Question

Question #9f76a

Answer 1

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, `"Kr"`, which has a total of `36` electrons surrounding its nucleus.

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

`Z` - the atomic number of the transition metal
`"O.N."` - its oxidation number
`n` - the 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, `x`, on the `"O"_2^(x)` ligand.

Your transition metal cation will be surrounded by

• two ammonia molecules, `"NH"_3`
• one ethylenediamine molecule, `"en"`
• one unknown ligand labeled `"O"_2^(x)`

Now, it is important to realize that ethylenediamine and the `"O"_2^x` ligand are 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 `6`, since the cobalt cation will form 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, `y+`, must be equal to the negative charge of the chloride anion, which is `1-`, which is equivalent to saying that

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

Now, ammonia and ethylenediamine are neutral molecules, which means that the overall charge on the `"O"_2^x` ligand 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 `"O"_2^(2-)` anion, also known as the peroxide anion.