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Schrodinger Equation

Boron is unab...

Question

Boron is unable to form $B F_{6}^{3 -}$ ion. Explain.

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Solution

Non existence of $B F_{6}^{3 -}$

Boron does not have $d$ -orbitals. Due to non-availability of $d$ orbitals, boron is unable to expand its octet. This causes the maximum covalence of boron to be restricted to $4$ and hence it can not form $B F_{6}^{3 -}$ ion.

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