The correct order of bond angles (smallest first) in H2S, NH3, BF3 and SiH4 is
H2S < SiH4 < NH3 < BF3
H2S < NH3 < BF3 < SiH4
H2S < NH3 < SiH4 < BF3
NH3 < H2S < SiH4 < BF3
1. In H2S molecule, central atom S is a 3rd period element and according to Dragos rule, when a 3rd period or higher period element overlap with a small element whose electronegetivity is low then that over lapping cannot be a effective overlapping, because of higher size of 3rd or higher periods element and smaller size of low electronegative element.
Here in H2S, H atom has very low electronegativity and smaller in size so it create more negative charge around sulphur(S) atom and size of S$$-$$2 ion increases, because of this higher size difference efficiency overlapping is not possible. So, any hybridization do not happen in between S and H atom. Lone pair electrons of S atom is present in the pure orbital like s, px, py, pz and it look like this :
As we know the angle between and py is 90o, so bond angle is 90o.
2. In NH3, 3 Bond pair(BP) + 1 lone pair (LP) present so angle between bond 107o.
3. $$\,\,\,\,$$ BF3 has sp2 hybridization. So bond angle is 120o.
4. Here Si is sp3 hybridised and shape is regular tetrahedral and bond angle 109o 28'
MCQ (Single Correct Answer)
The pair of species having identical shapes for molecules of both species is
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