# Question #8b3fa

May 25, 2015

In the case of $\left[C r {\left(N {H}_{3}\right)}_{6}\right] \left[C r {\left(C N\right)}_{6}\right]$, the first thing you need to notice is that neither the cation, nor the anion, have a sbubscript.

This means that the charges on the two complex ions will be equal. You can thus break up the coordination complex into a cation and an anion

${\left[C r {\left(N {H}_{3}\right)}_{6}\right]}^{\textcolor{red}{x +}}$ $\to$ cation

${\left[C r {\left(C N\right)}_{6}\right]}^{\textcolor{b l u e}{x -}}$ $\to$ anion.

Notice that the anion contains the cyanide anion, $C {N}^{-}$, which has an overall charge of -1. This means that the oxidation state of the metal plus the total negative charge of the six cyanide ions must be equal to -x, the overall charge of the anion.

Let $c$ be the charge of the metal. This means that

$c + 6 \cdot \left(- 1\right) = - x$ $\text{ } \textcolor{g r e e n}{\left(1\right)}$

Now look at the cation. Since ammonia is a neutral compound, the overall charge of the cation must be equal to the charge of the metal. This means that you have

$c = x$ $\text{ } \textcolor{g r e e n}{\left(2\right)}$

Use equations $\textcolor{g r e e n}{\left(1\right)}$ and $\textcolor{g r e e n}{\left(2\right)}$ to get the identity of $c$

$\left\{\begin{matrix}c - 6 = - x \\ c = x\end{matrix}\right.$ $\implies$ $x - 6 = - x \implies x = 3$

Therefore, the charge of the metal will be +3 and the two complex ions will be

${\left[C r {\left(N {H}_{3}\right)}_{6}\right]}^{3 +}$ and ${\left[C r {\left(C N\right)}_{6}\right]}^{3 -}$

You can use the same approach for the second coordination complex. This time, however, the two complex ions do have subscripts. This means that you have

${\left[C o {\left(e n\right)}_{3}\right]}_{\textcolor{red}{4}} {\left[M n {\left(C N\right)}_{5} I\right]}_{\textcolor{b l u e}{3}}$

${\left[C o {\left(e n\right)}_{3}\right]}^{\textcolor{b l u e}{3 +}}$ $\to$ cation

${\left[M n {\left(C N\right)}_{5} I\right]}^{\textcolor{red}{4 -}}$ $\to$ anion

So, ethylenediamine is a neutral compound, so the oxidation state of cobalt will match the overall charge of the complex ion. Thus, you're dealing with the $C {o}^{3 +}$ cation.

Once again, use a simple equation to find the charge of the manganese ion. This time, in addition to the negative charge that's coming from the cyanide anion, you get an extra -1 charge from the iodide anion, ${I}^{-}$. This means that you have ($c$ is the charge on the manganese ion)

$c + 5 \cdot \left(- 1\right) + \left(- 1\right) = - 4$

$c - 6 = - 4 \implies c = + 2$

Therefore, you're dealing with the $M {n}^{2 +}$ cation.