The number of molecules/ion/s having trigonal bipyramidal shape is: $$PF_5$$, $$BrF_5$$, $$PCl_5$$, $$[PtCl_4]^{2-}$$, $$BF_3$$, $$Fe(CO)_5$$

Find how many of the given species have trigonal bipyramidal (TBP) shape: $$PF_5, BrF_5, PCl_5, [PtCl_4]^{2-}, BF_3, Fe(CO)_5$$.

Method: For each species, count electron domains around the central atom and determine the geometry using VSEPR theory.

1. $$PF_5$$: P has 5 valence electrons, each F contributes 1 bond = 5 bond pairs, 0 lone pairs. Electron geometry: TBP. Molecular shape: Trigonal bipyramidal.

2. $$BrF_5$$: Br has 7 valence electrons, 5 bonds to F uses 5 electrons, leaving 2 electrons = 1 lone pair. Total electron domains = 6 (octahedral electron geometry). Shape: Square pyramidal (not TBP).

3. $$PCl_5$$: Same as $$PF_5$$. P has 5 bond pairs, 0 lone pairs. Shape: Trigonal bipyramidal.

4. $$[PtCl_4]^{2-}$$: Pt is a transition metal with d$$^8$$ configuration ($$Pt^{2+}$$). With 4 $$Cl^-$$ ligands, it adopts square planar geometry (not TBP).

5. $$BF_3$$: B has 3 valence electrons, 3 bonds = 3 bond pairs, 0 lone pairs. Shape: Trigonal planar (not TBP).

6. $$Fe(CO)_5$$: Fe has 8 valence electrons. In $$Fe(CO)_5$$, iron is $$Fe(0)$$ with 5 CO ligands forming 5 bond pairs. This follows the 18-electron rule (8 + 5×2 = 18). The geometry is Trigonal bipyramidal.

Count of TBP species: $$PF_5$$, $$PCl_5$$, $$Fe(CO)_5$$ = 3.

The answer is 3.