Draw the Lewis structure of bromoform (CHBr3). 2. What is the hybridization type of the carbon atom? 3. What orbitals are involved in the formation of the C-H bond? 4. What orbitals are involved in the formation of the C-Br bond? 5. What is the electron geometry around the carbon atom? 2. Corrected_text: 1. Draw the Lewis diagram of hydroxylamine (HNOH). 2. What is the hybridization type of the N and O atoms? 3. What is the electron geometry around the N and O atoms? 4. What orbitals are involved in the formation of the N-O bond? Title_with_topic: Lewis Structures and Hybridization in Bromoform and Hydroxylamine
The Correct Answer and Explanation is:
Title: Lewis Structures and Hybridization in Bromoform and Hydroxylamine
1. Bromoform (CHBr₃)
Lewis Structure of CHBr₃:
- Central atom: Carbon (C)
- Carbon makes four single bonds: one with hydrogen (H) and three with bromine (Br).
- Bromine atoms each have three lone pairs of electrons.
- Hydrogen has no lone pairs.
Structure:
markdownCopyEdit Br
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Br — C — H
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Br
2. Hybridization of the Carbon Atom:
The carbon in CHBr₃ forms four sigma bonds. This corresponds to sp³ hybridization.
3. Orbitals in the C–H Bond:
The C–H bond is formed from the overlap of an sp³ hybrid orbital on carbon and a 1s orbital on hydrogen.
4. Orbitals in the C–Br Bond:
Each C–Br bond is formed by the overlap of an sp³ hybrid orbital on carbon and a p orbital on bromine.
5. Electron Geometry around Carbon:
With four regions of electron density (all bonding), the electron geometry around carbon is tetrahedral.
2. Hydroxylamine (HNOH)
Lewis Structure of HNOH:
- Nitrogen (N) is the central atom, bonded to one hydrogen (H) and one hydroxyl group (OH), and has a lone pair.
- Oxygen (O) is bonded to one hydrogen and one nitrogen and also has two lone pairs.
Structure:
mathematicaCopyEditH – N – O – H
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lone pair on N
two lone pairs on O
Hybridization Types:
- Nitrogen (N): Forms two sigma bonds and has one lone pair ⇒ sp³ hybridized
- Oxygen (O): Forms two sigma bonds and has two lone pairs ⇒ sp³ hybridized
Electron Geometry:
- Nitrogen: 3 bonding + 1 lone pair = tetrahedral electron geometry
- Oxygen: 2 bonding + 2 lone pairs = tetrahedral electron geometry
Orbitals in the N–O Bond:
The N–O bond forms by the overlap of sp³ hybrid orbitals from both nitrogen and oxygen.
Explanation
The Lewis structures of both CHBr₃ and HNOH help visualize the bonding and electron arrangements, which are essential for understanding molecular geometry and hybridization. In CHBr₃, carbon is bonded to three bromine atoms and one hydrogen atom, resulting in four sigma bonds. This bonding pattern indicates sp³ hybridization, which gives a tetrahedral geometry. Each carbon–bromine bond results from the overlap of carbon’s sp³ orbital with bromine’s p orbital, while the carbon–hydrogen bond involves overlap with hydrogen’s 1s orbital.
In hydroxylamine (HNOH), nitrogen is central, bonded to a hydrogen atom and an OH group, and also has one lone pair. Both nitrogen and oxygen in this structure exhibit sp³ hybridization. Nitrogen has three electron domains (two bonds and one lone pair), and oxygen has four (two bonds and two lone pairs), giving both a tetrahedral electron geometry. The nitrogen–oxygen bond involves the overlap of sp³ hybrid orbitals from both atoms.
Understanding orbital overlaps and geometries provides insight into molecular shape, polarity, and reactivity. The sp³ hybridization in both molecules results in tetrahedral arrangements, which are commonly found in organic and biological molecules.
