QUESTION IMAGE
Question
chemistry 20: bonding unit. molecular shapes. choose model\ molecule shapes model 2. check the molecular geometry
ame\ tools on a test, you will be asked molecular geometry name, not electron geometry name. 3. check show lone pairs 4. (optional) ...and the \show bond angles\ box (bond angles, not tested in chem 20) 5. now build your models... using the phet animation, determine the shape of the molecule. put the name here and then try to draw a 3d shape. turn molecules around with mouse. molecule lewis dot diagram structural diagram nh₃ ch₄ bh₃
Response
For $\ce{NH_{3}}$:
Lewis Dot Diagram:
- Nitrogen ($\ce{N}$) has 5 valence electrons, and each hydrogen ($\ce{H}$) has 1.
- $\ce{N}$ forms 3 single bonds with 3 $\ce{H}$ atoms, using 3 electrons, and has 1 lone pair (2 electrons).
- The Lewis dot diagram: $\ce{H:\overset{..}{N}:H}$ (with a $\ce{H}$ on the bottom too, so $\ce{H - \overset{..}{N} - H}$ with a $\ce{H}$ below the $\ce{N}$).
Structural Diagram:
- Single bonds between $\ce{N}$ and each $\ce{H}$, with the lone pair on $\ce{N}$. Represented as a central $\ce{N}$ connected to 3 $\ce{H}$ atoms (like a pyramid base with $\ce{N}$ at the top, but in 2D, it's $\ce{H - N - H}$ with a $\ce{H}$ below the $\ce{N}$ and the lone pair above $\ce{N}$).
Molecular Shape:
- Using VSEPR theory, the electron geometry is tetrahedral (4 electron groups: 3 bonding, 1 lone pair), but molecular geometry (shape) is trigonal pyramidal.
For $\ce{CH_{4}}$:
Lewis Dot Diagram:
- Carbon ($\ce{C}$) has 4 valence electrons, each $\ce{H}$ has 1.
- $\ce{C}$ forms 4 single bonds with 4 $\ce{H}$ atoms, using all 4 valence electrons.
- Lewis dot diagram: $\ce{H:\overset{}{\underset{}{C}}:H}$ (with $\ce{H}$ atoms above and below too, so $\ce{H - C - H}$ with $\ce{H}$ above and below $\ce{C}$).
Structural Diagram:
- Four single bonds between $\ce{C}$ and each $\ce{H}$, forming a tetrahedral arrangement in 3D, but in 2D, it's a central $\ce{C}$ with 4 $\ce{H}$ atoms (one at each "corner" of a square, but actually tetrahedral).
Molecular Shape:
- Electron geometry and molecular geometry (since no lone pairs) is tetrahedral (4 bonding groups, 0 lone pairs).
For $\ce{BH_{3}}$:
Lewis Dot Diagram:
- Boron ($\ce{B}$) has 3 valence electrons, each $\ce{H}$ has 1.
- $\ce{B}$ forms 3 single bonds with 3 $\ce{H}$ atoms, using all 3 valence electrons (no lone pairs on $\ce{B}$).
- Lewis dot diagram: $\ce{H:\overset{}{B}:H}$ (with a $\ce{H}$ below the $\ce{B}$).
Structural Diagram:
- Three single bonds between $\ce{B}$ and each $\ce{H}$, forming a trigonal planar arrangement.
Molecular Shape:
- Electron geometry and molecular geometry (no lone pairs) is trigonal planar (3 bonding groups, 0 lone pairs).
Filling the Table:
| Molecule | Lewis Dot Diagram | Structural Diagram | Molecular Shape (3D Name) |
|---|---|---|---|
| $\ce{CH_{4}}$ | $\ce{H - C - H}$ (with $\ce{H}$ above and below $\ce{C}$) | Central $\ce{C}$ with 4 $\ce{H}$ (tetrahedral) | Tetrahedral |
| $\ce{BH_{3}}$ | $\ce{H - B - H}$ (with $\ce{H}$ below $\ce{B}$) | Central $\ce{B}$ with 3 $\ce{H}$ (trigonal planar) | Trigonal Planar |
To draw the 3D shapes:
- $\ce{NH_{3}}$: Imagine a pyramid with a triangular base (3 $\ce{H}$ atoms) and $\ce{N}$ at the apex, with the lone pair above $\ce{N}$.
