Problem 17.33

[Anna De Schutter - 1A]

Hi!

I'm a bit confused on how to do problem 17.33. For 17.33 a for example, I thought HN(CH2CH2NH2)2 would be tridentate (which it is!) because there are three available lone pairs, one on each nitrogen. But then for 17.33 c, where we are given H2O, the solutions manual says it is monodentate. I just thought that because oxygen has two lone pairs in this situation that it would be bidentate. Could someone explain to me how I can find out the maximum number of places on the ligand that can bind simultaneously to a single metal center?

Thank you so much!
Anna De Schutter - section 1A

[Tiffany Chen 1A]

Since the 2 lone pairs of H2O are on the same O atom, H2O can't bind to 2 different sites on a central metal atom. Therefore, H2O can bind to a maximum of only 1 site on a central metal atom, making it monodentate.

[Anna De Schutter - 1A]

Thank you!! :)