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'
Questions Asked from Chemical Bonding & Molecular Structure
On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions