Identification of Tetrahedral Geometry in Transition Metal Complexes

To determine which pairs of complex species exhibit tetrahedral stereochemistry, we analyze the oxidation state of the central metal ion, its $d$-electron configuration, the nature of the ligands (strong field or weak field), and the resulting hybridization.

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1. Analysis of Option (A): $[\text{FeCl}_4]^-$ and $[\text{Fe(CO)}_4]^{2-}$

• For $[\text{FeCl}_4]^-$:

  • Oxidation state of $\text{Fe}$: $x + 4(-1) = -1 \implies x = +3$.
  • Electronic configuration of $\text{Fe}^{3+}$: $3d^5 4s^0$.
  • Ligand field & Hybridization: $\text{Cl}^-$ is a weak field ligand, so no pairing of $3d$ electrons occurs. The 4 pairs of electrons from $\text{Cl}^-$ ligands occupy one $4s$ and three $4p$ orbitals.
  • Hybridization: $sp^3$
  • Geometry: Tetrahedral

• For $[\text{Fe(CO)}_4]^{2-}$:

  • Oxidation state of $\text{Fe}$: $x + 4(0) = -2 \implies x = -2$.
  • Electronic configuration of $\text{Fe}^{2-}$: Neutral $\text{Fe}$ ($Z=26$) has 8 valence electrons ($3d^6 4s^2$). With a $-2$ charge, $\text{Fe}^{2-}$ has 10 valence electrons ($3d^{10} 4s^0$).
  • Ligand field & Hybridization: Since the $3d$ subshell is completely filled ($3d^{10}$), inner $d$-orbitals cannot participate in bonding. The 4 $\text{CO}$ ligands donate electron pairs into the vacant $4s$ and $4p$ orbitals.
  • Hybridization: $sp^3$
  • Geometry: Tetrahedral

Since both species exhibit tetrahedral geometry, Option (A) is correct.

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2. Analysis of Option (B): $[\text{Co(CO)}_4]^-$ and $[\text{CoCl}_4]^{2-}$

• For $[\text{Co(CO)}_4]^-$:

  • Oxidation state of $\text{Co}$: $x + 4(0) = -1 \implies x = -1$.
  • Electronic configuration of $\text{Co}^-$: Neutral $\text{Co}$ ($Z=27$) has 9 valence electrons ($3d^7 4s^2$). With a $-1$ charge, $\text{Co}^-$ has 10 valence electrons ($3d^{10} 4s^0$).
  • Ligand field & Hybridization: The $3d$ subshell is fully filled ($3d^{10}$). Bonding occurs using empty $4s$ and $4p$ orbitals.
  • Hybridization: $sp^3$
  • Geometry: Tetrahedral

• For $[\text{CoCl}_4]^{2-}$:

  • Oxidation state of $\text{Co}$: $x + 4(-1) = -2 \implies x = +2$.
  • Electronic configuration of $\text{Co}^{2+}$: $3d^7 4s^0$.
  • Ligand field & Hybridization: $\text{Cl}^-$ is a weak field ligand, forming a high-spin complex with four coordination sites using $4s$ and $4p$ orbitals.
  • Hybridization: $sp^3$
  • Geometry: Tetrahedral

Since both species exhibit tetrahedral geometry, Option (B) is correct.

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3. Analysis of Option (C): $[\text{Ni(CO)}_4]$ and $[\text{Ni(CN)}_4]^{2-}$

• For $[\text{Ni(CO)}_4]$:

  • Oxidation state of $\text{Ni}$: $0$.
  • Electronic configuration of $\text{Ni}(0)$: $3d^8 4s^2$.
  • Ligand field & Hybridization: $\text{CO}$ is a strong field ligand, forcing the two $4s$ electrons into the $3d$ subshell, resulting in a $3d^{10} 4s^0$ configuration.
  • Hybridization: $sp^3$
  • Geometry: Tetrahedral

• For $[\text{Ni(CN)}_4]^{2-}$:

  • Oxidation state of $\text{Ni}$: $x + 4(-1) = -2 \implies x = +2$.
  • Electronic configuration of $\text{Ni}^{2+}$: $3d^8 4s^0$.
  • Ligand field & Hybridization: $\text{CN}^-$ is a strong field ligand, causing the 8 electrons in the $3d$ subshell to pair up completely into four $3d$ orbitals. This leaves one $3d$ orbital vacant for hybridization ($3d + 4s + 2\times 4p$).
  • Hybridization: $dsp^2$
  • Geometry: Square Planar

Since $[\text{Ni(CN)}_4]^{2-}$ is square planar, Option (C) is incorrect.

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4. Analysis of Option (D): $[\text{Cu(py)}_4]^+$ and $[\text{Cu(CN)}_4]^{3-}$

• For $[\text{Cu(py)}_4]^+$:

  • Oxidation state of $\text{Cu}$: $x + 4(0) = +1 \implies x = +1$.
  • Electronic configuration of $\text{Cu}^+$: $3d^{10} 4s^0$.
  • Ligand field & Hybridization: The $3d$ subshell is fully filled ($3d^{10}$); inner $d$-orbitals cannot participate in hybridization. Coordination occurs through $4s$ and $4p$ orbitals.
  • Hybridization: $sp^3$
  • Geometry: Tetrahedral

• For $[\text{Cu(CN)}_4]^{3-}$:

  • Oxidation state of $\text{Cu}$: $x + 4(-1) = -3 \implies x = +1$.
  • Electronic configuration of $\text{Cu}^+$: $3d^{10} 4s^0$.
  • Ligand field & Hybridization: The $3d$ subshell is fully filled ($3d^{10}$), leading to $sp^3$ hybridization.
  • Hybridization: $sp^3$
  • Geometry: Tetrahedral

Since both species exhibit tetrahedral geometry, Option (D) is correct.

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Conclusion:

The complex pairs in which both species exhibit tetrahedral stereochemistry are (A), (B), and (D).