# What is the chemical formula for diamminedichloroethylenediaminecobalt(III) bromide? What are its nine isomers?

May 25, 2015

The chemical formula for diamminedichloroethylenediaminecobalt(III) bromide is ["CoCl"_2("en")("NH"_3)_2]"Br"  , where (en) is the symbol for ethylenediamine, $\text{H"_2"N-CH"_2"-CH"_2"-NH"_2}$. I suppose it could be written as $\text{CoC"_2"H"_14"N"_4"Cl"_2"Br}$, but I didn't see that in my research. The name literally means ["two ammonia"-"two chlorine"-"ethylene two amine"-"cobalt (III)"] bromide.

I found four isomers of the diamminedichloroethylenediaminecobalt(III) ion on another pdf file at http://bchsapchemistry.wikispaces.com/file/detail/Chapter-24.Lectures+%2815%29.pdf . They are on page 5. You would just need to add the $\text{Br"^-}$ ion. The formula is written in a different order as "[Co(NH"_3)_2"Cl"_2("en")]^+, which I think is a better way to write it, though this was the only place I saw it written that way.

May 25, 2015

I'm not completely sure about the isomers, so I'll just help you with the formula now and maybe update the answer tomorrow after I'm sure I have all the 9 isomers you mentioned.

So, you're dealing with a coodination complex, which consists of a central metal atom, in your case cobalt, $C o$, surrounded by a number of ligands.

Notice that the name of the compound contains the cobalt (III) ion, or $C {o}^{3 +}$. This will help you confirm how many ligands surround the central cobalt atom.

The electron configuration of cobalt looks like this

"Co": ["Ar"] 4s^(2) 3d^(7)

Since cobalt is a transition metal, all the electrons located outside its noble gas core are considered to be valence electrons. The electron configuration of the $C {o}^{3 +}$ cation will be

"Co"^(3+): ["Ar"] 3d^(6)

A low spin complex such as this one will have these 6 valence electrons paired in three of the five 3d-orbitals, and use the remaining two 3d-orbitals, its 4s-orbital, and the three 4p-orbitals to form 6 ${d}^{2} s {p}^{3}$ hybrid orbitals.

These hybrid orbitals will accomodate the electron pairs that come from the ligands.

So, you need to add six ligands to your central cobalt atom. According to the written formula, these ligands will be

• Two chlorine atoms - this comes from dichloro
• Two ammine functional groups - this comes from diammine
• One ethylenediamine (abbreviated en).

You only need one ethylenediamine because its structure looks like this

The two ${\text{-NH}}_{2}$ groups will each bind to one of cobalt's six available electrons. Since the compound is a bromide, the charge on the coordination complex will be +1, in order to balance the -1 charge of the bromide anion, $B {r}^{-}$.

So, if you put all this together you'll get

$\left[C o {\underbrace{C {l}_{2}}}_{\textcolor{b l u e}{\text{dichloro")) (en) overbrace((NH_3)_2)^(color(green)("diammine}}}\right] B r$

Here's how one isomer would look like