The number of molecules/ions having trigonal bipyramidal shape is:
\(\text{PF}_5, \, \text{BrF}_5, \, \text{PCl}_5, \, [\text{PtCl}_4]^{2-}, \, \text{BF}_3, \, \text{Fe(CO)}_5\)
PF5, PCl5, and Fe(CO)5 have trigonal bipyramidal geometry.
BrF5: square pyramidal
[PtCl4]2−: square planar
BF3: trigonal planar
The problem asks to identify the number of molecules or ions from the given list that have a trigonal bipyramidal shape.
To determine the shape of a molecule or ion, we use the Valence Shell Electron Pair Repulsion (VSEPR) theory. The shape is determined by the number of bonding pairs and lone pairs of electrons around the central atom.
For coordination compounds like [PtCl₄]²⁻ and Fe(CO)₅, the shape is determined by the coordination number and hybridization of the central metal atom.
We will analyze each species individually to determine its shape.
1. PF₅ (Phosphorus Pentafluoride):
2. BrF₅ (Bromine Pentafluoride):
3. PCl₅ (Phosphorus Pentachloride):
4. [PtCl₄]²⁻ (Tetrachloroplatinate(II)):
5. BF₃ (Boron Trifluoride):
6. Fe(CO)₅ (Iron Pentacarbonyl):
Based on the analysis, the molecules/ions from the list that have a trigonal bipyramidal shape are:
The total number of species with a trigonal bipyramidal shape is 3.
From the given following (A to D) cyclic structures, those which will not react with Tollen's reagent are : 
Compound 'P' undergoes the following sequence of reactions : (i) NH₃ (ii) $\Delta$ $\rightarrow$ Q (i) KOH, Br₂ (ii) CHCl₃, KOH (alc), $\Delta$ $\rightarrow$ NC-CH₃. 'P' is : 

The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.