO_3 + 3HCl$
Aluminum Trihalides ($AlCl_3$):
In aqueous solution, Aluminum utilizes its empty 3d orbitals to expand its coordination number, forming a complex hydrated ion.
$AlCl_3 + 6H_2O \rightarrow [Al(H_2O)_6]^{3+} + 3Cl^-$
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