Question

Question #8b3fa

Answer 1

In the case of `[Cr(NH_3)_6][Cr(CN)_6]`, 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

`[Cr(NH_3)_6]^(color(red)(x+))` `->` cation

`[Cr(CN)_6]^(color(blue)(x-))` `->` anion.

Notice that the anion contains the cyanide anion, `CN^(-)`, 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 * (-1) = -x` `" "color(green)((1))`

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` `" "color(green)((2))`

Use equations `color(green)((1))` and `color(green)((2))` to get the identity of `c`

`{ (c -6 = -x), (c=x) :}` `=>` `x-6 = -x => x =3`

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

`[Cr(NH_3)_6]^(3+)` and `[Cr(CN)_6]^(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

`[Co(en)_3]_color(red)(4)[Mn(CN)_5I]_color(blue)(3)`

`[Co(en)_3]^(color(blue)(3+))` `->` cation

`[Mn(CN)_5I]^(color(red)(4-))` `->` 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 `Co^(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 * (-1) + (-1) = -4`

`c -6 = -4 => c = +2`

Therefore, you're dealing with the `Mn^(2+)` cation.