The hybridization of the nitrogen atom in H2C=NH is sp2.
3a) Reproduce Figure a) and replace carbon with nitrogen and add the correct number of electrons. 2p 11 2p forms 1 bond potential energy
Figure a) Electron configuration diagram of hybridization of carbon
1 mark for correct number of electrons in the ground state
1 mark for correct placement of electrons. Note lone pair can be in any of the sp2 orbitals
1 mark for identifying the correct number of pi and sigma bonds
3b) Reproduce Figure b) and replace one of the carbons with a nitrogen. Keep in mind the electronic configuration of nitrogen.
Left diagrams shows atomic orbitals on individual atoms. Right diagram shows the atomic orbitals coming together to form bonds
Figure b)
Left diagram: 1 mark for drawing orbitals correctly with an electron in each orbital. A dot instead of an arrow can be used for electrons.
Right diagram: 1 mark for correctly reproducing the combined orbitals that represent the sigma, pi bond and lone pair. Also for correctly changing the line diagram. (all correct or zero) 1 mark. Correctly adding labels showing p orbitals and pi bond. The other labels do not have to be reproduced.
The Correct Answer and Explanation is :
Let’s break down the problem step by step for a thorough explanation.
3a) Hybridization of the Nitrogen Atom in H2C=NH (sp2)
In the molecule H2C=NH, nitrogen is sp2 hybridized. To understand this, we need to look at nitrogen’s electron configuration and how it hybridizes.
Electron Configuration of Nitrogen (Ground State):
- Atomic number of nitrogen = 7
- Electron configuration: 1s² 2s² 2p³
For sp2 hybridization, the nitrogen atom needs to promote one of its 2s electrons to the empty 2p orbital. The resulting electron configuration for nitrogen in the sp2 hybridized state becomes:
- 2s¹ 2p² (with one electron in each of the three orbitals: two 2p orbitals and one 2s orbital).
Atomic Orbitals and Bonding in H2C=NH:
- The nitrogen atom uses one of its sp2 orbitals to form a sigma bond with the carbon in H2C=NH.
- The second sp2 orbital on nitrogen holds the lone pair of electrons.
- The third sp2 orbital participates in the formation of the C=N double bond, where one of the bonds is a sigma bond and the other is a pi bond.
Answer for 3a:
- Electron Configuration Diagram:
- Nitrogen has three hybridized sp2 orbitals and one unhybridized 2p orbital.
- The first sp2 orbital will overlap with the carbon to form a sigma bond.
- The second sp2 orbital will hold the lone pair of electrons.
- The third sp2 orbital will overlap with one of the carbon’s p orbitals to form the second sigma bond.
- The unhybridized 2p orbital on nitrogen will overlap with a p orbital on carbon to form the pi bond.
Pi bonds: 1 (from the C=N double bond)
Sigma bonds: 2 (one from the C-N single bond and one from the C=N double bond)
Lone pair: 1 (on nitrogen)
3b) Replace Carbon with Nitrogen in the Diagram (Electron Configuration of Nitrogen):
Now, in Figure b), we need to replace one of the carbons with nitrogen and keep in mind nitrogen’s electronic configuration.
Left Diagram (Atomic Orbitals):
- Nitrogen in sp2 hybridization:
- One sp2 orbital holds a lone pair.
- Two sp2 orbitals overlap with carbon to form sigma bonds.
- One unhybridized p orbital on nitrogen overlaps with one of carbon’s p orbitals to form the pi bond.
Right Diagram (Molecular Orbitals):
- The bonding of orbitals on the nitrogen atom will form sigma and pi bonds with carbon.
- The lone pair will be shown as a dot in the sp2 orbital.
Final Answer for 3b:
- Left Diagram:
- Draw the nitrogen atomic orbitals (1s, 2s, and 2p), with electrons filling the orbitals appropriately (2p² and 2s¹).
- Right Diagram:
- The atomic orbitals of nitrogen overlap with carbon’s orbitals to form sigma and pi bonds, with the lone pair in an sp2 orbital on nitrogen.
Image Generation:
I will now generate a diagram based on this explanation, following your request.
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Here is the diagram showing the hybridization of nitrogen in H2C=NH, with the correct placement of electrons and bonding. It includes the atomic orbitals of nitrogen, the sp2 hybridization, and the overlap of orbitals that form sigma and pi bonds with carbon. I hope this helps! Let me know if you’d like any adjustments or further details.
