Animated Solution for Chemistry - Coordination Compounds: Given below are two statements. Statement I $[\text{Mn}(\text{CN}{)}_6{]}^{3-}$, $[\text{Fe}(\text{CN}{)}_6{]}^{3-}$ and $[\text{Co}({\text{C}}_2{\text{O}}_4{)}_3{]}^{3-}$ are $d^{2}s{p}^3$ hybridised. Statement II $[{\text{MnCl}}_6{]}^{3-}$ and $[{\text{FeF}}_6{]}^{3-}$ are paramagnetic and have 4 and 5 unpaired electrons, respectively. In the light of the above statements, choose the correct answer from the options given below

Coordination chemistry is a fascinating world where the rules of classical bonding are beautifully bent. In this problem, we are tasked with verifying two statements regarding the hybridization and magnetic properties of five distinct coordination complexes. Let's embark on this journey by analyzing each complex through the lens of Valence Bond Theory (VBT) and Crystal Field Theory (CFT).

The Inner Orbital Complexes

Our first candidate is the hexacyanidomanganate(III) ion, $[\text{Mn}(\text{CN}{)}_6{]}^{3-}$. The central manganese atom is in a $+3$ oxidation state, which leaves it with a $3d^4$ electronic configuration. The cyanide ion (${\text{CN}}^{-}$) is a notoriously strong field ligand. It exerts a massive crystal field splitting energy (${\Delta }_o$), which forces the electrons to pair up in the lower energy $t_{2g}$ orbitals rather than jumping to the higher $e_g$ orbitals. This pairing leaves exactly two inner $3d$ orbitals vacant. These two $d$ orbitals mix with one $4s$ and three $4p$ orbitals to yield a $d^{2}s{p}^3$ hybridization.

Next, we look at the hexacyanidoferrate(III) ion, $[\text{Fe}(\text{CN}{)}_6{]}^{3-}$. Iron is also in a $+3$ state, giving it a $3d^5$ configuration. Just like before, the strong cyanide ligand forces maximum pairing. The five electrons occupy the inner orbitals, leaving two $3d$ orbitals empty. Consequently, it also undergoes $d^{2}s{p}^3$ hybridization.

Our third complex is $[\text{Co}({\text{C}}_2{\text{O}}_4{)}_3{]}^{3-}$. Cobalt is in a $+3$ state, meaning it has a $3d^6$ configuration. Here lies a classic trap! While the oxalate ion (${\text{C}}_2{\text{O}}_4^{2-}$) is generally considered a weak field ligand, it behaves as a strong field ligand when paired with ${\text{Co}}^{3+}$. This anomaly occurs because the high charge density of ${\text{Co}}^{3+}$ pulls the ligands closer, increasing the splitting energy ${\Delta }_o$ beyond the pairing energy. Thus, all six electrons pair up, leaving two inner $d$ orbitals empty, resulting in $d^{2}s{p}^3$ hybridization.

Since all three complexes are indeed $d^{2}s{p}^3$ hybridized, Statement I is absolutely true.

The Outer Orbital Complexes

Now, let's shift our focus to Statement II, starting with the hexafluoridoferrate(III) ion, $[{\text{FeF}}_6{]}^{3-}$. Iron is in a $+3$ state ($3d^5$). However, the fluoride ion (${\text{F}}^{-}$) is a weak field ligand. It cannot overcome the pairing energy, so the electrons remain unpaired, occupying all five $3d$ orbitals singly. Because no inner $d$ orbitals are available, the complex must utilize the outer $4d$ orbitals for bonding, leading to an $s{p}^3{d}^2$ hybridization. With 5 unpaired electrons, this complex is highly paramagnetic.

Finally, we examine the hexachloridomanganate(III) ion, $[{\text{MnCl}}_6{]}^{3-}$. Manganese is in a $+3$ state ($3d^4$). The chloride ion (${\text{Cl}}^{-}$) is another weak field ligand, so no pairing occurs. The four electrons remain unpaired. Although one inner $3d$ orbital is empty, an octahedral geometry strictly requires two empty $d$ orbitals to form a $d^{2}s{p}^3$ hybrid. Since it only has one, it is forced to use the outer $4d$ orbitals, resulting in an $s{p}^3{d}^2$ hybridization. It possesses exactly 4 unpaired electrons.

Since $[{\text{MnCl}}_6{]}^{3-}$ has 4 unpaired electrons and $[{\text{FeF}}_6{]}^{3-}$ has 5 unpaired electrons, Statement II is also perfectly true.

Both statements stand up to rigorous chemical scrutiny. The interplay between the metal's oxidation state and the ligand's field strength dictates the entire geometry and magnetic behavior of these beautiful molecules. Therefore, the correct option is (d).