Animated Solution for Chemistry - Coordination Compounds: Choose the correct statement(s) among the following :

Coordination chemistry is a beautiful puzzle where geometry, electronic configuration, and ligand behavior interlock to define the physical reality of a molecule. This JEE Advanced problem is a masterclass in testing these interconnected concepts. Let's break down each statement systematically.

The Tetrahedral Iron Complex

We begin with the complex ion $[FeC{l}_4{]}^{-}$. The first step is always to determine the oxidation state of the central metal. Here, iron is in the $+3$ oxidation state, giving it an electronic configuration of $[Ar]3d^5$.

The ligand involved is the chloride ion ($C{l}^{-}$), which is a classic weak field ligand. Because the coordination number is 4, the complex must adopt either a tetrahedral or a square planar geometry. A square planar geometry requires $ds{p}^2$ hybridization, which means an inner $3d$ orbital must be empty. However, the weak field nature of $C{l}^{-}$ cannot force the five unpaired $3d$ electrons to pair up. Consequently, the complex utilizes the outer $4s$ and $4p$ orbitals, resulting in $s{p}^3$ hybridization and a tetrahedral geometry. Statement (A) is absolutely correct.

The Art of Geometrical Isomerism

Next, we examine the octahedral complex $[Co(en)(N{H}_3{)}_{2}C{l}_2{]}^{+}$. This is a complex of the type $[M(AA)b_2{c}_2]$, where 'en' (ethylenediamine) is a symmetrical bidentate ligand.

To find the geometrical isomers, we must systematically arrange the ligands around the octahedral core. Remember, the bidentate 'en' ligand has a small "bite angle" and can only occupy cis positions relative to each other; it cannot span across trans positions.

Keeping this constraint in mind, we can form three distinct spatial arrangements: 1. Both $C{l}^{-}$ ligands are trans to each other, and both $N{H}_3$ ligands are cis. 2. Both $N{H}_3$ ligands are trans to each other, and both $C{l}^{-}$ ligands are cis. 3. All identical ligands ($C{l}^{-}$ and $N{H}_3$) are cis to each other.

Thus, there are 3 geometrical isomers, not 2. Statement (B) is incorrect.

The Battle of Magnetic Moments

Magnetic moment is a direct consequence of unpaired electrons. Let's compare our two complexes.

For $[FeC{l}_4{]}^{-}$, we established that it is a high-spin tetrahedral complex with a $d^5$ configuration. In a tetrahedral crystal field, the splitting energy (${\Delta }_t$) is less than the pairing energy ($P$). Thus, the configuration is $e^2{t}_2^3$, leaving 5 unpaired electrons ($n=5$). The spin-only magnetic moment is calculated as:

Now, for $[Co(en)(N{H}_3{)}_{2}C{l}_2{]}^{+}$, cobalt is in the $+3$ state, giving a $3d^6$ configuration. The presence of 'en' and $N{H}_3$ creates a strong ligand field. In this octahedral field, the splitting energy (${\Delta }_o$) overcomes the pairing energy ($P$). All six electrons pair up in the lower energy $t_{2g}$ orbitals ($t_{2g}^6{e}_g^0$). With 0 unpaired electrons ($n=0$), its magnetic moment is exactly $0\text{ B.M.}$

Clearly, $[FeC{l}_4{]}^{-}$ has a significantly higher magnetic moment. Statement (C) is correct.

The Hybridization Trap

Finally, let's look at the hybridization of the cobalt ion in $[Co(en)(N{H}_3{)}_{2}C{l}_2{]}^{+}$. As we just deduced, it is a low-spin $d^6$ complex where all electrons are paired in the $t_{2g}$ level.

This pairing leaves two inner $3d$ orbitals (specifically the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals of the $e_g$ set) completely empty. The metal ion will preferentially use these lower-energy inner orbitals for bonding. Therefore, the hybridization is $d^{2}s{p}^3$ (an inner orbital complex), not $s{p}^3{d}^2$ (which would imply an outer orbital complex). Statement (D) is incorrect.

By carefully applying the principles of Valence Bond Theory and Crystal Field Theory, we can confidently conclude that only statements (A) and (C) are correct.