# What is the likely charge on the metal ion with a complex whose formula is [M(SO_4)(NH_3)_5]^+Cl^(-)?

Mar 12, 2017

Well because the complex as written has a single chloride counterion, we can write..........

${\left[M \left(S {O}_{4}\right) {\left(N {H}_{3}\right)}_{5}\right]}^{+} C {l}^{-} \rightarrow {\left[M \left(S {O}_{4}\right) {\left(N {H}_{3}\right)}_{5}\right]}^{+} + C {l}^{-}$

#### Explanation:

${\left[M \left(S {O}_{4}\right) {\left(N {H}_{3}\right)}_{5}\right]}^{+} C {l}^{-} \rightarrow {\left[M \left(S {O}_{4}\right) {\left(N {H}_{3}\right)}_{5}\right]}^{+} + C {l}^{-}$

And the ammine ligands may be removed as neutral entities, whose removal causes no change to the charge of the complex.......

${\left[M \left(S {O}_{4}\right) {\left(N {H}_{3}\right)}_{5}\right]}^{+} \rightarrow {\left[M \left(S {O}_{4}\right)\right]}^{+} + 5 N {H}_{3}$

to leave ${\left[M \left(S {O}_{4}\right)\right]}^{+}$; but sulfato as a ligand is formulated as $S {O}_{4}^{2 -} :$

${\left[M \left(S {O}_{4}\right)\right]}^{+} \rightarrow S {O}_{4}^{2 -} + {M}^{3 +}$

So $M$ whatever it was (and it is likely to be a chromium or a cobalt salt) was ${M}^{3 +}$. And thus the oxidation state of the metal centre was $+ I I I$.

Note that all I have done here is to conserve CHARGE, and MASS. All of stoichiometry depends on these conservations.

Capisce?

Note, that looking at this question again, it was clearly formulated by someone who has not formally trained as a chemist. The $X$ symbol is reserved for the halides, and a chemist would have formulated the question of the basis of ${\left[M \left(S {O}_{4}\right) {\left(N {H}_{3}\right)}_{5}\right]}^{+}$.