To determine the geometry of the coordination complexes, we need to find the hybridization of the central metal atom. For a coordination number of 4, the geometry can be either tetrahedral ($sp^3$ hybridization) or square planar ($dsp^2$ hybridization).

General Rules:

1. Determine the oxidation state of the central metal ion.
2. Write down the electronic configuration of the central metal ion.
3. For coordination number 4:
   • If the central metal ion has a $d^{10}$ electronic configuration, the hybridization will be $sp^3$ and the geometry will be tetrahedral, as there are no vacant d-orbitals for $dsp^2$ hybridization.
   • If the central metal ion has a $d^8$ configuration with a strong field ligand (like $CN^-$, $CO$), electron pairing occurs, making one d-orbital vacant, leading to $dsp^2$ hybridization and square planar geometry.
   • If the ligands are weak field (like halides $Cl^-$, $Br^-$), they generally do not cause pairing. If no inner d-orbital is available, the hybridization will be $sp^3$, leading to tetrahedral geometry.

Let's analyze each option:

A: $[FeCl_4]^-$ and $[Fe(CO)_4]^{2-}$

1. $[FeCl_4]^-$:

   • Oxidation state of Fe: Let it be $x$. $x + 4(-1) = -1 \Rightarrow x = +3$.
   • Fe (Z=26) has the configuration $[Ar] 3d^6 4s^2$.
   • $Fe^{3+}$ has the configuration $[Ar] 3d^5$.
   • $Cl^-$ is a weak field ligand. It does not cause electron pairing.
   • For coordination number 4, the vacant orbitals available for hybridization are one 4s and three 4p orbitals.
   • Hybridization is $sp^3$, which corresponds to a tetrahedral geometry.

2. $[Fe(CO)_4]^{2-}$:

   • Oxidation state of Fe: Let it be $x$. $x + 4(0) = -2 \Rightarrow x = -2$.
   • Fe (Z=26) has the configuration $[Ar] 3d^6 4s^2$.
   • $Fe^{-2}$ has the electronic configuration $[Ar] 3d^8 4s^2$.
   • $CO$ is a strong field ligand, which causes the 4s electrons to shift to the 3d orbitals.
   • The resulting configuration of Fe is $[Ar] 3d^{10}$.
   • Since the 3d orbitals are completely filled, the hybridization must be $sp^3$.
   • The geometry is tetrahedral.

Since both complexes in pair A are tetrahedral, option A is correct.

B: $[Co(CO)_4]^-$ and $[CoCl_4]^{2-}$

1. $[Co(CO)_4]^-$:

   • Oxidation state of Co: Let it be $x$. $x + 4(0) = -1 \Rightarrow x = -1$.
   • Co (Z=27) has the configuration $[Ar] 3d^7 4s^2$.
   • $Co^{-1}$ has the configuration $[Ar] 3d^8 4s^2$.
   • CO is a strong field ligand, pushing the 4s electrons into the 3d orbitals.
   • The resulting configuration of Co is $[Ar] 3d^{10}$.
   • The 3d orbitals are full, so hybridization is $sp^3$.
   • The geometry is tetrahedral.

2. $[CoCl_4]^{2-}$:

   • Oxidation state of Co: Let it be $x$. $x + 4(-1) = -2 \Rightarrow x = +2$.
   • Co (Z=27) has the configuration $[Ar] 3d^7 4s^2$.
   • $Co^{2+}$ has the configuration $[Ar] 3d^7$.
   • $Cl^-$ is a weak field ligand. It does not cause electron pairing.
   • Hybridization involves the outer 4s and 4p orbitals, leading to $sp^3$ hybridization.
   • The geometry is tetrahedral.

Since both complexes in pair B are tetrahedral, option B is correct.

C: [Ni(CO)${}_4$] and [Ni(CN)${}_4$]${}^{2-}$

1. [Ni(CO)${}_4$]:

   • Oxidation state of Ni is 0.
   • Ni (Z=28) has the configuration $[Ar] 3d^8 4s^2$.
   • CO is a strong field ligand, so the 4s electrons are pushed into the 3d orbitals.
   • The resulting configuration of Ni is $[Ar] 3d^{10}$.
   • With a filled 3d subshell, the hybridization is $sp^3$.
   • The geometry is tetrahedral.

2. [Ni(CN)${}_4$]${}^{2-}$:

   • Oxidation state of Ni: Let it be $x$. $x + 4(-1) = -2 \Rightarrow x = +2$.
   • Ni (Z=28) has the configuration $[Ar] 3d^8 4s^2$.
   • $Ni^{2+}$ has the configuration $[Ar] 3d^8$.
   • $CN^-$ is a strong field ligand. It forces the pairing of the two unpaired 3d electrons.
   • This leaves one 3d orbital empty ($d_{x^2-y^2}$).
   • The hybridization is $dsp^2$ (using one 3d, one 4s, and two 4p orbitals).
   • The geometry is square planar.

Since one complex is tetrahedral and the other is square planar, option C is incorrect.

D: [Cu(py)${}_4$]${}^+$ and [Cu(CN)${}_4$]${}^{3-}$

1. [Cu(py)${}_4$]${}^+$:

   • Pyridine (py) is a neutral ligand.
   • Oxidation state of Cu is +1.
   • Cu (Z=29) has the configuration $[Ar] 3d^{10} 4s^1$.
   • $Cu^{+}$ has the configuration $[Ar] 3d^{10}$.
   • The 3d orbitals are completely filled. Hybridization must be $sp^3$.
   • The geometry is tetrahedral.

2. [Cu(CN)${}_4$]${}^{3-}$:

   • Oxidation state of Cu: Let it be $x$. $x + 4(-1) = -3 \Rightarrow x = +1$.
   • $Cu^{+}$ has the configuration $[Ar] 3d^{10}$.
   • The 3d orbitals are completely filled. Hybridization must be $sp^3$.
   • The geometry is tetrahedral.

Since both complexes in pair D are tetrahedral, option D is correct.