Question

In the structure of $H_{2}CS{F}_4$, the following bond angle values are given to decide the plane in which C = S is present:
Axial FSF angle (idealised = 180${}^o$) $\Rightarrow$ 170${}^o$
Equatorial FSF angle (idealised = 120${}^o$) $\Rightarrow$ 97${}^o$
After deciding the plane of double bond, which of the following statements is 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

Answer: (A)

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

The hydrogen atoms are in the same plane with the axial fluorine atoms.
Hydrogen atoms lie in the CSF${}_2$ axial plane. We know that the $\pi$ bond involving a p-orbital on the carbon atom must lie in the equatorial plane of the molecule and the resulting repulsion between the $\pi$ electrons and the electron pair bonding the equatorial fluorine atoms is dramatic. The F${}_{eq}$ - F${}_{eq}$ angle has been reduced to 97.5${}^o$.

Hence, option A is correct.