Match the following Complex ions with their respective hybridisation and choose the correct option.

(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

The correct answer is option 3. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).

The correct matching of the complex ions with their respective hybridizations is as follows:

(a) \([Ni(CN)_4]^{2-}\)(iv) \(dsp^2\):

In the complex ion \([Ni(CN)_4]^{2-}\), the central nickel ion \((Ni)\) is coordinated with four cyanide ions \((CN^-)\) as ligands. Each cyanide ion donates one electron pair to form a coordinate bond with the metal ion. Nickel \((Ni)\) has the electron configuration \([Ar] 3d^8 4s^2\). In the complex \([Ni(CN)_4]^{2-}\), the nickel ion loses two electrons to achieve a \(+2\) oxidation state. Hence, the electron configuration becomes \([Ar] 3d^8\). In \(dsp^2\) hybridization, the d orbitals \((3d)\) and the s orbital \((4s)\) undergo hybridization to form five hybrid orbitals: four \(sp^3\) hybrid orbitals and one unhybridized \(d\) orbital. The hybrid orbitals are arranged in a square planar geometry around the central nickel ion. In the case of \([Ni(CN)_4]^{2-}\), the four cyanide ligands occupy the four \(sp^3\) hybrid orbitals of nickel, forming four sigma \((\sigma )\) bonds. The remaining unhybridized \(d\) orbital of nickel can accept a pair of electrons from the cyanide ligands, forming \(d\pi -p\pi \) bonds, which contribute to the overall stability of the complex. Therefore, the correct hybridization of the central nickel ion in \([Ni(CN)_4]^{2-}\) is \(dsp^2\).

(b) \([Fe(CN)_6]^{4-}\)(ii) \(d^2sp^3\):

The \([Fe(CN)_6]^{4-}\) complex is an example of an octahedral coordination complex where the central iron ion is in the \(+2\) oxidation state (Fe^2+). In this case, iron has the electronic configuration of \( [Ar] 3d^6 4s^2 \). In the \([Fe(CN)_6]^{4-}\) complex, the iron ion uses six ligands (CN^-) arranged octahedrally around it. For octahedral complexes, the typical hybridization is \(d^2sp^3\). This means that the iron ion undergoes hybridization of one \(d\) orbital, one \(s\) orbital, and three \(p\) orbitals to form six equivalent \(sp^3d^2\) hybrid orbitals, which then overlap with the ligands.

The iron ion in the +2 oxidation state has the electronic configuration \( [Ar] 3d^6 \). It utilizes six orbitals (one \(d\) orbital, one \(s\) orbital, and three \(p\) orbitals) to form six \(d^2sp^3\) hybrid orbitals. These orbitals arrange themselves octahedrally, which is consistent with the geometry of the complex. The cyanide ligands \((CN^-)\) are strong field ligands and cause a large splitting of the d-orbitals, leading to a low-spin configuration in this complex.

(c) \([FeF_6]^{4-}\)(iii) \(sp^3d^2\):

In the complex ion \([FeF_6]^{4-}\), the central iron ion \((Fe)\) is coordinated with six fluoride ions \((F^-)\) as ligands. Each fluoride ion donates one electron pair to form a coordinate bond with the metal ion. Iron \((Fe)\) has the electron configuration \([Ar] 3d^6 4s^2\). In the complex \([FeF_6]^{4-}\), the iron ion loses two electrons to achieve a \(+2\) oxidation state. Hence, the electron configuration becomes \([Ar] 3d^6\).

In order to form six bonds, the electron configuration of iron can be represented as \([Ar] 3d^6 4s^0\). To accommodate the formation of six bonds, iron promotes one electron from the 4s orbital to one of the empty 3d orbitals. This results in the electron configuration \([Ar] 3d^5\).

In \(sp^3d^2\) hybridization, five hybrid orbitals are formed by hybridizing three \(3d\) orbitals, one \(4s\) orbital, and one 4p orbital. These five hybrid orbitals arrange themselves in an octahedral geometry around the central iron ion. Each hybrid orbital overlaps with the fluorine ligand's orbital to form five sigma \((\sigma )\) bonds. Therefore, the correct hybridization of the central iron ion in \([FeF_6]^{4-}\)is \(sp^3d^2\).

(d) \([Ni(CO)_4]\)(iv) \(sp^3\):

Nickel \((Ni)\) in its ground state has the electronic configuration \( [Ar] 3d^8 4s^2 \). For the \([Ni(CO)_4]\) complex, the nickel is in the zero oxidation state, so it retains its full \(3d\) and \(4s\) electrons. The complex \([Ni(CO)_4]\) contains four carbon monoxide (CO) ligands. Each \(CO\) ligand is a strong field ligand, meaning it has a significant ability to interact with the metal center through both sigma and pi bonding. In a tetrahedral coordination complex like \([Ni(CO)_4]\), the nickel atom undergoes \(sp^3\) hybridization. Here’s how this works:

The nickel atom mixes one \(s\) orbital and three \(p\) orbitals (from the 4s and 4p orbitals) to form four \(sp^3\) hybrid orbitals. These four \(sp^3\) hybrid orbitals arrange themselves in a tetrahedral geometry to minimize electron repulsion. In a tetrahedral arrangement, the four hybrid orbitals are oriented towards the corners of a tetrahedron. Each hybrid orbital forms a sigma bond with a \(CO\) ligand, resulting in a stable tetrahedral structure. Each \(CO\) ligand forms a sigma bond with one of the \(sp^3\) hybrid orbitals on the nickel atom. \(CO\) ligands also engage in back-donation, where the nickel's d orbitals interact with the CO's empty \( \pi^* \) orbitals. This interaction strengthens the bond but is not part of the primary hybridization. The result of this hybridization and bonding is a tetrahedral geometry around the nickel atom, with each corner of the tetrahedron occupied by a CO ligand. In summary, \([Ni(CO)_4]\) adopts a tetrahedral shape due to \(sp^3\) hybridization of the nickel atom, where four equivalent hybrid orbitals form sigma bonds with the CO ligands.