Animated Solution for Chemistry - Coordination Compounds: Match each set of hybrid orbitals from LIST-I with complex (es) given in LIST-II.

In this thrilling journey through coordination chemistry, we are tasked with determining the hybridization of six distinct coordination complexes and matching them to their corresponding orbital geometries. This is a classic test of your understanding of Valence Bond Theory (VBT) and Crystal Field Theory (CFT).

To conquer any hybridization problem in coordination chemistry, you must follow a strict, logical sequence: 1. Determine the Oxidation State: Find the charge on the central metal ion. 2. Write the Electronic Configuration: Know how many d-electrons you are working with. 3. Analyze the Ligand Strength: Is it a Strong Field Ligand (SFL) that forces pairing, or a Weak Field Ligand (WFL) that leaves electrons unpaired? 4. Assign the Hybridization: Based on the coordination number and the availability of inner d-orbitals, determine if the complex is inner orbital ($d^{2}s{p}^3$, $ds{p}^2$) or outer orbital ($s{p}^3{d}^2$, $s{p}^3$).

Let's break down each complex with surgical precision.

1. The Hexafluoroferrate Ion: $[Fe{F}_6{]}^{4-}$ Here, iron is in a $+2$ oxidation state, giving it a $3d^6$ configuration. Fluoride ($F^{-}$) is a notorious weak field ligand. It simply doesn't have the energetic muscle to force those d-electrons to pair up against their will. Since we need six empty orbitals for the octahedral geometry, and the inner $3d$ orbitals are occupied, we must look outward. We utilize one $4s$, three $4p$, and two $4d$ orbitals, resulting in an $s{p}^3{d}^2$ hybridization. This is an outer orbital complex.

2. The Titanium Complex: $[Ti(H_{2}O{)}_{3}C{l}_3]$ Titanium is in a $+3$ state, which means it has a highly sparse $3d^1$ configuration. With only one electron wandering around the $3d$ subshell, there are plenty of empty inner d-orbitals available. It effortlessly uses two of these inner $3d$ orbitals, leading to a $d^{2}s{p}^3$ hybridization.

3. The Hexaamminechromium Ion: $[Cr(N{H}_3{)}_6{]}^{3+}$ Chromium in the $+3$ state has a $3d^3$ configuration. According to Hund's rule, these three electrons occupy the $t_{2g}$ orbitals singly. This naturally leaves exactly two $e_g$ orbitals completely empty! Therefore, regardless of the ligand's strength, $C{r}^{3+}$ will always utilize these inner orbitals, resulting in a $d^{2}s{p}^3$ hybridization.

4. The Tetrachloroferrate Ion: $[FeC{l}_4{]}^{2-}$ Iron is again in a $+2$ state ($3d^6$). The coordination number is 4, and chloride ($C{l}^{-}$) is a weak field ligand. No pairing occurs. To accommodate four ligands, the metal uses one $4s$ and three $4p$ orbitals, giving us an $s{p}^3$ hybridization, which corresponds to a tetrahedral geometry.

5. Nickel Tetracarbonyl: $Ni(CO{)}_4$ This is a fascinating molecule! Nickel is in a rare $0$ oxidation state, possessing a full $3d^{8}4s^2$ valence shell. However, carbonyl ($CO$) is an exceptionally strong field ligand and a powerful $\pi$-acceptor. It aggressively forces the two $4s$ electrons to migrate into the $3d$ subshell, pairing up with the existing electrons to create a completely full $3d^{10}$ configuration. With the $4s$ orbital now vacant, the complex utilizes the $4s$ and three $4p$ orbitals, resulting in an $s{p}^3$ hybridization.

6. The Tetracyanonickelate Ion: $[Ni(CN{)}_4{]}^{2-}$ Here, nickel is in a $+2$ state, giving it a $3d^8$ configuration. Cyanide ($C{N}^{-}$) is a strong field ligand. It forces the two unpaired electrons in the $3d$ subshell to pair up into a single orbital. This strategic pairing liberates exactly one $3d$ orbital. The complex seizes this opportunity, utilizing one $3d$, one $4s$, and two $4p$ orbitals to achieve a $ds{p}^2$ hybridization, forming a beautiful square planar geometry.

With our analysis complete, the matching is straightforward: - P ($ds{p}^2$) perfectly matches with 6 ($[Ni(CN{)}_4{]}^{2-}$). - Q ($s{p}^3$) matches with both 4 ($[FeC{l}_4{]}^{2-}$) and 5 ($Ni(CO{)}_4$). - R ($s{p}^3{d}^2$) matches with 1 ($[Fe{F}_6{]}^{4-}$). - S ($d^{2}s{p}^3$) matches with both 2 ($[Ti(H_{2}O{)}_{3}C{l}_3]$) and 3 ($[Cr(N{H}_3{)}_6{]}^{3+}$).

Mastering these electronic transitions and ligand effects is the key to unlocking the beautiful structural logic of coordination chemistry!