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N76 [4]
3 years ago
12

Identify the hybridization of each carbon atom for the molecule above

Chemistry
1 answer:
ipn [44]3 years ago
8 0

Carbons starting from the left end:

  1. sp²
  2. sp²
  3. sp²
  4. sp
  5. sp

Refer to the sketch attached.

<h3>Explanation</h3>

The hybridization of a carbon atom depends on the number of electron domains that it has.

Each chemical bond counts as one single electron domain. This is the case for all chemical bonds: single, double, or triple. Each lone pair also counts as one electron domain. However, lone pairs are seldom seen on carbon atoms.

Each carbon atom has four valence electrons. It can form up to four chemical bonds. As a result, a carbon atom can have up to four electron domains. It has a minimum of two electron domains, with either two double bonds or one single bond and one triple bond.

  • A carbon atom with four electron domains is sp³ hybridized;
  • A carbon atom with three electron domains is sp² hybridized;
  • A carbon atom with two electron domains is sp hybridized.

Starting from the left end (H₂C=CH-) of the molecule:

  • The first carbon has three electron domains: two C-H single bonds and one C=C double bond; It is sp² hybridized.
  • The second carbon has three electron domains: one C-H single bond, one C-C single bond, and one C=C double bond; it is sp² hybridized.
  • The third carbon has three electron domains: two C-C single bonds and one C=O double bond; it is sp² hybridized.
  • The fourth carbon has two electron domains: one C-C single bond and one C≡C triple bond; it is sp hybridized.
  • The fifth carbon has two electron domains: one C-H single bond and one C≡C triple bond; it is sp hybridized.

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Assume that each atom is a sphere, and that the surface of each atom is in contact with its nearest neighbor. Determine the perc
tatiyna

Answer:

  • The percentage of unit cell volume that is occupied by atoms in a face- centered cubic lattice is 74.05%
  • The percentage of unit cell volume that is occupied by atoms in a body-centered cubic lattice is 68.03%  
  • The percentage of unit cell volume that is occupied by atoms in a diamond lattice is 34.01%

Explanation:

The percentage of unit cell volume = Volume of atoms/Volume of unit cell

Volume of sphere = \frac{4 }{3} \pi r^2

a) Percentage of unit cell volume occupied by atoms in face- centered cubic lattice:

let the side of each cube = a

Volume of unit cell = Volume of cube = a³

Radius of atoms = \frac{a\sqrt{2} }{4}

Volume of each atom = \frac{4 }{3} \pi (\frac{a\sqrt{2}}{4})^3 = \frac{\pi *a^3\sqrt{2}}{24}

Number of atoms/unit cell = 4

Total volume of the atoms = 4 X \frac{\pi *a^3\sqrt{2}}{24} = \frac{\pi *a^3\sqrt{2}}{6}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{2}}{6}}{a^3} =\frac{\pi *a^3\sqrt{2}}{6a^3} = \frac{\pi \sqrt{2}}{6} = 0.7405

= 0.7405 X 100% = 74.05%

b) Percentage of unit cell volume occupied by atoms in a body-centered cubic lattice

Radius of atoms = \frac{a\sqrt{3} }{4}

Volume of each atom =\frac{4 }{3} \pi (\frac{a\sqrt{3}}{4})^3 =\frac{\pi *a^3\sqrt{3}}{16}

Number of atoms/unit cell = 2

Total volume of the atoms = 2X \frac{\pi *a^3\sqrt{3}}{16} = \frac{\pi *a^3\sqrt{3}}{8}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{3}}{8}}{a^3} =\frac{\pi *a^3\sqrt{3}}{8a^3} = \frac{\pi \sqrt{3}}{8} = 0.6803

= 0.6803 X 100% = 68.03%

c) Percentage of unit cell volume occupied by atoms in a diamond lattice

Radius of atoms = \frac{a\sqrt{3} }{8}

Volume of each atom = \frac{4 }{3} \pi (\frac{a\sqrt{3}}{8})^3 = \frac{\pi *a^3\sqrt{3}}{128}

Number of atoms/unit cell = 8

Total volume of the atoms = 8X \frac{\pi *a^3\sqrt{3}}{128} = \frac{\pi *a^3\sqrt{3}}{16}

The percentage of unit cell volume = \frac{\frac{\pi *a^3\sqrt{3}}{16}}{a^3} =\frac{\pi *a^3\sqrt{3}}{16a^3} = \frac{\pi \sqrt{3}}{16} = 0.3401

= 0.3401  X 100% = 34.01%

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