Predict the hybridization of the oxygen atom in water, H 2 O. Draw a picture of its three-dimensional structure, and explain why its bond angle is 104.5°.

Welcome back everyone. What is the three dimensional structure of ammonia? What is the hybridization of its nitrogen atom and explain why its bond angle is 107 degrees. So we have a three part problem. Let's begin solving the first one. And if we want to draw a three dimensional structure, we want to recall the vs CPR theory. So let's count the number of the electrons for ammonia, nitrogen has five valence electrons. And we want to add three multiplied by one for each hydrogen. And this gives us a total of eight valence electrons. And what we want to do is assign a central atom which would be nitrogen because it is more electron and draw single bonds between nitrogen and every hydrogen. Since hygen follows the duet rule, we essentially can count the number of electrons that we have, we have six. So the two remaining electrons must belong to the central nitrogen such that nitrogen has an an octet. So now we have our Louis structure, we essentially want to predict its three dimensional structure based on the vs CPR theory. If the central atom is nitrogen and it has four electron regions we bonds in one lone pair. Then its electronic geometry is tetrahedral. We can understand this from the the vs CPR tables. And it's molecular geometry would be Trigon all we're a middle because there's one long pair present. So we essentially want to draw a trigonal pyramidal shape. Essentially, we are going to begin by drawing a blue nitrogen atom, we can color it slightly. And now we want to draw three hydrogens in a three dimensional plane. Two of them can be pointing towards us, right. So we are just going to draw to nitrogen hydrogen bonds and that one of them will be pointing away. And that's it. We have our three dimensional representation of the structure in this structure. The blue atom is nitrogen and the white one is hydrogen with a red outline. Now we have the answer for the first part, we can continue with the second part and we want to understand what's the hybridization of nitrogen. Now, if electronic geometry is tetrahedral, let's recall that we asteric number of four, which we get from three bonds in one long pair based on the C number of four. The hybridization is SP three S one P three, we can just add those superscripts and we will get four, right. So hybridization would be SP three. And now we essentially want to explain why the bond angle is 107 degrees for that purpose. We simply want to recall that if its electronic geometry is tetrahedral. Then the expected bond angle is 109.5 degrees. But due to the presence of long pair based on the vs cpr theory, that long pair is essentially decreasing the bond angle by about 2.5 degrees. So we can say the bun angle of ammonia is 107 degrees, which is less than the expected 109.5 degrees. DT repulsive force exerted buy the lamp pair on the bonding electrons. Right, based on the vs CPR theory L pairs are interacting with the bonding electrons. And that repulsive force is squeezing the angle, every lump pair is decreasing the angle by roughly 2.5 degrees. So if we add an additional lump pair such as in the water, we would have a an angle of 104.5 degrees. So those are our answers. Let's mark them and conclude the video. Thank you for watching.

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1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

1. A Review of General Chemistry

Molecular Geometry

Organic Chemistry

1. A Review of General Chemistry

Molecular Geometry

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6:24 minutes

Problem 16a

Textbook Question

Predict the hybridization of the oxygen atom in water, H₂O. Draw a picture of its three-dimensional structure, and explain why its bond angle is 104.5°.

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Identify the central atom in the molecule. In water (H2O), the central atom is oxygen.

Determine the number of valence electrons for the oxygen atom. Oxygen has 6 valence electrons.

Consider the bonding and lone pairs around the oxygen atom. In H2O, oxygen forms two sigma bonds with hydrogen atoms and has two lone pairs of electrons.

Use the VSEPR (Valence Shell Electron Pair Repulsion) theory to predict the hybridization. The steric number (number of sigma bonds + lone pairs) for oxygen in H2O is 4, indicating sp³ hybridization.

Explain the bond angle. The ideal bond angle for sp³ hybridization is 109.5°, but the presence of lone pairs, which exert greater repulsion than bonding pairs, reduces the bond angle to approximately 104.5°.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hybridization

Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals suitable for pairing electrons to form chemical bonds. In water (H2O), the oxygen atom undergoes sp3 hybridization, which involves the mixing of one s orbital and three p orbitals, resulting in four equivalent sp3 hybrid orbitals. This hybridization explains the tetrahedral arrangement of electron pairs around the oxygen atom.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. In water, the geometry is bent or V-shaped due to the two lone pairs on the oxygen atom, which repel the hydrogen atoms, resulting in a bond angle of 104.5°. This angle is less than the ideal tetrahedral angle of 109.5° due to the greater repulsion exerted by lone pairs compared to bonding pairs.

VSEPR Theory

Valence Shell Electron Pair Repulsion (VSEPR) theory is used to predict the shape of molecules based on electron pair repulsions. According to VSEPR theory, electron pairs around a central atom will arrange themselves to minimize repulsion. In H2O, the two lone pairs on oxygen push the hydrogen atoms closer together, resulting in a bond angle of 104.5°, which is characteristic of a bent molecular shape.

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Predict the hybridization of the oxygen atom in water, H2O. Draw a picture of its three-dimensional structure, and explain why its bond angle is 104.5°.

Key Concepts

Hybridization

Molecular Geometry

VSEPR Theory

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Predict the hybridization of the oxygen atom in water, H₂O. Draw a picture of its three-dimensional structure, and explain why its bond angle is 104.5°.