Animated Solution for Chemistry - Coordination Compounds: The correct option(s) regarding the complex $[Co(en)(N{H}_3{)}_3(H_{2}O){]}^{3+}$ :- $(en=H_{2}NC{H}_{2}C{H}_{2}N{H}_2)$ is (are)

The beauty of coordination chemistry lies in its ability to blend spatial geometry with quantum mechanics. In this thrilling JEE Advanced problem, we are tasked with dissecting the structural and electronic properties of the complex $[Co(en)(N{H}_3{)}_3(H_{2}O){]}^{3+}$. Let's embark on this journey step by step.

Geometrical Isomerism

Our first challenge is to determine the number of geometrical isomers for the given complex. We can classify this complex as a $[M(AA)b_{3}c]$ type, where 'AA' represents the bidentate ethylenediamine ($en$) ligand, 'b' represents the three ammonia ($N{H}_3$) ligands, and 'c' represents the single water ($H_{2}O$) molecule.

Because ethylenediamine has a relatively short carbon chain connecting its two nitrogen donor atoms, it is geometrically constrained. It cannot stretch across the central metal atom to occupy trans positions (180 degrees apart) without severe ring strain. Therefore, it must strictly occupy cis positions (90 degrees apart).

With the bidentate ligand locked in place, we have four remaining positions for the three $N{H}_3$ molecules and one $H_{2}O$ molecule. The position of the unique $H_{2}O$ molecule dictates the isomerism. We can place the $H_{2}O$ molecule either trans to one of the $N{H}_3$ ligands, or trans to one of the nitrogen atoms of the ethylenediamine ligand. This spatial arrangement yields exactly two geometrical isomers. Thus, Option A is correct.

Ligand Substitution

Option B presents a thought experiment: what happens if we replace the bidentate 'en' ligand with two monodentate cyanide ($C{N}^{-}$) ligands? The complex transforms into a $[M{a}_2{b}_{3}c]$ type system, specifically $[Co(CN{)}_2(N{H}_3{)}_3(H_{2}O){]}^{+}$.

To find the geometrical isomers here, we must focus on the three identical $N{H}_3$ ligands. They can arrange themselves in two fundamental ways:

1. Facial (fac): The three $N{H}_3$ ligands occupy the corners of one triangular face of the octahedron. In this highly symmetric arrangement, the remaining three positions are also facial. Placing the two $C{N}^{-}$ and one $H_{2}O$ in these slots yields only one unique isomer. 2. Meridional (mer): The three $N{H}_3$ ligands form a T-shape along a meridian of the octahedron. This leaves three positions where the two $C{N}^{-}$ ligands can be placed either cis or trans to each other, generating two distinct isomers.

Adding these up, we get a total of $1+2=3$ geometrical isomers. Therefore, Option B is also correct.

Magnetic Properties

Next, we dive into Crystal Field Theory to evaluate the magnetic nature of the complex. The central Cobalt atom is in a $+3$ oxidation state, giving it a $3d^6$ electron configuration.

The ligands surrounding the Cobalt ion dictate how these $d$-electrons behave. Both ethylenediamine and ammonia are strong field ligands. Their robust interaction with the metal creates a large crystal field splitting energy (${\Delta }_o$). This large energy gap forces all six $d$-electrons to pair up in the lower energy $t_{2g}$ orbitals, resulting in a $t_{2g}^6{e}_g^0$ configuration.

Since there are zero unpaired electrons, the complex is strictly diamagnetic, not paramagnetic. Consequently, Option C is incorrect.

Absorption Wavelength

Finally, we compare our complex, $[Co(en)(N{H}_3{)}_3(H_{2}O){]}^{3+}$, with a similar complex where the water molecule is replaced by an additional ammonia molecule: $[Co(en)(N{H}_3{)}_4{]}^{3+}$.

According to the spectrochemical series, water is a weaker field ligand than ammonia ($H_{2}O<N{H}_3$). Therefore, the overall ligand field strength of our original complex is weaker, leading to a smaller crystal field splitting energy (${\Delta }_o$).

The energy of the photon absorbed during a $d-d$ transition is directly equal to ${\Delta }_o$. According to the Planck-Einstein relation, $E=\frac{hc}{\lambda }$. Because energy and wavelength ($\lambda$) are inversely proportional, a smaller energy gap (${\Delta }_o$) means that the complex will absorb a photon of lower energy, which corresponds to a longer wavelength.

This confirms that our complex absorbs light at a longer wavelength compared to the all-nitrogen donor complex. Thus, Option D is correct.

By meticulously applying the principles of spatial geometry and Crystal Field Theory, we have successfully navigated this complex problem. The correct statements are indeed (A), (B), and (D).