Homework question 17.33

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Magdalena Palavecino 1A
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Homework question 17.33

Postby Magdalena Palavecino 1A » Mon Nov 20, 2017 6:30 pm

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?

Lucian1F
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Re: Homework question 17.33

Postby Lucian1F » Mon Nov 20, 2017 10:04 pm

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

Ya Gao
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Re: Homework question 17.33

Postby Ya Gao » Tue Nov 21, 2017 11:58 am

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.

Timothy Kim 1B
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Re: Homework question 17.33

Postby Timothy Kim 1B » Fri Nov 24, 2017 4:20 pm

If the atoms have a formal charge of 0, it will not want to bind with anything, correct?

Charles Ang 1E
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Re: Homework question 17.33

Postby Charles Ang 1E » Sat Nov 25, 2017 5:17 pm

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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