Problem 17.29

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Anna De Schutter - 1A
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Joined: Wed Feb 21, 2018 3:01 am

Problem 17.29

Postby Anna De Schutter - 1A » Mon Jun 04, 2018 10:55 pm

Hi!

In problem 17.29, we need to determine the oxidation number of each metal. In the solutions manual we are given the formula:
(#metal atoms)(oxidation number of the metal) + summation sign(# each ligand)(charge of each ligand)=charge of the ion

I was wondering, how to we determine the charge of each ligand? Do we draw the Lewis structure and then look at the formal charges?

Thank you! :)
Anna De Schutter - section 1A

Isabelle De Rego 1A
Posts: 40
Joined: Fri Apr 06, 2018 11:02 am

Re: Problem 17.29

Postby Isabelle De Rego 1A » Tue Jun 05, 2018 2:40 pm

For determining the charge of the ligand you should calculate the formal charge. The charges of the ligands + the oxidation state of the central atom should equal the overall charge. So if the overall charge is 1+ and you have 2 ligands with a -1 charge, your oxidation state should be 3+. Hope this helped!

Hannah Lee 1B
Posts: 26
Joined: Fri Apr 06, 2018 11:03 am

Re: Problem 17.29

Postby Hannah Lee 1B » Tue Jun 05, 2018 2:52 pm

my TA said we don't need to know that, however I think you're more or less correct and also using the charges of the other atoms/ions in the molecule can be used to find the ligand's charge.

Anna De Schutter - 1A
Posts: 66
Joined: Wed Feb 21, 2018 3:01 am

Re: Problem 17.29

Postby Anna De Schutter - 1A » Tue Jun 05, 2018 3:47 pm

I had my discussion section today, and my TA explained to me how to find the charge of a ligand :)

She told us that we needed to calculate the formal charge of the atom of the ligand bound to the transition metal. So for example for the coordination compound [Co(en)2(Cl)2)]*, we need to calculate the formal charge of N and Cl. The way to do that for N for example would be to give the two electrons of the bond between N and Co to N which would give it one extra lone pair (and N is already bound to two H and one C). As a result of this, the formal charge of N is 0. For Cl, we also give the two electrons of the bond between Cl and Co to Cl which would give it one extra lone pair (on top of the three lone pairs it already has). As a result of this, the formal charge of each Cl is -1. So the oxidation number of Co is +2.

*en refers to NH2(CH2)2NH2

Anna De Schutter - section 1A


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