Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: aficesk8ingal on September 07, 2004, 08:18:55 PM
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??? Hi! I was wondering if any of you could help me with the hybridization of diazomethane. I have a structure set up that is H2C=N=Nwith two pairs of electrons. I know how to find hybridization for the C and middle N by counting the pairs of electrons around it, but I don't know how to find the hybridization of the end N since it has only 1 double bond and 2 lone pairs. I'm confused if the lone pairs effect hybridization, or if the double bond just automatically indicates sp hybridization. Any insights would be amazing! Thanks! ;D
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The double bond doesn't automatically indicate sp. For eg, NO3- has a double bond and it's sp2.
As a guide, any atom with exactly two "groups" (a group is a lone pair of electrons or a bond regardless of bond order ie single double or triple) is sp hybridized. An atom with three groups is sp2 hybridized; one with four groups is sp3 hybridized.
So the end N has a one double bond (1 group) and 2 lone pairs (2 groups). Total is 3 groups, so sp2.
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Demotivator is right. The geometry of the atom will determine the hybridization. The N has 3 groups which define a plane and would have to indicate sp2 hybridized. It is tricky, so I understand your confusion, that is while most Chemist tend to avoid Valence Bond Theory.
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I have a structure set up that is H2C=N=N with two pairs of electrons.
Perhaps I'll describe H2C=N=N in terms of the number of electrons on each atom in the molecule.
Each H contain only a bonding pair (C-H) each.
The C atom contains 4 bonding pairs.
The N attach to C has 4 valence electron. It has donated its 5th electron to the other N atom and inherit a positive charge on it. It exhibits a valency of 4, thus contains 4 bonding pairs.
The last N has 6 valence electrons. It accepted it 6th electron from the othe N atom, thus inherit a negative charge. It exhibits a valency of 2, thus forming double bond with the N atom.
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Yeah, ok but there are two other contributing resonance structures of diazomethane. That means that in reality, the end N does not accept a full electron. In the other structures the end N looks like its sp hybridized and does not accept an electron. It just so happens that the other 2 structures are minor contributors but enough to make tne actual end N hybridization to be sp1.8. :)
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Thank you all so much for your replies! :) It makes a lot more sense now, and I really appreciate your *delete me*!! Thanks again!!!! ;D