QJEEM04Ch1fba: JEE Main 2004 Chemistry - Class 11 JEEM04Ch1fba Solution: The correct order of bond angles (smallest first) in H 2 S, NH…

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, p_x, p_y, p_z and it look like this :



As we know the angle between and p_y is 90^o, so bond angle is 90^o.

2. In NH₃, 3 Bond pair(BP) + 1 lone pair (LP) present so angle between bond 107^o.



3. $$$$ BF₃ has sp² hybridization. So bond angle is 120^o.



4.
Here Si is sp³ hybridised and shape is regular tetrahedral and bond angle 109^o 28'