(a) Compare the bond enthalpies (Table 8.4) of the...

(a) Compare the bond enthalpies (Table 8.4) of the carbon-carbon single, double, and triple bonds to deduce an average $\pi$-bond contribution to the enthalpy. What fraction of a single bond does this quantity represent? (b) Make a similar comparison of nitrogen-nitrogen bonds. What do you observe? (c) Write Lewis structures of $\mathrm{N}_{2} \mathrm{H}_{4}, \mathrm{~N}_{2} \mathrm{H}_{2}$, and $\mathrm{N}_{2}$, and determine the hybridization around nitrogen in each case. (d) Propose a reason for the large difference in your observations of parts (a) and (b).

Key Concepts

Lewis Structures

Lewis structures are diagrams that represent the valence electron arrangement in molecules. They show bonding electrons as lines and non-bonding electrons as dots, providing a simple visual method to understand the connectivity of atoms, the presence of multiple bonds, and the potential for lone pair repulsions that can influence molecular shape and reactivity.

Lone Pair Repulsion Effects

Lone pair repulsion is a key factor in determining bond energies and molecular geometries. In comparing bonds such as nitrogen-nitrogen versus carbon-carbon, differences in lone pair interactions can cause variations in bond strength and stability. This concept explains observations like the distinct energetic behavior of multiple bonds involving elements with different numbers of lone pairs.

Molecular Hybridization

Molecular hybridization is the process by which atomic orbitals mix to form new hybrid orbitals, which then dictate the geometry and bonding properties of a molecule. By analyzing the bonding and electron pair distribution in molecules, one can determine whether atoms adopt sp, sp2, sp3, or other hybridization states, aiding in the prediction of molecular structure and bond angles.

? Bond Contribution

The ? bond contribution refers to the additional stabilization that arises from the presence of pi bonds in a double or triple bond, above the base stabilization provided by a sigma bond. By comparing bond enthalpies of single, double, and triple bonds, one can estimate the average contribution of a pi bond and assess what fraction of a single bond’s energy this amount represents.

Sigma and Pi Bonds

Covalent bonds are often composed of a sigma (?) bond and one or more pi (?) bonds. The sigma bond forms from the head-on overlap of orbitals and is usually the strongest and most stable component, while pi bonds result from the lateral overlap of p-orbitals. In multiple bonds, the additional pi bonds contribute extra bonding interactions, but typically with different energetic significance compared to sigma bonds.

Bond Enthalpy

Bond enthalpy quantifies the energy required to break a chemical bond in a molecule. It is a measure of bond strength and is determined by the difference in energy between bonded atoms and the separate atoms. Comparisons of bond enthalpies provide insight into how different bonding types (single, double, triple) contribute to molecular stability.

Transcript

00:01 I thought this was a little strange.

00:02 I'm not exactly sure i did what they were looking for, but i got answers for everything.

00:08 In this problem, we're asked to compare bond entopies for carbon, and we're looking at the carbon single, the carbon, carbon double, and the carbon -carbon -triple bonds, to deduce the average pie bond contribution.

00:42 And i did, and what fraction of a single bond is? this represent.

01:00 Okay.

01:01 And to do this, i looked up in the table that they referenced, the bond enthelpes for each of these three bonds, and they were, as you see right here, and these are all kilojoules per mole.

01:28 And remember that this is one sigma bond.

01:34 This is one sigma and one pi, and this is one sigma and two pi.

01:42 So here's where, i hope i did this right, i found the difference, between my double and my single by taking 614.

01:50 I'm going to leave my units off.

01:51 Minus 348, and i got 266 kilojoules per mole.

02:00 That would be one of my pie bonds.

02:03 My pie bond in this in a double bond.

02:09 And for my triple bond, i took 839 minus 614, and i got to, let me see, 400, where is it now? 225 kilojoules per mole and that's for my second that would be for that second pie bond.

02:39 So to get the average of those i just took one half times 266 plus 225 and that gave me 245 i believe 245 kilojoules per mole.

02:59 And to figure my fraction i took for 240, whoops, 240.

03:03 Kilojoules per mole and i divided by that by 348 kilojoules per mole and it gave me about 0 .70 was the i guess the percent contribution and i wasn't that might have actually now that i'm looking at that i might take my 245 divided by 614 and my 245 times two would have been 491 divided by 839.

04:01 I'm looking at that.

04:02 That might be more appropriate.

04:04 That might be what they're asking for.

04:05 I didn't think of that until i was solving the problem.

04:10 Solving that now, that's 0 .40, close enough.

04:16 And 491 divided by 839 is 0 .59 for both of these...

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