Reaction Process

This question asks about the products formed when the complex ion [Co(NH3)5Cl]$^{2+}$ reacts with [Cr(OH2)6]$^{2+}$ in an acidic solution indicated by the presence of H3O$^{+}$. This is a typical inorganic redox reaction involving coordination complexes.

Metal Oxidation States

First, let's determine the oxidation states of the metal ions in the reactant complexes:

• In [Co(NH3)5Cl]$^{2+}$, the neutral ligand is NH3 (charge 0) and Cl has a charge of -1. Let the oxidation state of Co be $x$. So, $x + 5(0) + (-1) = +2$. This gives $x = +3$. The cobalt is in the +3 oxidation state, Co(III).
• In [Cr(OH2)6]$^{2+}$, the neutral ligand is H2O (charge 0). Let the oxidation state of Cr be $y$. So, $y + 6(0) = +2$. This gives $y = +2$. The chromium is in the +2 oxidation state, Cr(II).

We have Co(III) reacting with Cr(II). Co(III) is often an oxidizing agent, and Cr(II) is a strong reducing agent. Therefore, an electron transfer is expected where Co(III) is reduced to Co(II) and Cr(II) is oxidized to Cr(III). After the reaction, we expect the oxidation states of the metals in the products to be Co(II) and Cr(III).

Inner-Sphere Mechanism

Reactions between metal complexes like this often proceed via an inner-sphere electron transfer mechanism, especially when a bridging ligand is available. In the complex [Co(NH3)5Cl]$^{2+}$, the chloride ion (Cl$^{-}$) can act as a bridging ligand. The Cr(II) complex, [Cr(OH2)6]$^{2+}$, is labile, meaning its ligands can be easily exchanged. The Co(III) complex, [Co(NH3)5Cl]$^{2+}$, is inert. In an inner-sphere mechanism:

1. The labile Cr(II) complex replaces one of its water ligands with the chloride ligand from the Co(III) complex, forming a bridged intermediate: $[(NH3)5Co^{\text{III}}\text{-Cl-Cr}^{\text{II}}(OH2)5]^{4+}$.
2. An electron is transferred through the bridging chloride ligand from Cr(II) to Co(III), resulting in $[(NH3)5Co^{\text{II}}\text{-Cl-Cr}^{\text{III}}(OH2)5]^{4+}$.
3. The bridged intermediate then cleaves. Cleavage typically occurs on the Cr-Cl bond.

Final Products

When the bridge breaks at the Cr-Cl bond, we get:

• A Co(II) species: Initially, it is $[Co(NH3)5Cl]^{2+}$, but Co(II) complexes are very labile. In the acidic aqueous medium, the NH3 and Cl ligands are rapidly replaced by water molecules. The NH3 ligands are protonated by H3O$^{+}$ to form NH4$^{+}$. The final stable Co(II) species in water is the aqua complex, [Co(OH2)6]$^{2+}$.
• A Cr(III) species: The complex $[Cr(OH2)5Cl]^{2+}$ is formed. Cr(III) complexes are relatively inert, so this complex is stable and does not readily exchange its ligands.

Considering the overall reaction in acidic solution which also involves 5H3O$^{+}$ leading to the formation of 5NH4$^{+}$ ions (as per the balanced equation stoichiometry often associated with this reaction), the complete replacement of NH3 ligands by water on the Co center is expected. Thus, the products A and B are [Co(OH2)6]$^{2+}$ and [Cr(OH2)5Cl]$^{2+}$, respectively. The balanced chemical equation is: $[Co(NH3)5Cl]^{2+} + [Cr(OH2)6]^{2+} + 5H3O^+ \rightarrow [Co(OH2)6]^{2+} + [Cr(OH2)5Cl]^{2+} + 5NH4^+$ The products A and B correspond to $[Co(OH2)6]^{2+}$ and $[Cr(OH2)5Cl]^{2+}$.