Animated Solution for Chemistry - Coordination Compounds: The pair(s) of complexes wherein both exhibit tetrahedral geometry is(are) (Note: py = pyridine Given: Atomic numbers of Fe, Co, Ni and Cu are 26, 27, 28 and 29, respectively)

When a central metal ion is surrounded by exactly four ligands, the resulting coordination complex can adopt one of two primary geometries: tetrahedral or square planar. The deciding factor between these two shapes lies deep within the quantum mechanical realm of hybridization.

If the metal ion utilizes one $s$ and three $p$ orbitals from its outermost shell, it undergoes $s{p}^3$ hybridization, leading to a tetrahedral geometry. Conversely, if the metal ion has an empty inner $d$-orbital available, it can mix that $d$-orbital with one $s$ and two $p$ orbitals to form $ds{p}^2$ hybridization, resulting in a flat, square planar geometry. The availability of this inner $d$-orbital depends heavily on two things: the oxidation state of the metal and the strength of the approaching ligands.

Let's dive into the first pair of complexes: $[FeC{l}_4{]}^{-}$ and $[Fe(CO{)}_4{]}^{2-}$.

In the tetrachloroferrate(III) ion, $[FeC{l}_4{]}^{-}$, iron is in a $+3$ oxidation state. A neutral iron atom has the electron configuration $[Ar]3d^{6}4s^2$. Stripping away three electrons leaves us with a $3d^5$ configuration. The chloride ion ($C{l}^{-}$) is a classic weak field ligand. It lacks the energetic punch required to force those five unpaired $d$-electrons to pair up against their natural repulsion. Because no inner $3d$ orbitals are freed up, the iron ion must look outward, utilizing its empty $4s$ and $4p$ orbitals. This results in $s{p}^3$ hybridization, making the complex tetrahedral.

Now, look at the tetracarbonylferrate(-II) ion, $[Fe(CO{)}_4{]}^{2-}$. This is where things get fascinating. Iron here possesses a rare $-2$ oxidation state! Starting from the neutral $3d^{6}4s^2$, gaining two electrons pushes it to $3d^{8}4s^2$. Carbon monoxide ($CO$) is a notoriously strong field ligand and a powerful $\pi$-acceptor. It forces the two $4s$ electrons to demote and pair up within the $3d$ subshell, creating a completely filled $3d^{10}$ configuration. With the $3d$ subshell entirely packed, there are absolutely no inner $d$-orbitals available. The complex has no choice but to use the outer $4s$ and $4p$ orbitals, resulting in $s{p}^3$ hybridization. Thus, it is also tetrahedral. Option A is a match!

Moving to Option B, we encounter $[Co(CO{)}_4{]}^{-}$ and $[CoC{l}_4{]}^{2-}$.

The tetracarbonylcobaltate(-I) ion features cobalt in a $-1$ state. Neutral cobalt is $3d^{7}4s^2$, so $C{o}^{-}$ becomes $3d^{8}4s^2$. Just like the iron carbonyl complex, the strong $CO$ ligands force the $4s$ electrons into the $3d$ subshell, yielding a full $3d^{10}$ configuration. A full $d$-subshell guarantees $s{p}^3$ hybridization, making it tetrahedral.

Its partner, the tetrachlorocobaltate(II) ion, has cobalt in a $+2$ state, giving a $3d^7$ configuration. The weak field chloride ligands cannot force pairing. The inner $d$-orbitals remain occupied by unpaired electrons, forcing the use of outer orbitals. This leads to $s{p}^3$ hybridization and a tetrahedral geometry. Option B is also a perfect match!

In Option C, we compare $[Ni(CO{)}_4]$ and $[Ni(CN{)}_4{]}^{2-}$.

Tetracarbonylnickel(0) features neutral nickel ($3d^{8}4s^2$). The strong $CO$ ligands again force the $4s$ electrons into the $3d$ subshell, resulting in a $3d^{10}$ configuration. As we've established, $3d^{10}$ dictates $s{p}^3$ hybridization and a tetrahedral shape.

However, the tetracyanonickelate(II) ion breaks the pattern. Nickel is in a $+2$ state, meaning it has a $3d^8$ configuration. Cyanide ($C{N}^{-}$) is an exceptionally strong field ligand. It possesses enough energy to force the two unpaired electrons in the $3d$ subshell to pair up into a single orbital. This brilliant maneuver frees up exactly one inner $3d$ orbital! The nickel ion seizes this opportunity, mixing the newly available $3d$ orbital with one $4s$ and two $4p$ orbitals to achieve $ds{p}^2$ hybridization. This results in a square planar geometry. Because they do not share the same geometry, Option C is incorrect.

Finally, we examine Option D: $[Cu(py{)}_4{]}^{+}$ and $[Cu(CN{)}_4{]}^{3-}$.

Both of these complexes feature copper in a $+1$ oxidation state. Neutral copper is an anomaly with a $3d^{10}4s^1$ configuration. Removing one electron gives $C{u}^{+}$ a completely filled $3d^{10}$ configuration.

Here is the ultimate rule of thumb: If the $d$-subshell is completely full ($d^{10}$), the nature of the ligand is entirely irrelevant. Whether it is a weak field ligand like pyridine or a strong field ligand like cyanide, there are simply no inner $d$-orbitals available to be freed up. Both complexes are forced to use their outer $4s$ and $4p$ orbitals, resulting in $s{p}^3$ hybridization. Therefore, both complexes are unequivocally tetrahedral. Option D is correct!

By carefully tracking the oxidation states and understanding the profound influence of strong versus weak field ligands, we can confidently conclude that the pairs in options A, B, and D all exhibit tetrahedral geometry.