Finding Oxidation State
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Finding Oxidation State
For Week 9 HW, question #6 asks to give the oxidation state of the metal species in the complex [Co(NH3)4Cl2]Cl. I calculated +2, but it is incorrect. It says my "answer would be correct if the complex were neutral. Remember to account for the counterion charge." Could someone explain?
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Re: Finding Oxidation State
In the coordination compound [Co(NH3)4Cl2]Cl the two species are [Co(NH3)4Cl2] (the complex) and Cl (the counterion). The coordination compound's overall charge is 0, and since you know that the Cl counterion has a charge of -1, the [Co(NH3)4Cl2] must have a charge of +1. The two Cl's in [Co(NH3)4Cl2] have a total charge of -2, so the Co must have a charge of +3 to have the net charge be +1. Hope this helps!
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Re: Finding Oxidation State
You calculated the correct charge if the coordination compound was neutral, as the hint said, however, because there is a counterion of Cl present, which has a charge of -1, [Co(NH3)4Cl2] must therefore have a charge of +1. This means that whatever number you calculated for the oxidation number of Co in the neutral compound just needs to be increased by +1, giving the final answer of +3.
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Re: Finding Oxidation State
Since the Cl outside the brackets has a charge of -1, we know everything inside the brackets has to have a charge of +1.
To solve this,
we see that there are two Cls in the brackets so that gives a -2 charge. The (NH3)4 is neutral, meaning it has a charge of 0. This means that only the Co is left. To get the total charge to +1, we need to do x -2=1. We know here x=3, so the oxidation state for cobalt is 3.
I hope this helped.
To solve this,
we see that there are two Cls in the brackets so that gives a -2 charge. The (NH3)4 is neutral, meaning it has a charge of 0. This means that only the Co is left. To get the total charge to +1, we need to do x -2=1. We know here x=3, so the oxidation state for cobalt is 3.
I hope this helped.
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