Given the complex #[Cr(NH_3)_5X]^(2+)#, how would formulate its substitution reactions?

1 Answer
Dec 29, 2015

Answer:

Chromium #(III)# complexes are generally octahedral. We simply replace one of the primary ligands for one of the counterions and conserve mass and charge.

Explanation:

#[Cr(NH_3)_5Cl]^(2+)(aq) + NO_3^(-) rightleftharpoons [Cr(NH_3)_5(ONO_2)]^(2+)(aq) + Cl^-#

Alternatively:

#[Cr(NH_3)_5Cl]^(2+)(aq) + NO_3^(-) rightleftharpoons [Cr(NH_3)_4(ONO_2)Cl]^(+)(aq) + NH_3(aq)#

These equilibria would be going on all the time in aqueous solution; including the replacement of ammine or chloro ligands for water (below). What crystallizes out is the least soluble species; whatever this is.

#[Cr(NH_3)_5Cl]^(2+)(aq) + OH_2 rightleftharpoons [Cr(NH_3)_5(OH_2)]^(3+)(aq) + Cl^-#