Animated Solution for Chemistry - Coordination Compounds: The oxidation states of Cr, in $[Cr(H_{2}O{)}_6]C{l}_3$, $[Cr(C_6{H}_6{)}_2]$, and $K_2[Cr(CN{)}_2(O{)}_2(O_2)(N{H}_3)]$ respectively are

When dealing with coordination compounds, finding the oxidation state of the central metal ion is like solving a simple algebraic puzzle. The fundamental principle you must always remember is this: The sum of the oxidation states of all atoms (or ligands) in a neutral molecule is always equal to zero. If the complex is an ion, the sum equals the net charge of the ion.

Let's apply this golden rule to three fascinating chromium complexes and decode the oxidation state of chromium ($Cr$) in each one.

The Classic Aqua Complex

Our first candidate is $[Cr(H_{2}O{)}_6]C{l}_3$. Let's break it down into its constituent parts to find the oxidation state of chromium, which we will call $x$.

Inside the coordination sphere, we have six water ($H_{2}O$) molecules. Water is a neutral ligand, meaning it contributes a charge of $0$. Outside the coordination sphere, we have three chloride ($C{l}^{-}$) ions acting as counter ions. Each chloride ion carries a charge of $-1$.

Setting up our algebraic equation based on the golden rule:

Simplifying this, we get:

So, in this classic aqua complex, chromium exists in a $+3$ oxidation state.

The Zero-Valent Organometallic

Next up is $[Cr(C_6{H}_6{)}_2]$, known as bis(benzene)chromium. This is a beautiful example of an organometallic sandwich compound.

Here, the ligands are benzene ($C_6{H}_6$) rings. Benzene is a stable, neutral molecule, so its charge contribution is $0$. There are no counter ions present.

Let's set up the equation:

This might seem surprising at first, but yes, metals can absolutely have a zero oxidation state when they are coordinated exclusively to neutral ligands!

The Ligand Jungle

Finally, we face the most intimidating complex of the trio: $K_2[Cr(CN{)}_2(O{)}_2(O_2)(N{H}_3)]$. Don't let the long formula scare you; we just need to identify each piece carefully.

- Potassium ($K$): An alkali metal, always $+1$. We have two of them. - Cyanide ($C{N}^{-}$): A classic anionic ligand with a $-1$ charge. We have two. - Oxo ($(O{)}_2$): The formula writes $(O{)}_2$ to indicate two separate oxygen atoms acting as oxo ligands. Each oxo ligand ($O^{2-}$) carries a $-2$ charge. - Peroxo ($(O_2)$): The $(O_2)$ in parentheses represents a single diatomic peroxo group ($O_2^{2-}$), which carries an overall $-2$ charge. - Ammine ($N{H}_3$): Ammonia is a neutral ligand, so its charge is $0$.

Now, let's carefully plug all these values into our master equation:

Let's simplify the terms:

The $+2$ and $-2$ cancel each other out:

In this complex, chromium is pushed to its maximum common oxidation state of $+6$.

By systematically breaking down each complex, we found the oxidation states of chromium to be $+3$, $0$, and $+6$ respectively. This perfectly matches option (c). Mastering the charges of common ligands is the key to effortlessly solving these types of problems!