Determine Number of Chemical Species with Tetrahedral Geometry

To determine the total number of chemical species with tetrahedral geometry, we analyze each species individually:

1. $[\text{Co}(\text{CN})_4]^{4-}$:

   • Oxidation state of $\text{Co} = 0$.
   • Electronic configuration of $\text{Co}^0 = [\text{Ar}] 3d^7 4s^2 \equiv 3d^9$.
   • Cyanide ($\text{CN}^-$) is a strong field ligand. It causes pairing of electrons, making one $3d$ orbital available for $dsp^2$ hybridization.
   • Geometry: Square Planar

2. $[\text{Co}(\text{CO})_3(\text{NO})]$:

   • Nitrosyl ($\text{NO}$) acts as a 3-electron donor ($\text{NO}^+$).
   • The metal ion attains a pseudo-$d^{10}$ closed-shell electronic configuration ($\text{Co}^{-1}$).
   • The four ligands donate electron pairs into the vacant $4s$ and $4p$ orbitals, resulting in $sp^3$ hybridization.
   • Geometry: Tetrahedral

3. $\text{XeF}_4$:

   • Xenon ($\text{Xe}$) has 8 valence electrons.
   • It forms 4 single bonds with fluorine atoms and retains 2 lone pairs.
   • Steric number $= 4 \text{ (bond pairs)} + 2 \text{ (lone pairs)} = 6$, corresponding to $sp^3d^2$ hybridization.
   • Geometry (Shape): Square Planar

4. $[\text{PCl}_4]^+$:

   • Phosphorus ($\text{P}$) has $5 - 1 = 4$ valence electrons.
   • It forms 4 single bonds with chlorine atoms with 0 lone pairs.
   • Steric number $= 4$, corresponding to $sp^3$ hybridization.
   • Geometry: Tetrahedral

5. $[\text{PdCl}_4]^{2-}$:

   • Palladium is in the $+2$ oxidation state ($\text{Pd}^{2+}$), which is a $4d^8$ system.
   • For $4d$ and $5d$ transition series metals with $d^8$ configuration, all 4-coordinate complexes adopt $dsp^2$ hybridization regardless of ligand strength.
   • Geometry: Square Planar

6. $[\text{ICl}_4]^-$:

   • Iodine ($\text{I}$) has $7 + 1 = 8$ valence electrons.
   • It forms 4 single bonds with chlorine atoms and retains 2 lone pairs.
   • Steric number $= 6$, corresponding to $sp^3d^2$ hybridization.
   • Geometry (Shape): Square Planar

7. $[\text{Cu}(\text{CN})_4]^{3-}$:

   • Oxidation state of $\text{Cu} = +1$.
   • Electronic configuration of $\text{Cu}^+ = [\text{Ar}] 3d^{10}$.
   • Since the $3d$ subshell is completely filled, inner $d$-orbitals cannot participate in hybridization.
   • It uses empty $4s$ and $4p$ orbitals for $sp^3$ hybridization.
   • Geometry: Tetrahedral

8. $\text{P}_4$:

   • White phosphorus exists as a discrete $\text{P}_4$ molecule where each phosphorus atom is bonded to three other phosphorus atoms at the vertices of a tetrahedron.
   • Geometry: Tetrahedral

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Summary:

The species with tetrahedral geometry are:

1. $[\text{Co}(\text{CO})_3(\text{NO})]$
2. $[\text{PCl}_4]^+$
3. $[\text{Cu}(\text{CN})_4]^{3-}$
4. $\text{P}_4$

Thus, the total number of species with tetrahedral geometry is 4.