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Apparently it has $sp^3$ hybridization, but I don't understand why. Ammonia ($\ce{NH3}$) seems to me to not require $sp$ hybridization because all of its bond lengths are already equal. It has 3 hydrogens bonded to the $p$ orbitals. Why can't the lone electron pair in nitrogen's $2s$ shell stay put where it is?

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Well, I can think of these reasons that can justify the hybridization in ammonia molecule.

  1. Bond angle: If the molecule did not have hybrid orbits, and instead had unhybridized p-orbitals taking part in the bond formation, then the bond angle between the orbitals would be 90 degrees. And as for the real situation the bond angle is nearly 107 which makes the molecule more stable, decreasing the bond pair-bond pair and bond pair-lone pair repulsion.

  2. Energy: Recalling the definition of hybridization, it is the mixing of the atomic orbitals having slightly different energies to form new orbitals that have equal energies. This stabilizes the molecule. As for ammonia molecule due to hybridization the energies of the lone pair of electrons and the bond pair of electrons becomes almost equal, thereby increasing the stability of the molecule.

  3. Geometry: Thinking about the spatial arrangement of the atoms in the molecule of ammonia if there is no hybridization in the molecule then the size of the orbital containing lone pair of electrons would be different from that of the orbitals containing bond pair of electrons. Also if hybridization would not be taking place, explaining the geometry of the molecule, that is it's trigonal pyramidal shape, would not be possible.

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  • $\begingroup$ I'll mark this as the answer, I found the answer in the meantime after a lot of looking around. It seems that all atoms with 4 or more valence electrons will form sp3 hybrid orbitals (sp2 or sp if there is a double bond or triple bond respectively). I'm still not sure exactly why the s orbital can't stay as an s orbital, I guess because it's needed to shift everything to an sp so that the electrons as a group can be farther away from each other. In any case, thanks for a great answer. $\endgroup$ – Klik Jan 3 '14 at 7:52
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This question has been answered before. Look here: http://answers.yahoo.com/question/index?qid=20130803221208AAYOCBr If you look at the MO diagram for $\ce{NH_3}$ you'll find that the orbital which the lone pair resides in is higher in energy (in the $2s$ antibonding orbital http://www.d.umn.edu/~pkiprof/ChemWebV2/Bonding/MO-ammonia/index.html). Also, if I recall well, hybridization is not entirely correct. In hybridization, you typically assume that all of the hybridized molecular orbitals are degenerate. This is not always the case due to the different types overlap of the orbitals needed to create the molecular orbitals. Some hybridized molecular orbitals (even the ones of the same "type" such as $sp^3$ or $sp^2d$, etc) are higher in energy even though all of them are bonding molecular orbitals. You'll learn more of this in the second semester of inorganic chemistry. I think , ammonia is not really $sp^3$ due to the poor overlap (methane has a better overlap between the $s$ orbitals and $p$ oribitals). Take it with a grain of salt. It has been a while since I took anything related to MO's.

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  • $\begingroup$ I appreciate the help, this question was really giving me difficulty. I found out that Ammonia does in fact have sp3 hybridized orbitals. It seems that atoms with 4 or more valence electrons usually have sp3 hybridization. It allows the electron shape to become tetrahedral so the electrons can be farther apart. It also accounts for the bond angles in NH3 of 107 (slightly different than normal tetrahedral because of the lone pair of electrons). Cheers. $\endgroup$ – Klik Jan 3 '14 at 8:03
  • $\begingroup$ sp3 hybridization does NOT lead to bond angles of 107. Good sp3 hybridization leads to bond angles of 109.5 degrees. $\endgroup$ – CoffeeIsLife Jan 3 '14 at 8:07
  • $\begingroup$ It does in this case because of the lone electron pair. Really it does. $\endgroup$ – Klik Jan 3 '14 at 8:12
  • $\begingroup$ en.wikipedia.org/wiki/Ammonia#Structure $\endgroup$ – Klik Jan 3 '14 at 8:15
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    $\begingroup$ The bond angles are an indication of the type of hybridization. In this case, we say that the hybridization is sp3 since it resembles sp3. Ammonia has a poorer orbital overlap compared to methane. Also, MO theory and VSPER theory are not the same. They complement each other, but there are subtleties to it. If you are interested in MO theory, I recommend DekocK and Gray's "Chemical Structure and Bonding" $\endgroup$ – CoffeeIsLife Jan 3 '14 at 8:18

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