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In the structureof \( \boldsymbol{H}_{2} \boldsymbol{C} \boldsymbol{S} \boldsymbol{F}_{4}, \) to decide the plane in which \( C=S \) is present the following bond angle values are given: Axial \( \boldsymbol{F}-\boldsymbol{S}-\boldsymbol{F} \text { angle (idealised }=180) \) \( \Rightarrow 170 \) Equatorial \( \boldsymbol{F}-\boldsymbol{S}-\boldsymbol{F} \) angle (idealised \( = \) 120)\( \Rightarrow 97 \) After deciding the plane of double bond, which of the following statement is/are correct?

1. Two \( C-H \) bonds are in the same plane of axial \( S-F \) bonds
2. Two \( C-H \) bonds are in the same plane of equatorial \( S-F \) bonds
3. Total five atoms are in the same plane
4. Equatorial \( S-F \) bonds are perpendicular to plane of \( \pi \) - bond

Correct answer: Two \( C-H \) bonds are in the same plane of equatorial \( S-F \) bonds

Solution

The very small equatorial F–S–F angle shows strong repulsion in the equatorial plane, consistent with the C=S double bond occupying an equatorial position. Therefore the two C–H bonds lie in the same plane as the equatorial S–F bonds.

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