Question

Consider the structure of $A{l}_{2}M{e}_6$ compound and find the value of $\frac{x-y}{z}$, where:
$x=$ Maximum number of atoms that can lie in plane having terminal $(Al-Me)$ bonds.
$y=$ Total number of $3c-2e^{-}$ bonds.
$z=$ Total number of atoms that are $s{p}^3$ hybridized.

Answer 1

Have 2 $3C-2e$ bonds ( y = 2 )

Number of atoms lying on a plane having terminal are = ( x = 10 )

Number of $s{p}^3$ hybridised atoms are = ( z = 8 )

So value of $\frac{x-y}{z}=\frac{10-2}{8}=1$