Homework question 17.33
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Homework question 17.33
Hi, can someone please explain how to solve for questions b and c of this question? Why is b) a mono or bidentate when there are so many free electron pairs, and why in is it a monodentate in part c) if there are also multiple free electron pairs that could form bonds?
Re: Homework question 17.33
You need to take into account the molecular shapes. It needs to be possible for multiple electron pairs to bind to the transition metal with the molecular shape of the ligand
Re: Homework question 17.33
For ion CO3 2-, I think it is bidentate because if you draw the lewis structure of it, you will see that two out of the three oxygen have lone pairs around them and do not have a formal charge of 0. So for these two oxygen atoms, they would want to form bonds.
For water molecule, if you draw out the lewis structure, you'll see that there are lone pairs around the central oxygen atom, but even though there are two lone pairs around it, it can only form one bond as a ligand.
Hope it helps.
For water molecule, if you draw out the lewis structure, you'll see that there are lone pairs around the central oxygen atom, but even though there are two lone pairs around it, it can only form one bond as a ligand.
Hope it helps.
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Re: Homework question 17.33
If the atoms have a formal charge of 0, it will not want to bind with anything, correct?
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Re: Homework question 17.33
Timothy Kim 1F wrote:If the atoms have a formal charge of 0, it will not want to bind with anything, correct?
I may be mistaken, but I think that the potential to bind is not caused by the formal charge, but rather the presence of lone pairs on an atom.
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