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2 years ago

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 .

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Could a mixture be made up of only elements and no compounds. explain

lbvjy [14]

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3 years ago

Balance each chemical equation.<br> Hg2(C2H3O2)2(aq)+KCl(aq)→Hg2Cl2(s)+KC2H3O2(aq)

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Answer:

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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:

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

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

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

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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3 years ago

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3 years ago