Animated Solution for Chemistry - Coordination Compounds: Which of the following facts about the complex $[Cr(N{H}_3{)}_6]C{l}_3$ is wrong?

Have you ever looked at a coordination complex and wondered what secrets it holds within its brackets? Today, we are going to dissect the hexaamminechromium(III) chloride complex, $[Cr(N{H}_3{)}_6]C{l}_3$, piece by piece. This isn't just about finding the wrong statement; it's about understanding the beautiful dance of electrons and orbitals that gives this molecule its unique properties. Let's dive in!

When we look at the formula $[Cr(N{H}_3{)}_6]C{l}_3$, the first thing we need to do is separate the inside from the outside. The square brackets enclose the coordination sphere, which acts as a single, unbreakable unit in aqueous solutions. The chloride ions outside the brackets, however, are free to roam.

When this complex dissolves in water, it ionizes to give one $[Cr(N{H}_3{)}_6{]}^{3+}$ cation and three $C{l}^{-}$ anions.

Now, what happens if we introduce silver nitrate ($AgN{O}_3$) to this solution? Silver ions ($A{g}^{+}$) have a strong affinity for chloride ions ($C{l}^{-}$). Since we have three free chloride ions swimming around, they will immediately react with the silver ions to form a white precipitate of silver chloride ($AgCl$).

This confirms that the complex does indeed give a white precipitate with silver nitrate solution. So, the statement in option (d) is a perfectly correct fact!

Let's shift our focus to the central metal ion inside the coordination sphere. Ammonia ($N{H}_3$) is a neutral ligand, which means the entire $+3$ charge of the complex ion comes directly from the chromium atom.

A neutral chromium atom has the electronic configuration $[Ar]3d^{5}4s^1$. To form the $C{r}^{3+}$ ion, it loses three electrons—one from the $4s$ orbital and two from the $3d$ orbitals. This leaves us with:

Imagine those three electrons sitting in the five available $3d$ orbitals. According to Hund's rule, they will occupy three separate orbitals, leaving two $3d$ orbitals completely empty.

Now, six ammonia ligands are approaching, each carrying a lone pair of electrons to donate. They need six empty orbitals to form coordinate covalent bonds. Where will they go?

They will occupy the two empty $3d$ orbitals, the one empty $4s$ orbital, and the three empty $4p$ orbitals. This specific mixing of orbitals is known as $d^{2}s{p}^3$ hybridization.

Because the hybridization involves six orbitals, the resulting geometry is perfectly octahedral. This makes the statement in option (a) absolutely correct!

Here is where the catch lies. Notice that the complex used the inner $3d$ orbitals for its hybridization, rather than the outer $4d$ orbitals. Whenever a complex utilizes its inner $(n-1)d$ orbitals, we classify it as an inner orbital complex.

Option (c) claims that it is an outer orbital complex, which is a blatant lie! This is the wrong statement we were looking for.

But before we wrap up, let's quickly check option (b). If we look back at our $3d$ orbitals, we still have those three unpaired electrons sitting there. The presence of unpaired electrons means the complex will be weakly attracted to a magnetic field, making it paramagnetic. So, option (b) is also a correct fact.

By systematically breaking down the complex, we didn't just find the answer; we understood the why behind every single property. Keep this analytical mindset, and no coordination compound will ever intimidate you!