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Why is equitorial bond angle in sf4> ch2=sf4 (double bond in b/w), trying to understand it through bent rule but unable to so far...

Why is equitorial bond angle in sf4> ch2=sf4 (double bond in b/w), trying to understand it through bent rule but unable to so far...

Grade:12

3 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
6 years ago
SF4 has a trigonal bipyramidal structure with lone pair of electron in the equatorial position.The F-S-F bond angle is 101degree.
In the trigonal-bipyramidal species H2C= SF4 the H atoms are located in the plane of the axial F atoms.There are no lone pair in CH2=Sf4 molecule..The F-S-F bond angle is 97 degree.More s character of lone pair leads to increase concentration of p character on fluorine.Thus the bond angle is more as p character of fluorine atom is more as compared to SF4=CH2.
agam goel
35 Points
6 years ago
Sir, bond angle is directly proportional to s character and therefore inversely proportio al to p character? 
Sunil Kumar FP
askIITians Faculty 183 Points
6 years ago
Yes ,you are correct but actually what i want to tell is that greater s character of lone pair leads to lowering of equatorial bond angle(F-S-F) from 1200 to 101degree.This is due to increase p character on fluorine atom.
In case of CH2=SF4 the p character attained ie even more so the bond angle (F-S-F)decrease more from 120 to 97 degree.
Bent rule:-greater p character leads to less bond angle

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