# Question #9f76a

##### 1 Answer

The peroxide anion.

#### Explanation:

The idea here is that *transition metals* can obtain a stable **noble gas configuration** by forming *coordination complexes*.

For a given transition metal, its stable noble gas configuration will correspond to the electron configuration of the *noble gas* with which it **shares the period** in the periodic table.

Cobalt is located in period 4, so its noble gas electron configuration will match that of *krypton*,

So, the **effective atomic number** for a transition metal in a coordination complex is calculated like this

#color(blue)("EAN" = Z - "O.N." + n xx 2)" "# , where

**coordination number** of the metal cation

Now, your goal here is to use this equation to find the oxidation state of the transition metal. This will then allow you to find the charge,

Your transition metal cation will be surrounded by

,twoammonia molecules#"NH"_3# ,oneethylenediamine molecule#"en"# oneunknown ligand labeled#"O"_2^(x)#

Now, it is important to realize that ethylenediamine and the **bidentate ligands**, which means that they donate **two pairs of electrons** to the metal cation, not just one pair like you have for the *unidentate ligands*.

This means that the **coordination number** of the metal cation, which essentially tells you how many **atoms** are donating electrons to the metal cation, will be equal to **six** dative covalent bonds

#2 xx "NH"_3# #2 xx "en"# #2 xx "O"_2^(x)#

So, plug this into the equation for EAN and find the oxidation state of the metal

#"EAN" = 27 - "O.N." + 6 xx 2#

#"O.N." = 27 + 12 - 36 = +3#

Now, notice that you're dealing with a **neutral** coordination compound made up of

#["Co"("NH"_3)_2"O"_2^(x)("en")]^(y+)" "# and#" ""Cl"^(-)#

This tells you that the *overall charge* of the complex ion, **must be equal** to the negative charge of the chloride anion, which is

#(y+) + (1-) = 0 implies y = 1+#

Now, ammonia and ethylenediamine are **neutral molecules**, which means that the **overall charge** on the **must be equal to**

#"O.N." + overbrace((1 xx x))^(color(purple)("one O"_2color(white)(a)"ligand")) + overbrace(1 xx (1-))^(color(blue)("one Cl"^(-)"anion")) = 0#

This is equivalent to

#x = 1 - (+3) = -2#

Therefore, you're dealing with the **peroxide anion**.