Explain the following:
$(1)$ Aluminum forms $[AlF_6]^{3-}$ ion but boron does not form $[BF_6]^{3-}$ ion.
$(2)$ $PbX_2$ is more stable than $PbX_4$.
$(1)$ Aluminum forms $[AlF_6]^{3-}$ ion but boron does not form $[BF_6]^{3-}$ ion.
$(2)$ $PbX_2$ is more stable than $PbX_4$.
(N/A) $Al$ has vacant '$d$' orbitals and can expand its coordination number to form $[AlF_6]^{3-}$.
On the other hand,Boron does not have '$d$' orbitals and cannot expand its covalence beyond $4$,thus it forms $[BF_4]^{-}$ instead of $[BF_6]^{3-}$.
Due to the inert pair effect,the $+2$ oxidation state is more stable than the $+4$ oxidation state for lead $(Pb)$.
On the other hand,Boron does not have '$d$' orbitals and cannot expand its covalence beyond $4$,thus it forms $[BF_4]^{-}$ instead of $[BF_6]^{3-}$.
Due to the inert pair effect,the $+2$ oxidation state is more stable than the $+4$ oxidation state for lead $(Pb)$.
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