Homework 17.33

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Homework 17.33

Postby gracehart » Fri Nov 11, 2016 2:56 pm

Could you explain how to answer this homework question?

17.33 Which of the following ligands can be polydentate? If the ligand can be polydentate, give the maximum number of places on the ligand that can bind simultaneously to a single metal center: (a) chloride ion; (b) cyanide ion; (c) ethylenediaminetetraacetate; (d) N(CH2CH2NH2)3.

Thank you!

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Re: Homework 17.33

Postby NinaSheridan » Mon Nov 14, 2016 7:28 am

This is related, if someone can explain this, can they also explain why oxalato is only bidentate?

Omer Lavian 2K
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Re: Homework 17.33

Postby Omer Lavian 2K » Mon Nov 14, 2016 11:01 am

Nina, to answer your question, and maybe this will help with the original question as well, for a molecule to be polydentate, it needs to be able to contribute multiple lone pairs at the same time to a central metal atom. This means that a) there must be at least two atoms with available lone pairs on the ligand and b) the geometry of the molecule must be such that it is conducive to binding to a metal atom at multiple sites. The ideal geometry for a ligand to be polydentate is an atom with an available lone pair followed by two "spacer" atoms with no lone pairs followed by another atom with an available lone pair. Oxalato follows this model exactly. It generally donates lone pairs of electrons from its oxygen atoms possessing a -1 formal charge. These atoms are separated by two "spacer" carbon atoms, making the geometry of the molecule ideal for being a bidentate ligand. Therefore, oxalato is more often than not a bidentate ligand. Using these ideas, you can draw Lewis structures and see whether a molecule is polydentate and if so how many times it can bind to a central metal atom.

Also, certain ligands like sulfato or carbonato have lone pairs separated by one spacer atom, making them generally monodentate but occasionally bidentate given a small enough central atom.

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