Chemistry - CHEMICAL BONDING Question with Solution | TestHub
the number of planar molecule(s) in which orbital of the central atom participate in hybridisation.
Y = find the number of 90° angles in anionic part of solid
Find the value of
Answer:
Solution:
To find X, we analyze the given molecules for planarity and dz² orbital participation in hybridization. XeO₃: sp³ hybridization, trigonal pyramidal, non-planar. SF₄: sp³d hybridization, seesaw, non-planar. BrF₄⁻: sp³d² hybridization, square planar, planar. dz² participates. XeF₄: sp³d² hybridization, square planar, planar. dz² participates. IF₅: sp³d² hybridization, square pyramidal, non-planar. PCl₅: sp³d hybridization, trigonal bipyramidal, non-planar. So, X = 2 (BrF₄⁻, XeF₄).
Solid PCl₅ exists as [PCl₄]⁺[PCl₆]⁻. The anionic part is [PCl₆]⁻, which has sp³d² hybridization and an octahedral geometry. An octahedron has 12 angles of 90°. So, Y = 12.
Finally, calculate .
