Question

In the structureof $H_{2}CS{F}_4$, to decide the plane in which $C=S$ is present the following bond angle values are given :
Axial $F-S-F$ angle (idealised = 180) $\Rightarrow$ 170
Equatorial $F-S-F$ angle (idealised = 120) $\Rightarrow$ 97
After deciding the plane of double bond, which of the following statement is/are correct ?

• A. Two $C-H$ bonds are in the same plane of axial $S-F$ bonds
• B. Two $C-H$ bonds are in the same plane of equatorial $S-F$ bonds
• C. Total five atoms are in the same plane
• D. Equatorial $S-F$ bonds are perpendicular to plane of $\pi$- bond

Answer 1

The correct option is A Two $C-H$ bonds are in the same plane of axial $S-F$ bonds

Structure of $H_{2}CS{F}_4$ is triagonal bipyramidal. The plane of double bond is above and below the equatorial that cause repulsion and distortion of $F-S-F$ axial bond angle and $F-S-F$ equatorial bond angle. Two $C-H$ bonds are in the same plane of axial $S-F$ bond.