Out of the following, how many compounds have tetrahedral geometry?

NH₄⁺, XeF₄, [NiCl₄]²⁻, [PtCl₄]²⁻, [Cu(NH₃)₄]²⁺, BF₃ and [Ni(CO)₄]

The correct answer is 3.

To determine the number of compounds with tetrahedral geometry, we can analyze the hybridization and structure of each species one by one:

1. NH₄⁺ (Ammonium ion):
The central nitrogen atom has 5 valence electrons. In NH₄⁺, one electron is lost, leaving 4 valence electrons which form 4 sigma (σ) bonds with four hydrogen atoms. With 4 bond pairs and 0 lone pairs, the hybridization of nitrogen is sp³. Therefore, it has a tetrahedral geometry.

2. XeF₄ (Xenon tetrafluoride):
The central xenon atom has 8 valence electrons. It forms 4 single bonds with fluorine atoms, leaving 4 non-bonding electrons (2 lone pairs). With 4 bond pairs and 2 lone pairs (steric number = 6), the hybridization of xenon is sp³d², resulting in a square planar geometry.

3. [NiCl₄]^2- (Tetrachloronidcolate(II) ion):
The oxidation state of nickel is +2. Nickel has the electronic configuration [Ar] 3d⁸ 4s⁰. Since Cl^- is a weak field ligand, it does not force the pairing of unpaired 3d electrons. Thus, the empty 4s and three 4p orbitals hybridize to form sp³ hybrid orbitals. Therefore, the geometry is tetrahedral.

4. [PtCl₄]^2- (Tetrachloroplatinate(II) ion):
The oxidation state of platinum is +2. Platinum belongs to the 5d transition series. Because of the higher effective nuclear charge and larger size of 5d orbitals, the crystal field splitting energy is high enough to cause pairing of electrons even with weak field ligands like Cl^-. This results in dsp² hybridization and a square planar geometry.

5. [Cu(NH₃)₄]²⁺ (Tetraamminecopper(II) ion):
The oxidation state of copper is +2 with a 3d⁹ configuration. In the presence of NH₃ ligands, one 3d electron is excited/transferred, allowing the copper ion to undergo dsp² hybridization. Therefore, its geometry is square planar.

6. BF₃ (Boron trifluoride):
The central boron atom has 3 valence electrons, which form 3 sigma bonds with fluorine atoms. With 3 bond pairs and 0 lone pairs, the hybridization of boron is sp², giving it a trigonal planar geometry.

7. [Ni(CO)₄] (Tetracarbonylnickel(0)):
The oxidation state of nickel is 0, having the electronic configuration [Ar] 3d⁸ 4s². CO is a strong field ligand and causes the pairing of the 4s electrons into the 3d subshell, resulting in a fully filled 3d¹⁰ configuration. The empty 4s and 4p orbitals then hybridize to form sp³ hybrid orbitals. Therefore, the geometry is tetrahedral.

Conclusion:
Out of the given list, the 3 compounds with tetrahedral geometry are NH₄⁺, [NiCl₄]^2-, and [Ni(CO)₄].