AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

In the structure of $H_{2} {CSF}_{4}$ , to decide the plane in which $C = S$ is present the following bond angle values are given :
Axial FSF angle $(idealised$ $= 180^{\circ}) = 170^{\circ}$
Equatorial FSF angle $(idealised = 120^{\circ}) = 97^{\circ}$
After deciding the plane of double bond, which of the following statement is/are correct?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Equatorial S-F bonds are perpendicular to plane of $π$ -bond

Two $C - H$ bonds are in the same plane of equatorial S-F bonds

Total five atoms are in the same plane

Two $C - H$ bonds are in the same plane of axial S-F bonds

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Double bond lies in the equatorial plane.
Hence, two $C - H$ bonds lie in plane with axial S-F bonds.

Hence, the correct answer is A.

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!