For the following molecules PCl5, BrF3, ICl2−, XeF5−, NO3−,XeO2F2, PCl4+, CH3+Calculate the value of a+bca = Number of species having Sp3d hybridizationb = Number of species which are planar c = Number of species which are non-planar

Pcl5→sp3d, non planar Icl2−→sp3d, linear planar No3−→sp2, planar Pcl4+→sp3, tetrahedral, non− planar BrF3→sp3d, bent, T−shape, planar XeF5−→sp3d3, pentagonal, planar XeO2F2→sp3d, see−saw, non−planar CH3+→Sp2, Trigonal planar a=4, b=5, c=3 SO a+bc=3

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For the following molecules $P C l_{5}, B r F_{3}, I C {l_{2}}^{-}, X e {F_{5}}^{-}, N {O_{3}}^{-}, X e O_{2} F_{2}, P C {l_{4}}^{+}, C {H_{3}}^{+}$
Calculate the value of $\frac{a + b}{c}$
a = Number of species having $S p^{3} d$ hybridization
b = Number of species which are planar
c = Number of species which are non-planar

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answer is 3.

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Detailed Solution

$P c l_{5} \to s p^{3} d, n o n p l a n a r I c {l_{2}}^{-} \to s p^{3} d, l i n e a r p l a n a r N {o_{3}}^{-} \to s p^{2}, p l a n a r P c {l_{4}}^{+} \to s p^{3}, t e t r a h e d r a l, n o n - p l a n a r B r F_{3} \to s p^{3} d, b e n t, T - s h a p e, p l a n a r X e {F_{5}}^{-} \to s p^{3} d^{3}, p e n t a g o n a l, p l a n a r X e O_{2} F_{2} \to s p^{3} d, s e e - s a w, n o n - p l a n a r C {H_{3}}^{+} \to S p^{2}, T r i g o n a l p l a n a r a = 4, b = 5, c = 3 S O \frac{a + b}{c} = 3$