Question

For the following molecules :
$PCl_{5}, BrF_{3}, ICl^{-}_{2}, XeF^{-}_{5}, NO^{-}_{3}, XeO_{2}F_{2}, PCl^{+}_{4}, CH^{+}_{3}$ Calculate the value of $\frac{a+b}{c}$
a = Number of species having $sp^{3} \,d$-hybridisation
b = Number of species which are planar
c = Number of species which are non-planar

Answer 1

$PCl_{5} \to sp^{3}\,d$, non-planar
$BrF_{3} \to sp^{3}\,d$, bent, $T$-shape, planar
$ICl^{-}_{2} \to sp^{3}\,d$, linear, planar
$XeF^{-}_{5} \to sp^{3}\,d^{3}$, pentagonal planar
$NO^{-}_{3} \to sp^{2}$, planar
$XeO_{2}F_{2} \to sp^{3}\,d$, see-saw, non-planar
$PCl^{+}_{4} \to sp^{3}$, tetrahedral, non-planar
$CH^{+}_{3} \to sp^{2}$, Trigonal planar
$a=4, b=5, c=3$
So, $\frac{a+b}{c}=3$