Find the Shapes of molecule using the VSEPR Theory

BF₃:

                             Central atom is B. There are 3 fluorine atoms.

                    Valence electrons = 0

                            Negative Charge is = 0

                            Add all these, 3+3+0 = 6

                            Number of Bond (N/2) = 6/2 = 3

                            Table 1 tell us bond that it is a Triangular Planner.

BF₄^- :

                             Central atom is B. There are 4 fluorine atoms.

                  Valence electron = 3

                     Negative charge is = -1

                   Add all these, 3+4+1 = 8

                    Number of bond (N/2) = 8/2 = 4

Ammonia NH₃:

                                                 Central atom is N. There are 3 Hydrogen atoms.

                                                                         Valence electron = 5

                                           Add these, 5+3 = 8

                                           Number of bonds (N/2) = 8/2 = 2

                                  . _. .     Shape is Triangular pyramidal.

PCl₅:

                               Central atom is P. There are 5 Cl atoms.

                  Valence electron = 5 

                  Number of bond (N/2) = 10/2 = 5

                  Bonding electron = 5, Non Bonding electron = 0

           . _. .     Shape is Triangular bipyramidal.

CIF₃:

                       Central Atom is Cl.

                                       Valence electron = 7

   There three fluorine atoms, so the number of electron pairs shared is = 3

                       Sum is 10 No. of bond = 10/2 = 5

                         Shape is T-shaped.