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?

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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.

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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.

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