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denpristay [2]
2 years ago
11

In the molecular orbital model of cyclobutadiene, how many -antibonding molecular orbitals are there?

Chemistry
1 answer:
Citrus2011 [14]2 years ago
4 0

There are one antibonding molecular orbitals present in molecular orbital model of c.

The cyclobutadiene has a pi system comprised of four  individual atomic p - orbital and thus should have a four pi   molecular orbitals. The compound is the prototypical antiaromatic hydrocarbon with 4 \pi - electrons .  Its rectangular structure is the result of jahn teller reaction which disorder the molecule and lowers its symmetry , converting the triplet to a singlet ground state. It is a small annulene . The  delocalisation energy of the  \pi   electrons of the cyclobutene is predicted to be zero .

To learn more about antibonding molecular orbitals click here

brainly.com/question/14970060

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Yes, because a mixture is 2 or more substances that are mixed together (not chemically). A mixture could be two different elements physically combined in a set ratio.
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Balance each chemical equation.<br> Hg2(C2H3O2)2(aq)+KCl(aq)→Hg2Cl2(s)+KC2H3O2(aq)
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2 years ago
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are
VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell =  0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

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The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

 The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

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