ACHIEVE HW Week 9 #5
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ACHIEVE HW Week 9 #5
Hello, I was wondering if anyone could help me find the coordination number for this metal species, [Co(en)2(CO)2]Br I know to ignore what is outside of the brackets so to focus on what is bounded to Co, so I am confused why it would not be 4.
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Re: ACHIEVE HW Week 9 #5
The answer is 6, not 4 because (en) stands for ethylenediamine which is a bidentate ligand. In other words, you have it by 2 since it donates 2 electron pairs rather than one. So 2*2+2 would give you 6 instead of 4.
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Re: ACHIEVE HW Week 9 #5
Hi!
The coordination compound in the homework was [Cd(en)Br2]. In this case, the coordination number would be 4 because there are two bromines and an ethylenediamine attached to the cadmium. Bromine is a monodentate, meaning it bonds to the metal at only one site. Ethylenediamine is a bidentate, meaning it bonds to the metal at two sites. As a result, because there are two monodentates and one bidentate, the coordination number is 4.
In your case, for [Co(en)2(CO)2]Br, there are two monodentates (CO) and two bidentates (en). So, I believe the coordination number would be 6 because there are 6 sites at which ligands bind to the transition metal.
Hope that helps.
The coordination compound in the homework was [Cd(en)Br2]. In this case, the coordination number would be 4 because there are two bromines and an ethylenediamine attached to the cadmium. Bromine is a monodentate, meaning it bonds to the metal at only one site. Ethylenediamine is a bidentate, meaning it bonds to the metal at two sites. As a result, because there are two monodentates and one bidentate, the coordination number is 4.
In your case, for [Co(en)2(CO)2]Br, there are two monodentates (CO) and two bidentates (en). So, I believe the coordination number would be 6 because there are 6 sites at which ligands bind to the transition metal.
Hope that helps.
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Re: ACHIEVE HW Week 9 #5
The en is ethylenediamine (H2NCH2CH2NH2). This is bidentate, so it has two donor atoms. So these add 2 to the coordination number plus the 2 C and the 2 O, and you get 6 as the coordination number.
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Re: ACHIEVE HW Week 9 #5
Hi! The answer is 6 and not 4 because en is a bidentate ligand, therefore you would add 2 to your coordination number. I would memorize the ones like this that he went over in lecture.
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Re: ACHIEVE HW Week 9 #5
it's because the (en) is a bidentate ligand so both nitrogens bond to the metal so it by itself would have 4 bonding parts and then you still need to add the 2 from the CO, so the total would be 6.hope that helps
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Re: ACHIEVE HW Week 9 #5
So anytime you see "en" in a coordination compound you should remember that it stands for a bidentate ligand and add two to the coordination number if that's what the question is asking for.
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Re: ACHIEVE HW Week 9 #5
One (en) contains two NH2, so therefore Co binds to two of the two NH2 and two CO, so the coordination number is 2 x 2 + 2, which is 6.
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