Assuming that the following structures exist, draw the structures?

Consider then inorganic complexes, and now you are in the realm of inorganic coordination complexes, so you will have to work out hybridization.

So with a coordination number of 7, will it be d5sp, d3sp3 in a trigonal bipyramid or d4sp2 or d5p2 which will be trigonal prisim with an extra atom in one tetragonal face?

For instance, IF7 could be postulated as having a pentagon 'waist' with one fluorine above and below the plane. IF5, a square planar 'waist' with one fluorine above the plane, and a pair of electrons below the plane.

Consult the work of Gillespie and Nyholm for more details.

Sulfur has six valence electrons, nonetheless every fluorine actually has 7 valence electrons a piece (it's a team 7 atom). This means your whole electron rely is 22.... 2(7)+6+2(for the terrible cost)=?? So your structure could have the sulfur as the relevant atom with noooo pi (double) bonds. Believe about it, Fluorine is probably the most electronegative atom on the periodic desk. This implies it holds its electrons very tightly. The existence of a double bond would supply fluorine a +1 formal charge. That is sooo no longer taking place with the fact that fluorine has an electronegativity of 4 on the pauli scale. So, sulfur crucial atom, two sigma (single) bonds to fluorine, each fluorine has three units of lone pairs, that takes care of sixteen of our electrons. The opposite 6 make three lone pairs on the sulfur (which is able to manage the handed valence octet on the grounds that it's a significant atom).