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

Dec 29, 2015

Chromium $\left(I I I\right)$ complexes are generally octahedral. We simply replace one of the primary ligands for one of the counterions and conserve mass and charge.

#### Explanation:

${\left[C r {\left(N {H}_{3}\right)}_{5} C l\right]}^{2 +} \left(a q\right) + N {O}_{3}^{-} r i g h t \le f t h a r p \infty n s {\left[C r {\left(N {H}_{3}\right)}_{5} \left(O N {O}_{2}\right)\right]}^{2 +} \left(a q\right) + C {l}^{-}$

Alternatively:

${\left[C r {\left(N {H}_{3}\right)}_{5} C l\right]}^{2 +} \left(a q\right) + N {O}_{3}^{-} r i g h t \le f t h a r p \infty n s {\left[C r {\left(N {H}_{3}\right)}_{4} \left(O N {O}_{2}\right) C l\right]}^{+} \left(a q\right) + N {H}_{3} \left(a q\right)$

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.

${\left[C r {\left(N {H}_{3}\right)}_{5} C l\right]}^{2 +} \left(a q\right) + O {H}_{2} r i g h t \le f t h a r p \infty n s {\left[C r {\left(N {H}_{3}\right)}_{5} \left(O {H}_{2}\right)\right]}^{3 +} \left(a q\right) + C {l}^{-}$