Among the following, the correct statement(s) is are
(A) Al(CH3)3 has the three-centre two-electron

Q: Why doesn't carbon in the methyl group form a coordinate bond?

Carbon in a methyl group ($-\text{CH}_3$) has already used all its valence electrons to form bonds (three with hydrogen, one with aluminum). It has no lone pairs left to donate, so it must share its existing electron pair across three atoms.

Q: What exactly is a 3c-2e bond?

A 3-center-2-electron bond is an electron-deficient bond where three atoms share only two electrons. Instead of the usual two atoms sharing two electrons, the electron cloud is spread over three nuclei, holding them together.

Q: Why does AlCl3 form a 3c-4e bond instead of a 3c-2e bond?

Chlorine has three lone pairs of electrons. When it bridges two aluminum atoms, it forms a normal covalent bond with one Al (using one electron) and donates a full lone pair to the empty orbital of the other Al. This means 4 electrons are involved in the Al-Cl-Al bridge.

Q: I thought Lewis acidity decreases down the group because electronegativity decreases. Is that right?

While electronegativity plays a role, the primary factor here is atomic size. As the central atom gets larger (from B to Al), the empty p-orbital becomes more diffuse, and the nucleus is further away. This significantly weakens the attraction for an incoming electron pair, making the larger atom a weaker Lewis acid.

Q: Does back-bonding affect the Lewis acidity of BCl3?

Yes, back-bonding from Cl to B does reduce the Lewis acidity of BCl3 compared to BF3 or BBr3. However, when comparing BCl3 to AlCl3, the dominant factor is the much smaller size and higher charge density of Boron, which makes BCl3 a stronger Lewis acid overall despite the back-bonding.

Question

Among the following, the correct statement(s) is are

Options:

(A) $	ext{Al}(	ext{CH}_3)_3$ has the three-centre two-electron bonds in its dimeric structure

(B) $	ext{AlCl}_3$ has the three-centre two-electron bonds in its dimeric structure

(C) $	ext{BH}_3$ has the three-centre two-electron bonds in its dimeric structure

(D) The Lewis acidity of $	ext{BCl}_3$ is greater than that of $	ext{AlCl}_3$

Sigma Percentile · Accepted Answer

### The Mystery of Electron Deficiency
Group 13 elements like Boron and Aluminum have a classic problem: they are electron-deficient. With only three valence electrons, they form three bonds, ending up with an incomplete octet of six electrons.

To survive, they must find a way to share more electrons. This leads to the fascinating phenomenon of **dimerization**, where two molecules join forces to share their resources.

### The Case of Trimethylaluminium
Let's look at $	ext{Al}(	ext{CH}_3)_3$. When it dimerizes to form $	ext{Al}_2(	ext{CH}_3)_6$, the methyl groups act as bridges between the two aluminum atoms.

But there's a catch! Carbon in the methyl group has no lone pairs to donate. It has already used all its valence electrons.

So, what happens? The single electron pair that normally forms the $	ext{Al}-	ext{C}$ bond is forced to stretch and share itself across three atoms: two aluminums and one carbon. This creates a **3-center-2-electron (3c-2e) bond**. Therefore, statement (A) is perfectly correct.

### The Chlorine Bridge in Aluminum Chloride
Now, consider $	ext{AlCl}_3$ dimerizing to $	ext{Al}_2	ext{Cl}_6$. Here, the bridging atoms are chlorines.

Unlike the methyl group, chlorine is rich in electrons; it has three lone pairs!

When chlorine bridges two aluminum atoms, it forms a standard covalent bond with one aluminum and uses one of its lone pairs to form a **coordinate bond** with the empty orbital of the second aluminum.

Because there are two electrons from the covalent bond and two from the coordinate bond, the $	ext{Al}-	ext{Cl}-	ext{Al}$ bridge involves four electrons. This is a **3-center-4-electron (3c-4e) bond**. Thus, statement (B) is incorrect.

### The Famous Banana Bonds of Diborane
Next up is borane, $	ext{BH}_3$. It dimerizes to form the well-known diborane, $	ext{B}_2	ext{H}_6$.

The bridging hydrogen atoms have only one electron each. Just like the methyl groups, they cannot donate a lone pair.

They must share their single electron pair across two boron atoms and one hydrogen atom. This forms another **3-center-2-electron (3c-2e) bond**, famously known as a **banana bond** due to its curved electron cloud. Statement (C) is correct.

### The Battle of Lewis Acidity
Finally, let's compare the Lewis acidity of $	ext{BCl}_3$ and $	ext{AlCl}_3$. A Lewis acid's strength depends on its ability to attract and accept an incoming electron pair.

As we move down Group 13 from Boron to Aluminum, the atomic size increases significantly.

Because Aluminum is larger, its empty p-orbital is more diffuse, and its nucleus is further away from the surface. This drastically weakens the nucleus's ability to pull in an external electron pair.

Boron, being much smaller, has a higher charge density and attracts electron pairs much more aggressively. Therefore, the Lewis acidity of $	ext{BCl}_3$ is greater than that of $	ext{AlCl}_3$. Statement (D) is correct.

### Final Conclusion
By analyzing the bridging bonds and atomic sizes, we have successfully decoded the behavior of these Group 13 compounds. The correct statements are indeed **(A), (C), and (D)**.

Animated Solution for Chemistry - s and p-Block Elements: Among the following, the correct statement(s) is are

Select Answer:

Visualized Solution

$Analyzing Group 13 Dimers$

$Dimer of Al (CH_{3})_{3}$

$Dimer of AlCl_{3}$

$Dimer of BH_{3}$

$Lewis Acidity of BCl_{3} vs AlCl_{3}$

$Final Conclusion$

The Sigma Insight: Group 13 Elements

The Mystery of Electron Deficiency Group 13 elements like Boron and Aluminum have a classic problem: they are electron-deficient

The Case of Trimethylaluminium Let's look at $Al (CH_{3})_{3}$

The Chlorine Bridge in Aluminum Chloride Now, consider $AlCl_{3}$ dimerizing to $Al_{2} Cl_{6}$

The Famous Banana Bonds of Diborane Next up is borane, $BH_{3}$

The Battle of Lewis Acidity Finally, let's compare the Lewis acidity of $BCl_{3}$ and $AlCl_{3}$

Final Conclusion By analyzing the bridging bonds and atomic sizes, we have successfully decoded the behavior of these Group 13 compounds

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Animated Solution for Chemistry - s and p-Block Elements: Among the following, the correct statement(s) is are

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Group 13 Dimers

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dimer of Al(CH3​)3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dimer of AlCl3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dimer of BH3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lewis Acidity of BCl3​ vs AlCl3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion