Chemistry - CHEMICAL BONDING Question with Solution | TestHub
For the following molecules :
PCl₃, BrF₃, ICl₂⁻, XeF₅⁻, NO₃⁻, XeO₂F₂, PCl₄⁺, CH₃⁺, NO₂⁻
Calculate the value of
= Number of species having sp³d-hybridization
= Number of species which are planar
= Number of species which are non-planar
Answer:
Solution:
To determine a, b, and c, we analyze each species:
PCl₃: sp³ hybridization, pyramidal (non-planar) BrF₃: sp³d hybridization, T-shaped (planar) ICl₂⁻: sp³d hybridization, linear (planar) XeF₅⁻: sp³d³ hybridization, pentagonal planar (planar) NO₃⁻: sp² hybridization, trigonal planar (planar) XeO₂F₂: sp³d hybridization, seesaw (non-planar) PCl₄⁺: sp³ hybridization, tetrahedral (non-planar) CH₃⁺: sp² hybridization, trigonal planar (planar) NO₂⁻: sp² hybridization, bent (planar)
a = 3 (BrF₃, ICl₂⁻, XeO₂F₂) b = 6 (BrF₃, ICl₂⁻, XeF₅⁻, NO₃⁻, CH₃⁺, NO₂⁻) c = 3 (PCl₃, XeO₂F₂, PCl₄⁺)
Value =
