Energy level Diagram relating to SP3 hybridization

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Physics345
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Homework Statement



Experimental evidence suggests that the nitrogen atom in ammonia, NH3, has four identical orbitals in the shape of a pyramid or tetrahedron.

a) Draw an energy level diagram to show the formation of these hybrid orbitals.

(hint: No electron promotion is required)

b) Name the type of Hybrid orbitals found in NH3. Of the four hybrid orbitals on the N atom, how many will take part in bonding?

c)Draw for yourself the energy – level diagram showing the hybrid orbitals formed in the C atom when it bonds. Now look at those hybrid orbitals and those of the N atom, and describe how the bonding with a N atom will di²er with the bonding that occurs with a C atom, even though both atoms have four hybrid orbitals oriented in a tetrahedral shape

Homework Equations



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The Attempt at a Solution


a)

Energy level diagram of orbital hybridization: 2p, 2s and 1s orbitals on the left with electron arrows, and an arrow labelled Orbital Hybridization leading to four sp3 hybrid orbitals labelled s and P, plus 1s.


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b) sp^3 Is the hybrid orbital found in NH_3 in this case Nitrogen wants to complete its valence shell becoming like the noble gas Neon, thus it takes 3 electrons from each 1s orbital from the three hydrogen atoms and completes its 2p orbital forming a complete shell and a sp^3 hybridization. Therefore, it uses three hybrid orbitals.

c) The bonding with the nitrogen atom is different compared to the carbon atom. The carbon atom has 4 incomplete orbitals and only 2 electrons in its 2p orbitals while the third 2p orbital remains empty, while the nitrogen atom has 4 incomplete orbitals as well but the difference is that it has 3 electrons in each of its 2p orbitals.

Energy level diagram: on the left, 2p (three orbitals), 2s and 1s boxes drawn as circles with arrows; an arrow labelled Orbital Hybridization leads to four sp3 hybrid orbitals labelled s, P, P, P, with a 1s orbital shown below.

I was just wondering if I did these correctly. Thanks in advance.
 
Attachments
  • Energy level diagram: on the left, 2p (three orbitals), 2s and 1s boxes drawn as circles with arrows; an arrow labelled Orbital Hybridization leads to four sp3 hybrid orbitals labelled s, P, P, P, with a 1s orbital shown below.
    IO7JmdG.png
    5.5 KB · Views: 2,696
  • Energy level diagram of orbital hybridization: 2p, 2s and 1s orbitals on the left with electron arrows, and an arrow labelled Orbital Hybridization leading to four sp3 hybrid orbitals labelled s and P, plus 1s.
    KuUvJNv.png
    6.8 KB · Views: 6,940
The explanation looks basically right. I would not label the individual hybrid orbitals as s and p. They would all be called sp3 orbitals, as they all contain some commbination of characteristics from s and p orbitals. So each hydrogen's 1s orbital mixes with one sp3-orbital (not a p-orbital) to form a N-H bond.

In answer (b), I would clarify that three sp3 orbital are used for bonding while the fourth sp3 orbital contains a lone pair (it is a non-bonding orbital).
 
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Based on your comments, I have rewrote my answers: Let me know what you think :)
Energy level diagrams of sp3 orbital hybridization: 2p, 2s and 1s orbitals with electron arrows on the left, an arrow labelled Orbital Hybridization, and four sp3 hybrid orbitals on the right, shown twice with paragraph text between.

Ignore the attached file I accidentally posted that, now I can't figure out how to remove it, oh well.
 
Attachments
  • Handwritten-style chemistry answers showing Lewis structures and 3D sketches: an iodate ion IO3- with three bonding pairs and one lone pair on iodine, and a CHCl3 molecule with delta plus and minus charges marked on C-H and C-Cl bonds.
    GstDhwh.png
    18.6 KB · Views: 1,210
  • Energy level diagrams of sp3 orbital hybridization: 2p, 2s and 1s orbitals with electron arrows on the left, an arrow labelled Orbital Hybridization, and four sp3 hybrid orbitals on the right, shown twice with paragraph text between.
    p639acy.png
    26.9 KB · Views: 2,242
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