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oee [108]
3 years ago
14

What atomic or hybrid orbitals make up the sigma bond between c2 and c3 in propyne, chcch3 ? (c2 is the second carbon in the str

ucture as written.) orbital on c2 + orbital on c3 what is the approximate c1-c2-c3 bond angle ? ... °?

Chemistry
1 answer:
Anvisha [2.4K]3 years ago
4 0

Answer:

\boxed{\text{$\sigma$(sp-sp$^{3}$); 180$^{\circ}$}}

Explanation:

(a) Hybridization of orbitals

See the Lewis structure of propyne in the first diagram below.

C1 is directly bonded to two other atoms (H and C2) so it is sp hybridized.

C2 is directly bonded to two other atoms (C1 and C3) so it is sp hybridized.

C3 is directly bonded to four other atoms (C2 and 3 H) so it is sp³ hybridized.

(b) C2-C3 Sigma Bond

See the atomic and molecular orbitals in the second picture below.

C3 is using its hybrid atomic orbitals to form sigma molecular bonds. The C2-C3 sigma bond is formed by the overlap of the C2 sp atomic orbital with the C3 sp³ atomic orbital to make a σ(sp-sp³) molecular orbital.

(c) Bond angles

C1 and C2 are sp hybridized. Since the angle between sp orbitals is 180°, all atoms directly attached the C1 and C2 must be in a straight line. The C-C-C bond angle is 180°.

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The equilibrium constant for the gas-phase isomerization of borneol (c10h17oh) to isoborneol at 503 k is 0.106. a mixture consis
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K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106

The molar mass of isoborneol/borneol is 154.25 g/mol

Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol

Use the ICE approach

        borneol  →  isoborneol
I         0.0972           0.0486
C         -x                     +x
E     0.0972 - x        0.0486 + x

Total moles = 0.1458

Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P

0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832 

Thus,
<em>Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904</em>
<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>
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