Question

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

• A. Two $C-H$ bonds are in the same plane of equatorial $S-F$ bonds
• B. Total six atoms are in the same plane
• C. Axial $S-F$ bonds are perpendicular to plane of $=$ bond
• D. Both a and b

Answer 1

The correct option is D Both a and b



Equatorial bonds of $S$ are $s{p}^2$ hybridized. Hence, both the equatorial $F$ atoms and $H$ atoms of $C$ are in the same plane. Hence to total of six atoms in the same plane.
Two axial $F$ atom will be at perpendicular to the equatorial bonds according to trigonal bipyramidal geometry. But due to greater repulsion between double bond -single bond, the bong angle will be greater than ${90}^{\circ }$.
Hence (a) and (b) alone are correct.