## Homework Problem 17.33.)

Jesus A Cuevas - 1E
Posts: 29
Joined: Fri Apr 06, 2018 11:02 am

### Homework Problem 17.33.)

Could someone help me with the following problem, thanks in advance!

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)HN(CH2CH2NH2)2; (b)CO3 2- (c)H2O (d)Oxalate (C2O4-2)

sharonvivianv
Posts: 63
Joined: Fri Apr 06, 2018 11:05 am

### Re: Homework Problem 17.33.)

a) tridentate, you can see that there are three nitrogens meaning there are three lone pairs.
b) bidentate, usually if there is a charge on an oxygen, a metal will be able to bind. There is a minus 2 charge.
c) monodentate, honestly I've asked around and no one has been able to give a good explanation as to why H20 is monodentate so you should just memorize that
d) bidentate, again there is a minus 2 charge.

These aren't set answers that can help you reason through every problem like this because there are always exceptions. Even the UAs have trouble telling sometimes but sometimes the charge on the oxygens and the lone pairs on the nitrogen are helpful while identifying them as polydentate.

LilianKhosravi_1H
Posts: 32
Joined: Fri Apr 06, 2018 11:03 am

### Re: Homework Problem 17.33.)

H2O is a monodentate because even though it has two lone pairs, only one of those will bind and the other one won't because it has no way of biding to the central atom since it's on the oxygen with the other lone pair that is already bound. Basically, the positions of the lone pairs won't allow it to bind to the central atom twice.

nikitasridhar_1b
Posts: 42
Joined: Fri Apr 06, 2018 11:05 am

### Re: Homework Problem 17.33.)

what is the relation btwn charge and number of binding sites? i thought the no. of binding sites had to do wiht the number of lone pairs

Chem_Mod
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### Re: Homework Problem 17.33.)

Yes, number of atoms with lone pairs that can bind is what matters. Sometimes the charge may have the same value.