Chemistry · Covalent Bonding: Concept of electronegativity, Fajan's rule, dipole moment, Valence Shell Electron Pair Repulsion (VSEPR) theory and shapes of simple molecules
In the structureof to decide the plane in which is present the following bond an
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?
- 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
Step-by-step solution
The molecule H2CSF4 has sulfur as the central atom with four fluorine atoms and one double-bonded carbon (C=S). Using VSEPR theory, the geometry around sulfur is trigonal bipyramidal with the double bond occupying an equatorial position due to its higher repulsive effect, as evidenced by the reduced equatorial F-S-F angle (97°). The carbon in C=S is sp2 hybridized, so the two C-H bonds lie in the same sigma plane as the C=S bond. This sigma plane corresponds to the equatorial plane of sulfur, which also contains the two equatorial S-F bonds. Therefore, the two C-H bonds are coplanar with the equatorial S-F bonds.