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HW 17.33

Posted: Sun Jun 10, 2018 11:53 pm
by Anna De Schutter - 1A
Hi!

I'm a bit confused about the answers given in the solutions manual regarding problem 17.33. For part a, we consider HN(CH2CH2NH2)2 and the solutions manual says it is tridentate. For part b however (CO3 2-), the solutions manual says it can be monodentate or bidentate. I understand why CO3 2- can be monodentate or bidentate, but doesn't that mean that we can say that HN(CH2CH2NH2)2 is monodentate, bidentate or tridentate as well? Similarly for part d (oxalate, C2O4 2-), wouldn't we be able to say it is monodentate or bidentate instead of just bidentate (as written in the solutions manual)? Or am I understanding this wrongly?

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

Re: HW 17.33

Posted: Mon Jun 11, 2018 4:17 pm
by Tarek Abushamma
I think it could be monodentate or bidentate as well, it would depend on which would produce the most stable structure.

Re: HW 17.33

Posted: Mon Jun 11, 2018 6:05 pm
by Chem_Mod
Almost any ligand can be monodentate, but for some it is so rare that we only consider them to be polydentate. CO32– sometimes only binds once because there is only one atom between the two atoms with lone pairs. This sometimes causes more strain if both atoms bind to the metal. Fortunately, the question only asks which ligands can be polydentate; you do not need to determine whether they are commonly monodentate.

Re: HW 17.33

Posted: Tue Jun 12, 2018 9:57 am
by Anna De Schutter - 1A
Thank you! :)

Re: HW 17.33

Posted: Tue Jun 12, 2018 10:39 am
by Emma Leshan 1B
Ligands are more likely to be polydentate if there are many atoms between the ones with lone pairs. This allows the molecule to bend, interacting with the metal in more than one place. Also, I think one reason CO3 2- can't be tridentate is due to its double bond, which would make it harder for the molecule to bend.

Re: HW 17.33

Posted: Wed Jun 13, 2018 12:53 am
by nikitasridhar_1b
can't co32- technically have 8 binding sites since two of the oxygen atoms have 3 pairs of electrons and one has 2 pairs of electrons?

Re: HW 17.33

Posted: Thu Jun 14, 2018 10:30 am
by Anna De Schutter - 1A
No the lone pairs on the double bonded O won't bind with the metal because that would give O three bonds (one double with C and one single with the metal) and one lone pair, and thus a formal charge of :
FC=6-2-3=+1 which is less stable than it was when O was just bound to the C with a double bond and two lone pairs --> FC=0 in that case.

Also, only one lone pair of the single bonded O's will bind with a metal because that would give the O's a formal charge of 0 (two single bonds, one with C and one with the metal + 2 lone pairs). If more than one of the lone pairs binds to the metal, the formal charge won't be 0 anymore, thus unfavorable.

I hope this helps! :)
Anna De Schutter - section 1A