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I don't understand this question. 

 

Give the formula of the complex ion formed between

 

a. iron (III) ions and six cyanide ions

b. copper (II) ions and four chloride ions

c. cobalt (III) ions and three 1,2-ethanediamine bidentate ligands. 

 

I don't think I have learned this yet... but it was hw. 

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I'll be honest - I have no idea what c. is talking about. Never seen ethanediamine bidentate before. I can't even break down those words. Maybe, maybe once I cover organic chemistry I'll recognize it?

 

Anyway, for a and b, I think it is simply asking for a balanced chemical formula. You should be able to draw the way those ions bond (keeping in mind that the metals are ligands).

 

b., for example, is copper tetrachloride. Complex ion suggests ligands - a topic 13 subject. Thornley usually explains things pretty well, although this is one of his strangest videos: https://www.youtube.com/watch?v=9_5wd_m8N3M

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Hiya, 

 

a would be [Fe(CN)6]3-

b would be [CuCl4]2-

c would be [Co(C2H4(NH2)2))]3+ 

 

You're basically supposed to put the cation and the anion next to each other, and work out the overall charge. So for Fe3+ and 6 cyanide

I know that Fe is 3+ and CN is -, so Fe(CN)6 is (3) + (-6) = -3

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c is [Co(C2H4(NH2)2)3]3+

A bidentate ligand has 2 lone pairs of electrons to form 2 coordinate bonds with the Co3+ ion.

[Co(C2H4(NH2)3)]3+ is a octahedral complex ion with a coordination number of 6 (it has 6 coordinate bonds (2 per ligand in this case).

Here is a video that covers the structure of a complex ion;

http://youtu.be/HiVH7L4vi2I

 

Complete playlist for topic 3, new syllabus:

https://www.youtube.com/playlist?list=PLluIsqNl4jcr7cYXf4590-WBw9GqEcv7m

Edited by Msj Chem
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