- $\ce{CH_{4}}$: A tetrahedron (4 triangular faces), with $\ce{C}$ at the center and $\ce{H}$ at each vertex.
- $\ce{BH_{3}}$: A flat triangle (trigonal planar), with $\ce{B}$ at the center and $\ce{H}$ at each corner of the triangle.
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For $\ce{NH_{3}}$:
Lewis Dot Diagram:
- Nitrogen ($\ce{N}$) has 5 valence electrons, and each hydrogen ($\ce{H}$) has 1.
- $\ce{N}$ forms 3 single bonds with 3 $\ce{H}$ atoms, using 3 electrons, and has 1 lone pair (2 electrons).
- The Lewis dot diagram: $\ce{H:\overset{..}{N}:H}$ (with a $\ce{H}$ on the bottom too, so $\ce{H - \overset{..}{N} - H}$ with a $\ce{H}$ below the $\ce{N}$).
Structural Diagram:
- Single bonds between $\ce{N}$ and each $\ce{H}$, with the lone pair on $\ce{N}$. Represented as a central $\ce{N}$ connected to 3 $\ce{H}$ atoms (like a pyramid base with $\ce{N}$ at the top, but in 2D, it's $\ce{H - N - H}$ with a $\ce{H}$ below the $\ce{N}$ and the lone pair above $\ce{N}$).
Molecular Shape:
- Using VSEPR theory, the electron geometry is tetrahedral (4 electron groups: 3 bonding, 1 lone pair), but molecular geometry (shape) is trigonal pyramidal.
For $\ce{CH_{4}}$:
Lewis Dot Diagram:
- Carbon ($\ce{C}$) has 4 valence electrons, each $\ce{H}$ has 1.
- $\ce{C}$ forms 4 single bonds with 4 $\ce{H}$ atoms, using all 4 valence electrons.
- Lewis dot diagram: $\ce{H:\overset{}{\underset{}{C}}:H}$ (with $\ce{H}$ atoms above and below too, so $\ce{H - C - H}$ with $\ce{H}$ above and below $\ce{C}$).
Structural Diagram:
- Four single bonds between $\ce{C}$ and each $\ce{H}$, forming a tetrahedral arrangement in 3D, but in 2D, it's a central $\ce{C}$ with 4 $\ce{H}$ atoms (one at each "corner" of a square, but actually tetrahedral).
Molecular Shape:
- Electron geometry and molecular geometry (since no lone pairs) is tetrahedral (4 bonding groups, 0 lone pairs).
For $\ce{BH_{3}}$:
Lewis Dot Diagram:
- Boron ($\ce{B}$) has 3 valence electrons, each $\ce{H}$ has 1.
- $\ce{B}$ forms 3 single bonds with 3 $\ce{H}$ atoms, using all 3 valence electrons (no lone pairs on $\ce{B}$).
- Lewis dot diagram: $\ce{H:\overset{}{B}:H}$ (with a $\ce{H}$ below the $\ce{B}$).
Structural Diagram:
- Three single bonds between $\ce{B}$ and each $\ce{H}$, forming a trigonal planar arrangement.
Molecular Shape:
- Electron geometry and molecular geometry (no lone pairs) is trigonal planar (3 bonding groups, 0 lone pairs).
Filling the Table:
| Molecule | Lewis Dot Diagram | Structural Diagram | Molecular Shape (3D Name) |
|---|---|---|---|
| $\ce{CH_{4}}$ | $\ce{H - C - H}$ (with $\ce{H}$ above and below $\ce{C}$) | Central $\ce{C}$ with 4 $\ce{H}$ (tetrahedral) | Tetrahedral |
| $\ce{BH_{3}}$ | $\ce{H - B - H}$ (with $\ce{H}$ below $\ce{B}$) | Central $\ce{B}$ with 3 $\ce{H}$ (trigonal planar) | Trigonal Planar |
To draw the 3D shapes:
- $\ce{NH_{3}}$: Imagine a pyramid with a triangular base (3 $\ce{H}$ atoms) and $\ce{N}$ at the apex, with the lone pair above $\ce{N}$.
- $\ce{CH_{4}}$: A tetrahedron (4 triangular faces), with $\ce{C}$ at the center and $\ce{H}$ at each vertex.
- $\ce{BH_{3}}$: A flat triangle (trigonal planar), with $\ce{B}$ at the center and $\ce{H}$ at each corner of the triangle.