Question

Predict hybridization and shape of $SF$${}_4$ molecule.

• A. sp${}^3$d and see saw
• B. sp${}^3$ and tetrahedral
• C. sp and T-shape
• D. sp${}^3$d${}^2$ and octahedral

Answer 1

The correct option is A sp${}^3$d and see saw

Sulphur in $S{F}_4$ is in the formal $+4$ oxidation state. Out of sulphur's total of six valence electrons, two form a lone pair. The structure of $S{F}_4$ can, therefore, be anticipated using the principles of VSEPR theory: it is a see-saw shape, with $S$ at the centre. One of the three equatorial positions is occupied by a nonbonding lone pair of electrons.

The d orbital comes into play when the central atom has more than 8 valence electrons. That's called an expanded octet and can happen when the central atom is in the third row of the periodic table or lower (as sulphur is). When you draw the Lewis structure for $S{F}_4$ you see that it has four single sulfur-fluorine bonds, but then you have an additional pair of electrons that must go on the central sulphur atoms as a lone pair. That's five electron groups surrounding the sulphur atom, The fifth hybridised orbitally comes from the sulphur atom's potential (but unoccupied in the unbonded atom) 3d sublevel.