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Aluminium forms AlF(6)^(3-) but boron ...

Aluminium forms `AlF_(6)^(3-)` but boron does not form `BF_(6)^(3-)`. Why so ?

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To understand why aluminum can form \( \text{AlF}_6^{3-} \) while boron cannot form \( \text{BF}_6^{3-} \), we can break down the explanation into several steps: ### Step 1: Understanding the Electronic Configuration - Aluminum (Al) has the electronic configuration of \( [Ne] 3s^2 3p^1 \). - Boron (B) has the electronic configuration of \( [He] 2s^2 2p^1 \). ### Step 2: Valence Shell and Covalency - The valence shell of aluminum is the third shell, which includes the 3s and 3p orbitals, and it also has access to the 3d orbitals. ...
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Explain the following (a) Gallium has higher ionisation enthalpy than aluminium (b) Boron does not exist as B^(3+) ion (c) Aluminium forms [AlF_(6)]^(3-) ion but boron does not form [BF_(6)]^(3-) ion.

Explain the following : (a) Boron has high melting and boiling points. (b) The p pi - p pi back bonding occurs in the halides of boron and not in those of aluminimum. ( c) Boron and aluminium halides behave as Lewis acids. (d) Aluminium forms [AlF_(6)]^(3-) ion, but boron does not form [BF_(6)]^(3-) ion.

Aluminium forms [AIF_(6)]^(3-) ion but boron does not form [BF_(6)]^(3-) ion. Explain.

Why boron does not form B^(3+) ion?

Assertion : All forms [Alf_(6)]^(3-) but B does not form [BF_(6)]^(3-) . Reason : B does not react with fluorine.